Draft
DRAFT
Lecture Notes in:
STRUCTURAL ENGINEERING
Analysis and Design
Victor E. Saouma
Dept. of Civil Environmental and Architectural Engineering
University of Colorado, Boulder, CO 80309-0428
Draft
DRAFT
Lecture Notes in:
STRUCTURAL ENGINEERING
Analysis and Design
Victor E. Saouma
Dept. of Civil Environmental and Architectural Engineering
University of Colorado, Boulder, CO 80309-0428
Draft
3
PREFACE
Whereas there are numerous excellent textbooks covering Structural Analysis, or Structural
Design, I felt that there was a need for a single reference which
•
Provides a
succinct, yet rigorous
, coverage of Structural Engineering.
•
Combines
, as much as possible, Analysis with Design.
•
Presents numerous,
carefully selected, example problems
.
in a properly type set document.
As such, and given the reluctance of undergraduate students to go through extensive verbage
in order to capture a key concept, I have opted for an unusual format, one in which each key
idea is clearly distinguishable. In addition, such a format will hopefully foster group learning
among students who can easily reference misunderstood points.
Finally, whereas all problems have been taken from a variety of references, I have been
very careful in not only properly selecting them, but also in enhancing their solution through
appropriate figures and L
A
TEX typesetting macros.
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1 A BRIEF HISTORY OF STRUCTURAL ARCHITECTURE
31
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
The Medieval Period (477-1492)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
. . . . . . . . . . . . . . . . . . . . . . . . . 36
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
Pre Modern Period, Seventeenth Century
. . . . . . . . . . . . . . . . . . . . . . 42
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
. . . . . . . . . . . . . . . . . . . . . . . . . . 45
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
The pre-Modern Period; Coulomb and Navier
. . . . . . . . . . . . . . . . . . . . 47
The Modern Period (1857-Present)
. . . . . . . . . . . . . . . . . . . . . . . . . . 48
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
55
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Structures and their Surroundings
. . . . . . . . . . . . . . . . . . . . . . . . . . 55
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
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6 INTERNAL FORCES IN STRUCTURES
113
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
E 6-1 Simple Shear and Moment Diagram
. . . . . . . . . . . . . . . . . . . . . 117
E 6-2 Sketches of Shear and Moment Diagrams
. . . . . . . . . . . . . . . . . . 119
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
E 6-3 Frame Shear and Moment Diagram
. . . . . . . . . . . . . . . . . . . . . . 120
E 6-4 Frame Shear and Moment Diagram; Hydrostatic Load
. . . . . . . . . . . 123
E 6-5 Shear Moment Diagrams for Frame
. . . . . . . . . . . . . . . . . . . . . . 125
E 6-6 Shear Moment Diagrams for Inclined Frame
. . . . . . . . . . . . . . . . . 127
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
7 ARCHES and CURVED STRUCTURES
131
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134
E 7-1 Three Hinged Arch, Point Loads. (Gerstle 1974)
. . . . . . . . . . . . . . 134
E 7-2 Semi-Circular Arch, (Gerstle 1974)
. . . . . . . . . . . . . . . . . . . . . . 135
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
E 7-3 Statically Indeterminate Arch, (Kinney 1957)
. . . . . . . . . . . . . . . . 137
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
E 7-4 Semi-Circular Box Girder, (Gerstle 1974)
. . . . . . . . . . . . . . . . . . 140
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
E 7-5 Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974)
. . . 144
8 DEFLECTION of STRUCTRES; Geometric Methods
149
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149
Differential Equation of the Elastic Curve
. . . . . . . . . . . . . . . . . . 151
Moment Temperature Curvature Relation
. . . . . . . . . . . . . . . . . . 152
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
. . . . . . . . . . . . . . . . . . . . . . . . . . . 152
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152
Curvature Area Method (Moment Area)
. . . . . . . . . . . . . . . . . . . 154
. . . . . . . . . . . . . . . . . . . . 154
. . . . . . . . . . . . . . . . . . . 154
E 8-2 Moment Area, Cantilevered Beam
. . . . . . . . . . . . . . . . . . . . . . 157
E 8-3 Moment Area, Simply Supported Beam
. . . . . . . . . . . . . . . . . . . 157
. . . . . . . . . . . . . . . . . . . . . . . . 159
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
E 8-6 Frame Subjected to Temperature Loading
. . . . . . . . . . . . . . . . . . 162
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
E 11-1 Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads
236
12 KINEMATIC INDETERMINANCY; STIFFNESS METHOD
253
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
12.1.1 Stiffness vs Flexibility
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257
12.3.1 Force-Displacement Relations
. . . . . . . . . . . . . . . . . . . . . . . . . 257
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259
12.3.2.1 Uniformly Distributed Loads
. . . . . . . . . . . . . . . . . . . . 260
. . . . . . . . . . . . . . . . . . . . . . . . . 260
12.4 Slope Deflection; Direct Solution
. . . . . . . . . . . . . . . . . . . . . . . . . . . 261
12.4.1 Slope Deflection Equations
. . . . . . . . . . . . . . . . . . . . . . . . . . 261
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
E 12-1 Propped Cantilever Beam, (Arbabi 1991)
. . . . . . . . . . . . . . . . . . 263
E 12-2 Two-Span Beam, Slope Deflection, (Arbabi 1991)
. . . . . . . . . . . . . . 264
E 12-3 Two-Span Beam, Slope Deflection, Initial Deflection, (Arbabi 1991)
. . . 265
Frames, Slope Deflection, (Arbabi 1991)
. . . . . . . . . . . . . . . 267
12.5 Moment Distribution; Indirect Solution
. . . . . . . . . . . . . . . . . . . . . . . 269
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
. . . . . . . . . . . . . . . . . . . . . . . . . . . 269
. . . . . . . . . . . . . . . . . . . . . . . . . 269
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
12.5.1.4 Distribution Factor (DF)
. . . . . . . . . . . . . . . . . . . . . . 270
. . . . . . . . . . . . . . . . . . . . . . . . . . 271
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272
E 12-5 Continuous Beam, (Kinney 1957)
. . . . . . . . . . . . . . . . . . . . . . . 272
E 12-6 Continuous Beam, Simplified Method, (Kinney 1957)
. . . . . . . . . . . . 275
E 12-7 Continuous Beam, Initial Settlement, (Kinney 1957)
. . . . . . . . . . . . 277
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
E 12-9 Frame with Side Load, (Kinney 1957)
. . . . . . . . . . . . . . . . . . . . 283
E 12-10Moment Distribution on a Spread-Sheet
. . . . . . . . . . . . . . . . . . . 285
287
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
13.1.1 Structural Idealization
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287
13.1.2 Structural Discretization
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
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14 DESIGN PHILOSOPHIES of ACI and AISC CODES
349
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 350
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351
14.3.1 The Normal Distribution
. . . . . . . . . . . . . . . . . . . . . . . . . . . 351
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
359
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 370
E 15-3 Earthquake Load on a Frame
. . . . . . . . . . . . . . . . . . . . . . . . . 374
E 15-4 Earthquake Load on a Tall Building, (Schueller 1996)
. . . . . . . . . . . 375
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378
E 15-6 Thermal Expansion/Stress (Schueller 1996)
. . . . . . . . . . . . . . . . . 378
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
15.5 Other Important Considerations
. . . . . . . . . . . . . . . . . . . . . . . . . . . 379
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
387
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394
16.6.1 ASCII File with Steel Section Properties
. . . . . . . . . . . . . . . . . . . 404
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . 464
E 19-7 Column Design Using AISC Charts, (
. . . . . . . . . . . . . . . . . . . 465
467
20.1 Review from Strength of Materials
. . . . . . . . . . . . . . . . . . . . . . . . . . 467
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 470
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471
20.3.1 Failure Modes and Classification of Steel Beams
. . . . . . . . . . . . . . 471
. . . . . . . . . . . . . . . . . . . . . . . . . . 472
20.3.1.2 Partially Compact Section
. . . . . . . . . . . . . . . . . . . . . 473
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 474
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 475
E 20-1 Shape Factors, Rectangular Section
. . . . . . . . . . . . . . . . . . . . . 475
E 20-2 Shape Factors, T Section
. . . . . . . . . . . . . . . . . . . . . . . . . . . 475
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 477
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 479
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 480
21 UNBRACED ROLLED STEEL BEAMS
483
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 486
E 21-1 Adequacy of an unbraced beam, (
. . . . . . . . . . . . . . . . . . . . . 486
E 21-2 Adequacy of an unbraced beam, II (
. . . . . . . . . . . . . . . . . . . . 488
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489
E 21-3 Design of Laterally Unsupported Beam, (
. . . . . . . . . . . . . . . . . 490
21.5 Summary of AISC Governing Equations
. . . . . . . . . . . . . . . . . . . . . . . 495
497
22.1 Potential Modes of Failures
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 497
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
E 22-3 Design of Steel Beam-Column, (
. . . . . . . . . . . . . . . . . . . . . . 505
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548
E 26-1 Prestressed Concrete I Beam
. . . . . . . . . . . . . . . . . . . . . . . . . 550
26.3 Case Study: Walnut Lane Bridge
. . . . . . . . . . . . . . . . . . . . . . . . . . . 552
26.3.1 Cross-Section Properties
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 552
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 554
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555
557
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 557
27.1.2 Possible Arrangement of Bars
. . . . . . . . . . . . . . . . . . . . . . . . . 558
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 558
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 559
. . . . . . . . . . . . . . . . . . . . . . . . . 559
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 562
. . . . . . . . . . . . . . . . . . . . . . . . . 563
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 566
E 27-3 R/C Column, Using Design Charts
. . . . . . . . . . . . . . . . . . . . . . 571
28 ELEMENTS of STRUCTURAL RELIABILITY
573
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573
28.3 Distributions of Random Variables
. . . . . . . . . . . . . . . . . . . . . . . . . . 575
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 575
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 577
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577
28.4.1 Performance Function Identification
. . . . . . . . . . . . . . . . . . . . . 577
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577
28.4.3 Mean and Standard Deviation of a Performance Function
. . . . . . . . . 578
. . . . . . . . . . . . . . . . . . . . . . . . . . 579
28.4.3.2 Monte Carlo Simulation
. . . . . . . . . . . . . . . . . . . . . . . 579
28.4.3.3 Point Estimate Method
. . . . . . . . . . . . . . . . . . . . . . . 582
28.4.3.4 Taylor’s Series-Finite Difference Estimation
. . . . . . . . . . . . 583
28.4.4 Overall System Reliability
. . . . . . . . . . . . . . . . . . . . . . . . . . . 584
28.4.5 Target Reliability Factors
. . . . . . . . . . . . . . . . . . . . . . . . . . . 584
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 584
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CONTENTS
17
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 639
E 33-1 Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads
641
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 651
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 654
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656
33.4.6 Example of Transverse Deflection
. . . . . . . . . . . . . . . . . . . . . . . 657
33.4.7 Effect of Bracing Trusses
. . . . . . . . . . . . . . . . . . . . . . . . . . . 661
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List of Figures
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
From Vitruvius Ten Books on Architecture, (Vitruvius 1960)
. . . . . . . . . . . 34
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
Discourses Concerning Two New Sciences
. . . . . . . . . . . . . . . 41
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
1.12 Experimental Set Up Used by Hooke
. . . . . . . . . . . . . . . . . . . . . . . . . 43
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Philosophiae Naturalis Principia Mathematica
. . . . . . . . . . . . 44
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
1.17 Nervi’s Palazetto Dello Sport
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
Types of Forces in Structural Elements (1D)
. . . . . . . . . . . . . . . . . . . . . 58
Basic Aspects of Cable Systems
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
Variations in Post and Beams Configurations
. . . . . . . . . . . . . . . . . . . . 62
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
Examples of Air Supported Structures
. . . . . . . . . . . . . . . . . . . . . . . . 65
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
2.10 Sequence of Structural Engineering Courses
. . . . . . . . . . . . . . . . . . . . . 67
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
Examples of Static Determinate and Indeterminate Structures
. . . . . . . . . . . 72
Geometric Instability Caused by Concurrent Reactions
. . . . . . . . . . . . . . . 74
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
A Statically Indeterminate Truss
. . . . . . . . . . . . . . . . . . . . . . . . . . . 84
X and Y Components of Truss Forces
. . . . . . . . . . . . . . . . . . . . . . . . 85
Draft
LIST OF FIGURES
21
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
10.1 Statically Indeterminate 3 Cable Structure
. . . . . . . . . . . . . . . . . . . . . . 200
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
10.15Definition of Flexibility Terms for a Rigid Frame
. . . . . . . . . . . . . . . . . . 220
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
11.1 Uniformly Loaded Beam and Frame with Free or Fixed Beam Restraint
. . . . . 228
11.2 Uniformly Loaded Frame, Approximate Location of Inflection Points
. . . . . . . 229
11.3 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments
. 230
11.4 Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces
231
11.5 Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments
232
11.6 Horizontal Force Acting on a Frame, Approximate Location of Inflection Points
. 233
11.7 Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear
. . . 235
11.8 ***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment
235
11.9 Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force
236
11.10Example; Approximate Analysis of a Building
. . . . . . . . . . . . . . . . . . . . 237
11.11Free Body Diagram for the Approximate Analysis of a Frame Subjected to Vertical Loads
237
11.12Approximate Analysis of a Building; Moments Due to Vertical Loads
. . . . . . . 239
11.13Approximate Analysis of a Building; Shears Due to Vertical Loads
. . . . . . . . 241
11.14Approximate Analysis for Vertical Loads; Spread-Sheet Format
. . . . . . . . . . 242
11.15Approximate Analysis for Vertical Loads; Equations in Spread-Sheet
. . . . . . . 243
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LIST OF FIGURES
23
15.13Load Life of a Structure, (Lin and Stotesbury 1981)
. . . . . . . . . . . . . . . . 381
15.14Concept of Tributary Areas for Structural Member Loading
. . . . . . . . . . . . 382
15.15One or Two Way actions in Slabs
. . . . . . . . . . . . . . . . . . . . . . . . . . . 382
15.16Load Transfer in R/C Buildings
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 384
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386
16.1 Stress Strain Curves of Concrete and Steel
. . . . . . . . . . . . . . . . . . . . . . 388
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388
16.3 Residual Stresses in Rolled Sections
. . . . . . . . . . . . . . . . . . . . . . . . . 390
16.4 Residual Stresses in Welded Sections
. . . . . . . . . . . . . . . . . . . . . . . . . 390
16.5 Influence of Residual Stress on Average Stress-Strain Curve of a Rolled Section
. 391
16.6 Concrete Stress-Strain curve
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395
16.9 prefabricated Steel Joists
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407
17.1 Stress Concentration Around Circular Hole
. . . . . . . . . . . . . . . . . . . . . 410
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411
17.3 Effect of Staggered Holes on Net Area
. . . . . . . . . . . . . . . . . . . . . . . . 411
17.4 Gage Distances for an Angle
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417
17.6 Tearing Failure Limit State
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
18.2 Stability of a Rigid Bar with Initial Imperfection
. . . . . . . . . . . . . . . . . . 439
18.3 Stability of a Two Rigid Bars System
. . . . . . . . . . . . . . . . . . . . . . . . 439
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442
18.6 Simply Supported Beam Column; Differential Segment; Effect of Axial Force P
. 444
18.7 Solution of the Tanscendental Equation for the Buckling Load of a Fixed-Hinged Column
448
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 449
18.10Column Effective Length in a Frame
. . . . . . . . . . . . . . . . . . . . . . . . . 450
18.11Standard Alignment Chart (AISC)
. . . . . . . . . . . . . . . . . . . . . . . . . . 451
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452
18.13Euler Buckling, and SSRC Column Curve
. . . . . . . . . . . . . . . . . . . . . . 453
19.1 SSRC Column Curve and AISC Critical Stresses
. . . . . . . . . . . . . . . . . . 456
According to LRFD, for Various
. . . . . . . . . . . . . . . . 459
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 468
20.2 Stress distribution at different stages of loading
. . . . . . . . . . . . . . . . . . . 469
20.3 Stress-strain diagram for most structural steels
. . . . . . . . . . . . . . . . . . . 469
20.4 Flexural and Shear Stress Distribution in a Rectangular Beam
. . . . . . . . . . 470
472
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473
20.7 Nominal Moments for Compact and Partially Compact Sections
. . . . . . . . . . 474
20.8 AISC Requirements for Shear Design
. . . . . . . . . . . . . . . . . . . . . . . . . 479
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Draft
LIST OF FIGURES
25
589
29.4 Axial and Flexural Stresses
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 590
29.5 Design of a Statically Indeterminate Arch
. . . . . . . . . . . . . . . . . . . . . . 591
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 594
30.1 Eiffel Tower (Billington and Mark 1983)
. . . . . . . . . . . . . . . . . . . . . . . 603
30.2 Eiffel Tower Idealization, (Billington and Mark 1983)
. . . . . . . . . . . . . . . . 605
30.3 Eiffel Tower, Dead Load Idealization; (Billington and Mark 1983)
. . . . . . . . . 605
30.4 Eiffel Tower, Wind Load Idealization; (Billington and Mark 1983)
. . . . . . . . . 606
30.5 Eiffel Tower, Wind Loads, (Billington and Mark 1983)
. . . . . . . . . . . . . . . 607
30.6 Eiffel Tower, Reactions; (Billington and Mark 1983)
. . . . . . . . . . . . . . . . 607
30.7 Eiffel Tower, Internal Gravity Forces; (Billington and Mark 1983)
. . . . . . . . . 609
30.8 Eiffel Tower, Horizontal Reactions; (Billington and Mark 1983)
. . . . . . . . . . 609
30.9 Eiffel Tower, Internal Wind Forces; (Billington and Mark 1983)
. . . . . . . . . . 610
31.1 Cable Structure Subjected to
. . . . . . . . . . . . . . . . . . . . . . . . . . 612
31.2 Longitudinal and Plan Elevation of the George Washington Bridge
. . . . . . . . 614
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616
31.5 Location of Cable Reactions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617
31.6 Vertical Reactions in Columns Due to Central Span Load
. . . . . . . . . . . . . 617
31.7 Cable Reactions in Side Span
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 618
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 619
31.9 Deck Idealization, Shear and Moment Diagrams
. . . . . . . . . . . . . . . . . . . 620
32.1 Magazzini Generali; Overall Dimensions, (Billington and Mark 1983)
. . . . . . . 622
32.2 Magazzini Generali; Support System, (Billington and Mark 1983)
. . . . . . . . . 622
32.3 Magazzini Generali; Loads (Billington and Mark 1983)
. . . . . . . . . . . . . . . 623
32.4 Magazzini Generali; Beam Reactions, (Billington and Mark 1983)
. . . . . . . . . 623
32.5 Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983)
. . 624
32.6 Magazzini Generali; Internal Moment, (Billington and Mark 1983)
. . . . . . . . 624
625
32.8 Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark 1983)
625
32.9 Magazzini Generali; Effect of Lateral Supports, (Billington and Mark 1983)
. . . 626
33.1 Flexible, Rigid, and Semi-Flexible Joints
. . . . . . . . . . . . . . . . . . . . . . . 627
628
629
33.4 Axial and Flexural Stresses
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 630
33.5 Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981)
. . . . . . . . . . . 632
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634
33.7 Design Example of a Tubular Structure, (Lin and Stotesbury 1981)
. . . . . . . . 635
33.8 A Basic Portal Frame, (Lin and Stotesbury 1981)
. . . . . . . . . . . . . . . . . . 636
33.9 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments
. 637
33.10Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces
638
33.11Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments
638
33.12Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear
. . . 640
33.13***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment
640
33.14Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force
641
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Structural Engineering
Draft
List of Tables
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
Static Determinacy and Stability of Trusses
. . . . . . . . . . . . . . . . . . . . . 83
Conjugate Beam Boundary Conditions
. . . . . . . . . . . . . . . . . . . . . . . . 165
Possible Combinations of Real and Hypothetical Formulations
. . . . . . . . . . . 175
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180
Summary of Expressions for the Internal Strain Energy and External Work
. . . 198
Z
L
0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
10.3 Displacement Computations for a Rectangular Frame
. . . . . . . . . . . . . . . 219
11.1 Columns Combined Approximate Vertical and Horizontal Loads
. . . . . . . . . 250
11.2 Girders Combined Approximate Vertical and Horizontal Loads
. . . . . . . . . . 251
12.1 Stiffness vs Flexibility Methods
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
12.2 Degrees of Freedom of Different Structure Types Systems
. . . . . . . . . . . . . 255
13.1 Example of Nodal Definition
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
13.2 Example of Element Definition
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 288
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
13.4 Degrees of Freedom of Different Structure Types Systems
. . . . . . . . . . . . . 293
14.1 Allowable Stresses for Steel and Concrete
. . . . . . . . . . . . . . . . . . . . . . 351
values for Steel and Concrete Structures
. . . . . . . . . . . . . . . . . 355
14.3 Strength Reduction Factors, Φ
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 355
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360
15.2 Weights of Building Materials
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
15.3 Average Gross Dead Load in Buildings
. . . . . . . . . . . . . . . . . . . . . . . . 361
15.4 Minimum Uniformly Distributed Live Loads, (UBC 1995)
. . . . . . . . . . . . . 362
15.5 Wind Velocity Variation above Ground
. . . . . . . . . . . . . . . . . . . . . . . . 366
Coefficients for Wind Load, (UBC 1995)
. . . . . . . . . . . . . . . . . . . . . 367
15.7 Wind Pressure Coefficients
. . . . . . . . . . . . . . . . . . . . . 367
15.8 Importance Factors for Wind and Earthquake Load, (UBC 1995)
. . . . . . . . . 368
15.9 Approximate Design Wind Pressure
for Ordinary Wind Force Resisting Building Structures
368
Factors for Different Seismic Zones, ubc
. . . . . . . . . . . . . . . . . . . . . . 372
Draft
Part I
ANALYSIS
Draft
Chapter 1
A BRIEF HISTORY OF
STRUCTURAL ARCHITECTURE
If I have been able to see a little farther than some others,
it was because I stood on the shoulders of giants
.
Sir Isaac Newton
1
More than any other engineering discipline, Architecture/Mechanics/Structures is the proud
outcome of a of a long and distinguished history. Our profession, second oldest, would be better
appreciated if we were to develop a sense of our evolution.
1.1
Before the Greeks
2
Throughout antiquity, structural engineering existing as an art rather than a science. No
record exists of any rational consideration, either as to the strength of structural members or
as to the behavior of structural materials. The builders were guided by rules of thumbs and
experience, which were passed from generation to generation, guarded by secrets of the guild,
and seldom supplemented by new knowledge. Despite this, structures erected before Galileo are
by modern standards quite phenomenal (pyramids, Via Appia, aqueducs, Colisseums, Gothic
cathedrals to name a few).
3
The first structural engineer in history seems to have been
Imhotep
, one of only two com-
moners to be deified. He was the builder of the step pyramid of Sakkara about 3,000 B.C., and
yielded great influence over ancient Egypt.
4
Hamurrabi’s code in Babylonia (1750 BC) included among its 282 laws penalties for those
“architects†whose houses collapsed, Fig.
1.2
Greeks
5
The greek philosopher
Pythagoras
(born around 582 B.C.) founded his famous school, which
was primarily a secret religious society, at Crotona in southern Italy. At his school he allowed
Draft
1.3 Romans
33
Figure 1.2: Archimed
conqueror of Syracuse.
1.3
Romans
10
Science made much less progress under the Romans than under the Greeks. The Romans
apparently were more practical, and were not as interested in abstract thinking though they
were excellent fighters and builders.
11
As the roman empire expanded, the Romans built great roads (some of them still in use)
such as the Via Appia, Cassia, Aurelia; Also they built great bridges (such as the third of a
mile bridge over the Rhine built by Caesars), and stadium (Colliseum).
12
One of the most notable Roman construction was the
Pantheon
, Fig.
. It is the best-
Figure 1.3: Pantheon
preserved major edifice of ancient Rome and one of the most significant buildings in architectural
history. In shape it is an immense cylinder concealing eight piers, topped with a dome and
fronted by a rectangular colonnaded porch. The great vaulted dome is 43 m (142 ft) in diameter,
and the entire structure is lighted through one aperture, called an
oculus
, in the center of the
dome. The Pantheon was erected by the Roman emperor Hadrian between AD 118 and 128.
Victor Saouma
Structural Engineering
Draft
Chapter 2
INTRODUCTION
2.1
Structural Engineering
1
Structural engineers are responsible for the detailed analysis and design of:
Architectural structures:
Buildings, houses, factories. They must work in close cooperation
with an architect who will ultimately be responsible for the design.
Civil Infrastructures:
Bridges, dams, pipelines, offshore structures. They work with trans-
portation, hydraulic, nuclear and other engineers. For those structures they play the
leading role.
Aerospace, Mechanical, Naval structures:
aeroplanes, spacecrafts, cars, ships, submarines
to ensure the structural safety of those important structures.
2.2
Structures and their Surroundings
2
Structural design is affected by various environmental constraints:
1. Major movements: For example, elevator shafts are usually shear walls good at resisting
lateral load (wind, earthquake).
2. Sound and structure interact:
•
A
dome
roof will concentrate the sound
•
A
dish
roof will diffuse the sound
3. Natural light:
•
A flat roof in a building may not provide adequate light.
•
A Folded plate will provide adequate lighting (analysis more complex).
•
A bearing and shear wall building may not have enough openings for daylight.
•
A Frame design will allow more light in (analysis more complex).
4. Conduits for cables (electric, telephone, computer), HVAC ducts, may dictate type of
floor system.
5. Net clearance between columns (unobstructed surface) will dictate type of framing.
Draft
2.6 Structural Analysis
57
2.6
Structural Analysis
12
Given an
existing
structure subjected to a certain load determine internal forces (axial,
shear, flexural, torsional; or stresses), deflections, and verify that no unstable failure can occur.
13
Thus the basic structural requirements are:
Strength:
stresses should not exceed critical values:
σ < σ
f
Stiffness:
deflections should be controlled: ∆
<
∆
max
Stability:
buckling or cracking should also be prevented
2.7
Structural Design
14
Given a set of forces,
dimension
the structural element.
Steel/wood Structures
Select appropriate section.
Reinforced Concrete:
Determine dimensions of the element and internal reinforcement (num-
ber and sizes of reinforcing bars).
15
For
new structures
,
iterative
process between analysis and design. A preliminary design is
made using
rules of thumbs
(best known to Engineers with design experience) and analyzed.
Following design, we check for
Serviceability:
deflections, crack widths under the applied load. Compare with acceptable
values specified in the design code.
Failure:
and compare the failure load with the applied load times the appropriate factors of
safety.
If the design is found not to be acceptable, then it must be modified and reanalyzed.
16
For
existing structures rehabilitation
, or verification of an old infrastructure, analysis is
the most important component.
17
In summary, analysis is always required.
2.8
Load Transfer Elements
18
From Strength of Materials, Fig.
Axial:
cables, truss elements, arches, membrane, shells
Flexural:
Beams, frames, grids, plates
Torsional:
Grids, 3D frames
Shear:
Frames, grids, shear walls.
Victor Saouma
Structural Engineering
Draft
Chapter 3
EQUILIBRIUM & REACTIONS
To every action there is an equal and opposite reaction.
Newton’s third law of motion
3.1
Introduction
1
In the analysis of structures (hand calculations), it is often easier (but not always necessary)
to start by determining the reactions.
2
Once the reactions are determined, internal forces are determined next; finally, deformations
(deflections and rotations) are determined last
.
3
Reactions are necessary to determine
foundation load
.
4
Depending on the type of structures, there can be different types of support conditions, Fig.
Roller:
provides a restraint in only one direction in a 2D structure, in 3D structures a roller
may provide restraint in one or two directions. A roller will allow rotation.
Hinge:
allows rotation but no displacements.
Fixed Support:
will prevent rotation and displacements in all directions.
3.2
Equilibrium
5
Reactions are determined from the appropriate equations of static equilibrium.
6
Summation of forces and moments,
in a static system
must be equal to zero
.
1
This is the sequence of operations in the
flexibility
method which lends itself to hand calculation. In the
stiffness
method, we determine displacements firsts, then internal forces and reactions. This method is most
suitable to computer implementation.
2
In a dynamic system Σ
F
=
ma
where
m
is the mass and
a
is the acceleration.
Draft
3.3 Equations of Conditions
71
Structure Type
Equations
Beam, no axial forces
Σ
F
y
Σ
M
z
2D Truss, Frame, Beam
Σ
F
x
Σ
F
y
Σ
M
z
Grid
Σ
F
z
Σ
M
x
Σ
M
y
3D Truss, Frame
Σ
F
x
Σ
F
y
Σ
F
z
Σ
M
x
Σ
M
y
Σ
M
z
Alternate Set
Beams, no axial Force
Σ
M
A
z
Σ
M
B
z
2 D Truss, Frame, Beam
Σ
F
x
Σ
M
A
z
Σ
M
B
z
Σ
M
A
z
Σ
M
B
z
Σ
M
C
z
Table 3.1: Equations of Equilibrium
3. The right hand side of the equation should be zero
If your reaction is negative, then it will be in a direction opposite from the one assumed.
16
Summation of all external forces (including reactions) is not necessarily zero (except at hinges
and at points outside the structure).
17
Summation of external forces is equal and
opposite
to the internal ones. Thus the net
force/moment is equal to zero.
18
The external forces give rise to the (non-zero) shear and moment diagram.
3.3
Equations of Conditions
19
If a structure has an
internal hinge
(which may connect two or more substructures), then
this will provide an additional equation (Σ
M
= 0 at the hinge) which can be exploited to
determine the reactions.
20
Those equations are often exploited in trusses (where each connection is a hinge) to determine
reactions.
21
In an
inclined roller
support with
S
x
and
S
y
horizontal and vertical projection, then the
reaction R would have, Fig.
R
x
R
y
=
S
y
S
x
(3.3)
3.4
Static Determinacy
22
In statically determinate structures, reactions depend only on the geometry, boundary con-
ditions and loads.
23
If the reactions can not be determined simply from the equations of static equilibrium (and
equations of conditions if present), then the reactions of the structure are said to be
statically
indeterminate
.
Victor Saouma
Structural Engineering
Draft
3.4 Static Determinacy
73
A rigid plate is supported by two aluminum cables and a steel one. Determine the force in
each cable
.
If the rigid plate supports a load P, determine the stress in each of the three cables.
Solution:
1. We have three unknowns and only two independent equations of equilibrium. Hence the
problem is statically indeterminate to the first degree.
Σ
M
z
= 0;
⇒
P
left
Al
=
P
right
Al
Σ
F
y
= 0;
⇒
2
P
Al
+
P
St
=
P
Thus we effectively have two unknowns and one equation.
2. We need to have a third equation to solve for the three unknowns. This will be de-
rived from the
compatibility of the displacements
in all three cables, i.e. all three
displacements must be equal:
σ
=
P
A
ε
=
∆
L
L
ε
=
σ
E





⇒
∆
L
=
P L
AE
P
Al
L
E
Al
A
Al
|
{z
}
∆
Al
=
P
St
L
E
St
A
St
|
{z
}
∆
St
⇒
P
Al
P
St
=
(
EA
)
Al
(
EA
)
St
or
−
(
EA
)
St
P
Al
+ (
EA
)
Al
P
St
= 0
3. Solution of this system of two equations with two unknowns yield:
"
2
1
−
(
EA
)
St
(
EA
)
Al
# (
P
Al
P
St
)
=
(
P
0
)
⇒
(
P
Al
P
St
)
=
"
2
1
−
(
EA
)
St
(
EA
)
Al
#
−
1
(
P
0
)
=
1
2(
EA
)
Al
+ (
EA
)
St
|
{z
}
Determinant
"
(
EA
)
Al
−
1
(
EA
)
St
2
# (
P
0
)
5
This example problem will be the only statically indeterminate problem analyzed in CVEN3525.
Victor Saouma
Structural Engineering
Draft
Chapter 4
TRUSSES
4.1
Introduction
4.1.1
Assumptions
1
Cables and trusses are 2D or 3D structures composed of an assemblage of simple one dimen-
sional components which transfer only
axial
forces along their axis.
2
Cables can carry only tensile forces, trusses can carry tensile and compressive forces.
3
Cables tend to be
flexible
, and hence, they tend to oscillate and therefore must be stiffened.
4
Trusses are extensively used for bridges, long span roofs, electric tower, space structures.
5
For trusses, it is assumed that
1. Bars are
pin-connected
2. Joints are frictionless hinges
3. Loads are applied at the
joints only
.
6
A truss would typically be composed of triangular elements with the bars on the
upper
chord
under compression and those along the
lower chord
under tension. Depending on the
orientation of the diagonals
, they can be under either tension or compression.
7
Fig.
illustrates some of the most common types of trusses.
8
It can be easily determined that in a Pratt truss, the diagonal members are under tension,
while in a Howe truss, they are in compression. Thus, the Pratt design is an excellent choice
for steel whose members are slender and long diagonal member being in tension are not prone
to buckling. The vertical members are less likely to buckle because they are shorter. On the
other hand the Howe truss is often preferred for for heavy timber trusses.
9
In a truss analysis or design, we seek to determine the internal force along each member, Fig.
1
In practice the bars are riveted, bolted, or welded directly to each other or to gusset plates, thus the bars
are not free to rotate and so-called
secondary bending moments
are developed at the bars. Another source
of secondary moments is the dead weight of the element.
Draft
4.2 Trusses
83
4.1.2
Basic Relations
Sign Convention:
Tension positive, compression negative. On a truss the axial forces are
indicated as forces acting on the joints.
Stress-Force:
σ
=
P
A
Stress-Strain:
σ
=
Eε
Force-Displacement:
ε
=
∆
L
L
Equilibrium:
Σ
F
= 0
4.2
Trusses
4.2.1
Determinacy and Stability
10
Trusses are
statically determinate
when all the bar forces can be determined from the
equations of
statics
alone. Otherwise the truss is
statically indeterminate
.
11
A truss may be statically/externally determinate or indeterminate with respect to the reac-
tions (more than 3 or 6 reactions in 2D or 3D problems respectively).
12
A truss may be internally determinate or indeterminate, Table
13
If we refer to
j
as the number of joints,
R
the number of reactions and
m
the number of
members, then we would have a total of
m
+
R
unknowns and 2
j
(or 3
j
) equations of statics
(2D or 3D at each joint). If we do not have enough equations of statics then the problem is
indeterminate, if we have too many equations then the truss is unstable, Table
2D
3D
Static Indeterminacy
External
R >
3
R >
6
Internal
m
+
R >
2
j
m
+
R >
3
j
Unstable
m
+
R <
2
j
m
+
R <
3
j
Table 4.1: Static Determinacy and Stability of Trusses
14
If
m <
2
j
−
3 (in 2D) the truss is not internally stable, and it will not remain a rigid body
when it is detached from its supports. However, when attached to the supports, the truss will
be rigid.
15
Since each joint is pin-connected, we can apply Σ
M
= 0 at each one of them. Furthermore,
summation of forces applied on a joint must be equal to zero.
16
For 2D trusses the external equations of equilibrium which can be used to determine the
reactions are Σ
F
X
= 0, Σ
F
Y
= 0 and Σ
M
Z
= 0. For 3D trusses the available equations are
Σ
F
X
= 0, Σ
F
Y
= 0, Σ
F
Z
= 0 and Σ
M
X
= 0, Σ
M
Y
= 0, Σ
M
Z
= 0.
17
For a 2D truss we have 2 equations of equilibrium Σ
F
X
= 0 and Σ
F
Y
= 0 which can be
applied at each joint. For 3D trusses we would have three equations: Σ
F
X
= 0, Σ
F
Y
= 0 and
Σ
F
Z
= 0.
Victor Saouma
Structural Engineering
Draft
4.2 Trusses
85
Figure 4.4: X and Y Components of Truss Forces
23
In truss analysis, there is
no sign convention
. A member is
assumed
to be under tension
(or compression). If after analysis, the force is found to be negative, then this would imply that
the wrong assumption was made, and that the member should have been under compression
(or tension).
24
On a
free body diagram
, the internal forces are represented by arrow acting
on the joints
and not as end forces on the element itself. That is for tension, the arrow is pointing away from
the joint, and for compression toward the joint, Fig.
Figure 4.5: Sign Convention for Truss Element Forces
Example 4-1: Truss, Method of Joints
Using the method of joints, analyze the following truss
Victor Saouma
Structural Engineering
Draft
4.2 Trusses
87
Node B:
(+
-
) Σ
F
x
= 0;
⇒
F
BC
=
43
.
5
k
Tension
(+
6
) Σ
F
y
= 0;
⇒
F
BH
=
20
k
Tension
Node H:
(+
-
) Σ
F
x
= 0;
⇒
F
AH
x
−
F
HC
x
−
F
HG
x
= 0
43
.
5
−
24
√
24
2
+32
2
(
F
HC
)
−
24
√
24
2
+10
2
(
F
HG
) = 0
(+
6
) Σ
F
y
= 0;
⇒
F
AH
y
+
F
HC
y
−
12
−
F
HG
y
−
20 = 0
58 +
32
√
24
2
+32
2
(
F
HC
)
−
12
−
10
√
24
2
+10
2
(
F
HG
)
−
20 = 0
This can be most conveniently written as
"
0
.
6
0
.
921
−
0
.
8 0
.
385
# (
F
HC
F
HG
)
=
(
−
7
.
5
52
)
(4.2)
Solving we obtain
F
HC
=
−
7
.
5
k
Tension
F
HG
=
52
k
Compression
Node E:
Σ
F
y
= 0;
⇒
F
EF
y
= 62
⇒
F
EF
=
√
24
2
+32
2
32
(62)
= 77
.
5
k
Σ
F
x
= 0;
⇒
F
ED
=
F
EF
x
⇒
F
ED
=
24
32
(
F
EF
y
) =
24
32
(62) = 46
.
5
k
Victor Saouma
Structural Engineering
Draft
Chapter 5
CABLES
5.1
Funicular Polygons
1
A cable is a slender flexible member with zero or negligible flexural stiffness, thus it can only
transmit
tensile
2
The tensile force at any point acts in the direction of the tangent to the cable (as any other
component will cause bending).
3
Its strength stems from its ability to undergo extensive changes in slope at the point of load
application.
4
Cables resist vertical forces by undergoing
sag
(
h
) and thus developing tensile forces. The
horizontal component of this force (
H
) is called
thrust
.
5
The distance between the cable supports is called the
chord
.
6
The sag to span ratio is denoted by
r
=
h
l
(5.1)
7
When a set of concentrated loads is applied to a cable of negligible weight, then the cable
deflects into a series of linear segments and the resulting shape is called the
funicular polygon
.
8
If a cable supports vertical forces only, then the horizontal component
H
of the cable tension
T
remains constant.
Example 5-1: Funicular Cable Structure
Determine the reactions and the tensions for the cable structure shown below.
1
Due to the zero flexural rigidity it will buckle under axial compressive forces.
Draft
5.2 Uniform Load
99
=
H
cos
θ
B
=
51
0
.
999
= 51
.
03
k
(5.6-d)
T
CD
;
tan
θ
C
=
4
.
6
30
= 0
.
153
⇒
θ
C
= 8
.
7 deg
(5.6-e)
=
H
cos
θ
C
=
51
0
.
988
= 51
.
62
k
(5.6-f)
5.2
Uniform Load
5.2.1
qdx
; Parabola
9
Whereas the forces in a cable can be determined from statics alone, its configuration must
be derived from its deformation. Let us consider a cable with distributed load
p
(
x
)
per unit
horizontal projection
of the cable length
. An infinitesimal portion of that cable can be
assumed to be a straight line, Fig.
and in the absence of any horizontal load we have
H
T+dT
T
H
V
V+dV
H
θ
θ
q(x)
dx
dy
ds
ds
L
x
V
q(x)
y
y(x)
dx
h
y’
x’
x
y
L/2
Figure 5.1: Cable Structure Subjected to
q
(
x
)
H
=constant. Summation of the vertical forces yields
(+
?
) Σ
F
y
= 0
⇒ −
V
+
qdx
+ (
V
+
dV
) = 0
(5.7-a)
dV
+
qdx
= 0
(5.7-b)
where
V
is the vertical component of the cable tension at
. Because the cable must be
tangent to
T
, we have
tan
θ
=
V
H
(5.8)
2
Thus neglecting the weight of the cable
3
Note that if the cable was subjected to its own weight then we would have
qds
instead of
pdx
.
Victor Saouma
Structural Engineering
Draft
5.2 Uniform Load
101
15
Combining Eq.
and
we obtain
y
=
4
hx
L
2
(
L
−
x
)
(5.18)
16
If we shift the origin to midspan, and reverse
y
, then
y
=
4
h
L
2
x
2
(5.19)
Thus the cable assumes a parabolic shape (as the moment diagram of the applied load).
17
The maximum tension occurs at the support where the vertical component is equal to
V
=
qL
2
and the horizontal one to
H
, thus
T
max
=
p
V
2
+
H
2
=
s
qL
2
2
+
H
2
=
H
s
1 +
qL/
2
H
2
(5.20)
Combining this with Eq.
we obtain
.
T
max
=
H
p
1 + 16
r
2
≈
H
(1 + 8
r
2
)
(5.21)
5.2.2
â€
qds
; Catenary
18
Let us consider now the case where the cable is subjected to its own weight (plus ice and
wind if any). We would have to replace
qdx
by
qds
in Eq.
dV
+
qds
= 0
(5.22)
The differential equation for this new case will be derived exactly as before, but we substitute
qdx
by
qds
, thus Eq.
becomes
d
2
y
dx
2
=
−
q
H
ds
dx
(5.23)
19
But
ds
2
=
dx
2
+
dy
2
, hence:
d
2
y
dx
2
=
−
q
H
s
1 +
dy
dx
2
(5.24)
solution of this differential equation is considerably more complicated than for a parabola.
20
We let
dy/dx
=
p
, then
dp
dx
=
−
q
H
q
1 +
p
2
(5.25)
5
Recalling that (
a
+
b
)
n
=
a
n
+
na
n
−
1
b
+
n
(
n
−
1)
2!
a
n
−
2
b
2
+
·
or (1 +
b
)
n
= 1 +
nb
+
n
(
n
−
1)
b
2
2!
+
n
(
n
−
1)(
n
−
2)
b
3
3!
+
· · ·
;
Thus for
b
2
<<
1,
√
1 +
b
= (1 +
b
)
1
2
≈
1 +
b
2
Victor Saouma
Structural Engineering
Draft
Chapter 6
INTERNAL FORCES IN
STRUCTURES
1
This chapter will start as a review of shear and moment diagrams which you have studied
in both
Statics
and
Strength of Materials
, and will proceed with the analysis of statically
determinate frames, arches and grids.
2
By the end of this lecture, you should be able to draw the shear, moment and torsion (when
applicable) diagrams for each member of a structure.
3
Those diagrams will subsequently be used for member design. For instance, for flexural
design, we will consider the section subjected to the highest moment, and make sure that the
internal moment is equal and opposite to the external one. For the ASD method, the basic
beam equation (derived in Strength of Materials)
σ
=
M C
I
, (where
M
would be the design
moment obtained from the moment diagram) would have to be satisfied.
4
Some of the examples first analyzed in chapter
(Reactions), will be revisited here. Later
on, we will determine the deflections of those same problems.
6.1
Design Sign Conventions
5
Before we (re)derive the Shear-Moment relations, let us
arbitrarily
define a sign convention.
6
The sign convention adopted here, is the one commonly used for design purposes
7
With reference to Fig.
2D:
Load
Positive along the beam’s local y axis (assuming a right hand side convention),
that is positive upward.
Axial:
tension positive.
Flexure
A positive moment is one which causes tension in the lower fibers, and com-
pression in the upper ones. Alternatively, moments are drawn on the compression
side (useful to keep in mind for frames).
1
Later on, in more advanced analysis courses we will use a different one.
Draft
6.2 Load, Shear, Moment Relations
115
6.2
Load, Shear, Moment Relations
8
Let us (re)derive the basic relations between load, shear and moment. Considering an in-
finitesimal length
dx
of a beam subjected to a positive load
w
(
x
), Fig.
. The infinitesimal
Figure 6.3: Free Body Diagram of an Infinitesimal Beam Segment
section must also be in equilibrium.
9
There are no axial forces, thus we only have two equations of equilibrium to satisfy Σ
F
y
= 0
and Σ
M
z
= 0.
10
Since
dx
is infinitesimally small, the small variation in load along it can be neglected, therefore
we assume
w
(
x
) to be constant along
dx
.
11
To denote that a small change in shear and moment occurs over the length
dx
of the element,
we add the differential quantities
dV
x
and
dM
x
to
V
x
and
M
x
on the right face.
12
Next considering the first equation of equilibrium
(+
6
) Σ
F
y
= 0
⇒
V
x
+
w
x
dx
−
(
V
x
+
dV
x
) = 0
or
dV
dx
=
w
(
x
)
(6.1)
The slope of the shear curve at any point along the axis of a member is given by
the load curve at that point.
13
Similarly
(+
) Σ
M
O
= 0
⇒
M
x
+
V
x
dx
−
w
x
dx
dx
2
−
(
M
x
+
dM
x
) = 0
Neglecting the
dx
2
term, this simplifies to
dM
dx
=
V
(
x
)
(6.2)
The slope of the moment curve at any point along the axis of a member is given by
the shear at that point.
2
In this derivation, as in all other ones we should assume all quantities to be positive.
Victor Saouma
Structural Engineering
Draft
6.4 Examples
117
Shear
Moment
Load
Shear
Positive Constant
Negative Constant
Positive Constant
Negative Constant
Negative Increasing Negative Decreasing
Negative Increasing Negative Decreasing
Positive Increasing Positive Decreasing
Positive Increasing Positive Decreasing
Figure 6.5: Slope Relations Between Load Intensity and Shear, or Between Shear and Moment
6.4
Examples
6.4.1
Beams
Example 6-1: Simple Shear and Moment Diagram
Draw the shear and moment diagram for the beam shown below
Solution:
The free body diagram is drawn below
Victor Saouma
Structural Engineering
Draft
6.4 Examples
119
Example 6-2: Sketches of Shear and Moment Diagrams
For each of the following examples, sketch the shear and moment diagrams.
Victor Saouma
Structural Engineering
Draft
6.4 Examples
121
Solution:
Victor Saouma
Structural Engineering
Draft
6.4 Examples
129
B-C
Σ
F
x
0
= 0
⇒
N
B
x
0
=
V
C
z
0
=
−
60kN
Σ
F
y
0
= 0
⇒
V
B
y
0
=
V
C
y
0
= +40kN
Σ
M
y
0
= 0
⇒
M
B
y
0
=
M
C
y
0
=
−
120kN.m
Σ
M
z
0
= 0
⇒
M
B
z
0
=
V
0
y
C
(4) = (40)(4) = +160kN.m
Σ
T
x
0
= 0
⇒
T
B
x
0
=
−
M
C
z
0
=
−
40kN.m
A-B
Σ
F
x
0
= 0
⇒
N
A
x
0
=
V
B
y
0
= +40kN
Σ
F
y
0
= 0
⇒
V
A
y
0
=
N
B
x
0
= +60kN
Σ
M
y
0
= 0
⇒
M
A
y
0
=
T
B
x
0
= +40kN.m
Σ
M
z
0
= 0
⇒
M
A
z
0
=
M
B
z
0
+
N
B
x
0
(4) = 160 + (60)(4) = +400kN.m
Σ
T
x
0
= 0
⇒
T
A
x
0
=
M
B
y
0
=
−
120kN.m
The interaction between axial forces
N
and shear
V
as well as between moments
M
and
torsion
T
is clearly highlighted by this example.
120 kN-m
20 kN/m
120 kN-m
40 kN
40 kN
120 kN-m
40 kN
40 kN
120 kN-m
40 kN
40 kN
C
B
120 kN-m
120 kN-m
40 kN
120 kN-m
40 kN
x’
y’
z’
x’
y’
120 kN-m
120 kN-m
x’
y’
60 kN
40 kN-m
40 kN-m
60 kN
160 kN-m
40 kN-m
60 kN
60 kN
60 kN
40 kN-m
60 kN
40 kN-m
40 kN-m
60 kN
40 kN-m
60 kN
160 kN-m
160 kN-m
40 kN-m
60 kN
60 kN
160 kN-m
400 kN-m
40 kN-m
z’
y’
Victor Saouma
Structural Engineering
Draft
Chapter 7
ARCHES and CURVED
STRUCTURES
1
This chapter will concentrate on the analysis of arches.
2
The concepts used are identical to the ones previously seen, however the major (and only)
difference is that equations will be written in polar coordinates.
3
Like cables, arches can be used to reduce the bending moment in long span structures. Es-
sentially, an arch can be considered as an inverted cable, and is transmits the load primarily
through axial compression, but can also resist flexure through its flexural rigidity.
4
A parabolic arch uniformly loaded will be loaded in compression only.
5
A semi-circular arch unifirmly loaded will have some flexural stresses in addition to the
compressive ones.
7.1
Arches
6
In order to optimize dead-load efficiency, long span structures should have their shapes ap-
proximate the coresponding moment diagram, hence an arch, suspended cable, or tendon con-
figuration in a prestressed concrete beam all are nearly parabolic, Fig.
7
Long span structures can be built using flat construction such as girders or trusses. However,
for spans in excess of 100 ft, it is often more economical to build a curved structure such as an
arch, suspended cable or thin shells.
8
Since the dawn of history, mankind has tried to span distances using arch construction.
Essentially this was because an arch required materials to resist compression only (such as
stone, masonary, bricks), and labour was not an issue.
9
The basic issues of static in arch design are illustrated in Fig.
where the vertical load is per
unit horizontal projection (such as an external load but not a self-weight). Due to symmetry,
the vertical reaction is simply
V
=
wL
2
, and there is no shear across the midspan of the arch
(nor a moment). Taking moment about the crown,
M
=
Hh
−
wL
2
L
2
−
L
4
= 0
(7.1)
Draft
7.1 Arches
133
Solving for
H
H
=
wL
2
8
h
(7.2)
We recall that a similar equation was derived for arches., and
H
is analogous to the
C
−
T
forces in a beam, and
h
is the overall height of the arch, Since
h
is much larger than
d
,
H
will
be much smaller than
C
−
T
in a beam.
10
Since equilibrium requires
H
to remain constant across thee arch, a parabolic curve would
theoretically result in no moment on the arch section.
11
Three-hinged arches are statically determinate structures which shape can acomodate sup-
port settlements and thermal expansion without secondary internal stresses. They are also easy
to analyse through statics.
12
An arch carries the vertical load across the span through a combination of axial forces and
flexural ones. A well dimensioned arch will have a small to negligible moment, and relatively
high normal compressive stresses.
13
An arch is far more efficient than a beam, and possibly more economical and aesthetic than
a truss in carrying loads over long spans.
14
If the arch has only two hinges, Fig.
, or if it has no hinges, then bending moments may
exist either at the crown or at the supports or at both places.
APPARENT LINE
OF PRESSURE WITH
ARCH BENDING
EXCEPT AT THE BASE
h
h’
V
V
M
w
h
H’=wl /8h’<
2
wl /8h
2
H’
H’<H
APPARENT LINE OF
PRESSURE WITH
ARCH BENDING
INCLUDING BASE
V
V
H’<H
M
w
h
L
crown
M
M
base
base
h’
H’<H
Figure 7.3: Two Hinged Arch, (Lin and Stotesbury 1981)
15
Since
H
varies inversely to the rise
h
, it is obvious that one should use as high a rise as
possible. For a combination of aesthetic and practical considerations, a span/rise ratio ranging
from 5 to 8 or perhaps as much as 12, is frequently used. However, as the ratio goes higher, we
may have buckling problems, and the section would then have a higher section depth, and the
arch advantage diminishes.
16
In a parabolic arch subjected to a uniform horizontal load there is no moment. However, in
practice an arch is not subjected to uniform horizontal load. First, the depth (and thus the
weight) of an arch is not usually constant, then due to the inclination of the arch the actual
self weight is not constant. Finally, live loads may act on portion of the arch, thus the line of
action will not necessarily follow the arch centroid. This last effect can be neglected if the live
load is small in comparison with the dead load.
Victor Saouma
Structural Engineering
Draft
7.1 Arches
135
Solving those four equations simultaneously we have:





140 26
.
25 0 0
0
1
0 1
1
0
1 0
80
60
0 0














R
Ay
R
Ax
R
Cy
R
Cx









=









2
,
900
80
50
3
,
000









⇒









R
Ay
R
Ax
R
Cy
R
Cx









=









15
.
1
k
29
.
8
k
34
.
9
k
50
.
2
k









(7.4)
We can check our results by considering the summation with respect to b from the right:
(+
) Σ
M
B
z
= 0;
−
(20)(20)
−
(50
.
2)(33
.
75) + (34
.
9)(60) = 0
√
(7.5)
Example 7-2: Semi-Circular Arch, (Gerstle 1974)
Determine the reactions of the three hinged statically determined semi-circular arch under
its own dead weight
w
(per unit arc length
s
, where
ds
=
rdθ
).
R cos
θ
R
A
B
C
R
A
B
θ
dP=wRd
θ
θ
r
θ
Figure 7.6: Semi-Circular three hinged arch
Solution:
I Reactions
The reactions can be determined by
integrating
the load over the entire struc-
ture
1. Vertical Reaction
is determined first:
(+
) Σ
M
A
= 0;
−
(
C
y
)(2
R
) +
Z
θ
=
Ï€
θ
=0
wRdθ
| {z }
dP
R
(1 + cos
θ
)
|
{z
}
moment arm
= 0
(7.6-a)
⇒
C
y
=
wR
2
Z
θ
=
Ï€
θ
=0
(1 + cos
θ
)
dθ
=
wR
2
[
θ
−
sin
θ
]
|
θ
=
Ï€
θ
=0
=
wR
2
[(
Ï€
−
sin
Ï€
)
−
(0
−
sin 0)]
=
Ï€
2
wR
(7.6-b)
2. Horizontal Reactions
are determined next
(+
) Σ
M
B
= 0;
−
(
C
x
)(
R
) + (
C
y
)(
R
)
−
Z
θ
=
Ï€
2
θ
=0
wRdθ
| {z }
dP
R
cos
θ
| {z }
moment arm
= 0
(7.7-a)
Victor Saouma
Structural Engineering
Draft
Chapter 8
DEFLECTION of STRUCTRES;
Geometric Methods
1
Deflections of structures must be determined in order to satisfy serviceability requirements
i.e. limit deflections under
service
loads to acceptable values (such as
∆
L
≤
360).
2
Later on, we will see that deflection calculations play an important role in the analysis of
statically indeterminate structures.
3
We shall focus on flexural deformation, however the end of this chapter will review axial and
torsional deformations as well.
4
Most of this chapter will be a
review
of subjects covered in
Strength of Materials
.
5
This chapter will examine deflections of structures based on geometric considerations. Later
on, we will present a more pwerful method based on energy considerations.
8.1
Flexural Deformation
8.1.1
Curvature Equation
6
Let us consider a segment (between point 1 and point 2), Fig.
of a beam subjected to
flexural loading.
7
The
slope
is denoted by
θ
, the change in slope per unit length is the
curvature
κ
, the
radius
of curvature
is
Ï
.
8
From
Strength of Materials
we have the following relations
ds
=
Ïdθ
⇒
dθ
ds
=
1
Ï
(8.1)
9
We also note by extension that ∆
s
=
Ï
∆
θ
10
As a first order approximation, and with
ds
≈
dx
and
dy
dx
=
θ
Eq.
becomes
κ
=
1
Ï
=
dθ
dx
=
d
2
y
dx
2
(8.2)
Draft
8.1 Flexural Deformation
151
14
Thus the slope
θ
, curvature
κ
, radius of curvature
Ï
are related to the
y
displacement at a
point
x
along a flexural member by
κ
=
d
2
y
dx
2
1 +
dy
dx
2
3
2
(8.9)
15
If the displacements are very small, we will have
dy
dx
<<
1, thus Eq.
reduces to
κ
=
d
2
y
dx
2
=
1
Ï
(8.10)
8.1.2
Differential Equation of the Elastic Curve
16
Again with reference to Figure
a positive
dθ
at a positive
y
(upper fibers) will cause a
shortening
of the upper fibers
∆
u
=
−
y
∆
θ
(8.11)
17
This equation can be rewritten as
lim
∆
s
→
0
∆
u
∆
s
=
−
y
lim
∆
s
→
0
∆
θ
∆
s
(8.12)
and since ∆
s
≈
∆
x
du
dx
|{z}
ε
=
−
y
dθ
dx
(8.13)
Combining this with Eq.
1
Ï
=
κ
=
−
ε
y
(8.14)
This is the fundamental relationship between curvature (
κ
), elastic curve (
y
), and linear strain
(
ε
).
18
Note that so far we made no assumptions about material properties, i.e. it can be elastic or
inelastic.
19
For the elastic case:
ε
x
=
σ
E
σ
=
−
M y
I
)
ε
=
−
M y
EI
(8.15)
Combining this last equation with Eq.
yields
1
Ï
=
dθ
dx
=
d
2
y
dx
2
=
M
EI
(8.16)
This fundamental equation relates moment to curvature.
Victor Saouma
Structural Engineering
Draft
8.2 Flexural Deformations
153
Solution:
At:
0
≤
x
≤
2
L
3
1. Moment Equation
EI
d
2
y
dx
2
=
M
x
=
wL
3
x
−
5
18
wL
2
(8.22)
2. Integrate once
EI
dy
dx
=
wL
6
x
2
−
5
18
wL
2
x
+
C
1
(8.23)
However we have at
x
= 0,
dy
dx
= 0,
⇒
C
1
= 0
3. Integrate twice
EIy
=
wL
18
x
3
−
5
wL
2
36
x
2
+
C
2
(8.24)
Again we have at
x
= 0,
y
= 0,
⇒
C
2
= 0
At:
2
L
3
≤
x
≤
L
1. Moment equation
EI
d
2
y
dx
2
=
M
x
=
wL
3
x
−
5
18
wL
2
−
w
(
x
−
2
L
3
)(
x
−
2
L
3
2
)
(8.25)
2. Integrate once
EI
dy
dx
=
wL
6
x
2
−
5
18
wL
2
x
−
w
6
(
x
−
2
L
3
)
3
+
C
3
(8.26)
Applying the boundary condition at
x
=
2
L
3
, we must have
dy
dx
equal to the value
coming from the left,
⇒
C
3
= 0
Victor Saouma
Structural Engineering
Draft
8.2 Flexural Deformations
155
Figure 8.2: Moment Area Theorems
Victor Saouma
Structural Engineering
Draft
Chapter 9
ENERGY METHODS; Part I
9.1
Introduction
1
Energy methods are powerful techniques for both formulation (of the stiffness matrix of an
element
) and for the analysis (i.e. deflection) of structural problems.
2
We shall explore two techniques:
1. Real Work
2. Virtual Work (Virtual force)
9.2
Real Work
3
We start by revisiting the first law of thermodynamics:
The time-rate of change of the total energy (i.e., sum of the kinetic energy and the
internal energy) is equal to the sum of the rate of work done by the external forces
and the change of heat content per unit time.
d
dt
(
K
+
U
) =
W
e
+
H
(9.1)
where
K
is the kinetic energy,
U
the internal strain energy,
W
e
the external work, and
H
the
heat input to the system.
4
For an adiabatic system (no heat exchange) and if loads are applied in a quasi static manner
(no kinetic energy), the above relation simplifies to:
W
e
=
U
(9.2)
5
Simply stated, the first law stipulates that the external work must be equal to the internal
strain energy due to the external load.
1
More about this in
Matrix Structural Analysis
.
Draft
9.2 Real Work
173
Figure 9.2: Strain Energy Definition
9.2.2
Internal Work
9
Considering an infinitesimal element from an arbitrary structure subjected to uniaxial state
of stress, the strain energy can be determined with reference to Fig.
. The net force acting
on the element while deformation is taking place is
P
=
σ
x
dydz
. The element will undergo a
displacement
u
=
ε
x
dx
. Thus, for a linear elastic system, the strain energy density is
dU
=
1
2
σε
.
And the total strain energy will thus be
U
=
1
2
Z
Vol
ε Eε
|{z}
σ
d
Vol
(9.7)
10
When this relation is applied to various structural members it would yield:
Axial Members:
U
=
Z
Vol
εσ
2
d
Vol
σ
=
P
A
ε
=
P
AE
dV
=
Adx











U
=
Z
L
0
P
2
2
AE
dx
(9.8)
Torsional Members:
U
=
1
2
Z
Vol
ε Eε
|{z}
σ
d
Vol
U
=
1
2
Z
Vol
γ
xy
Gγ
xy
| {z }
Ï„
xy
d
Vol
Ï„
xy
=
T r
J
γ
xy
=
Ï„
xy
G
d
Vol
=
rdθdrdx
J
=
Z
r
o
Z
2
Ï€
0
r
2
dθ dr

































U
=
Z
L
0
T
2
2
GJ
dx
(9.9)
Victor Saouma
Structural Engineering
Draft
9.3 Virtual Work
175
9.3
Virtual Work
11
A severe limitation of the method of real work is that only deflection along the externally
applied load can be determined.
12
A more powerful method is the virtual work method.
13
The principle of Virtual Force (VF) relates
force
systems which satisfy the requirements of
equilibrium
, and
deformation
systems which satisfy the requirement of
compatibility
14
In any application the force system could either be the actual set of
external
loads
d
p
or
some
virtual
force system which happens to satisfy the condition of
equilibrium
δ
p
. This set
of external forces will induce internal actual forces
d
σ
or internal virtual forces
δ
σ
compatible
with the externally applied load.
15
Similarly the deformation could consist of either the actual joint deflections
d
u
and compati-
ble internal deformations
d
ε
of the structure, or some
virtual
external and internal deformation
δ
u
and
δ
ε
which satisfy the conditions of
compatibility
.
16
It is often simplest to assume that the
virtual load is a unit load
.
17
Thus we may have 4 possible combinations, Table
: where:
d
corresponds to the actual,
Force
Deformation
IVW
Formulation
External
Internal
External
Internal
1
d
p
d
σ
d
u
d
ε
2
δ
p
δ
σ
d
u
d
ε
δU
∗
Flexibility
3
d
p
d
σ
δ
u
δ
ε
δU
Stiffness
4
δ
p
δ
σ
δ
u
δ
ε
Table 9.1: Possible Combinations of Real and Hypothetical Formulations
and
δ
(with an overbar) to the hypothetical values.
This table calls for the following observations
1. The second approach is the same one on which the method of virtual or unit load is based.
It is simpler to use than the third as a internal force distribution compatible with the
assumed virtual force can be easily obtained for statically determinate structures. This
approach will yield exact solutions for statically determinate structures.
2. The third approach is favored for statically indeterminate problems or in conjunction with
approximate solution. It requires a proper “guess†of a displacement shape and is the
basis of the stiffness method.
18
Let us consider an arbitrary structure and load it with both real and virtual loads in the
following sequence, Fig.
. For the sake of simplicity, let us assume (or consider) that this
structure develops only axial stresses.
1. If we apply the virtual load, then
1
2
δP δ
∆ =
1
2
Z
d
Vol
δσδεd
Vol
(9.12)
Victor Saouma
Structural Engineering
Draft
Chapter 10
STATIC INDETERMINANCY;
FLEXIBILITY METHOD
All the examples in this chapter are taken verbatim from White, Gergely and Sexmith
10.1
Introduction
1
A statically indeterminate structure has more unknowns than equations of equilibrium (and
equations of conditions if applicable).
2
The advantages of a statically indeterminate structures are:
1. Lower internal forces
2. Safety in redundancy, i.e. if a support or members fails, the structure can
redistribute
its
internal forces to accomodate the changing B.C. without resulting in a sudden failure.
3
Only disadvantage is that it is more complicated to analyse.
4
Analysis mehtods of statically indeterminate structures
must satisfy
three requirements
Equilibrium
Force-displacement
(or stress-strain) relations (linear elastic in this course).
Compatibility
of displacements (i.e. no discontinuity)
5
This can be achieved through two classes of solution
Force or Flexibility
method;
Displacement or Stiffness
method
6
The flexibility method is first illustrated by the following problem of a statically indeterminate
cable structure in which a rigid plate is supported by two aluminum cables and a steel one. We
seek to determine the force in each cable, Fig.
1. We have three unknowns and only two independent equations of equilibrium. Hence the
problem is statically indeterminate to the first degree.
Draft
10.1 Introduction
201
6. We observe that the solution of this sproblem, contrarily to statically determinate ones,
depends on the elastic properties.
7
Another example is the propped cantiliver beam of length
L
, Fig.
x
x
f
BB
P
B
A
C
L/2
L/2
P
D
1
QL/2
PL/4
-
+
-
-
-PL
PL/2
-(1)L/2
Primary Structure Under Actual Load
Primary Structure Under Redundant Loading
Bending Moment Diagram
Figure 10.2: Propped Cantilever Beam
1. First we remove the roller support, and are left with the cantilever as a primary structure.
2. We then determine the deflection at point
B
due to the applied load
P
using the virtual
force method
1
.D
=
Z
δM
M
EI
dx
(10.6-a)
=
Z
L/
2
0
0
−
px
EI
dx
+
Z
L/
2
0
−
P L
2
+
P x
(
−
x
)
dx
(10.6-b)
=
1
EI
Z
L/
2
0
P L
2
x
+
P x
2
dx
(10.6-c)
=
1
EI
"
P Lx
2
4
+
P x
3
3
#
L/
2
0
(10.6-d)
=
5
48
P L
3
EI
(10.6-e)
Victor Saouma
Structural Engineering
Draft
10.3 Short-Cut for Displacement Evaluation
203
Note that
D
0
i
is the vector of initial displacements, which is usually zero unless we have
an initial displacement of the support (such as support settlement).
7. The reactions are then obtained by simply inverting the flexibility matrix.
9
Note that from Maxwell-Betti’s reciprocal theorem, the flexibility matrix [
f
] is always sym-
metric.
10.3
Short-Cut for Displacement Evaluation
10
Since deflections due to flexural effects must be determined numerous times in the flexibility
method, Table
may simplify some of the evaluation of the internal strain energy.
You are
strongly discouraged to use this table while still at school!
.
g
2
(
x
)
g
1
(
x
)
L
a
H
H
H
L
a
L
a
b
L
c
Lac
Lac
2
Lc
(
a
+
b
)
2
H
H
H
L
c
Lac
2
Lac
3
Lc
(2
a
+
b
)
6
L
c
Lac
2
Lac
6
Lc
(
a
+2
b
)
6
L
c
d
La
(
c
+
d
)
2
La
(2
c
+
d
)
6
La
(2
c
+
d
)+
Lb
(
c
+2
d
)
6
L
c
d e
La
(
c
+4
d
+
e
)
6
La
(
c
+2
d
)
6
La
(
c
+2
d
)+
Lb
(2
d
+
e
)
6
Table 10.1: Table of
Z
L
0
g
1
(
x
)
g
2
(
x
)
dx
10.4
Examples
Example 10-1: Steel Building Frame Analysis, (White et al. 1976)
A small, mass-produced industrial building, Fig.
, is to be framed in structural steel
with a typical cross section as shown below. The engineer is considering three different designs
for the frame: (a) for poor or unknown soil conditions, the foundations for the frame may not
be able to develop any dependable horizontal forces at its bases. In this case the idealized
base conditions are a hinge at one of the bases and a roller at the other; (b) for excellent
Victor Saouma
Structural Engineering
Draft
10.4
Examples
219
D
δM
M
R
δM M dx
R
δM
M
EI
dx
AB
BC
CD
Total
f
11
A
A
A
A
+
h
3
3
+
Lh
2
+
h
3
3
+
2
h
3
3
EI
c
+
Lh
2
EI
b
f
12
A
A
+
h
2
L
2
+
L
2
h
2
0
+
h
2
L
2
EI
c
+
L
2
h
2
EI
b
f
13
A
A
−
h
2
2
−
Lh
−
h
2
2
−
h
2
EI
c
−
Lh
EI
b
f
21
A
A
+
h
2
L
2
+
L
2
h
2
0
+
h
2
L
2
EI
c
+
L
2
h
2
EI
b
f
22
+
L
2
h
+
L
3
3
0
+
L
2
h
EI
c
+
L
3
3
EI
b
f
23
−
hL
−
L
2
2
0
−
hL
EI
c
−
L
2
2
EI
b
f
31
A
A
−
h
2
2
−
Lh
−
h
2
2
−
h
2
EI
c
−
Lh
EI
b
f
32
−
hL
−
L
2
2
0
−
hL
EI
c
−
L
2
2
EI
b
f
33
+
h
+
L
+
h
+
2
h
EI
c
+
L
EI
b
D
1
Q
A
A
aaa
−
h
2
(2
h
+15
L
−
30)
6
+
Lh
(20
−
5
L
)
2
0
−
h
2
(2
h
+15
L
−
30)
6
EI
c
+
Lh
(20
−
5
L
)
2
EI
b
D
2
Q
aaa
−
Lh
(2
h
+10
L
−
20)
2
+
L
2
(30
−
10
L
)
6
0
−
Lh
(2
h
+10
L
−
20)
2
EI
c
+
L
2
(30
−
10
L
)
6
EI
b
D
3
Q
aaa
+
h
(2
h
+10
L
−
20)
2
−
L
(20
−
5
L
)
2
0
−
h
(2
h
+10
L
−
20)
2
EI
c
−
L
(20
−
5
L
)
2
EI
b
Table 10.3: Displacement Computations for a Rectangular Frame
Victor
Saouma
Structural
Engineering
Draft
10.4 Examples
221
3. In the following discussion the contributions to displacements due to axial strain are denoted
with a single prime (
0
) and those due to curvature by a double prime (
00
).
4. Consider the axial strain first. A unit length of frame member shortens as a result of the
temperature decrease from 85
â—¦
F to 45
â—¦
F at the middepth of the member. The strain is therefore
α
∆
T
= (0
.
0000055)(40) = 0
.
00022
(10.56)
5. The effect of axial strain on the relative displacements needs little analysis. The horizontal
member shortens by an amount (0.00022)(20) = 0.0044 ft. The shortening of the vertical
members results in no relative displacement in the vertical direction 2. No rotation occurs.
6. We therefore have
D
0
1∆
=
−
0
.
0044 ft,
D
0
2∆
= 0, and
D
0
3∆
= 0
.
Figure 10.16:
7. The effect of curvature must also be considered. A frame element of length
dx
undergoes an
angular strain as a result of the temperature gradient as indicated in Figure
. The change
in length at an extreme fiber is
=
α
∆
T dx
= 0
.
0000055(25)
dx
= 0
.
000138
dx
(10.57)
8. with the resulting real rotation of the cross section
dφ
=
/
0
.
5 = 0
.
000138
dx/
0
.
5 = 0
.
000276
dx
radians
(10.58)
9. The relative displacements of the primary structure at
D
are found by the virtual force
method.
10. A virtual force
δQ
is applied in the direction of the desired displacement and the resulting
moment diagram
δM
determined.
11. The virtual work equation
δQD
=
Z
δM dφ
(10.59)
is used to obtain each of the desired displacements
D
.
12. The results, which you should verify, are
D
00
1∆
=
0
.
0828
ft
(10.60-a)
D
00
2∆
=
0
.
1104
ft
(10.60-b)
D
00
3∆
=
−
0
.
01104
radians
(10.60-c)
Victor Saouma
Structural Engineering
Draft
10.4 Examples
223
with units in kips and feet.
21. A moment diagram may now be constructed, and other internal force quantities computed
from the now known values of the redundants. The redundants have been valuated separately
for effects of temperature and foundation settlement. These effects may be combined with those
due to loading using the principle of superposition.
Example 10-9: Braced Bent with Loads and Temperature Change, (White et al. 1976)
The truss shown in Figure
reperesents an internal braced bent in an enclosed shed,
with lateral loads of 20 kN at the panel points. A temperature drop of 30
â—¦
C may occur on
the outer members (members 1-2, 2-3, 3-4, 4-5, and 5-6). We wish to analyze the truss for the
loading and for the temperature effect.
Solution:
1. The first step in the analysis is the definition of the two redundants. The choice of forces
in diagonals 2-4 and 1-5 as redundants facilitates the computations because some of the load
effects are easy to analyze. Figure
??
-b shows the definition of
R
1
and
R
2
.
2. The computations are organized in tabular form in Table
. The first column gives the
bar forces
P
in the primary structure caused by the actual loads. Forces are in kN. Column
2 gives the force in each bar caused by a unit load (1 kN) corresponding to release 1. These
are denoted
p
1
and also represent the bar force ¯
q
1
/δQ
1
caused by a virtual force
δQ
1
applied
Figure 10.17:
at the same location. Column 3 lists the same quantity for a unit load and for a virtual force
δQ
2
applied at release 2. These three columns constitute a record of the truss analysis needed
for this problem.
Victor Saouma
Structural Engineering
Draft
10.4
Examples
225
D
1
Q
D
2
Q
f
11
f
21
f
12
f
22
D
1∆
D
2∆
P
p
1
p
2
L/EA
¯
q
1
P L/EA
¯
q
2
P L/EA
¯
q
1
p
1
L/EA
¯
q
2
p
2
L/EA
¯
qp
2
L/EA
¯
q
2
p
2
L/EA
∆
/
temp
¯
q
1
∆
/
¯
q
2
∆
/
multiply by
L
c
/EA
c
L
c
/EA
c
L
c
/EA
c
L
c
/EA
c
L
c
/EA
c
L
c
/EA
c
L
c
/EA
c
L
c
10
−
4
L
c
10
−
4
L
c
1-2
60.0
0
−
0.707
1
0
−
42.42
0
0
0
0.50
−
0.0003
0
2.12
2-3
20.00
−
0.707
0
1
−
14.14
0
0.50
0
0
0
−
0.0003
2.12
0
3-4
0
−
0.707
0
2
0
0
1.00
0
0
0
−
0.0003
2.12
0
4-5
0
−
0.707
0
1
0
0
0.50
0
0
0
−
0.0003
2.12
0
5-6
−
20.00
0
−
0.707
1
0
14.14
0
0
0
0.50
−
0.0003
0
2.12
6-1
40.00
0
−
0.707
2
0
−
56.56
0
0
0
1.00
0
0
0
2-5
20.00
−
0.707
−
0.707
2
−
28.28
−
28.28
1.00
1.00
1.00
1.00
0
0
0
1-5
0
0
1.00
2.828
0
0
0
0
0
2.83
0
0
0
2-6
−
56.56
0
1.00
2.828
0
−
160.00
0
0
0
2.83
0
0
0
2-4
0
1.00
0
2.828
0
0
2.83
0
0
0
0
0
0
3-5
−
28.28
1.00
0
2.838
−
80.00
0
2.83
0
0
0
0
0
0
−
122.42
−
273.12
8.66
1.00
1.00
8.66
6.36
4.24
Victor
Saouma
Structural
Engineering
Draft
Chapter 11
APPROXIMATE FRAME
ANALYSIS
1
Despite the widespread availability of computers, approximate methods of analysis are justi-
fied by
1. Inherent assumption made regarding the validity of a linear elastic analysis
vis a vis
of
an ultimate failure design.
2. Ability of structures to redistribute internal forces.
3. Uncertainties in load and material properties
2
Vertical loads are treated separately from the horizontal ones.
3
We use the design sign convention for moments (+ve tension below), and for shear (ccw +ve).
4
Assume girders to be numbered from left to right.
5
In all free body diagrams assume positivee forces/moments, and take algeebraic sums.
6
The key to the approximate analysis method is our ability to sketch the deflected shape of a
structure and identify inflection points.
7
We begin by considering a uniformly loaded beam and frame. In each case we consider an
extreme end of the restraint: a) free or b) restrained. For the frame a relativly flexible or stiff
column would be analogous to a free or fixed restrain on the beam, Fig.
11.1
Vertical Loads
8
With reference to Fig.
, we now consider an
intermediary
case as shown in Fig.
9
With the location of the inflection points identified, we may now determine all the reactions
and internal forces from statics.
10
If we now consider a multi-bay/multi-storey frame, the girders at each floor are assumed to
be continuous beams, and columns are assumed to resist the resulting unbalanced moments
from the girders, we may make the following assumptions
Draft
11.1 Vertical Loads
229
0.5H
0.5H
0.1 L
0.1 L
Figure 11.2: Uniformly Loaded Frame, Approximate Location of Inflection Points
Victor Saouma
Structural Engineering
Draft
11.1 Vertical Loads
231
M
+
=
1
8
wL
2
s
=
w
1
8
(0
.
8)
2
L
2
= 0
.
08
wL
2
(11.1)
Maximum negative moment
at each end of the girder is given by, Fig.
M
lef t
=
M
rgt
=
−
w
2
(0
.
1
L
)
2
−
w
2
(0
.
8
L
)(0
.
1
L
) =
−
0
.
045
wL
2
(11.2)
Girder Shear
are obtained from the free body diagram, Fig.
V
rgt
i-1
V
lft
i
P
above
P
below
Figure 11.4: Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces
V
lf t
=
wL
2
V
rgt
=
−
wL
2
(11.3)
Column axial force
is obtained by summing all the girder shears to the axial force transmit-
ted by the column above it. Fig.
P
dwn
=
P
up
+
V
rgt
i
−
1
−
V
lf t
i
(11.4)
Column Moment
are obtained by considering the free body diagram of columns Fig.
M
top
=
M
bot
above
−
M
rgt
i
−
1
+
M
lf t
i
M
bot
=
−
M
top
(11.5)
Victor Saouma
Structural Engineering
Draft
11.2 Horizontal Loads
245
Horizontal Loads, Portal Method
1. Column Shears
V
5
=
15
(2)(3)
= 2
.
5
k
V
6
= 2(
V
5
) = (2)(2
.
5) = 5
k
V
7
= 2(
V
5
) = (2)(2
.
5) = 5
k
V
8
=
V
5
= 2
.
5
k
V
1
=
15+30
(2)(3)
= 7
.
5
k
V
2
= 2(
V
1
) = (2)(7
.
5) = 15
k
V
3
= 2(
V
1
) = (2)(2
.
5) = 15
k
V
4
=
V
1
= 7
.
5
k
2. Top Column Moments
M
top
5
=
V
1
H
5
2
=
(2
.
5)(14)
2
=
17
.
5
k.ft
M
bot
5
=
−
M
top
5
=
−
17
.
5
k.ft
M
top
6
=
V
6
H
6
2
=
(5)(14)
2
=
35
.
0
k.ft
M
bot
6
=
−
M
top
6
=
−
35
.
0
k.ft
M
top
7
=
V
up
7
H
7
2
=
(5)(14)
2
=
35
.
0
k.ft
M
bot
7
=
−
M
top
7
=
−
35
.
0
k.ft
M
top
8
=
V
up
8
H
8
2
=
(2
.
5)(14)
2
=
17
.
5
k.ft
M
bot
8
=
−
M
top
8
=
−
17
.
5
k.ft
3. Bottom Column Moments
M
top
1
=
V
dwn
1
H
1
2
=
(7
.
5)(16)
2
=
60
k.ft
M
bot
1
=
−
M
top
1
=
−
60
k.ft
M
top
2
=
V
dwn
2
H
2
2
=
(15)(16)
2
=
120
k.ft
M
bot
2
=
−
M
top
2
=
−
120
k.ft
M
top
3
=
V
dwn
3
H
3
2
=
(15)(16)
2
=
120
k.ft
M
bot
3
=
−
M
top
3
=
−
120
k.ft
M
top
4
=
V
dwn
4
H
4
2
=
(7
.
5)(16)
2
=
60
k.ft
M
bot
4
=
−
M
top
4
=
−
60
k.ft
4. Top Girder Moments
M
lf t
12
=
M
top
5
=
17
.
5
k.ft
M
rgt
12
=
−
M
lf t
12
=
−
17
.
5
k.ft
M
lf t
13
=
M
rgt
12
+
M
top
6
=
−
17
.
5 + 35 =
17
.
5
k.ft
M
rgt
13
=
−
M
lf t
13
=
−
17
.
5
k.ft
M
lf t
14
=
M
rgt
13
+
M
top
7
=
−
17
.
5 + 35 =
17
.
5
k.ft
M
rgt
14
=
−
M
lf t
14
=
−
17
.
5
k.ft
Victor Saouma
Structural Engineering
Draft
11.2 Horizontal Loads
247
6. Top Girder Shear
V
lf t
12
=
−
2
M
lf t
12
L
12
=
−
(2)(17
.
5)
20
=
−
1
.
75
k
V
rgt
12
= +
V
lf t
12
=
−
1
.
75
k
V
lf t
13
=
−
2
M
lf t
13
L
13
=
−
(2)(17
.
5)
30
=
−
1
.
17
k
V
rgt
13
= +
V
lf t
13
=
−
1
.
17
k
V
lf t
14
=
−
2
M
lf t
14
L
14
=
−
(2)(17
.
5)
24
=
−
1
.
46
k
V
rgt
14
= +
V
lf t
14
=
−
1
.
46
k
7. Bottom Girder Shear
V
lf t
9
=
−
2
M
lf t
12
L
9
=
−
(2)(77
.
5)
20
=
−
7
.
75
k
V
rgt
9
= +
V
lf t
9
=
−
7
.
75
k
V
lf t
10
=
−
2
M
lf t
10
L
10
=
−
(2)(77
.
5)
30
=
−
5
.
17
k
V
rgt
10
= +
V
lf t
10
=
−
5
.
17
k
V
lf t
11
=
−
2
M
lf t
11
L
11
=
−
(2)(77
.
5)
24
=
−
6
.
46
k
V
rgt
11
= +
V
lf t
11
=
−
6
.
46
k
8. Top Column Axial Forces (+ve tension, -ve compression)
P
5
=
−
V
lf t
12
=
−
(
−
1
.
75)
k
P
6
= +
V
rgt
12
−
V
lf t
13
=
−
1
.
75
−
(
−
1
.
17) =
−
0
.
58
k
P
7
= +
V
rgt
13
−
V
lf t
14
=
−
1
.
17
−
(
−
1
.
46) = 0
.
29
k
P
8
=
V
rgt
14
=
−
1
.
46
k
9. Bottom Column Axial Forces (+ve tension, -ve compression)
P
1
=
P
5
+
V
lf t
9
= 1
.
75
−
(
−
7
.
75)
= 9
.
5
k
P
2
=
P
6
+
V
rgt
10
+
V
lf t
9
=
−
0
.
58
−
7
.
75
−
(
−
5
.
17) =
−
3
.
16
k
P
3
=
P
7
+
V
rgt
11
+
V
lf t
10
= 0
.
29
−
5
.
17
−
(
−
6
.
46)
= 1
.
58
k
P
4
=
P
8
+
V
rgt
11
=
−
1
.
46
−
6
.
46
=
−
7
.
66
k
Design Parameters
On the basis of the two approximate analyses, vertical and lateral load,
we now seek the design parameters for the frame, Table
Structural Engineering
Draft
11.2 Horizontal Loads
249
Portal Method
PORTAL.XLS
Victor E. Saouma
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
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X
X
X
X
X
X
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X
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X
X
X
X
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X
X
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X
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X
X
X
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X
X
X
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X
X
X
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X
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X
X
X
X
X
X
X
Y
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Y
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Y
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Y
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Y
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Y
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Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
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Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
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1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
A
B
C
D
E
F
G
H
I
J
K
L
M
N
O
P
Q
R
S
PORTAL METHOD
# of Bays
3
L1
L2
L3
Ë
Ë
Ë
Ë
Ë
Ì
Ì
Ì
Ì
Ì
Ã
Ã
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Ã
Ã
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ÃŽ
ÃŽ
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Ñ
Ñ
Ñ
Ñ
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Ó
Ó
Ó
Ó
Ó
Ô
Ô
Ô
Ô
Ô
Õ
Õ
Õ
Õ
Õ
Ö
Ö
Ö
Ö
Ö
×
×
×
×
×
Ø
Ø
Ø
Ø
Ø
Ù
Ù
Ù
Ù
Ù
Ú
Ú
Ú
Ú
Ú
Û
Û
Û
Û
Û
Ü
Ü
Ü
Ü
Ü
20
30
24
MOMENTS
Ã
Ã
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Þ
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ß
ß
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ß
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á
á
á
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ä
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ä
ä
ä
ä
ä
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Ã¥
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Ã¥
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æ
æ
æ
æ
æ
æ
æ
æ
ç
ç
ç
ç
ç
ç
ç
ç
è
è
è
è
è
è
è
è
é
é
é
é
é
é
é
é
# of Storeys
2
Bay 1
Bay 2
Bay 3
ê
ê
ê
ê
ë
ë
ë
ë
ì
ì
ì
ì
Ã
Ã
Ã
Ã
î
î
î
î
ï
ï
ï
ï
ð
ð
ð
ð
ñ
ñ
ñ
ñ
ò
ò
ò
ò
ó
ó
ó
ó
ô
ô
ô
ô
õ
õ
õ
õ
ö
ö
ö
ö
÷
÷
÷
÷
ø
ø
ø
ø
ù
ù
ù
ù
ú
ú
ú
ú
û
û
û
û
Force
Shear
Col
Beam
Column
Beam
Column
Beam
Col
ü
ü
ü
ü
ý
ý
ý
ý
þ
þ
þ
þ
ÿ
ÿ
ÿ
ÿ
H
Lat.
Tot
Ext
Int
Lft
Rgt
Lft
Rgt
Lft
Rgt
=+H9
=-I8
=+J8+K9
=-M8
=+N8+O9
=-Q8
H1
14 15
=+C9
=+D9/(2*$F$2) =2*E9
=+E9*B9/2
=+F9*B9/2
=+K9
=+H9
=-H9
=-K9
=+K10
=+H10
=+H12-H10 =-I11
=+K12-K10+J11 =-M11
=+O12-O10+N11 =-Q11
H2
16 30
=SUM($C$9:C12) =+D12/(2*$F$2) =2*E12
=+E12*B12/2
=+F12*B12/2
=+K12
=+H12
=-H12
=-K12
=+K13
=+H13
SHEAR
Bay 1
Bay 2
Bay 3
Col
Beam
Column
Beam
Column
Beam
Col
!
!
!
!
"
"
"
"
#
#
#
#
Lft
Rgt
Lft
Rgt
Lft
Rgt
=-2*I8/I$3
=+I18
=-2*M8/M$3
=+M18
=-2*Q8/Q$3
=+Q18
$
$
$
$
=+E9
=+F9
=+F9
=+E9
=+H19
=+K19
=+O19
=+S19
=-2*I11/I$3 =+I21
=-2*M11/M$3
=+M21
=-2*Q11/Q$3
=+Q21
=+E12
=+F12
=+F12
=+E12
=+H22
=+K22
=+O22
=+S22
AXIAL FORCE
Bay 1
Bay 2
Bay 3
%
%
%
%
&
&
&
&
'
'
'
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(
(
(
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+
+
,
,
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.
.
.
/
/
/
/
0
0
0
0
Col
Beam
Column
Beam
Column
Beam
Col
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
5
5
5
5
0
0
0
=-I18
=+J18-M18
=+N18-Q18
=+R18
0
0
0
=+H28-I21
=+K28+J21-M21
=+O28+N21-Q21
=+S28+R21
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
:
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
;
Figure 11.19: Portal Method; Equations in Spread-Sheet
Victor Saouma
Structural Engineering
Draft
11.2 Horizontal Loads
251
Mem.
Vert.
Hor.
Design
Values
-ve Moment
9.00
77.50
86.50
9
+ve Moment
16.00
0.00
16.00
Shear
5.00
7.75
12.75
-ve Moment
20.20
77.50
97.70
10
+ve Moment
36.00
0.00
36.00
Shear
7.50
5.17
12.67
-ve Moment
13.0
77.50
90.50
11
+ve Moment
23.00
0.00
23.00
Shear
6.00
6.46
12.46
-ve Moment
4.50
17.50
22.00
12
+ve Moment
8.00
0.00
8.00
Shear
2.50
1.75
4.25
-ve Moment
10.10
17.50
27.60
13
+ve Moment
18.00
0.00
18.00
Shear
3.75
1.17
4.92
-ve Moment
6.50
17.50
24.00
14
+ve Moment
11.50
0.00
11.50
Shear
3.00
1.46
4.46
Table 11.2: Girders Combined Approximate Vertical and Horizontal Loads
Victor Saouma
Structural Engineering
Draft
Chapter 12
KINEMATIC
INDETERMINANCY; STIFFNESS
METHOD
12.1
Introduction
12.1.1
Stiffness vs Flexibility
1
There are two classes of structural analysis methods, Table
Flexibility:
where the primary unknown is a force, where equations of equilibrium are the
starting point, static indeterminancy occurs if there are more unknowns than equations,
and displacements of the entire structure (usually from virtual work) are used to write an
equation of compatibility of displacements in order to solve for the redundant forces.
Stiffness:
method is the counterpart of the flexibility one. Primary unknowns are displace-
ments, and we start from expressions for the forces written in terms of the displacements
(at the element level) and then apply the equations of equilibrium. The structure is con-
sidered to be
kinematically indeterminate
to the
n
th degree where
n
is the total number
of independent displacements. From the displacements, we then compute the internal
forces.
Flexibility
Stiffness
Primary Variable (d.o.f.)
Forces
Displacements
Indeterminancy
Static
Kinematic
Force-Displacement
Displacement(Force)/Structure
Force(Displacement)/Element
Governing Relations
Compatibility of displacement
Equilibrium
Methods of analysis
“Consistent Deformationâ€
Slope Deflection; Moment Distribution
Table 12.1: Stiffness vs Flexibility Methods
2
In the flexibility method, we started by releasing as many redundant forces as possible in
order to render the structure
statically determinate
, and this made it quite flexible. We then
Draft
12.2 Degrees of Freedom
255
Figure 12.2: Independent Displacements
Type
Node 1
Node 2
1 Dimensional
{
p
}
F
y
1
,
M
z
2
F
y
3
,
M
z
4
Beam
{
δ
}
v
1
,
θ
2
v
3
,
θ
4
2 Dimensional
{
p
}
F
x
1
F
x
2
Truss
{
δ
}
u
1
u
2
{
p
}
F
x
1
,
F
y
2
,
M
z
3
F
x
4
,
F
y
5
,
M
z
6
Frame
{
δ
}
u
1
,
v
2
,
θ
3
u
4
,
v
5
,
θ
6
3 Dimensional
{
p
}
F
x
1
,
F
x
2
Truss
{
δ
}
u
1
,
u
2
{
p
}
F
x
1
,
F
y
2
,
F
y
3
,
F
x
7
,
F
y
8
,
F
y
9
,
T
x
4
M
y
5
,
M
z
6
T
x
10
M
y
11
,
M
z
12
Frame
{
δ
}
u
1
,
v
2
,
w
3
,
u
7
,
v
8
,
w
9
,
θ
4
,
θ
5
θ
6
θ
10
,
θ
11
θ
12
Table 12.2: Degrees of Freedom of Different Structure Types Systems
Victor Saouma
Structural Engineering
Draft
12.3 Kinematic Relations
257
12.2.1
Methods of Analysis
12
There are three methods for the stiffness based analysis of a structure
Slope Deflection:
(Mohr, 1892) Which results in a system of
n
linear equations with
n
un-
knowns, where
n
is the degree of kinematic indeterminancy (i.e. total number of inde-
pendent displacements/rotation).
Moment Distribution:
(Cross, 1930) which is an iterative method to solve for the
n
dis-
placements and corresponding internal forces in flexural structures.
Direct Stiffness method:
(˜1960) which is a formal statement of the stiffness method and
cast in matrix form is by far the most powerful method of structural analysis.
The first two methods lend themselves to hand calculation, and the third to a computer based
analysis.
12.3
Kinematic Relations
12.3.1
Force-Displacement Relations
13
Whereas in the flexibility method we sought to obtain a displacement in terms of the forces
(through virtual work) for an entire structure, our starting point in the stiffness method is to
develop a set of relationship for the
force in terms of the displacements for a single element
.









V
1
M
1
V
2
M
2









=





− − − −
− − − −
− − − −
− − − −














v
1
θ
1
v
2
θ
2









(12.1)
14
We start from the differential equation of a beam, Fig.
in which we have all positive
known displacements, we have from strength of materials
M
=
−
EI
d
2
v
dx
2
=
M
1
−
V
1
x
+
m
(
x
)
(12.2)
where
m
(
x
) is the moment applied due to the applied load only. It is positive when counter-
clockwise.
15
Integrating twice
−
EIv
0
=
M
1
x
−
1
2
V
1
x
2
+
f
(
x
) +
C
1
(12.3)
−
EIv
=
1
2
M
1
x
2
−
1
6
V
1
x
3
+
g
(
x
) +
C
1
x
+
C
2
(12.4)
where
f
(
x
) =
R
m
(
x
)
dx
, and
g
(
x
) =
R
f
(
x
)
dx
.
16
Applying the boundary conditions at
x
= 0
v
0
=
θ
1
v
=
v
1
)
⇒
(
C
1
=
−
EIθ
1
C
2
=
−
EIv
1
(12.5)
Victor Saouma
Structural Engineering
Draft
12.5 Moment Distribution; Indirect Solution
277
Example 12-7: Continuous Beam, Initial Settlement, (Kinney 1957)
For the following beam find the moments at
A, B,
and
C
by moment distribution. The
support at
C
settles by 0.1 in. Use
E
= 30
,
000 k/in
2
.
Solution:
1. Fixed-end moments: Uniform load:
M
F
AB
=
wL
2
12
=
(5)(20
2
)
12
= +167
k.ft
(12.68-a)
M
F
BA
=
−
167
k.ft
(12.68-b)
Victor Saouma
Structural Engineering
Draft
12.5 Moment Distribution; Indirect Solution
279
Solution:
1. The first step is to perform the usual moment distribution. The reader should fully under-
stand that this balancing operation adjusts the internal moments at the ends of the members
by a series of corrections as the joints are considered to rotate, until Σ
M
= 0 at each joint.
The reader should also realize that
during this balancing operation no translation of any joint
is permitted
.
2. The fixed-end moments are
M
F
BC
=
(18)(12)(6
2
)
18
2
= +24
k.ft
(12.71-a)
M
F
CB
=
(18)(6)(12
2
)
18
2
=
−
48
k.ft
(12.71-b)
3. Moment distribution
Joint
A
B
C
D
Balance
CO
Member
AB
BA
BC
CB
CD
DC
K
10
10
20
20
15
15
DF
0
0.333
0.667
0.571
0.429
0
FEM
+24.0
-48.0
FEM
+
+13.7
?
+27.4
+20.6
-
+10.3
C
DC; BC
-6.3
-12.6
-25.1
-
?
-12.5
s
B
AB; CB
+
+3.6
?
+7.1
+5.4
-
+2.7
C
BC; DC
-0.6
-1.2
-2.4
-
?
-1.2
s
B
AB; CB
+
+0.3
?
+0.7
+0.5
-
+0.02
C
BC; DC
-0.1
-0.2
B
Total
-6.9
-13.9
+13.9
-26.5
+26.5
+13.2
4. The
final moments listed in the table are correct only if there is no translation of any joint
.
It is therefore necessary to determine whether or not, with the above moments existing, there
is any tendency for side lurch of the top of the frame.
5. If the frame is divided into three free bodies, the result will be as shown below.
Victor Saouma
Structural Engineering
Draft
12.5 Moment Distribution; Indirect Solution
281
62
Recalling that the fixed end moment is
M
F
= 6
EI
∆
L
2
= 6
EK
m
∆, where
K
m
=
I
L
2
=
K
L
we
can write
∆ =
M
F
AB
6
EK
m
=
M
F
DC
6
EK
m
(12.72-a)
⇒
M
F
AB
M
F
DC
=
K
AB
m
K
DC
m
=
10
15
(12.72-b)
63
These fixed-end moments could, for example, have the values of
−
10 and
−
15
k.ft
or
−
20 and
−
30, or
−
30 and
−
45, or any other combination so long as the above ratio is maintained. The
proper procedure is to choose values for these fixed-end moments of approximately the same
order of magnitude as the original fixed-end moments due to the real loads. This will result in
the same accuracy for the results of the balance for the side-sway correction that was realized
in the first balance for the real loads. Accordingly, it will be assumed that
P
, and the resulting
∆, are of such magnitudes as to result in the fixed-end moments shown below
8. Obviously, Σ
M
= 0 is not satisfied for joints B and C in this deflected frame. Therefore
these joints must rotate until equilibrium is reached. The effect of this rotation is determined
in the distribution below
Joint
A
B
C
D
Balance
CO
Member
AB
BA
BC
CB
CD
DC
K
10
10
20
20
15
15
DF
0
0.333
0.667
0.571
0.429
0
FEM
-30.0
-30.0
+
-45.0
?
-45.0
+
+12.9
?
+25.8
+19.2
-
+9.6
C
BC; DC
+2.8
+5.7
+11.4
-
?
+5.7
s
B
AB; CB
+
-1.6
?
-3.3
-2.4
-
-1.2
C
BC; DC
+0.2
+0.5
+1.1
-
?
+0.5
s
B
AB; CB
-0.3
-0.2
-
-0.1
C
Total
-27.0
-23.8
+23.8
+28.4
-28.4
-36.7
9. During the rotation of joints
B
and
C
, as represented by the above distribution,
the value
of
∆
has remained constant, with
P
varying in magnitude as required to maintain
∆.
10. It is now possible to determine the final value of
P
simply by adding the shears in the
columns. The shear in any member, without external loads applied along its length, is obtained
by adding the end moments algebraically and dividing by the length of the member. The final
Victor Saouma
Structural Engineering
Draft
Chapter 13
DIRECT STIFFNESS METHOD
13.1
Introduction
13.1.1
Structural Idealization
1
Prior to analysis, a structure must be idealized for a suitable mathematical representation.
Since it is practically impossible (and most often unnecessary) to model every single detail,
assumptions must be made. Hence, structural idealization is as much an art as a science.
Some
of the questions confronting the analyst include:
1. Two dimensional versus three dimensional; Should we model a single bay of a building,
or the entire structure?
2. Frame or truss, can we neglect flexural stiffness?
3. Rigid or semi-rigid connections (most important in steel structures)
4. Rigid supports or elastic foundations (are the foundations over solid rock, or over clay
which may consolidate over time)
5. Include or not secondary members (such as diagonal braces in a three dimensional anal-
ysis).
6. Include or not axial deformation (can we neglect the axial stiffness of a beam in a build-
ing?)
7. Cross sectional properties (what is the moment of inertia of a reinforced concrete beam?)
8. Neglect or not haunches (those are usually present in zones of high negative moments)
9. Linear or nonlinear analysis (linear analysis can not predict the peak or failure load, and
will underestimate the deformations).
10. Small or large deformations (In the analysis of a high rise building subjected to wind
load, the moments should be amplified by the product of the axial load times the lateral
deformation,
P
−
∆ effects).
11. Time dependent effects (such as creep, which is extremely important in prestressed con-
crete, or cable stayed concrete bridges).
Draft
13.1 Introduction
289
(such as beam element), while element 2 has a code 2 (such as a truss element). Material group
1 would have different elastic/geometric properties than material group 2.
Group
Element
Material
No.
Type
Group
1
1
1
2
2
1
3
1
2
Table 13.3: Example of Group Number
6
From the analysis, we first obtain the nodal displacements, and then the element internal
forces. Those internal forces vary according to the element type. For a two dimensional frame,
those are the axial and shear forces, and moment at each node.
7
Hence, the need to define two coordinate systems (one for the entire structure, and one for
each element), and a sign convention become apparent.
13.1.3
Coordinate Systems
8
We should differentiate between 2 coordinate systems:
Global:
to describe the structure nodal coordinates. This system can be arbitrarily selected
provided it is a Right Hand Side (RHS) one, and we will associate with it upper case axis
labels,
X, Y, Z
, Fig.
or 1,2,3 (running indeces within a computer program).
Local:
system is associated with each element and is used to describe the element internal
forces. We will associate with it lower case axis labels,
x, y, z
(or 1,2,3), Fig.
9
The
x
-axis is assumed to be along the member, and the direction is chosen such that it points
from the 1st node to the 2nd node, Fig.
10
Two dimensional structures will be defined in the X-Y plane.
13.1.4
Sign Convention
11
The sign convention in structural analysis is completely different than the one previously
adopted in structural analysis/design, Fig.
(where we focused mostly on flexure and defined
a positive moment as one causing “tension belowâ€. This would be awkward to program!).
12
In matrix structural analysis the sign convention adopted is consistent with the prevailing
coordinate system. Hence, we define a positive moment as one which is counter-clockwise, Fig.
13
Fig.
illustrates the sign convention associated with each type of element.
14
Fig.
also shows the geometric (upper left) and elastic material (upper right) properties
associated with each type of element.
Victor Saouma
Structural Engineering
Draft
13.1 Introduction
291
Figure 13.3: Sign Convention, Design and Analysis
Figure 13.4: Total Degrees of Freedom for various Type of Elements
Victor Saouma
Structural Engineering
Draft
13.1 Introduction
293
Type
Node 1
Node 2
[
k
]
[
K
]
(Local)
(Global)
1 Dimensional
{
p
}
F
y
1
,
M
z
2
F
y
3
,
M
z
4
Beam
4
×
4
4
×
4
{
δ
}
v
1
,
θ
2
v
3
,
θ
4
2 Dimensional
{
p
}
F
x
1
F
x
2
Truss
2
×
2
4
×
4
{
δ
}
u
1
u
2
{
p
}
F
x
1
,
F
y
2
,
M
z
3
F
x
4
,
F
y
5
,
M
z
6
Frame
6
×
6
6
×
6
{
δ
}
u
1
,
v
2
,
θ
3
u
4
,
v
5
,
θ
6
{
p
}
T
x
1
,
F
y
2
,
M
z
3
T
x
4
,
F
y
5
,
M
z
6
Grid
6
×
6
6
×
6
{
δ
}
θ
1
,
v
2
,
θ
3
θ
4
,
v
5
,
θ
6
3 Dimensional
{
p
}
F
x
1
,
F
x
2
Truss
2
×
2
6
×
6
{
δ
}
u
1
,
u
2
{
p
}
F
x
1
,
F
y
2
,
F
y
3
,
F
x
7
,
F
y
8
,
F
y
9
,
T
x
4
M
y
5
,
M
z
6
T
x
10
M
y
11
,
M
z
12
Frame
12
×
12
12
×
12
{
δ
}
u
1
,
v
2
,
w
3
,
u
7
,
v
8
,
w
9
,
θ
4
,
θ
5
θ
6
θ
10
,
θ
11
θ
12
Table 13.4: Degrees of Freedom of Different Structure Types Systems
Victor Saouma
Structural Engineering
Draft
13.2 Stiffness Matrices
295
13.2
Stiffness Matrices
13.2.1
Truss Element
22
From strength of materials, the force/displacement relation in axial members is
σ
=
E
Aσ
|{z}
P
=
AE
L
∆
|{z}
1
(13.1)
Hence, for a unit displacement, the applied force should be equal to
AE
L
. From statics, the force
at the other end must be equal and opposite.
23
The truss element (whether in 2D or 3D) has only one degree of freedom associated with
each node. Hence, from Eq.
, we have
[
k
t
] =
AE
L


u
1
u
2
p
1
1
−
1
p
2
−
1
1


(13.2)
13.2.2
Beam Element
24
Using Equations
and
we can determine the forces associated with
each unit displacement.
[
k
b
] =





v
1
θ
1
v
2
θ
2
V
1
Eq.
v
1
= 1)
Eq.
θ
1
= 1)
Eq.
v
2
= 1)
Eq.
θ
2
= 1)
M
1
Eq.
v
1
= 1)
Eq.
θ
1
= 1)
Eq.
v
2
= 1)
Eq.
θ
2
= 1)
V
2
Eq.
v
1
= 1)
Eq.
θ
1
= 1)
Eq.
v
2
= 1)
Eq.
θ
2
= 1)
M
2
Eq.
v
1
= 1)
Eq.
θ
1
= 1)
Eq.
v
2
= 1)
Eq.
θ
2
= 1)





(13.3)
25
The stiffness matrix of the beam element (neglecting shear and axial deformation) will thus
be
[
k
b
] =









v
1
θ
1
v
2
θ
2
V
1
12
EI
z
L
3
6
EI
z
L
2
−
12
EI
z
L
3
6
EI
z
L
2
M
1
6
EI
z
L
2
4
EI
z
L
−
6
EI
z
L
2
2
EI
z
L
V
2
−
12
EI
z
L
3
−
6
EI
z
L
2
12
EI
z
L
3
−
6
EI
z
L
2
M
2
6
EI
z
L
2
2
EI
z
L
−
6
EI
z
L
2
4
EI
z
L









(13.4)
26
We note that this is identical to Eq.









V
1
M
1
V
2
M
2









=





v
1
θ
1
v
2
θ
2
V
1
12
EI
z
L
3
6
EI
z
L
2
−
12
EI
z
L
3
6
EI
z
L
2
M
1
6
EI
z
L
2
4
EI
z
L
−
6
EI
z
L
2
2
EI
z
L
V
2
−
12
EI
z
L
3
−
6
EI
z
L
2
12
EI
z
L
3
−
6
EI
z
L
2
M
2
6
EI
z
L
2
2
EI
z
L
−
6
EI
z
L
2
4
EI
z
L





|
{z
}
k
(
e
)









v
1
θ
1
v
2
θ
2









(13.5)
Victor Saouma
Structural Engineering
Draft
13.3 Direct Stiffness Method
297
Figure 13.7: Problem with 2 Global d.o.f.
θ
1
and
θ
2
Victor Saouma
Structural Engineering
Draft
13.3 Direct Stiffness Method
315
% Solve for the Displacements in meters and radians
Displacements=inv(Ktt)*P’
% Extract Kut
Kut=Kaug(4:9,1:3);
% Compute the Reactions and do not forget to add fixed end actions
Reactions=Kut*Displacements+FEA_Rest’
% Solve for the internal forces and do not forget to include the fixed end actions
dis_global(:,:,1)=[0,0,0,Displacements(1:3)’];
dis_global(:,:,2)=[Displacements(1:3)’,0,0,0];
for elem=1:2
dis_local=Gamma(:,:,elem)*dis_global(:,:,elem)’;
int_forces=k(:,:,elem)*dis_local+fea(1:6,elem)
end
function [k,K,Gamma]=stiff(EE,II,A,i,j)
% Determine the length
L=sqrt((j(2)-i(2))^2+(j(1)-i(1))^2);
% Compute the angle theta (carefull with vertical members!)
if(j(1)-i(1))~=0
alpha=atan((j(2)-i(2))/(j(1)-i(1)));
else
alpha=-pi/2;
end
% form rotation matrix Gamma
Gamma=[
cos(alpha)
sin(alpha)
0
0
0
0;
-sin(alpha) cos(alpha)
0
0
0
0;
0
0
1
0
0
0;
0
0
0
cos(alpha) sin(alpha) 0;
0
0
0 -sin(alpha) cos(alpha) 0;
0
0
0
0
0
1];
% form element stiffness matrix in local coordinate system
EI=EE*II;
EA=EE*A;
k=[EA/L,
0,
0, -EA/L,
0,
0;
0,
12*EI/L^3,
6*EI/L^2,
0, -12*EI/L^3,
6*EI/L^2;
0,
6*EI/L^2,
4*EI/L,
0,
-6*EI/L^2,
2*EI/L;
-EA/L,
0,
0,
EA/L,
0,
0;
0, -12*EI/L^3, -6*EI/L^2,
0,
12*EI/L^3, -6*EI/L^2;
0,
6*EI/L^2,
2*EI/L,
0,
-6*EI/L^2,
4*EI/L];
% Element stiffness matrix in global coordinate system
K=Gamma’*k*Gamma;
This simple proigram will produce the following results:
Displacements =
0.0010
-0.0050
Victor Saouma
Structural Engineering
Draft
13.3 Direct Stiffness Method
317
Figure 13.14: ID Values for Simple Beam
2. The
structure
stiffness matrix is assembled
K
=



1
2
−
3
−
4
1
12
EI/L
2
−
6
EI/L
2
−
12
EI/L
3
−
6
EI/L
2
2
−
6
EI/L
2
4
EI/L
6
EI/L
2
2
EI/L
−
3
−
12
EI/L
3
6
EI/L
2
12
EI/L
3
6
EI/L
2
−
4
−
6
EI/L
2
2
EI/L
6
EI/L
2
4
EI/L



3. The global matrix can be rewritten as





−
P
√
0
√
R
3
?
R
4
?





=



12
EI/L
2
−
6
EI/L
2
−
12
EI/L
3
−
6
EI/L
2
−
6
EI/L
2
4
EI/L
6
EI/L
2
2
EI/L
−
12
EI/L
3
6
EI/L
2
12
EI/L
3
6
EI/L
2
−
6
EI/L
2
2
EI/L
6
EI/L
2
4
EI/L








∆
1
?
θ
2
?
∆
3
√
θ
4
√





4.
K
tt
is inverted (or actually decomposed) and stored in the same global matrix





L
3
/
3
EI
L
2
/
2
EI
−
12
EI/L
3
−
6
EI/L
2
L
2
/
2
EI
L/EI
6
EI/L
2
2
EI/L
−
12
EI/L
3
6
EI/L
2
12
EI/L
3
6
EI/L
2
−
6
EI/L
2
2
EI/L
6
EI/L
2
4
EI/L





5. Next we compute the equivalent load,
P
0
t
=
P
t
−
K
tu
∆
u
, and overwrite
P
t
by
P
0
t
P
t
−
K
tu
∆
u
=







−
P
0
0
0







−





L
3
/
3
EI
L
2
/
2
EI
−
12
EI/L
3
−
6
EI/L
2
L
2
/
2
EI
L/EI
6
EI/L
2
2
EI/L
−
12
EI/L
3
6
EI/L
2
12
EI/L
3
6
EI/L
2
−
6
EI/L
2
2
EI/L
6
EI/L
2
4
EI/L












−
P
0
0
0







=







−
P
0
0
0







Victor Saouma
Structural Engineering
Draft
13.3 Direct Stiffness Method
319
5. We compute the equivalent load,
P
0
t
=
P
t
−
K
tu
∆
u
, and overwrite
P
t
by
P
0
t
P
t
−
K
tu
∆
u
=







0
M
0
0







−





L
3
/
3
EI
−
L/
6
EI
6
EI/L
2
−
6
EI/L
2
−
L/
6
EI
L/
3
EI
6
EI/L
2
−
6
EI/L
2
6
EI/L
2
6
EI/L
2
12
EI/L
3
−
12
EI/L
3
−
6
EI/L
2
−
6
EI/L
2
−
12
EI/L
3
12
EI/L
3












0
M
0
0







=







0
M
0
0







6. Solve for the displacements,
∆
t
=
K
−
1
tt
P
0
t
, and overwrite
P
t
by
∆
t







θ
1
θ
2
0
0







=





L
3
/
3
EI
−
L/
6
EI
6
EI/L
2
−
6
EI/L
2
−
L/
6
EI
L/
3
EI
6
EI/L
2
−
6
EI/L
2
6
EI/L
2
6
EI/L
2
12
EI/L
3
−
12
EI/L
3
−
6
EI/L
2
−
6
EI/L
2
−
12
EI/L
3
12
EI/L
3












0
M
0
0







=









−
M L/
6
EI
M L/
3
EI
0
0









7. Solve for the reactions,
R
t
=
K
ut
∆
tt
+
K
uu
∆
u
, and overwrite
∆
u
by
R
u







−
M L/
6
EI
M L/
3
EI
R
1
R
2







=





L
3
/
3
EI
−
L/
6
EI
6
EI/L
2
−
6
EI/L
2
−
L/
6
EI
L/
3
EI
6
EI/L
2
−
6
EI/L
2
6
EI/L
2
6
EI/L
2
12
EI/L
3
−
12
EI/L
3
−
6
EI/L
2
−
6
EI/L
2
−
12
EI/L
3
12
EI/L
3














−
M L/
6
EI
M L/
3
EI
0
0









=









−
M L/
6
EI
M L/
3
EI
M/L
−
M/L









Cantilivered Beam/Initial Displacement and Concentrated Moment
1. The
element
stiffness matrix is
k
=



−
2
−
3
−
4
1
−
2
12
EI/L
3
6
EI/L
2
−
12
EI/L
3
6
EI/L
2
−
3
6
EI/L
2
4
EI/L
−
6
EI/L
2
2
EI/L
−
4
−
12
EI/L
3
−
6
EI/L
2
12
EI/L
3
−
6
EI/L
2
1
6
EI/L
2
2
EI/L
−
6
EI/L
2
4
EI/L



2. The
structure
stiffness matrix is assembled
K
=



1
−
2
−
3
−
4
1
4
EI/L
6
EI/L
2
2
EI/L
−
6
EI/L
2
−
2
6
EI/L
2
12
EI/L
3
6
EI/L
2
−
12
EI/L
3
−
3
2
EI/L
6
EI/L
2
4
EI/L
−
6
EI/L
2
−
4
−
6
EI/L
2
−
12
EI/L
3
−
6
EI/L
2
12
EI/L
3



3. The global matrix can be rewritten as





M
√
R
2
?
R
3
?
R
4
?





=



4
EI/L
6
EI/L
2
2
EI/L
−
6
EI/L
2
6
EI/L
2
12
EI/L
3
6
EI/L
2
−
12
EI/L
3
2
EI/L
6
EI/L
2
4
EI/L
−
6
EI/L
2
−
6
EI/L
2
−
12
EI/L
3
−
6
EI/L
2
12
EI/L
3








θ
1
?
∆
2
√
θ
3
√
∆
4
√





Victor Saouma
Structural Engineering
Draft
Chapter 14
DESIGN PHILOSOPHIES of ACI
and AISC CODES
14.1
Safety Provisions
1
Structures and structural members must always be designed to carry some reserve load above
what is expected under normal use. This is to account for
Variability in Resistance:
The
actual
strengths (resistance) of structural elements will dif-
fer from those
assumed
by the designer due to:
1. Variability in the strength of the material (greater variability in concrete strength
than in steel strength).
2. Differences between the actual dimensions and those specified (mostly in placement
of steel rebars in R/C).
3. Effect of simplifying assumptions made in the derivation of certain formulas.
Variability in Loadings:
All loadings are variable. There is a greater variation in the live
loads than in the dead loads. Some types of loadings are very difficult to quantify (wind,
earthquakes).
Consequences of Failure:
The consequence of a structural component failure must be care-
fully assessed. The collapse of a beam is likely to cause a localized failure. Alternatively
the failure of a column is likely to trigger the failure of the whole structure. Alternatively,
the failure of certain components can be preceded by warnings (such as excessive defor-
mation), whereas other are sudden and catastrophic. Finally, if no redistribution of load
is possible (as would be the case in a statically determinate structure), a higher safety
factor must be adopted.
2
The purpose of safety provisions is to limit the
probability of failure
and yet permit
economical structures.
3
The following items must be considered in determining safety provisions:
1. Seriousness of a failure, either to humans or goods.
Draft
14.3 Ultimate Strength Method
351
σ < σ
all
=
σ
yld
F.S.
(14.1)
where
F.S.
is the factor of safety.
10
Major limitations of this method
1. An elastic analysis can not easily account for creep and shrinkage of concrete.
2. For concrete structures, stresses are not linearly proportional to strain beyond 0
.
45
f
0
c
.
3. Safety factors are not rigorously determined from a probabilistic approach, but are the
result of experience and judgment.
11
Allowable strengths are given in Table
Steel, AISC/ASD
Tension, Gross Area
F
t
= 0
.
6
F
y
Tension, Effective Net Area
∗
F
t
= 0
.
5
F
u
Bending
F
b
= 0
.
66
F
y
Shear
F
v
= 0
.
40
F
y
Concrete, ACI/WSD
Tension
0
Compression
0
.
45
f
0
c
∗
Effective net area will be defined in section
Table 14.1: Allowable Stresses for Steel and Concrete
14.3
Ultimate Strength Method
14.3.1
The Normal Distribution
12
The normal distribution has been found to be an excellent approximation to a large class of
distributions, and has some very desirable mathematical properties:
1.
f
(
x
) is symmetric with respect to the mean
µ
.
2.
f
(
x
) is a “bell curve†with inflection points at
x
=
µ
±
σ
.
3.
f
(
x
) is a valid
probability distribution function
as:
Z
∞
−∞
f
(
x
) = 1
(14.2)
4. The
probability
that
x
min
< x < x
max
is given by:
P
(
x
min
< x < x
max
) =
Z
x
max
x
min
f
(
x
)
dx
(14.3)
Victor Saouma
Structural Engineering
Draft
14.3 Ultimate Strength Method
353
Figure 14.3: Frequency Distributions of Load
Q
and Resistance
R
Failure would occur for negative values of
X
19
The
probability of failure
P
f
is equal to the ratio of the shaded area to the total area
under the curve in Fig.
Figure 14.4: Definition of Reliability Index
20
If
X
is assumed to follow a
Normal Distribution
than it has a mean value
X
=
ln
R
Q
m
and a standard deviation
σ
.
21
We define the
safety index
(or
reliability index
) as
β
=
X
σ
22
For standard distributions and for
β
= 3
.
5, it can be shown that the probability of failure is
P
f
=
1
9
,
091
or 1
.
1
×
10
−
4
. That is 1 in every 10,000 structural members designed with
β
= 3
.
5
will fail because of either excessive load or understrength sometime in its lifetime.
23
Reliability indices are a relative measure of the current condition and provide a qualitative
estimate of the structural performance.
24
Structures with relatively high reliable indices will be expected to perform well. If the value
is too low, then the structure may be classified as a hazard.
25
Target values for
β
are shown in Table
, and in Fig.
Victor Saouma
Structural Engineering
Draft
Chapter 15
LOADS
15.1
Introduction
1
The main purpose of a structure is to transfer load from one point to another: bridge deck to
pier; slab to beam; beam to girder; girder to column; column to foundation; foundation to soil.
2
There can also be secondary loads such as thermal (in restrained structures), differential
settlement of foundations, P-Delta effects (additional moment caused by the product of the
vertical force and the lateral displacement caused by lateral load in a high rise building).
3
Loads are generally subdivided into two categories
Vertical Loads
or gravity load
1.
dead load
(DL)
2.
live load
(LL)
also included are snow loads.
Lateral Loads
which act horizontally on the structure
1.
Wind load
(WL)
2.
Earthquake load
(EL)
this also includes hydrostatic and earth loads.
4
This distinction is helpful not only to compute a structure’s load, but also to assign different
factor of safety to each one.
5
For a detailed coverage of loads, refer to the Universal Building Code (UBC), (UBC 1995).
15.2
Vertical Loads
6
For closely spaced identical loads (such as joist loads), it is customary to treat them as a
uniformly distributed load rather than as discrete loads, Fig.
Draft
15.2 Vertical Loads
361
Material
lb
/
ft
2
Ceilings
Channel suspended system
1
Acoustical fiber tile
1
Floors
Steel deck
2-10
Concrete-plain 1 in.
12
Linoleum 1/4 in.
1
Hardwood
4
Roofs
Copper or tin
1-5
5 ply felt and gravel
6
Shingles asphalt
3
Clay tiles
9-14
Sheathing wood
3
Insulation 1 in. poured in place
2
Partitions
Clay tile 3 in.
17
Clay tile 10 in.
40
Gypsum Block 5 in.
14
Wood studs 2x4 (12-16 in. o.c.)
2
Plaster 1 in. cement
10
Plaster 1 in. gypsum
5
Walls
Bricks 4 in.
40
Bricks 12 in.
120
Hollow concrete block (heavy aggregate)
4 in.
30
8 in.
55
12 in.
80
Hollow concrete block (light aggregate)
4 in.
21
8 in.
38
12 in.
55
Table 15.2: Weights of Building Materials
Material
lb
/
ft
2
Timber
40-50
Steel
50-80
Reinforced concrete
100-150
Table 15.3: Average Gross Dead Load in Buildings
Victor Saouma
Structural Engineering
Draft
15.3 Lateral Loads
363
Floor
Roof
10
9
8
7
6
5
4
3
2
Total
Cumulative R (%)
8.48
16.96
25.44
33.92
42.4
51.32
59.8
60
60
60
Cumulative LL
20
80
80
80
80
80
80
80
80
80
740
Cumulative R
×
LL
18.3
66.4
59.6
52.9
46.08
38.9
32.2
32
32
32
410
The resulting design live load for the bottom column has been reduced from 740 Kips to
410 Kips .
5. The total dead load is
DL
= (10)(60) = 600 Kips, thus the total reduction in load is
740
−
410
740+600
×
100= 25% .
15.2.3
Snow
19
Roof snow load vary greatly depending on
geographic location and elevation
. They
range from 20 to 45 psf, Fig.
Figure 15.2: Snow Map of the United States, ubc
20
Snow loads are always given on the projected length or area on a slope, Fig.
21
The steeper the roof, the lower the snow retention. For snow loads greater than 20 psf and
roof pitches
α
more than 20
â—¦
the snow load
p
may be reduced by
R
= (
α
−
20)
p
40
−
0
.
5
(psf)
(15.2)
22
Other examples of loads acting on inclined surfaces are shown in Fig.
15.3
Lateral Loads
15.3.1
Wind
23
Wind load depend on:
velocity of the wind, shape of the building, height, geograph-
ical location, texture of the building surface and stiffness of the structure
.
Victor Saouma
Structural Engineering
Draft
Chapter 16
STRUCTURAL MATERIALS
1
Proper understanding of structural materials is essential to both structural analysis and to
structural design.
2
Characteristics of the most commonly used structural materials will be highlighted.
16.1
Steel
16.1.1
Structural Steel
3
Steel is an
alloy
of iron and carbon. Its properties can be greatly varied by altering the
carbon content (always less than 0.5%) or by adding other elements such as silicon, nickel,
manganese and copper.
4
Practically all grades of steel have a Young Modulus equal to
29,000 ksi
, density of 490
lb/cu ft, and a coefficient of thermal expansion equal to 0
.
65
×
10
−
5
/deg F.
5
The yield stress of steel can vary from 40 ksi to 250 ksi. Most commonly used structural steel
are
A36
(
σ
yld
= 36 ksi) and A572 (
σ
yld
= 50 ksi), Fig.
6
Structural steel can be rolled into a wide variety of shapes and sizes. Usually the most
desirable members are those which have a large section moduli (
S
) in proportion to their area
(
A
), Fig.
7
Steel can be bolted, riveted or welded.
8
Sections are designated by the shape of their cross section, their depth and their weight. For
example W 27
×
114 is a W section, 27 in. deep weighing 114 lb/ft.
9
Common sections are:
S
sections were the first ones rolled in America and have a slope on their inside flange surfaces
of 1 to 6.
W
or wide flange sections have a much smaller inner slope which facilitates connections and
rivetting. W sections constitute about 50% of the tonnage of rolled structural steel.
C
are channel sections
MC
Miscellaneous channel which can not be classified as a C shape by dimensions.
Draft
16.1 Steel
389
HP
is a bearing pile section.
M
is a miscellaneous section.
L
are angle sections which may have equal or unequal sides.
WT
is a T section cut from a W section in two.
Properties of structural steel are tabulated in Table
ASTM
Desig.
Shapes Available
Use
σ
y
(ksi)
σ
u
(ksi)
A36
Shapes and bars
Riveted,
bolted,
welded; Buildings and
bridges
36 up through 8 in.
(32
above 8.)
A500
Cold formed welded and
seamless sections;
General
structural
purpose
Riveted,
welded or bolted;
Grade A: 33; Grade B: 42;
Grade C: 46
A501
Hot formed welded and
seamless sections;
Bolted and welded
36
A529
Plates and bars
1
2
in and
less thick;
Building frames and
trusses;
Bolted and
welded
42
A606
Hot and cold rolled sheets;
Atmospheric corrosion
resistant
45-50
A611
Cold rolled sheet in cut
lengths
Cold formed sections
Grade C 33; Grade D 40;
Grade E 80
A 709
Structural shapes, plates
and bars
Bridges
Grade 36: 36 (to 4 in.);
Grade 50: 50; Grade 100:
100 (to 2.5in.) and 90 (over
2.5 to 4 in.)
Table 16.1: Properties of Major Structural Steels
10
Rolled sections, Fig.
and welded ones, Fig
have
residual stresses
. Those originate
during the rolling or fabrication of a member. The member is hot just after rolling or welding,
it cools unevenly because of varying exposure. The area that cool first become stiffer, resist
contraction, and develop compressive stresses. The remaining regions continue to cool and
contract in the plastic condition and develop tensile stresses.
11
Due to those residual stresses, the stress-strain curve of a rolled section exhibits a non-linear
segment prior to the theoretical yielding, Fig.
. This would have important implications
on the flexural and axial strength of beams and columns.
16.1.2
Reinforcing Steel
12
Steel is also used as reinforcing bars in concrete, Table
. Those bars have a deformation
on their surface to increase the bond with concrete, and usually have a yield stress of 60 ksi
.
1
Stirrups which are used as vertical reinforcement to resist shear usually have a yield stress of only 40 ksi.
Victor Saouma
Structural Engineering
Draft
16.1 Steel
391
.
.
.
Maximum
residual
compressive
stress
no residual stress
Ideal coupon containing
Members with
residual stress
F
F
y
p
2
1
3
1
2
3
Average copressive strain
Average stress P/A
Shaded portion indicates area
which has achieved a stress F
y
Figure 16.5: Influence of Residual Stress on Average Stress-Strain Curve of a Rolled Section
Bar Designation
Diameter
Area
Perimeter
Weight
(in.)
(
in
2
)
in
lb/ft
No. 2
2/8=0.250
0.05
0.79
0.167
No. 3
3/8=0.375
0.11
1.18
0.376
No. 4
4/8=0.500
0.20
1.57
0.668
No. 5
5/8=0.625
0.31
1.96
1.043
No. 6
6/8=0.750
0.44
2.36
1.5202
No. 7
7/8=0.875
0.60
2.75
2.044
No. 8
8/8=1.000
0.79
3.14
2.670
No. 9
9/8=1.128
1.00
3.54
3.400
No. 10
10/8=1.270
1.27
3.99
4.303
No. 11
11/8=1.410
1.56
4.43
5.313
No. 14
14/8=1.693
2.25
5.32
7.650
No. 18
18/8=2.257
4.00
7.09
13.60
Table 16.2: Properties of Reinforcing Bars
Victor Saouma
Structural Engineering
Draft
16.3 Concrete
393
or
E
= 33
γ
1
.
5
q
f
0
c
(16.2)
where both
f
0
c
and
E
are in psi and
γ
is in
lbs
/
ft
3
.
24
Typical concrete (compressive) strengths range from 3,000 to 6,000 psi; However
high
strength concrete
can go up to 14,000 psi.
25
All concrete fail at an ultimate strain of 0.003, Fig.
Figure 16.6: Concrete Stress-Strain curve
26
Pre-peak nonlinearity is caused by micro-cracking Fig.
ε
u
f’
c
.5f’
c
linear
non-linear
Figure 16.7: Concrete microcracking
27
The tensile strength of concrete
f
0
t
is about 10% of the compressive strength.
28
Density of normal weight concrete is 145
lbs
/
ft
3
and 100
lbs
/
ft
3
for lightweight concrete.
Victor Saouma
Structural Engineering
Draft
16.6 Steel Section Properties
401
Designation
A
d
t
w
b
f
t
f
bf
2
tf
hc
tw
I
x
S
x
r
x
I
y
S
y
r
y
Z
x
Z
y
J
in
2
in
in
in
in
in
4
in
3
in
in
4
in
3
in
in
3
in
3
in
4
WT 8.x 50.
14.7
8.48
0.585
10.425
0.985
0
12.1
76.8
11.4
2.28
93.10
17.90
2.51
20.70
27.40
3.85
WT 8.x 45.
13.1
8.38
0.525
10.365
0.875
0
13.5
67.2
10.1
2.27
81.30
15.70
2.49
18.10
24.00
2.72
WT 8.x 39.
11.3
8.26
0.455
10.295
0.760
0
15.6
56.9
8.6
2.24
69.20
13.40
2.47
15.30
20.50
1.78
WT 8.x 34.
9.8
8.16
0.395
10.235
0.665
0
18.0
48.6
7.4
2.22
59.50
11.60
2.46
13.00
17.70
1.19
WT 8.x 29.
8.4
8.22
0.430
7.120
0.715
0
16.5
48.7
7.8
2.41
21.60
6.06
1.60
13.80
9.43
1.10
WT 8.x 25.
7.4
8.13
0.380
7.070
0.630
0
18.7
42.3
6.8
2.40
18.60
5.26
1.59
12.00
8.16
0.76
WT 8.x 23.
6.6
8.06
0.345
7.035
0.565
0
20.6
37.8
6.1
2.39
16.40
4.67
1.57
10.80
7.23
0.65
WT 8.x 20.
5.9
8.01
0.305
6.995
0.505
0
23.3
33.1
5.3
2.37
14.40
4.12
1.57
9.43
6.37
0.40
WT 8.x 18.
5.3
7.93
0.295
6.985
0.430
0
24.1
30.6
5.1
2.41
12.20
3.50
1.52
8.93
5.42
0.27
WT 8.x 16.
4.6
7.94
0.275
5.525
0.440
0
25.8
27.4
4.6
2.45
6.20
2.24
1.17
8.27
3.52
0.23
WT 8.x 13.
3.8
7.84
0.250
5.500
0.345
0
28.4
23.5
4.1
2.47
4.80
1.74
1.12
8.12
2.74
0.13
WT 7.x365.
107.0
11.21
3.070
17.890
4.910
0
1.9
739.0
95.4
2.62
2360.00
264.00
4.69
211.00
408.00
714.00
WT 7.x333.
97.8
10.82
2.830
17.650
4.520
0
2.0
622.0
82.1
2.52
2080.00
236.00
4.62
182.00
365.00
555.00
WT 7.x303.
88.9
10.46
2.595
17.415
4.160
0
2.2
524.0
70.6
2.43
1840.00
211.00
4.55
157.00
326.00
430.00
WT 7.x275.
80.9
10.12
2.380
17.200
3.820
0
2.4
442.0
60.9
2.34
1630.00
189.00
4.49
136.00
292.00
331.00
WT 7.x250.
73.5
9.80
2.190
17.010
3.500
0
2.6
375.0
52.7
2.26
1440.00
169.00
4.43
117.00
261.00
255.00
WT 7.x228.
66.9
9.51
2.015
16.835
3.210
0
2.8
321.0
45.9
2.19
1280.00
152.00
4.38
102.00
234.00
196.00
WT 7.x213.
62.6
9.34
1.875
16.695
3.035
0
3.0
287.0
41.4
2.14
1180.00
141.00
4.34
91.70
217.00
164.00
WT 7.x199.
58.5
9.15
1.770
16.590
2.845
0
3.2
257.0
37.6
2.10
1090.00
131.00
4.31
82.90
201.00
135.00
WT 7.x185.
54.4
8.96
1.655
16.475
2.660
0
3.4
229.0
33.9
2.05
994.00
121.00
4.27
74.40
185.00
110.00
WT 7.x171.
50.3
8.77
1.540
16.360
2.470
0
3.7
203.0
30.4
2.01
903.00
110.00
4.24
66.20
169.00
88.30
WT 7.x156.
45.7
8.56
1.410
16.230
2.260
0
4.0
176.0
26.7
1.96
807.00
99.40
4.20
57.70
152.00
67.50
WT 7.x142.
41.6
8.37
1.290
16.110
2.070
0
4.4
153.0
23.5
1.92
722.00
89.70
4.17
50.40
137.00
51.80
WT 7.x129.
37.8
8.19
1.175
15.995
1.890
0
4.9
133.0
20.7
1.88
645.00
80.70
4.13
43.90
123.00
39.30
WT 7.x117.
34.2
8.02
1.070
15.890
1.720
0
5.3
116.0
18.2
1.84
576.00
72.50
4.10
38.20
110.00
29.60
WT 7.x106.
31.0
7.86
0.980
15.800
1.560
0
5.8
102.0
16.2
1.81
513.00
65.00
4.07
33.40
99.00
22.20
WT 7.x 97.
28.4
7.74
0.890
15.710
1.440
0
6.4
89.8
14.4
1.78
466.00
59.30
4.05
29.40
90.20
17.30
WT 7.x 88.
25.9
7.61
0.830
15.650
1.310
0
6.9
80.5
13.0
1.76
419.00
53.50
4.02
26.30
81.40
13.20
WT 7.x 80.
23.4
7.49
0.745
15.565
1.190
0
7.7
70.2
11.4
1.73
374.00
48.10
4.00
22.80
73.00
9.84
WT 7.x 73.
21.3
7.39
0.680
15.500
1.090
0
8.4
62.5
10.2
1.71
338.00
43.70
3.98
20.20
66.30
7.56
WT 7.x 66.
19.4
7.33
0.645
14.725
1.030
0
8.8
57.8
9.6
1.73
274.00
37.20
3.76
18.60
56.60
6.13
WT 7.x 60.
17.7
7.24
0.590
14.670
0.940
0
9.7
51.7
8.6
1.71
247.00
33.70
3.74
16.50
51.20
4.67
WT 7.x 55.
16.0
7.16
0.525
14.605
0.860
0
10.9
45.3
7.6
1.68
223.00
30.60
3.73
14.40
46.40
3.55
WT 7.x 50.
14.6
7.08
0.485
14.565
0.780
0
11.8
40.9
6.9
1.67
201.00
27.60
3.71
12.90
41.80
2.68
WT 7.x 45.
13.2
7.01
0.440
14.520
0.710
0
13.0
36.4
6.2
1.66
181.00
25.00
3.70
11.50
37.80
2.03
WT 7.x 41.
12.0
7.16
0.510
10.130
0.855
0
11.2
41.2
7.1
1.85
74.20
14.60
2.48
13.20
22.40
2.53
WT 7.x 37.
10.9
7.09
0.450
10.070
0.785
0
12.7
36.0
6.3
1.82
66.90
13.30
2.48
11.50
20.30
1.94
WT 7.x 34.
10.0
7.02
0.415
10.035
0.720
0
13.7
32.6
5.7
1.81
60.70
12.10
2.46
10.40
18.50
1.51
WT 7.x 31.
9.0
6.95
0.375
9.995
0.645
0
15.2
28.9
5.1
1.80
53.70
10.70
2.45
9.16
16.40
1.10
WT 7.x 27.
7.8
6.96
0.370
8.060
0.660
0
15.4
27.6
4.9
1.88
28.80
7.16
1.92
8.87
11.00
0.97
WT 7.x 24.
7.1
6.89
0.340
8.030
0.595
0
16.8
24.9
4.5
1.87
25.70
6.40
1.91
8.00
9.82
0.73
WT 7.x 22.
6.3
6.83
0.305
7.995
0.530
0
18.7
21.9
4.0
1.86
22.60
5.65
1.89
7.05
8.66
0.52
WT 7.x 19.
5.6
7.05
0.310
6.770
0.515
0
19.8
23.3
4.2
2.04
13.30
3.94
1.55
7.45
6.07
0.40
WT 7.x 17.
5.0
6.99
0.285
6.745
0.455
0
21.5
20.9
3.8
2.04
11.70
3.45
1.53
6.74
5.32
0.28
WT 7.x 15.
4.4
6.92
0.270
6.730
0.385
0
22.7
19.0
3.5
2.07
9.79
2.91
1.49
6.25
4.49
0.19
WT 7.x 13.
3.8
6.95
0.255
5.025
0.420
0
24.1
17.3
3.3
2.12
4.45
1.77
1.08
5.89
2.77
0.18
WT 7.x 11.
3.3
6.87
0.230
5.000
0.335
0
26.7
14.8
2.9
2.14
3.50
1.40
1.04
5.20
2.19
0.10
WT 6.x168.
49.4
8.41
1.775
13.385
2.955
0
2.7
190.0
31.2
1.96
593.00
88.60
3.47
68.40
137.00
120.00
WT 6.x153.
44.8
8.16
1.625
13.235
2.705
0
3.0
162.0
27.0
1.90
525.00
79.30
3.42
59.10
122.00
92.00
WT 6.x140.
41.0
7.93
1.530
13.140
2.470
0
3.2
141.0
24.1
1.86
469.00
71.30
3.38
51.90
110.00
70.90
WT 6.x126.
37.0
7.70
1.395
13.005
2.250
0
3.5
121.0
20.9
1.81
414.00
63.60
3.34
44.80
97.90
53.50
WT 6.x115.
33.9
7.53
1.285
12.895
2.070
0
3.8
106.0
18.5
1.77
371.00
57.50
3.31
39.40
88.40
41.60
WT 6.x105.
30.9
7.36
1.180
12.790
1.900
0
4.1
92.1
16.4
1.73
332.00
51.90
3.28
34.50
79.70
32.20
WT 6.x 95.
27.9
7.19
1.060
12.670
1.735
0
4.6
79.0
14.2
1.68
295.00
46.50
3.25
29.80
71.30
24.40
WT 6.x 85.
25.0
7.01
0.960
12.570
1.560
0
5.1
67.8
12.3
1.65
259.00
41.20
3.22
25.60
63.00
17.70
WT 6.x 76.
22.4
6.86
0.870
12.480
1.400
0
5.6
58.5
10.8
1.62
227.00
36.40
3.19
22.00
55.60
12.80
WT 6.x 68.
20.0
6.70
0.790
12.400
1.250
0
6.1
50.6
9.5
1.59
199.00
32.10
3.16
19.00
49.00
9.22
WT 6.x 60.
17.6
6.56
0.710
12.320
1.105
0
6.8
43.4
8.2
1.57
172.00
28.00
3.13
16.20
42.70
6.43
WT 6.x 53.
15.6
6.45
0.610
12.220
0.990
0
8.0
36.3
6.9
1.53
151.00
24.70
3.11
13.60
37.50
4.55
WT 6.x 48.
14.1
6.36
0.550
12.160
0.900
0
8.8
32.0
6.1
1.51
135.00
22.20
3.09
11.90
33.70
3.42
WT 6.x 44.
12.8
6.26
0.515
12.125
0.810
0
9.4
28.9
5.6
1.50
120.00
19.90
3.07
10.70
30.20
2.54
WT 6.x 40.
11.6
6.19
0.470
12.080
0.735
0
10.3
25.8
5.0
1.49
108.00
17.90
3.05
9.49
27.20
1.92
WT 6.x 36.
10.6
6.13
0.430
12.040
0.670
0
11.3
23.2
4.5
1.48
97.50
16.20
3.04
8.48
24.60
1.46
WT 6.x 33.
9.5
6.06
0.390
12.000
0.605
0
12.4
20.6
4.1
1.47
87.20
14.50
3.02
7.50
22.00
1.09
WT 6.x 29.
8.5
6.09
0.360
10.010
0.640
0
13.5
19.1
3.8
1.50
53.50
10.70
2.51
6.97
16.30
1.05
WT 6.x 27.
7.8
6.03
0.345
9.995
0.575
0
14.1
17.7
3.5
1.51
47.90
9.58
2.48
6.46
14.60
0.79
WT 6.x 25.
7.3
6.09
0.370
8.080
0.640
0
13.1
18.7
3.8
1.60
28.20
6.97
1.96
6.90
10.70
0.89
WT 6.x 23.
6.6
6.03
0.335
8.045
0.575
0
14.5
16.6
3.4
1.58
25.00
6.21
1.94
6.12
9.50
0.66
WT 6.x 20.
5.9
5.97
0.295
8.005
0.515
0
16.5
14.4
3.0
1.57
22.00
5.51
1.93
5.30
8.41
0.48
WT 6.x 18.
5.2
6.25
0.300
6.560
0.520
0
18.1
16.0
3.2
1.76
12.20
3.73
1.54
5.71
5.73
0.37
WT 6.x 15.
4.4
6.17
0.260
6.520
0.440
0
20.9
13.5
2.8
1.75
10.20
3.12
1.52
4.83
4.78
0.23
WT 6.x 13.
3.8
6.11
0.230
6.490
0.380
0
23.6
11.7
2.4
1.75
8.66
2.67
1.51
4.20
4.08
0.15
WT 6.x 11.
3.2
6.16
0.260
4.030
0.425
0
20.9
11.7
2.6
1.90
2.33
1.16
0.85
4.63
1.83
0.15
WT 6.x 10.
2.8
6.08
0.235
4.005
0.350
0
23.1
10.1
2.3
1.90
1.88
0.94
0.82
4.11
1.49
0.09
WT 6.x 8.
2.4
5.99
0.220
3.990
0.265
0
24.7
8.7
2.0
1.92
1.41
0.71
0.77
3.72
1.13
0.05
WT 6.x 7.
2.1
5.95
0.200
3.970
0.225
0
27.2
7.7
1.8
1.92
1.18
0.59
0.75
3.32
0.95
0.04
WT 5.x 56.
16.5
5.68
0.755
10.415
1.250
0
5.2
28.6
6.4
1.32
118.00
22.60
2.68
13.40
34.60
7.50
WT 5.x 50.
14.7
5.55
0.680
10.340
1.120
0
5.8
24.5
5.6
1.29
103.00
20.00
2.65
11.40
30.50
5.41
WT 5.x 44.
12.9
5.42
0.605
10.265
0.990
0
6.5
20.8
4.8
1.27
89.30
17.40
2.63
9.65
26.50
3.75
WT 5.x 39.
11.3
5.30
0.530
10.190
0.870
0
7.4
17.4
4.0
1.24
76.80
15.10
2.60
8.06
22.90
2.55
WT 5.x 34.
10.0
5.20
0.470
10.130
0.770
0
8.4
14.9
3.5
1.22
66.80
13.20
2.59
6.85
20.00
1.78
WT 5.x 30.
8.8
5.11
0.420
10.080
0.680
0
9.4
12.9
3.0
1.21
58.10
11.50
2.57
5.87
17.50
1.23
WT 5.x 27.
7.9
5.05
0.370
10.030
0.615
0
10.6
11.1
2.6
1.19
51.70
10.30
2.56
5.05
15.70
0.91
WT 5.x 25.
7.2
4.99
0.340
10.000
0.560
0
11.6
10.0
2.4
1.18
46.70
9.34
2.54
4.52
14.20
0.69
WT 5.x 23.
6.6
5.05
0.350
8.020
0.620
0
11.2
10.2
2.5
1.24
26.70
6.65
2.01
4.65
10.10
0.75
WT 5.x 20.
5.7
4.96
0.315
7.985
0.530
0
12.5
8.8
2.2
1.24
22.50
5.64
1.98
3.99
8.59
0.49
WT 5.x 17.
4.8
4.86
0.290
7.960
0.435
0
13.6
7.7
1.9
1.26
18.30
4.60
1.94
3.48
7.01
0.29
WT 5.x 15.
4.4
5.24
0.300
5.810
0.510
0
14.8
9.3
2.2
1.45
8.35
2.87
1.37
4.01
4.42
0.31
WT 5.x 13.
3.8
5.16
0.260
5.770
0.440
0
17.0
7.9
1.9
1.44
7.05
2.44
1.36
3.39
3.75
0.20
WT 5.x 11.
3.2
5.09
0.240
5.750
0.360
0
18.4
6.9
1.7
1.46
5.71
1.99
1.33
3.02
3.05
0.12
WT 5.x 10.
2.8
5.12
0.250
4.020
0.395
0
17.7
6.7
1.7
1.54
2.15
1.07
0.87
3.10
1.68
0.12
WT 5.x 9.
2.5
5.05
0.240
4.010
0.330
0
18.4
6.1
1.6
1.56
1.78
0.89
0.84
2.90
1.40
0.08
WT 5.x 8.
2.2
4.99
0.230
4.000
0.270
0
19.2
5.4
1.5
1.57
1.45
0.72
0.81
3.03
1.15
0.05
WT 5.x 6.
1.8
4.93
0.190
3.960
0.210
0
23.3
4.3
1.2
1.57
1.09
0.55
0.79
2.50
0.87
0.03
Victor Saouma
Structural Engineering
Draft
17.2 Geometric Considerations
411
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
HIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIHIH
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KIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIK
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KIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIKIK
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LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
LILILILILILILILILILILILILILILILILILILILILILIL
Effective hole size
Bolt size
Hole size
D
D +
D +
16
1
8
1
Figure 17.2: Hole Sizes
17
Accordingly, for a plate of width
w
and thickness
t
, and with a single hole which accomodates
a bolt of diameter
D
, the net area would be
A
n
=
wt
−
D
+
1
8
t
(17.2)
18
Whenever there is a need for more than two rivets or bolts, the holes are not aligned but
rather
staggered
. Thus there is more than one potential failure line which can be used to
determine the net area.
19
For staggered holes, we define the net are in terms of, Fig.
Figure 17.3: Effect of Staggered Holes on Net Area
pitch
s
, longitudinal center to center spacing between two consecutive holes.
gage
g
, transverse center to center spacing between two adjacent holes.
Victor Saouma
Structural Engineering
Draft
17.2 Geometric Considerations
413
Leg
8
7
6
5
4
3
1
2
3
2
1
2
2
1
3
4
1
1
2
1
3
8
1
1
4
1
g
4
1
2
4
3
1
2
3
2
1
2
2
1
3
4
1
3
8
1
1
8
1
7
8
7
8
3
4
5
8
g
1
3
2
1
2
2
1
4
2
g
2
3
3
2
1
2
1
3
4
Table 17.1: Usual Gage Lengths
g
,
g
1
,
g
2
Solution:
We consider various paths:
A
−
D
:
h
12
−
2
15
16
+
1
16
i
(0
.
25)
= 2
.
50
in
2
(17.5-a)
A
−
B
−
D
:
h
12
−
3
15
16
+
1
16
+
(2
.
125)
2
4(2
.
5)
+
(2
.
125)
2
4(4)
i
(0
.
25) = 2
.
43
in
2
(17.5-b)
A
−
B
−
C
:
h
12
−
3
15
16
+
1
16
+
(2
.
125)
2
4(2
.
5)
+
(1
.
875)
2
4(4)
i
(0
.
25) = 2
.
4
in
2
(17.5-c)
Thus path A-B-C controls.
Example 17-2: Net Area, Angle
Determine the net area
A
n
for the angle shown below if
15
16
in. diameter holes are used.
Victor Saouma
Structural Engineering
Draft
17.2 Geometric Considerations
415
Type of members
Minimum Number
Special
A
e
of Fasteners/line
Requirements
Fastened Rolled Sections
Members having all cross sectional elements
connected to transmit tensile force
1 (or welds)
A
n
W, M, or S shapes with flange widths not less
than 2/3 the depth, and structural tees cut
from these shapes, provided the connection is
to the flanges. Bolted or riveted connections
shall have no fewer than three fasteners per
line in the direction of stress
3 (or welds)
b
d
≥
0
.
67
0
.
9
A
n
W, M, or S not meeting above conditions,
structural tees cut from these shapes and
all other shapes, including built-up cross-
sections. Bolted or riveted connections shall
have no fewer than three fasteners per line in
the direction of stress
3 (or welds)
0
.
85
A
n
Structural tees cut from W, M, or S con-
nected at flanges only
3 (or welds)
b
d
≥
0
.
67
0
.
9
A
n
All members with bolted or riveted connec-
tions having only two fasteners per line in the
direction of stress
2
0
.
75
A
n
Welded Plates
Welds
l >
2
w
A
g
Welds
2
w > l >
1
.
5
w
0
.
87
A
g
Welds
l/w.
1
0
.
75
A
g
b
Flange width;
d
section depth;
l
Weld length;
w
Plate width;
Table 17.2: Effective Net Area
A
e
for Bolted and Welded Connections
Victor Saouma
Structural Engineering
Draft
17.3 LRFD Design of Tension Members
417
where
Φ
t
=
resistance factor relating to tensile strength
T
n
=
nominal strength of a tension member
T
u
=
factored load on a tension member
30
From Table
, Φ
t
is 0.75 and 0.9 for fracture and yielding failure respectively.
17.3.1
Tension Failure
31
The design strength Φ
t
T
n
is the smaller of that based on, Fig.
1.
Yielding in the gross section:
We can not allow yielding of the gross section, because
this will result in unacceptable elongation of the entire member.
Φ
t
T
n
= Φ
t
F
y
A
g
= 0
.
90
F
y
A
g
(17.12)
or
2.
Fracture in the net section:
Yielding is locally allowed, because
A
e
is applicable only
on a small portion of the element. Local excessive elongation is allowed, however fracture
must be prevented. This mode of failure usually occurs if there is insufficient distance
behind the pin.
Φ
t
T
n
= Φ
t
F
u
A
e
= 0
.
75
F
u
A
e
(17.13)
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o
o o
o o
o
0.75F A
0.9F A
0.75F A
u e
u e
y g
Figure 17.5: Net and Gross Areas
17.3.2
Block Shear Failure
32
For bolted connections,
tearing failure
may occur and control the strength of the tension
member.
33
For instance, with reference to Fig.
the angle tension member attached to the
gusset
Victor Saouma
Structural Engineering
Draft
Chapter 18
COLUMN STABILITY
18.1
Introduction; Discrete Rigid Bars
18.1.1
Single Bar System
1
Let us begin by considering a rigid bar connected to the support by a spring and axially
loaded at the other end, Fig.
. Taking moments about point
A
:
M
M
M
M
M
M
M
M
M
N
N
N
N
N
N
N
N
N
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
P
QRQRQRQRQRQ
QRQRQRQRQRQ
QRQRQRQRQRQ
QRQRQRQRQRQ
SRSRSRSRSRS
SRSRSRSRSRS
SRSRSRSRSRS
SRSRSRSRSRS
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
TRTRTRTRT
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
URURURURU
VRVRVRVRV
VRVRVRVRV
VRVRVRVRV
VRVRVRVRV
WRWRWRWRW
WRWRWRWRW
WRWRWRWRW
WRWRWRWRW
P
B
k
A
B
P
∆
θ
A
θ
P
Stable Equilibrium
Neutral Equilibrium
Unstable Equilibrium
(stable)
L
L
k
sin
θ
θ
L
k
XYXYX
XYXYX
XYXYX
ZYZYZ
ZYZYZ
ZYZYZ
[Y[Y[
[Y[Y[
[Y[Y[
\Y\Y\
\Y\Y\
\Y\Y\
]Y]Y]
]Y]Y]
]Y]Y]
^Y^Y^
^Y^Y^
^Y^Y^
Stable
Neutral
Unstable
Figure 18.1: Stability of a Rigid Bar
Σ
M
A
=
P
∆
−
kθ
= 0
(18.1-a)
∆ =
Lθ
for small rotation
(18.1-b)
P θL
−
kθ
= 0
(18.1-c)
(
P
−
k
L
) = 0
(18.1-d)
Draft
18.1 Introduction; Discrete Rigid Bars
439
θ
0
θ
0
θ
0
_`_`_`_`_`_
_`_`_`_`_`_
_`_`_`_`_`_
_`_`_`_`_`_
_`_`_`_`_`_
_`_`_`_`_`_
_`_`_`_`_`_
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d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
d
L
B
P
∆
θ
A
θ
P
Perfect System
L
(1-
θ
)
L
k
k
k
Figure 18.2: Stability of a Rigid Bar with Initial Imperfection
θ
2
θ
1
k( - )
θ
2
θ
1
k( - )
θ
1
θ
2
θ
1
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l
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l
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m
m
m
m
m
m
m
m
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m
m
m
m
m
m
m
m
m
m
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m
m
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zfz
L
k
L
k
A
B
C
P
C
A
B
P
P
θ
2
B
k
θ
1
L
B
C
1
.618
P
L
P
P
P
P
Figure 18.3: Stability of a Two Rigid Bars System
Victor Saouma
Structural Engineering
Draft
18.1 Introduction; Discrete Rigid Bars
441
"
1
−
√
5
2
−
1
−
1
1
−
−
1
−
√
5
2
# (
θ
1
θ
2
)
=
(
0
0
)
(18.12-c)
"
−
0
.
618
−
1
−
1
−
1
.
618
# (
θ
1
θ
2
)
=
(
0
0
)
(18.12-d)
we now arbitrarily set
θ
1
= 1, then
θ
2
=
−
1
/
1
.
618 =
−
0
.
618, thus the second eigenmode is
(
θ
1
θ
2
)
=
(
1
−
0
.
618
)
(18.13)
18.1.3
‡
Analogy with Free Vibration
12
The problem just considered bears great ressemblence with the vibration of a two degree of
freedom mass spring system, Fig.
m = m
1
2
m = 2m
1
k = k
u
2
1
u
1
u
u
2
u
2
k u
2
k(u -u )
2
1
2m u
2
..
m u
1
..
k u
1
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
{|{
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
}|}
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~|~
~|~
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
k = k
2
k = k
3
Figure 18.4: Two DOF Dynamic System
13
Each mass is subjected to an inertial force equals to the mass times the acceleration, and
the spring force:
2
m
¨
u
2
+
k
u
2
+
k
(
u
2
−
u
1
) = 0
(18.14-a)
m
¨
u
1
+
k
u
1
+
k
9
u
2
−
u
1
0 = 0
(18.14-b)
or in matrix form
"
m
0
0
2
m
#
|
{z
}
M
(
¨
u
1
¨
u
2
)
|
{z
}
¨
U
+
"
2
k
−
k
−
k
2
k
#
|
{z
}
K
(
u
1
u
2
)
|
{z
}
U
(18.15)
14
The characteristic equation is
|
K
−
λ
M
|
where
λ
=
ω
2
, and
ω
is the natural frequency.
2
h
−
λ
−
h
−
h
2
h
−
2
λ
= 0
(18.16)
Victor Saouma
Structural Engineering
Draft
Chapter 19
STEEL COMPRESSION
MEMBERS
1
This chapter will cover the elementary analysis and design of steel column according to the
LRFD provisions.
19.1
AISC Equations
2
By introducing a
slenderness parameter
λ
c
(not to be confused with the slenderness ratio)
defined as
λ
2
c
=
F
y
F
Euler
cr
=
F
y
Ï€
2
E
KL
rmin
2
(19.1-a)
λ
c
=
KL
r
min
s
F
y
Ï€
2
E
(19.1-b)
Hence, this parameter accounts for both the slenderness ratio as well as steel material prop-
erties. This new parameter is a more suitable one than the slenderness ratio (which was a
delimeter for elastic buckling) for the inelastic buckling.
3
Equation
becomes
F
cr
F
y
= 1
−
λ
2
c
4
for
λ
c
≤
√
2
(19.2)
4
When
λ
c
>
√
2, then Euler equation applies
F
cr
F
y
=
1
λ
2
c
for
λ
c
≥
√
2
(19.3)
Hence the first equation is based on inelastic buckling (with gross yielding as a limiting case),
and the second one on elastic buckling. The two curves are tangent to each other.
5
The above equations are valid for concentrically straight members.
Draft
19.1 AISC Equations
457
Kl
r
min
λ
c
φ
c
F
cr
Kl
r
min
λ
c
φ
c
F
cr
Kl
r
min
λ
c
φ
c
F
cr
Kl
r
min
λ
c
φ
c
F
cr
Kl
r
min
λ
c
φ
c
F
cr
1
0.01
30.60
41
0.46
28.01
81
0.91
21.66
121
1.36
14.16
161
1.81
8.23
2
0.02
30.59
42
0.47
27.89
82
0.92
21.48
122
1.37
13.98
162
1.82
8.13
3
0.03
30.59
43
0.48
27.76
83
0.93
21.29
123
1.38
13.80
163
1.83
8.03
4
0.04
30.57
44
0.49
27.63
84
0.94
21.11
124
1.39
13.62
164
1.84
7.93
5
0.06
30.56
45
0.50
27.51
85
0.95
20.92
125
1.40
13.44
165
1.85
7.84
6
0.07
30.54
46
0.52
27.37
86
0.96
20.73
126
1.41
13.27
166
1.86
7.74
7
0.08
30.52
47
0.53
27.24
87
0.98
20.54
127
1.42
13.09
167
1.87
7.65
8
0.09
30.50
48
0.54
27.10
88
0.99
20.35
128
1.44
12.92
168
1.88
7.56
9
0.10
30.47
49
0.55
26.97
89
1.00
20.17
129
1.45
12.74
169
1.90
7.47
10
0.11
30.44
50
0.56
26.83
90
1.01
19.98
130
1.46
12.57
170
1.91
7.38
11
0.12
30.41
51
0.57
26.68
91
1.02
19.79
131
1.47
12.40
171
1.92
7.30
12
0.13
30.37
52
0.58
26.54
92
1.03
19.60
132
1.48
12.23
172
1.93
7.21
13
0.15
30.33
53
0.59
26.39
93
1.04
19.41
133
1.49
12.06
173
1.94
7.13
14
0.16
30.29
54
0.61
26.25
94
1.05
19.22
134
1.50
11.88
174
1.95
7.05
15
0.17
30.24
55
0.62
26.10
95
1.07
19.03
135
1.51
11.71
175
1.96
6.97
16
0.18
30.19
56
0.63
25.94
96
1.08
18.84
136
1.53
11.54
176
1.97
6.89
17
0.19
30.14
57
0.64
25.79
97
1.09
18.65
137
1.54
11.37
177
1.99
6.81
18
0.20
30.08
58
0.65
25.63
98
1.10
18.46
138
1.55
11.20
178
2.00
6.73
19
0.21
30.02
59
0.66
25.48
99
1.11
18.27
139
1.56
11.04
179
2.01
6.66
20
0.22
29.96
60
0.67
25.32
100
1.12
18.08
140
1.57
10.89
180
2.02
6.59
21
0.24
29.90
61
0.68
25.16
101
1.13
17.89
141
1.58
10.73
181
2.03
6.51
22
0.25
29.83
62
0.70
24.99
102
1.14
17.70
142
1.59
10.58
182
2.04
6.44
23
0.26
29.76
63
0.71
24.83
103
1.16
17.51
143
1.60
10.43
183
2.05
6.37
24
0.27
29.69
64
0.72
24.66
104
1.17
17.32
144
1.61
10.29
184
2.06
6.30
25
0.28
29.61
65
0.73
24.50
105
1.18
17.13
145
1.63
10.15
185
2.07
6.23
26
0.29
29.53
66
0.74
24.33
106
1.19
16.94
146
1.64
10.01
186
2.09
6.17
27
0.30
29.45
67
0.75
24.16
107
1.20
16.75
147
1.65
9.87
187
2.10
6.10
28
0.31
29.36
68
0.76
23.99
108
1.21
16.56
148
1.66
9.74
188
2.11
6.04
29
0.33
29.27
69
0.77
23.82
109
1.22
16.37
149
1.67
9.61
189
2.12
5.97
30
0.34
29.18
70
0.79
23.64
110
1.23
16.18
150
1.68
9.48
190
2.13
5.91
31
0.35
29.09
71
0.80
23.47
111
1.24
16.00
151
1.69
9.36
191
2.14
5.85
32
0.36
28.99
72
0.81
23.29
112
1.26
15.81
152
1.70
9.23
192
2.15
5.79
33
0.37
28.90
73
0.82
23.11
113
1.27
15.62
153
1.72
9.11
193
2.16
5.73
34
0.38
28.79
74
0.83
22.94
114
1.28
15.44
154
1.73
9.00
194
2.18
5.67
35
0.39
28.69
75
0.84
22.76
115
1.29
15.25
155
1.74
8.88
195
2.19
5.61
36
0.40
28.58
76
0.85
22.58
116
1.30
15.07
156
1.75
8.77
196
2.20
5.55
37
0.41
28.47
77
0.86
22.40
117
1.31
14.88
157
1.76
8.66
197
2.21
5.50
38
0.43
28.36
78
0.87
22.21
118
1.32
14.70
158
1.77
8.55
198
2.22
5.44
39
0.44
28.25
79
0.89
22.03
119
1.33
14.52
159
1.78
8.44
199
2.23
5.39
40
0.45
28.13
80
0.90
21.85
120
1.35
14.34
160
1.79
8.33
200
2.24
5.33
Table 19.1: Design Stress
φF
cr
for
F
y
=36 ksi in Terms of
KL
r
min
Victor Saouma
Structural Engineering
Draft
Chapter 20
BRACED ROLLED STEEL BEAMS
1
This chapter deals with the behavior and design of
laterally supported
steel beams accord-
ing to the LRFD provisions.
2
If a beam is not laterally supported, we will have a failure mode governed by lateral torsional
buckling.
3
By the end of this lecture you should be able to select the most efficient section (light weight
with adequate strength) for a given bending moment and also be able to determine the flexural
strength of a given beam.
20.1
Review from Strength of Materials
20.1.1
Flexure
4
shows portion of an originally straight beam which has been bent to the radius
Ï
by end couples
M
, thus the segment is subjected to pure bending. It is assumed that plane
cross-sections normal to the length of the unbent beam remain plane after the beam is bent.
Therefore, considering the cross-sections
AB
and
CD
a unit distance apart, the similar sectors
Oab
and
bcd
give
ε
=
y
Ï
(20.1)
where
y
is measured from the axis of rotation (neutral axis). Thus strains are proportional to
the distance from the neutral axis.
5
The corresponding variation in stress over the cross-section is given by the stress-strain dia-
gram of the material rotated 90
o
from the conventional orientation, provided the strain axis
ε
is scaled with the distance
y
). The bending moment
M
is given by
M
=
Z
yσdA
(20.2)
where
dA
is an differential area a distance
y
) from the neutral axis. Thus the moment
M
can be determined if the stress-strain relation is known.
Draft
20.1 Review from Strength of Materials
469
Figure 20.2: Stress distribution at different stages of loading
Elastic Region
Plastic Region
Stress
Strain
ε
F
F
y
Figure 20.3: Stress-strain diagram for most structural steels
Victor Saouma
Structural Engineering
Draft
20.2 Nominal Strength
471
F
2
=
Z
A
(
M
+
dM
)
y
I
dA
(20.8-c)
Ï„ bdx
=
F
2
−
F
1
(20.8-d)
=
Z
A
dM y
I
dA
(20.8-e)
Ï„
=
dM
dx
1
Ib
Z
A
ydA
(20.8-f)
(20.8-g)
substituting
V
=
dM/dx
we now obtain
Ï„
=
V Q
Ib
Q
=
R
A
ydA
(20.9)
this is the shear formula, and
Q
is the first moment of the area.
12
For a rectangular beam, we will have a parabolic shear stress distribution.
13
For a W section, it can be easily shown that about 95% of the shear force is carried by the
web, and that the average shear stress is within 10% of the actual maximum shear stress.
20.2
Nominal Strength
The strength requirement for beams in load and resistance factor design is stated as
φ
b
M
n
≥
M
u
(20.10)
where:
φ
b
strength reduction factor; for flexure 0
.
90
M
n
nominal moment strength
M
u
factored service load moment.
14
The equations given in this chapter are valid for flexural members with the following kinds
of cross section and loading:
1. Doubly symmetric (such as W sections) and loaded in Plane of symmetry
2. Singly symmetric (channels and angles) loaded in plane of symmetry or through the shear
center parallel to the web
20.3
Flexural Design
20.3.1
Failure Modes and Classification of Steel Beams
15
The strength of flexural members is limited by:
Plastic Hinge:
at a particular cross section.
1
More about shear centers in
Mechanics of Materials II
.
Victor Saouma
Structural Engineering
Draft
20.3 Flexural Design
473
19
The nominal strength
M
n
for laterally stable compact sections according to LRFD is
M
n
=
M
p
(20.12)
where:
M
p
plastic moment strength =
ZF
y
Z
plastic section modulus
F
y
specified minimum yield strength
20
Note that section properties, including
Z
values are tabulated in Section
20.3.1.2
Partially Compact Section
21
If the width to thickness ratios of the compression elements exceed the
λ
p
values mentioned
in Eq.
but do not exceed the following
λ
r
, the section is partially compact and we can
have local buckling.
Flange:
λ
p
<
b
f
2
t
f
≤
λ
r
λ
p
=
65
√
F
y
λ
r
=
141
√
F
y
−
F
r
Web:
λ
p
<
h
c
t
w
≤
λ
r
λ
p
=
640
√
F
y
λ
r
=
970
√
F
y
(20.13)
where, Fig.
F
y
specified minimum yield stress in
ksi
b
f
width of the flange
t
f
thickness of the flange
h
c
unsupported height of the web which is twice the distance from the neutral axis
to the inside face of the compression flange less the fillet or corner radius.
t
w
thickness of the web.
F
r
residual stress = 10
.
0 ksi for rolled sections and 16
.
5 ksi for welded sections.
Figure 20.6: W Section
22
The nominal strength of partially compact sections according to LRFD is, Fig.
Victor Saouma
Structural Engineering
Draft
Chapter 21
UNBRACED ROLLED STEEL
BEAMS
21.1
Introduction
1
In a previous chapter we have examined the behavior of laterally supported beams. Under
those conditions, the potential modes of failures were either the formation of a plastic hinge (if
the section is compact), or local buckling of the flange or the web (partially compact section).
2
Rarely are the compression flange of beams entirely free of all restraint, and in general there
are two types of lateral supports:
1. Continuous lateral support by embedment of the compression flange in a concrete slab.
2. Lateral support at intervals through cross beams, cross frames, ties, or struts.
3
Now that the beam is not laterally supported, we ought to consider a third potential mode
of failure, lateral torsional buckling.
21.2
Background
4
Whereas it is beyond the scope of this course to derive the governing differential equation
for flexural torsional buckling (which is covered in either
Mechanics of Materials II
or in
Steel
Structures
), we shall review some related topics in order to understand the AISC equations
later on
5
There are two types of torsion:
Saint-Venant’s torsion:
or pure torsion (torsion is constant throughout the length) where it
is assumed that the cross-sectional plane prior to the application of torsion remains plane,
and only rotation occurs.
Warping torsion:
out of plane effects arise when the flanges are laterally displaced during
twisting. Compression flange will bend in one direction laterally while its tension flange
will bend in another. In this case part of the torque is resisted by bending and the rest
by Saint-Venant’s torsion.
Draft
21.3 AISC Equations
485
X
1
=
Ï€
S
x
s
EGJA
2
(21.4-a)
X
2
= 4
C
w
I
y
S
x
GJ
2
(21.4-b)
are not really physical properties but a combination of cross-sectional ones which simplifies
the writing of Eq.
. Those values are tabulated in the AISC manual.
9
The flexural efficiency of the member increases when
X
1
decreases and/or
X
2
increases.
21.3.2
Governing Moments
1.
L
b
< L
p
: “very short†Plastic hinge
M
n
=
M
p
=
Z
x
F
y
(21.5)
2.
L
p
< L
b
< L
r
: “short†inelastic lateral torsional buckling
M
n
=
C
b
"
M
p
−
(
M
p
−
M
r
)
L
b
−
L
p
L
r
−
L
p
!#
≤
M
p
(21.6)
and
C
b
= 1
.
75 + 1
.
05
M
1
M
2
+ 0
.
3
M
1
M
2
2
≤
2
.
3
(21.7)
where
M
1
is the smaller and
M
2
is the larger end moment
in the unbraced segment
M
1
M
2
is negative when the moments cause single curvature (i.e. one of them is clockwise,
the other counterclockwise), hence the most severe loading case with constant
M
gives
C
b
= 1
.
75
−
1
.
05 + 0
.
3 = 1
.
0.
M
r
is the moment strength available for service load when extreme fiber reach the yield
stress
F
y
;
M
r
= (
F
y
−
F
r
)
S
x
(21.8)
3.
L
r
< L
b
“long†elastic lateral torsional buckling, and the critical moment is the same as
in Eq.
M
cr
=
C
b
Ï€
L
b
s
Ï€E
L
b
2
C
w
I
y
+
EI
y
GJ
≤
C
b
M
r
and
M
p
(21.9)
or if expressed in terms of
X
1
and
X
2
M
cr
=
C
b
S
x
X
1
√
2
L
b
r
y
v
u
u
u
t
1 +
X
2
1
X
2
2
L
b
r
y
2
≤
C
b
M
r
(21.10)
10
Fig.
summarizes the governing equations.
Victor Saouma
Structural Engineering
Draft
Chapter 22
Beam Columns, (Unedited)
UNEDITED
Examples taken from
Structural Steel Design; LRFD
, by T. Burns, Delmar Publishers,
1995
22.1
Potential Modes of Failures
22.2
AISC Specifications
P
u
φ
c
P
n
+
8
9
M
ux
φ
b
M
nx
+
M
uy
φ
b
M
ny
!
≤
1
.
0 if
P
u
φ
c
P
n
≥
.
20
P
u
2
φ
c
P
n
+
M
ux
φ
b
M
nx
≤
1
.
0 if
P
u
φ
c
P
n
≤
.
20
(22.1)
22.3
Examples
22.3.1
Verification
Example 22-1: Verification, (?)
A W 12
×
120 is used to support the loads and moments as shown below, and is not subjected
to sidesway. Determine if the member is adequate and if the factored bending moment occurs
about the weak axis. The column is assumed to be perfectly pinned (
K
= 1
.
0) in both the
strong and weak directions and no bracing is supplied. Steel is A36 and assumed
C
b
= 1
.
0.
Draft
22.3 Examples
499
be developed or whether buckling behavior controls the bending behavior. The values for
L
p
and
L
r
can be calculated using Eq. 6-1 and 6-2 respectively or they can be found in
the beam section of the LRFD manual. In either case these values for a W 12
×
120 are:
L
p
= 13
ft
L
r
= 75
.
5
ft
5. Since our unbraced length falls between these two values, the beam will be controlled
by inelastic buckling, and the nominal moment capacity
M
n
can be calculated from Eq.
6-5. Using this equation we must first calculate the plastic and elastic moment capacity
values,
M
p
and
M
r
.
M
p
=
F
y
Z
x
= (36)
ksi
(186)
in
3
ft
(12)
in
= 558
k.ft
M
r
= (
F
yw
−
10)
ksi
S
x
(36
−
10)
ksi
(163)
in
3
= 353
.
2
k.ft
6. Using Eq. 6-5 (assuming
C
b
is equal to 1.0) we find:
M
n
=
C
b
[
M
p
−
(
M
p
−
M
r
)]
l
b
−
L
p
L
r
−
L
p
M
n
= 1
.
0[558
−
(558
−
453
.
2)]
k.ft
(15
−
13)
ft
(75
.
5
−
15)
ft
= 551
.
4
k.ft
7. Therefore the design moment capacity is as follows:
φ
b
M
n
= 0
.
90(551
.
4)
k.ft
= 496
.
3
k.ft
8. Now consider the effects of moment magnification on this section. Based on the alternative
method and since the member is not subjected to sidesway (
M
lt
= 0)
M
u
=
B
1
M
nt
B
1
=
C
m
1
−
Pu
Pe
C
m
=
.
60
−
.
4
M
1
M
2
C
m
=
.
60
−
.
4
−
50
50
= 1
.
0
P
u
= 400
k
P
e
=
Ï€
2
EA
g
Kl
r
2
(Euler’s Buckling Load Equation)
P
e
=
Ï€
2
(29
,
000
ksi
)(35
.
3
in
2
)
32
.
67
2
= 9
,
456
k
9. Therefore, calculating the
B
1
magnifier we find:
B
1
=
1
1
−
(400)
k
(9
,
456)
k
= 1
.
044
Calculating the amplified moment as follows:
M
u
=
B
1
M
nt
M
u
= 1
.
044(50)
k.ft
= 52
.
2
k.ft
Therefore the adequacy of the section is calculated from Eq. as follows:
P
u
φ
c
P
n
+
8
9
M
u
x
φ
b
M
n
x
≤
1
.
0
94000
k
(907
.
6)
k
+
8
9
(52
.
2)
k.ft
(496
.
3)
k.ft
=
.
53
<
1
.
0
√
Victor Saouma
Structural Engineering
Draft
Chapter 23
STEEL CONNECTIONS
23.1
Bolted Connections
1
Bolted connections, Fig.
are increasingly used instead of rivets (too labor intensive) and
more often than welds (may cause secondary cracks if not properly performed).
23.1.1
Types of Bolts
2
Most common high strength bolts are the A325 and A490.
3
A325 is made from heat-treated medium carbon steel, min.
F
b
y
≈
81
−
92 ksi,
F
b
u
= 120 ksi
4
A490 is a higher strength manufactured from an alloy of steel, min.
F
b
y
≈
115
−
130 ksi, and
F
b
u
= 150 ksi
5
Most common diameters are 3/4â€, 7/8†for building constructions; 7/8†and 1†for bridges,
Table
Bolt Diameter (in.)
Nominal Area (in
2
)
5/8
0.3068
3/4
0.4418
7/8
0.6013
1
0.7854
1 1/8
0.9940
1 1/4
1.2272
Table 23.1: Nominal Areas of Standard Bolts
23.1.2
Types of Bolted Connections
6
There are two types of bolted connections:
Bearing type
which transmits the load by a combination of shear and bearing on the bolt,
Fig.
Draft
23.1 Bolted Connections
509
Slip-critical
transmits load by friction, Fig.
. In addition of providing adequate at ulti-
mate load, it must not slip during service loads.
€R€
€R€
€R€
€R€
€R€
€R€
ÂRÂ
ÂRÂ
ÂRÂ
ÂRÂ
ÂRÂ
ÂRÂ
P
‚R‚
‚R‚
‚R‚
ƒRƒ
ƒRƒ
ƒRƒ
P
P
P
P
„R„
„R„
„R„
Â…RÂ…
Â…RÂ…
Â…RÂ…
†R†
†R†
†R†
†R†
†R†
‡R‡
‡R‡
‡R‡
‡R‡
‡R‡
P
P
P
P
P
Figure 23.2: Stress Transfer by Shear and Bearing in a Bolted Connection
µ
T
µ
T
µ
T
P
P
T
T
T
T
P=
High-strength bolt
P
P
Figure 23.3: Stress Transfer in a Friction Type Bolted Connection
7
Possible failure modes (or “limit statesâ€) which may control the strength of a bolted connec-
tion are shown in Fig.
23.1.3
Nominal Strength of Individual Bolts
Tensile Strength
The nominal strength
R
n
of one fastener in tension is
R
n
=
F
b
u
A
n
(23.1)
where
F
b
u
is the tensile strength of the bolt, and
A
n
the area through the threaded portion,
also known as “tensile stress areaâ€. The ratio of the tensile stress area to the gross area
A
g
ranges from 0.75 to 0.79.
Victor Saouma
Structural Engineering
Draft
Chapter 24
REINFORCED CONCRETE
BEAMS; Part I
24.1
Introduction
1
Recalling that concrete has a tensile strength (
f
0
t
) about one tenth its compressive strength
(
f
0
c
), concrete by itself is a very poor material for flexural members.
2
To provide tensile resistance to concrete beams, a reinforcement must be added. Steel is
almost universally used as reinforcement (longitudinal or as fibers), but in poorer countries
other indigenous materials have been used (such as bamboos).
3
The following lectures will focus exclusively on the flexural design and analysis of reinforced
concrete rectangular sections. Other concerns, such as shear, torsion, cracking, and deflections
are left for subsequent ones.
4
Design of reinforced concrete structures is governed in most cases by the
Building Code
Requirements for Reinforced Concrete
, of the American Concrete Institute (ACI-318). Some of
the most relevant provisions of this code are enclosed in this set of notes.
5
We will focus on determining the amount of flexural (that is longitudinal) reinforcement
required at a
given section
. For that section, the moment which should be considered for
design is the one obtained from the
moment envelope
at that particular point.
24.1.1
Notation
6
In R/C design, it is customary to use the following notation
Draft
24.2 Cracked Section, Ultimate Strength Design Method
519
24.1.4
Basic Relations and Assumptions
13
In developing a design/analysis method for reinforced concrete, the following
basic relations
will be used:
1. Equilibrium: of forces and moment at the cross section. 1) Σ
F
x
= 0 or Tension in the
reinforcement = Compression in concrete; and 2) Σ
M
= 0 or external moment (that is the
one obtained from the moment envelope) equal and opposite to the internal one (tension
in steel and compression of the concrete).
2. Material Stress Strain: We recall that all normal strength concrete have a failure strain
u
=
.
003 in compression irrespective of
f
0
c
.
14
Basic
assumptions
used:
Compatibility of Displacements:
Perfect bond between steel and concrete (no slip). Note
that those two materials do also have very close coefficients of thermal expansion under
normal temperature.
Plane section remain plane
⇒
strain is proportional to distance from neutral axis.
24.1.5
ACI Code
15
The ACI code is based on limit strength, or Φ
M
n
≥
M
u
thus a similar design philosophy
is used as the one adopted by the LRFD method of the AISC code,
ACI-318: 8.1.1; 9.3.1;
9.3.2
16
The required strength is based on (
ACI-318: 9.2
)
U
= 1
.
4
D
+ 1
.
7
L
= 0
.
75(1
.
4
D
+ 1
.
7
L
+ 1
.
7
W
)
(24.1)
17
We should consider the behaviors of a reinforced concrete section under increasing load:
1. Section uncracked
2. Section cracked, elastic
3. Section cracked, limit state
The second analysis gives rise to the Working Stress Design (WSD) method (to be covered in
Structural Engineering II), and the third one to the Ultimate Strength Design (USD) method.
24.2
Cracked Section, Ultimate Strength Design Method
24.2.1
Equivalent Stress Block
18
In determining the limit state moment of a cross section, we consider Fig.
. Whereas
the strain distribution is linear (
ACI-318 10.2.2
), the stress distribution is non-linear because
the stress-strain curve of concrete is itself non-linear beyond 0
.
5
f
0
c
.
19
Thus we have two alternatives to investigate the moment carrying capacity of the section,
ACI-318: 10.2.6
Victor Saouma
Structural Engineering
Draft
24.2 Cracked Section, Ultimate Strength Design Method
521
24
We have two equations and three unknowns (
α
,
β
1
, and
β
). Thus we need to use test data
to solve this problem
. From
experimental tests
, the following relations are obtained
f
0
c
(ppsi)
<
4,000
5,000
6,000
7,000
8,000
α
.72
.68
.64
.60
.56
β
.425
.400
.375
.350
.325
β
1
.85
.80
.75
.70
.65
Thus we have a general equation for
β
1
(
ACI-318 10.2.7.3
):
β
1
=
.
85
if
f
0
c
≤
4
,
000
=
.
85
−
(
.
05)(
f
0
c
−
4
,
000)
1
1
,
000
if 4
,
000
< f
0
c
<
8
,
000
(24.2)
Figure 24.2: Whitney Stress Block
24.2.2
Balanced Steel Ratio
25
Next we seek to determine the
balanced steel ratio
Ï
b
such that failure occurs by simulta-
neous yielding of the steel
f
s
=
f
y
and crushing of the concrete
ε
c
= 0
.
003,
ACI-318: 10.3.2
We will separately consider the two failure possibilities:
Tension Failure:
we stipulate that the steel stress is equal to
f
y
:
Ï
=
A
s
bd
A
s
f
y
=
.
85
f
0
c
ab
=
.
85
f
0
c
bβ
1
c
)
⇒
c
=
Ïf
y
0
.
85
f
0
c
β
1
d
(24.3)
Compression Failure:
where the concrete strain is equal to the ultimate strain; From the
strain diagram
ε
c
= 0
.
003
c
d
=
.
003
.
003+
ε
s
)
⇒
c
=
.
003
f
s
E
s
+
.
003
d
(24.4)
1
This approach is often used in Structural Engineering. Perform an analytical derivation, if the number of
unknowns then exceeds the number of equations, use experimental data.
Victor Saouma
Structural Engineering
Draft
25.1 T Beams, (ACI 8.10)
531
d
h
f
b
c
ε
s
A
s
f
y
a=
β
1
c
.85f’
c
ε
u
Figure 25.3: T Beam Strain and Stress Diagram
= +
A
s
A
sf
A
s
-A
sf
Figure 25.4: Decomposition of Steel Reinforcement for T Beams
Victor Saouma
Structural Engineering
Draft
25.1 T Beams, (ACI 8.10)
533
Solution:
1. Check requirements for isolated T sections
(a)
b
e
= 30
in
should not exceed 4
b
w
= 4(14) = 56
in
√
(b)
h
f
≥
b
u
2
⇒
7
≥
14
2
√
2. Assume Rectangular section
a
=
(12
.
48)(50)
(0
.
85)(3)(30)
= 8
.
16
in
> h
f
3. For a T section
A
sf
=
.
85
f
0
c
h
f
(
b
−
b
w
)
f
y
=
(
.
85)(3)(7)(30
−
14)
50
= 5
.
71
in
2
Ï
f
=
5
.
71
(14)(36)
=
.
0113
A
sw
= 12
.
48
−
5
.
71 = 6
.
77
in
2
Ï
w
=
12
.
48
(14)(36)
=
.
025
Ï
b
=
.
85
β
1
f
0
c
f
y
87
87+
f
y
= (
.
85)(
.
85)
3
50
87
87+50
=
.
0275
4. Maximum permissible ratio
Ï
max
=
.
75(
Ï
b
+
Ï
f
)
=
.
75(
.
0275 +
.
0113) =
.
029
> Ï
w
√
5. The design moment is then obtained from
M
n
1
= (5
.
71)(50)(36
−
7
2
) = 9
,
280
k.in
a
=
(6
.
77)(50)
(
.
85)(3)(14)
= 9
.
48
in
M
n
2
= (6
.
77)(50)(36
−
9
.
48
2
) = 10
,
580
k.in
M
d
= (
.
9)(9
,
280 + 10
,
580) = 17
,
890
k.in
→
17
,
900
k.in
Example 25-2: T Beam; Moment Capacity II
Determine the moment capacity of the following section, assume flange dimensions to satisfy
ACI requirements;
A
s
= 6#10 = 7
.
59
in
2
;
f
0
c
= 3 ksi;
f
y
=60 ksi.
Victor Saouma
Structural Engineering
Draft
25.1 T Beams, (ACI 8.10)
535
20"
3"
11"
47"
Solution:
1. Determine effective flange width:
1
2
(
b
−
b
w
)
≤
8
h
f
16
h
f
+
b
w
= (16)(3) + 11 = 59
in
L
4
=
24
4
12
= 72
in
Center Line spacing
= 47
in









b
= 47
in
2. Assume
a
= 3
in
A
s
=
M
d
φf
y
(
d
−
a
2
)
=
6
,
400
0
.
9)(60)(20
−
3
2
)
= 6
.
40
in
2
a
=
A
s
f
y
(
.
85)
f
0
c
b
=
(6
.
4)(60)
(
.
85)(3)(47)
= 3
.
20
in
> h
f
3. Thus a T beam analysis is required.
A
sf
=
.
85
f
0
c
(
b
−
b
w
)
h
f
f
y
=
(
.
85)(3)(47
−
11)(3)
60
= 4
.
58
in
2
M
d
1
=
φA
sf
f
y
(
d
−
h
f
2
) = (
.
90)(4
.
58)(60)(20
−
3
2
) = 4
,
570
k.in
M
d
2
=
M
d
−
M
d
1
= 6
,
400
−
4
,
570 = 1
,
830
k.in
4. Now, this is similar to the design of a rectangular section. Assume
a
=
d
5
=
20
5
= 4
.
in
A
s
−
A
sf
=
1
,
830
(
.
90)(60)
20
−
4
2
= 1
.
88
in
2
5. check
a
=
1
.
88)(60)
(
.
85)(3)(11)
= 4
.
02
in
≈
4
.
00
A
s
= 4
.
58 + 1
.
88 = 6
.
46
in
2
Ï
w
=
6
.
46
(11)(20)
=
.
0294
Ï
f
=
4
.
58
(11)(20)
=
.
0208
Ï
b
= (
.
85)(
.
85)
3
60
87
87+60
=
.
0214
Ï
max
=
.
75(
.
0214 +
.
0208) =
.
0316
> Ï
w
√
6. Note that 6.46
in
2
(T beam) is close to
A
s
= 6
.
40
in
2
if rectangular section was assumed.
Victor Saouma
Structural Engineering
Draft
Chapter 26
PRESTRESSED CONCRETE
26.1
Introduction
1
Beams with longer spans are architecturally more appealing than those with short ones.
However, for a reinforced concrete beam to span long distances, it would have to have to be
relatively deep (and at some point the self weight may become too large relative to the live
load), or higher grade steel and concrete must be used.
2
However, if we were to use a steel with
f
y
much higher than
≈
60 ksi in reinforced concrete
(R/C), then to take full advantage of this higher yield stress while maintaining full bond between
concrete and steel, will result in unacceptably wide crack widths. Large crack widths will in
turn result in corrosion of the rebars and poor protection against fire.
3
One way to control the concrete cracking and reduce the tensile stresses in a beam is to
prestress the beam by applying an initial state of stress which is opposite to the one which will
be induced by the load.
4
For a simply supported beam, we would then seek to apply an initial tensile stress at the
top and compressive stress at the bottom. In prestressed concrete (P/C) this can be achieved
through prestressing of a tendon placed below the elastic neutral axis.
5
Main advantages of P/C: Economy, deflection & crack control, durability, fatigue strength,
longer spans.
6
There two type of Prestressed Concrete beams:
Pretensioning:
Steel is first stressed, concrete is then poured around the stressed bars. When
enough concrete strength has been reached the steel restraints are released, Fig.
Postensioning:
Concrete is first poured, then when enough strength has been reached a steel
cable is passed thru a hollow core inside and stressed, Fig.
26.1.1
Materials
7
P/C beams usually have higher compressive strength than R/C. Prestressed beams can have
f
0
c
as high as 8,000 psi.
8
The importance of high yield stress for the steel is illustrated by the following simple example.
Draft
26.1 Introduction
545
If we consider the following:
1. An unstressed steel cable of length
l
s
2. A concrete beam of length
l
c
3. Prestress the beam with the cable, resulting in a stressed length of concrete and steel
equal to
l
0
s
=
l
0
c
.
4. Due to shrinkage and creep, there will be a change in length
∆
l
c
= (
ε
sh
+
ε
cr
)
l
c
(26.1)
we want to make sure that this amout of deformation is substantially smaller than the
stretch of the steel (for prestressing to be effective).
5. Assuming ordinary steel:
f
s
= 30
ksi
,
E
s
= 29
,
000
ksi
,
ε
s
=
30
29
,
000
= 1
.
03
×
10
−
3
in
/
in
6. The total steel elongation is
ε
s
l
s
= 1
.
03
×
10
−
3
l
s
7. The creep and shrinkage strains are about
ε
cr
+
ε
sh
'
.
9
×
10
−
3
8. The residual stres which is left in the steel after creep and shrinkage took place is thus
(1
.
03
−
.
90)
×
10
−
3
(29
×
10
3
) = 4
ksi
(26.2)
Thus the total loss is
30
−
4
30
= 87% which is unacceptably too high.
9. Alternatively if initial stress was 150
ksi
after losses we would be left with 124
ksi
or a
17% loss.
10. Note that the actual loss is (
.
90
×
10
−
3
)(29
×
10
3
) = 26
ksi
in each case
9
Having shown that losses would be too high for low strength steel, we will use
Strands
usually composed of 7 wires. Grade 250 or 270 ksi, Fig.
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Figure 26.3: 7 Wire Prestressing Tendon
Tendon
have diameters ranging from 1/2 to 1 3/8 of an inch. Grade 145 or 160 ksi.
Wires
come in bundles of 8 to 52.
Note that yield stress is not well defined for steel used in prestressed concrete, usually we take
1% strain as effective yield.
10
Steel relaxation is the reduction in stress at constant strain (as opposed to creep which
is reduction of strain at constant stress) occrs. Relaxation occurs indefinitely and produces
Victor Saouma
Structural Engineering
Draft
26.1 Introduction
547
Q
P
P
h/2
h/3
2Q
P
P
h/2
h/3
2Q
P
P
2h/3
P
P
h/2
ž|ž|ž
ž|ž|ž
ž|ž|ž
ž|ž|ž
ž|ž|ž
Ÿ|Ÿ|Ÿ
Ÿ|Ÿ|Ÿ
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Ÿ|Ÿ|Ÿ
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 |Â
 |Â
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¡|¡
¡|¡
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¡|¡
¡|¡
¢|¢
¢|¢
¢|¢
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£|£
£|£
£|£
£|£
¤|¤|¤
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¤|¤|¤
Â¥|Â¥|Â¥
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Â¥|Â¥|Â¥
Â¥|Â¥|Â¥
Â¥|Â¥|Â¥
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§|§
§|§
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©|©
©|©
©|©
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ª|ª
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¯|¯
¯|¯
¯|¯
¯|¯
°|°|°
°|°|°
°|°|°
°|°|°
°|°|°
±|±
±|±
±|±
±|±
±|±
²|²|²
²|²|²
²|²|²
²|²|²
²|²|²
³|³
³|³
³|³
³|³
³|³
´|´|´
´|´|´
´|´|´
´|´|´
´|´|´
´|´|´
µ|µ
µ|µ
µ|µ
µ|µ
µ|µ
µ|µ
¶|¶|¶
¶|¶|¶
¶|¶|¶
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¼|¼|¼|¼|¼|¼|¼
¼|¼|¼|¼|¼|¼|¼
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½|½|½|½|½|½|½
½|½|½|½|½|½|½
½|½|½|½|½|½|½
½|½|½|½|½|½|½
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¾|¾
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¿|¿
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W
h
f’
f
f
y
fc
c
c
f =f
c t
2fc
2f
2fc
0
2f =2f
t
c
c
2fc
0
0
2fc
2f =2f
t
c
2fc
0
2fc
f
fc
c
c
c
f
f
0
2fc
f =f
c
t
c
f
fc
c
fc
c
c
c
f
f
f
+
+
+
+
+
+
=
=
=
=
=
=
0
0
Midspan
Ends
Midspan
Q
Ends
f
Figure 26.4: Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978)
P sin
θ
P sin
θ
P cos
θ
P cos
θ
P
P cos
θ
P cos
θ
P sin
θ
P sin
θ
P cos
θ
P sin
θ
P
2
P sin
θ
P sin
θ
P cos
θ
M
P cos
θ
P sin
θ
P sin
θ
P cos
θ
P sin
θ
P
θ
θ
P
P
e
P
P
P
P e
P e
e
P
θ
M
P
P
None
P
P
P
2
None
P
P
(a)
(b)
(d)
(f)
(g)
(e)
P
(c)
Equivalent load on concrete from tendon Moment from prestressing
Member
Figure 26.5: Determination of Equivalent Loads
Victor Saouma
Structural Engineering
Draft
26.2 Flexural Stresses
549
4.
P
e
and
M
0
+
M
DL
+
M
LL
f
1
=
−
P
e
A
c
1
−
ec
1
r
2
−
M
0
+
M
DL
+
M
LL
S
1
f
2
=
−
P
e
A
c
1 +
ec
2
r
2
+
M
0
+
M
DL
+
M
LL
S
2
(26.7)
The internal stress distribution at each one of those four stages is illustrated by Fig.
S
2
Mo
+
S
2
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e
c
2
c
1
Pi e c
Ic
2
Stage 1
Stage 2
Stage 4
r
e c
2
)
2
( 1 +
P i
Ac
Mo
+
S
2
r
e c
2
)
2
( 1 +
P i
Ac
P i
Ac
P i
Ac
Pi e c
1
Ic
P i
Ac
( 1 -
r
e c
2
1
)
r
e c
2
)
2
( 1 +
P i
Ac
P i
Ac
( 1 -
r
e c
2
1
) - Mo
S
1
- Mo
S
1
( 1 -
r
e c
2
1
)
P i
Ac
Ac
P e
- Mo
S
( 1 -
r
e c
2
1
)
- Md + Ml
S
-
S
1
Mt
Ac
P e ( 1 -
r
e c
2
1
)
Md + Ml
S
+
+ Mt
r
e c
2
)
2
( 1 +
Ac
P e
r
e c
2
)
2
( 1 +
Ac
P e
Mo
+
S
2
2
1
1
Figure 26.7: Flexural Stress Distribution for a Beam with Variable Eccentricity; Maximum
Moment Section and Support Section, (Nilson 1978)
17
Those (service) flexural stresses must be below those specified by the ACI code (where the
subscripts
c
,
t
,
i
and
s
refer to compression, tension, initial and service respectively):
f
ci
permitted concrete compression stress at initial stage
.
60
f
0
ci
f
ti
permitted concrete tensile stress at initial stage
<
3
q
f
0
ci
f
cs
permitted concrete compressive stress at service stage
.
45
f
0
c
f
ts
permitted concrete tensile stress at initial stage
6
p
f
0
c
or 12
p
f
0
c
Note that
f
ts
can reach 12
p
f
0
c
only if appropriate deflection analysis is done, because section
would be cracked.
18
Based on the above, we identify two types of prestressing:
Victor Saouma
Structural Engineering
Draft
26.2 Flexural Stresses
551
M
0
=
(
.
183)(40)
2
8
= 36
.
6
k.ft
(26.9-b)
The flexural stresses will thus be equal to:
f
w
0
1
,
2
=
∓
M
0
S
1
,
2
=
∓
(36
.
6)(12
,
000)
1
,
000
=
∓
439
psi
(26.10)
f
1
=
−
P
i
A
c
1
−
ec
1
r
2
−
M
0
S
1
(26.11-a)
=
−
83
−
439 =
−
522
psi
(26.11-b)
f
ti
= 3
q
f
0
c
= +190
√
(26.11-c)
f
2
=
−
P
i
A
c
1 +
ec
2
r
2
+
M
0
S
2
(26.11-d)
=
−
1
,
837 + 439 =
−
1
,
398
psi
(26.11-e)
f
ci
=
.
6
f
0
c
=
−
2
,
400
√
(26.11-f)
3.
P
e
and
M
0
. If we have 15% losses, then the effective force
P
e
is equal to (1
−
0
.
15)169 =
144
k
f
1
=
−
P
e
A
c
1
−
ec
1
r
2
−
M
0
S
1
(26.12-a)
=
−
144
,
000
176
1
−
(5
.
19)(12)
68
.
2
−
439
(26.12-b)
=
−
71
−
439 =
−
510
psi
(26.12-c)
f
2
=
−
P
e
A
c
1 +
ec
2
r
2
+
M
0
S
2
(26.12-d)
=
−
144
,
000
176
1 +
(5
.
19)(12)
68
.
2
+ 439
(26.12-e)
=
−
1
,
561 + 439 =
−
1
,
122
psi
(26.12-f)
note that
−
71 and
−
1
,
561 are respectively equal to (0
.
85)(
−
83) and (0
.
85)(
−
1
,
837)
respectively.
4.
P
e
and
M
0
+
M
DL
+
M
LL
M
DL
+
M
LL
=
(0
.
55)(40)
2
8
= 110
k.ft
(26.13)
and corresponding stresses
f
1
,
2
=
∓
(110)(12
,
000)
1
,
000
=
∓
1
,
320
psi
(26.14)
Thus,
f
1
=
−
P
e
A
c
1
−
ec
1
r
2
−
M
0
+
M
DL
+
M
LL
S
1
(26.15-a)
Victor Saouma
Structural Engineering
Draft
26.3 Case Study: Walnut Lane Bridge
553
80 ft
CENTER
LINE
ELEVATION OF BEAM HALF
9.25’
44 ’
9.25’
BEAM CROSS SECTIONS
TRANSVERSE DIAPHRAGMS
ROAD
SIDEWALK
CROSS - SECTION OF BRIDGE
CROSS - SECTION OF BEAM
6’-7"
3’-3"
7"
10"
3"
6 "
3 "
7"
1/2
1/2
30"
52"
10"
7"
TRANSVERSE DIAPHRAGM
SLOTS FOR CABLES
Figure 26.8: Walnut Lane Bridge, Plan View
Victor Saouma
Structural Engineering
Draft
Chapter 27
COLUMNS
27.1
Introduction
1
Columns resist a combination of axial
P
and flexural load
M
, (or
M
=
P e
for eccentrically
applied load).
27.1.1
Types of Columns
Types of columns, Fig.
Tied column
Spiral column
Composite column
Pipe column
tie steel
main longitudinal steel reinforcement
Figure 27.1: Types of columns
2
Lateral reinforcement, Fig.
1. Restrains longitudinal steel from outward buckling
2. Restrains Poisson’s expansion of concrete
3. Acts as shear reinforcement for horizontal (wind & earthquake) load
4. Provide ductility
very important to resist earthquake load.
Draft
27.2 Short Columns
559
5
note:
1. 0.85 is obtained also from test data
2. Matches with beam theory using rect. stress block
3. Provides an adequate factor of safety
27.2.2
Eccentric Columns
5
Sources of flexure, Fig.
M
L
M
R
P
e
Figure 27.4: Sources of Bending
1. Unsymmetric moments
M
L
6
=
M
R
2. uncertainty of loads (must assume a minimum eccentricity)
3. unsymmetrical reinforcement
6
Types of Failure, Fig.
1. large eccentricity of load
⇒
failure by yielding of steel
2. small eccentricity of load
⇒
failure by crushing of concrete
3. balanced condition
7
Assumptions
A
0
s
=
A
s
;
Ï
=
A
s
bd
=
A
0
s
bd
=
f
0
s
=
f
y
27.2.2.1
Balanced Condition
8
There is one specific eccentricity
e
b
=
M
P
such that failure will be triggered by simultaneous
1. steel yielding
2. concrete crushing
Victor Saouma
Structural Engineering
Draft
27.2 Short Columns
561
A f
s s
ε
s
A f
s
’
s
d
d’
h/2
A’
A
s
s
b
ε
’
s
a
e
e’
0.85f’
c
P
n
c
A f
s s
c
ε
s
c
A f
s
A f
s
’
y
y
Figure 27.6: Strain and Stress Diagram of a R/C Column
Victor Saouma
Structural Engineering
Draft
27.2 Short Columns
571
error=0.1805
a= 20.7445
iter= 2
c=24.4053
epsi_sc=0.0015
f_sc=44.2224
f_sc=40
M_n=1.6860e+04
P_n=1.4789e+03
epsi_s=-4.1859e-04
f_s=12.1392
Failure by Tension
e=24
error=1
a=10
iter=0
iter=1
c=11.7647
epsi_sc=-6.0000e-
f_sc=-1.7400
P_n_1=504.5712
P_n_2=381.9376
a_new=7.5954
error=-0.3166
a=7.5954
iter=2
c=8.9358
epsi_sc=-0.0010
f_sc=-29.8336
P_n_1=294.2857
P_n_2=312.6867
a_new=7.9562
error=0.0453
a=7.9562
iter=2
P_n_1=294.2857
Example 27-3: R/C Column, Using Design Charts
Design the reinforcement for a column with
h
= 20
in
,
b
= 12
in
,
d
0
= 2
.
5
in
,
f
0
c
= 4
,
000
psi
,
f
y
= 60
,
000
psi
, to support
P
DL
= 56
k
,
P
LL
= 72
k
,
M
DL
= 88
k.ft
,
M
LL
= 75
k.ft
,
Solution:
1. Ultimate loads
P
u
= (1
.
4)(56) + (1
.
7)(72) = 201
k
⇒
P
n
=
201
0
.
7
= 287
k
M
u
= (1
.
4)(88) + (1
.
7)(75) = 251
k.ft
⇒
M
n
=
251
0
.
7
= 358
k.ft
(27.26)
Victor Saouma
Structural Engineering
Draft
Chapter 28
ELEMENTS of STRUCTURAL
RELIABILITY
28.1
Introduction
1
Traditionally, evaluations of structural adequacy have been expressed by safety factors
SF
=
C
D
, where
C
is the
capacity
(i.e. strength) and
D
is the demand (i.e. load). Whereas this
evaluation is quite simple to understand, it suffers from many limitations: it 1) treats all loads
equally; 2) does not differentiate between capacity and demands respective uncertainties; 3) is
restricted to service loads; and last but not least 4) does not allow comparison of relative reli-
abilities among different structures for different performance modes. Another major deficiency
is that all parameters are assigned a single value in an analysis which is then
deterministic
.
2
Another approach, a
probabilistic
one, extends the factor of safety concept to explicitly in-
corporate uncertainties in the parameters. The uncertainties are quantified through statistical
analysis of existing data or judgmentally assigned.
3
This chapter will thus develop a procedure which will enable the Engineer to perform a
reliability
based analysis of a structure, which will ultimately yield a
reliability index
. This is
turn is a “universal†indicator on the adequacy of a structure, and can be used as a metric to
1) assess the health of a structure, and 2) compare different structures targeted for possible
remediation.
28.2
Elements of Statistics
4
Elementary statistics formulaes will be reviewed, as they are needed to properly understand
structural reliability.
5
When a set of
N
values
x
i
is clustered around a particular one, then it may be useful to
characterize the set by a few numbers that are related to its
moments
(the sums of integer
powers of the values):
Mean:
estimates the value around which the data clusters.
µ
=
1
N
N
X
i
=1
x
i
(28.1)
Draft
28.3 Distributions of Random Variables
575
Skewness:
characterizes the degree of asymmetry of a distribution around its mean. It is
defined in a non-dimensional value. A positive one signifies a distribution with an asym-
metric tail extending out toward more positive
x
Skew =
1
N
Σ
N
i
=1
x
i
−
µ
σ
3
(28.8)
Kurtosis:
is a nondimensional quantity which measures the “flatness†or “peakedness†of a
distribution. It is normalized with respect to the curvature of a normal distribution.
Hence a negative value would result from a distribution resembling a loaf of bread, while
a positive one would be induced by a sharp peak:
Kurt =
1
N
Σ
N
i
=1
x
i
−
µ
σ
4
−
3
(28.9)
the
−
3 term makes the value zero for a normal distribution.
6
The expected value (or mean), standard deviation and coefficient of variation are interdepen-
dent: knowing any two, we can determine the third.
28.3
Distributions of Random Variables
7
Distribution of variables can be mathematically represented.
28.3.1
Uniform Distribution
8
Uniform distribution implies that any value between
x
min
and
x
max
is equaly likely to occur.
28.3.2
Normal Distribution
9
The general normal (or Gauss) distribution is given by, Fig.
φ
(
x
) =
1
√
2
πσ
e
−
1
2
[
x
−
µ
σ
]
2
(28.10)
10
A normal distribution
N
(
µ, σ
2
) can be normalized by defining
y
=
x
−
µ
σ
(28.11)
and
y
would have a distribution
N
(0
,
1):
φ
(
y
) =
1
√
2
Ï€
e
−
y
2
2
(28.12)
11
The normal distribution has been found to be an excellent approximation to a large class of
distributions, and has some very desirable mathematical properties:
Victor Saouma
Structural Engineering
Draft
28.4 Reliability Index
577
28.3.3
Lognormal Distribution
12
A random variable is lognormally distributed if the natural logarithm of the variable is
normally distributed.
13
It provides a reasonable shape when the coefficient of variation is large.
28.3.4
Beta Distribution
14
Beta distributions are very flexible and can assume a variety of shapes including the normal
and uniform distributions as special cases.
15
On the other hand, the beta distribution requires four parameters.
16
Beta distributions are selected when a particular shape for the probability density function
is desired.
28.3.5
BiNormal distribution
28.4
Reliability Index
28.4.1
Performance Function Identification
17
Designating
F
the capacity to demand ratio
C/D
(or
performance function
), in general
F
is a function of one or more variables
x
i
which describe the geometry, material, loads, and
boundary conditions
F
=
F
(
x
i
)
(28.17)
and thus
F
is in turn a random variable with its own probability distribution function, Fig.
18
A performance function evaluation typically require a structural analysis, this may range
from a simple calculation to a detailled finite element study.
28.4.2
Definitions
19
Reliability indices,
β
are used as a relative measure of the reliability or confidence in the
ability of a structure to perform its function in a satisfactory manner. In other words they are
a measure of the performance function.
20
Probabilistic methods are used to systematically evaluate uncertainties in parameters that
affect structural performance, and there is a relation between the reliability index and risk.
21
Reliability index is defined in terms of the performance function capacity
C
, and the applied
load or demand
D
. It is assumed that both
C
and
D
are
random variables
.
22
The
safety margin
is defined as
Y
=
C
−
D
. Failure would occur if
Y <
0 Next,
C
and
D
can be combined and the result expressed logarithmically, Fig.
X
= ln
C
D
(28.18)
Victor Saouma
Structural Engineering
Draft
28.4 Reliability Index
579
30
The objective is to determine the mean and standard deviation of the performance function
defined in terms of
C/D
.
31
Those two parameters, in turn, will later be required to compute the reliability index.
28.4.3.1
Direct Integration
32
Given a function random variable
x
, the mean value of the function is obtained by integrating
the function over the probability distribution function of the random variable
µ
[
F
(
x
)] =
Z
∞
−∞
g
(
x
)
f
(
x
)
dx
(28.21)
33
For more than one variable,
µ
[
F
(
x
)] =
Z
∞
−∞
Z
∞
−∞
· · ·
Z
∞
−∞
F
(
x
1
, x
2
,
· · ·
, x
n
)
F
(
x
1
, x
2
,
· · ·
, x
n
)
dx
1
dx
2
· · ·
dx
n
(28.22)
34
Note that in practice, the function
F
(
x
) is very rarely available for practical problems, and
hence this method is seldom used.
28.4.3.2
Monte Carlo Simulation
35
The performance function is evaluated for many possible values of the random variables.
36
Assuming that all variables have a normal distribution, then this is done through the following
algorithm
1. initialize random number generators
2. Perform
n
analysis, for each one:
(a) For each variable, determine a random number for the given distribution
(b) Transform the random number
(c) Analyse
(d) Determine the performance function, and store the results
3. From all the analyses, determine the mean and the standard deviation, compute the
reliability index.
4. Count the number of analyses,
n
f
which performance function indicate failure, the likeli-
hood of structural failure will be
p
(
f
) =
n
f
/n
.
37
A sample program (all subroutines are taken from (Press, Flannery, Teukolvsky and Vetterling
1988) which generates
n
normally distributed data points, and then analyze the results, deter-
mines mean and standard deviation, and sort them (for histogram plotting), is shown below:
program nice
parameter(ns=100000)
real x(ns), mean, sd
write(*,*)’enter mean, standard-deviation and n ’
read(*,*)mean,sd,n
Victor Saouma
Structural Engineering
Draft
28.4 Reliability Index
581
ix3=mod(ia3*ix3+ic3,m3)
j=1+(97*ix3)/m3
if(j.gt.97.or.j.lt.1)pause
ran1=r(j)
r(j)=(float(ix1)+float(ix2)*rm2)*rm1
return
end
c-----------------------------------------------------------------------------
subroutine moment(data,n,ave,adev,sdev,var,skew,curt)
c given array of data of length N, returns the mean AVE, average
c
deviation ADEV, standard deviation SDEV, variance VAR, skewness SKEW,
c
and kurtosis CURT
dimension data(n)
if(n.le.1)pause ’n must be at least 2’
s=0.
do 11 j=1,n
s=s+data(j)
11
continue
ave=s/n
adev=0.
var=0.
skew=0.
curt=0.
do 12 j=1,n
s=data(j)-ave
adev=adev+abs(s)
p=s*s
var=var+p
p=p*s
skew=skew+p
p=p*s
curt=curt+p
12
continue
adev=adev/n
var=var/(n-1)
sdev=sqrt(var)
if(var.ne.0.)then
skew=skew/(n*sdev**3)
curt=curt/(n*var**2)-3.
else
pause ’no skew or kurtosis when zero variance’
endif
return
end
c------------------------------------------------------------------------------
subroutine sort(n,ra)
dimension ra(n)
l=n/2+1
ir=n
10
continue
if(l.gt.1)then
l=l-1
rra=ra(l)
else
rra=ra(ir)
ra(ir)=ra(1)
ir=ir-1
if(ir.eq.1)then
ra(1)=rra
Victor Saouma
Structural Engineering
Draft
28.4 Reliability Index
583
42
This is accomplished by limiting ourselves to all possible combinations of
µ
i
±
σ
i
.
43
For each analysis we determine SFF, as well as its logarithm.
44
Mean and standard deviation of the logarithmic values are then determined from the 2
n
analyses, and then
β
is the ratio of the mean to the standard deviation.
28.4.3.4
Taylor’s Series-Finite Difference Estimation
45
In the previous method, we have cut down the number of deterministic analyses to 2
n
, in the
following method, we reduce it even further to 2
n
+ 1, (US Army Corps of Engineers 1992, US
Army Corps of Engineers 1993, Bryant, Brokaw and Mlakar 1993).
46
This simplified approach starts with the first order Taylor series expansion of Eq.
about the mean and limited to linear terms, (Benjamin and Cornell 1970).
µ
F
=
F
(
µ
i
)
(28.23)
where
µ
i
is the mean for all random variables.
47
For independent random variables, the variance can be approximated by
V ar
(
F
) =
σ
2
F
=
X
∂F
∂x
i
σ
i
2
(28.24-a)
∂F
∂x
i
≈
F
+
i
−
F
−
i
2
σ
i
(28.24-b)
F
+
i
=
F
(
µ
1
,
· · ·
, µ
i
+
σ
i
,
· · ·
, µ
n
)
(28.24-c)
F
−
i
=
F
(
µ
1
,
· · ·
, µ
i
−
σ
i
,
· · ·
, µ
n
)
(28.24-d)
where
σ
i
are the standard deviations of the variables. Hence,
σ
F
=
X
F
+
i
−
F
−
i
2
!
(28.25)
Finally, the reliability index is given by
β
=
ln
µ
F
σ
F
(28.26)
48
The procedure can be summarized as follows:
1. Perform an initial analysis in which all variables are set equal to their mean value. This
analysis provides the mean
µ
.
2. Perform 2
n
analysis, in which all variables are set equal to their mean values, except
variable
i
, which assumes a value equal to
µ
i
+
σ
i
, and then
µ
i
−
σ
i
.
3. For each pair of analysis in which variable
x
i
is modified, determine
Victor Saouma
Structural Engineering
Draft
28.5 Reliability Analysis
585
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
β
10
0
10
1
10
2
10
3
10
4
10
5
10
6
10
7
1/(Probability of Failure)
Probability of Failure in terms of
β
Good
Hazardous
Unsatisfactory
Poor
Below Average
Above Average
Figure 28.3: Probability of Failure in terms of
β
Victor Saouma
Structural Engineering
Draft
Chapter 29
DESIGN II
29.1
Frames
29.1.1
Beam Column Connections
1
The connection between the beam and the column can be, Fig.
θ
b
θ
c
θ
b
θ
c
θ
b
θ
c
=
θ
b
θ
c
=
θ
b
θ
c
=
θ
θ
θ
θ
b
b
c
c
M=K(
-
)
s
s
Flexible
Rigid
Semi-Flexible
Figure 29.1: Flexible, Rigid, and Semi-Flexible Joints
Flexible
that is a hinge which can transfer forces only. In this case we really have cantiliver
action only. In a flexible connection the column and beam end moments are both equal
to zero,
M
col
=
M
beam
= 0. The end rotation are not equal,
θ
col
6
=
θ
beam
.
Rigid:
The connection is such that
θ
beam
=
θ
col
and moment can be transmitted through the
connection. In a rigid connection, the end moments and rotations are equal (unless there
is an externally applied moment at the node),
M
col
=
M
beam
6
= 0,
θ
col
=
θ
beam
.
Semi-Rigid:
The end moments are equal and not equal to zero, but the rotation are different.
θ
beam
6
=
θ
col
,
M
col
=
M
beam
6
= 0. Furthermore, the difference in rotation is resisted by
the spring
M
spring
=
K
spring
(
θ
col
−
θ
beam
).
29.1.2
Behavior of Simple Frames
2
For vertical load across the beam rigid connection will reduce the maximum moment in the
beam (at the expense of a negative moment at the ends which will in turn be transferred to
Draft
29.1 Frames
589
w
P
w/2
-w/2
w/2
-w/2
w/2
-w/2
w/2
M’
M’
M’/L
-M’/L
M’/L
-M’/L
a
b
h
L
c
d
e
f
g
h
i
j
k
l
p
M
w/2
w/2
M’
-M’/L
p
-M’/L
p
w/2
-w/2
M/h
-M/h
p/2
p/2
-M/L
w/2
-w/2
0.36M/h
-0.36M/h
-M/L
p/2
p/2
-w/2
0.68M/h
-0.5M’/L
p/2
p/2
M
M
w/2
w/2
w/2
w/2
M
M
M/h
w/2
w/2
M’/2
M’/2
p/2
M/L
-M/L
0.4M
0.4M
0.64M
0.4M/h
w/2
w/2
M’/2
p/2
-M’/L
M’/L
0.55M
0.45M
M’/4
M’/4
M’/4
0.68M/h
p/2
w/2
w/2
M’/2L
-M’/2L
POST AND BEAM STRUCTURE
SIMPLE BENT FRAME
THREE-HINGE PORTAL
THREE-HINGE PORTAL
TWO-HINGE FRAME
RIGID FRAME
-0.68M/h
M’/4
0.45M
M’/2
Deformation
Shear
Moment
Axial
Frame Type
W=wL, M=wL/8, M’=Ph
2
Figure 29.3: Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal
Frames Subjected to Vertical and Horizontal Loads
Victor Saouma
Structural Engineering
Draft
29.1 Frames
591
29.1.4
Design of a Statically Indeterminate Arch
Adapted from (Kinney 1957)
Design a two-hinged, solid welded-steel arch rib for a hangar. The moment of inertia of the
rib is to vary as necessary. The span, center to center of hinges, is to be 200 ft. Ribs are to be
placed 35 ft center to center, with a rise of 35 ft. Roof deck, purlins and rib will be assumed to
weight 25 lb/ft
2
on roof surface, and snow will be assumed at 40 lb/ft
2
of this surface. Twenty
purlins will be equally spaced around the rib.
Figure 29.5: Design of a Statically Indeterminate Arch
1. The center line of the rib will be taken as the segment of a circle. By computation the
radius of this circle is found to be 160.357 ft, and the length of the
arc AB
to be 107.984
ft.
2. For the analysis the arc
AB
will be considered to be divided into ten segments, each with
a length of 10.798 ft. Thus a concentrated load is applied to the rib by the purlins framing
at the center of each segment. (The numbered segments are indicated in Fig.
??
.
3. Since the total dead and snow load is 65 lb/ft
2
of roof surface, the value of each concen-
trated force will be
P
= 10
.
798
×
35
×
65 = 24
.
565
k
(29.8)
4. The computations necessary to evaluate the coodinates of the centers of the various seg-
ments, referred to the hinge at
A
, are shown in Table
. Also shown are the values of
∆
x
, the horizontal projection of the distance between the centers of the several segments.
5. If experience is lacking and the designing engineer is therefore at a loss as to the initial
assumptions regarding the sectional variation along the rib, it is recommended that the
Victor Saouma
Structural Engineering
Draft
29.1 Frames
593
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
Shear
M
to
M
simple
Segment
x
y
=
δM
right
∆
x
increment
beam
M δM
δM M
(ft)
(ft
·
k)
(k)
(ft)
(ft
·
k)
(ft
·
k)
A
0.0
0.0
245.650
1
4.275
3.29
221.085
4.275
1,050
1,050
3,500
10.9
2
13.149
9.44
196.520
8.874
1,962
3,010
28,400
89.2
3
22.416
14.98
171.955
9.267
1,821
4,830
72,400
224.4
4
32.034
19.88
147.390
9.618
1,654
6,490
129,100
395.4
5
41.960
24.13
122.825
9.926
1,463
7,950
191,800
582.2
6
52.149
27.69
98.260
10.189
1,251
9,200
254,800
767.0
7
62.557
30.57
73.695
10.408
1,023
10,220
312,400
934.4
8
73.134
32.73
49.130
10.577
779
11,000
360,000
1,071.4
9
83.831
34.18
24.565
10.697
526
11,530
394,100
1,168.4
10
94.601
34.91
0.000
10.770
265
11,790
411,600
1,218.6
Crown
100.00
35.00
5.399
0
11,790
P
2,158,100
6,461.9
Table 29.2: Calculation of Horizontal Force
M
simple
Total
M
Segment
y
beam
H
A
y
at segment
(ft)
(ft
·
k)
(ft
·
k)
(ft
·
k)
A
0
1
3.29
1,050
–1,100
–50
2
9.44
3,010
–3,150
–140
3
14.98
4,830
–5,000
–170
4
19.88
6,490
–6,640
–150
5
24.13
7,950
–8,060
–110
6
27.69
9,200
–9,250
–50
7
30.57
10,220
–10,210
+10
8
32.73
11,000
–10,930
+70
9
34.18
11,530
–11,420
+110
10
34.19
11,790
–11,660
+130
Crown
35.00
11,790
–11,690
+100
Table 29.3: Moment at the Centers of the Ribs
Victor Saouma
Structural Engineering
Draft
29.1 Frames
595
Segment
V
(k)
V
cos
α
−
H
sin
α
=
S
(k)
V
sin
α
+
H
cos
α
=
N
(k)
A
245.6
192
-208
= -16
153
+ 261
= 414
1
221.1
177
-199
= -22
132
+268
= 400
2
196.5
165
-181
= -16
106
+281
= 387
3
172.0
150
-162
= -12
83
+292
= 375
4
147.4
133
-142
= -9
62
+303
= 365
5
122.8
114
-121
= -7
44
+311
= 355
6
98.3
94
-100
= -6
29
+319
= 348
7
73.7
72
-78
= -6
17
+325
= 342
8
49.1
48
-56
= -8
8
+329
= 337
9
24.6
244
-34
= -10
2
+332
= 334
10
0.0
0
-11
= -11
0
+334
= 334
Crown
+334
= 334
Table 29.4: Values of Normal and Shear Forces
3 to the crown is made to vary linerarly. The adequacies of the sections thus determined
for the centers of the several segments are checked in Table
17. It is necessary to recompute the value of
H
A
because the rib now has a varying moment
of inertia. Equation
must be altered to include the
I
of each segment and is now
written as
H
A
=
−
P
M δM /I
P
δM M /I
(29.14)
18. The revised value for
H
A
is easily determined as shown in Table
. Note that the
values in column (2) are found by dividing the values of
M δM
for the corresponding
segments in column (8) of Table
by the total
I
for each segment as shown in Table
. The values in column (3) of Table
are found in a similar manner from the
values in column (9) of Table
. The simple beam moments in column (5) of Table
are taken directly from column (7) of Table
19. Thus the revised
H
A
is
H
A
=
−
2
P
M δM /I
2
P
δM M /I
=
−
f rac
2
×
1
,
010
.
852
×
3
.
0192 =
−
334
.
81
k
(29.15)
20. The revised values for the axial thrust
N
at the centers of the various segments are
computed in Table
21. The sections previously designed at the centers of the segments are checked for adequacy
in Table
. From this table it appears that all sections of the rib are satisfactory.
This cannot be definitely concluded, however, until the secondary stresses caused by the
deflection of the rib are investigated.
29.1.5
Temperature Changes in an Arch
Adapted from (Kinney 1957)
Determine the effects of rib shortening and temperature changes in the arch rib of Fig.
Consider a temperature drop of 100
â—¦
F.
Victor Saouma
Structural Engineering
Draft
Chapter 30
Case Study I: EIFFEL TOWER
Adapted from (Billington and Mark 1983)
30.1
Materials, & Geometry
1
The tower was built out of wrought iron, less expensive than steel,and Eiffel had more ex-
pereince with this material, Fig.
Figure 30.1: Eiffel Tower (Billington and Mark 1983)
Draft
30.2 Loads
605
Width
Location
Height
Width/2
Estimated
Actual
dy
dx
β
Support
0
164
328
.333
18.4
o
First platform
186
108
216
240
.270
15.1
o
second platform
380
62
123
110
.205
11.6
o
Intermediate platform
644
20
40
.115
6.6
o
Top platform
906
1
2
.0264
1.5
o
Top
984
0
0
0.000
0
o
4
The tower is supported by four inclined supports, each with a cross section of 800
in
2
. An
idealization of the tower is shown in Fig.
ACTUAL
CONTINUOUS
CONNECTION
ACTUAL
POINTS OF
CONNECTION
CONTINUOUS
CONNECTION
IDEALIZED
Figure 30.2: Eiffel Tower Idealization, (Billington and Mark 1983)
30.2
Loads
5
The total weight of the tower is 18
,
800
k
.
6
The dead load is not uniformly distributed, and is approximated as follows, Fig.
Figure 30.3: Eiffel Tower, Dead Load Idealization; (Billington and Mark 1983)
Victor Saouma
Structural Engineering
Draft
Chapter 31
Case Study II: GEORGE
WASHINGTON BRIDGE
31.1
Theory
1
Whereas the forces in a cable can be determined from statics alone, its configuration must
be derived from its deformation. Let us consider a cable with distributed load
p
(
x
)
per unit
horizontal projection
of the cable length (thus neglecting the weight of the cable). An
infinitesimal portion of that cable can be assumed to be a straight line, Fig.
and in the
absence of any horizontal load we have
H
=constant. Summation of the vertical forces yields
(+
?
) Σ
F
y
= 0
⇒ −
V
+
wdx
+ (
V
+
dV
) = 0
(31.1-a)
dV
+
wdx
= 0
(31.1-b)
where
V
is the vertical component of the cable tension at
x
(Note that if the cable was
subjected to its own weight then we would have
wds
instead of
wdx
). Because the cable must
be tangent to
T
, we have
tan
θ
=
V
H
(31.2)
Substituting into Eq.
yields
d
(
H
tan
θ
) +
wdx
= 0
⇒ −
d
dx
(
H
tan
θ
) =
w
(31.3)
2
But
H
is constant (no horizontal load is applied), thus, this last equation can be rewritten as
−
H
d
dx
(tan
θ
) =
w
(31.4)
3
Written in terms of the vertical displacement
v
, tan
θ
=
dv
dx
which when substituted in Eq.
yields the
governing equation for cables
−
Hv
00
=
w
(31.5)
4
For a cable subjected to a uniform load
w
, we can determine its shape by double integration
of Eq.
−
Hv
0
=
wx
+
C
1
(31.6-a)
Draft
31.1 Theory
613
−
Hv
=
wx
2
2
+
C
1
x
+
C
2
(31.6-b)
and the constants of integrations
C
1
and
C
2
can be obtained from the boundary conditions:
v
= 0 at
x
= 0 and at
x
=
L
⇒
C
2
= 0 and
C
1
=
−
wL
2
. Thus
v
=
w
2
H
x
(
L
−
x
)
(31.7)
This equation gives the shape
v
(
x
) in terms of the horizontal force
H
,
5
Since the maximum sag
h
occurs at midspan (
x
=
L
2
) we can solve for the horizontal force
H
=
wL
2
8
h
(31.8)
we note the analogy with the maximum moment in a simply supported uniformly loaded beam
M
=
Hh
=
wL
2
8
. Furthermore, this relation clearly shows that the horizontal force is inversely
proportional to the sag
h
, as
h
&
H
%
. Finally, we can rewrite this equation as
r
def
=
h
L
(31.9-a)
wL
H
=
8
r
(31.9-b)
6
Eliminating
H
from Eq.
and
we obtain
v
= 4
h
−
x
2
L
2
+
x
L
!
(31.10)
Thus the cable assumes a parabolic shape (as the moment diagram of the applied load).
7
Whereas the horizontal force
H
is constant throughout the cable, the tension
T
is not. The
maximum tension occurs at the support where the vertical component is equal to
V
=
wL
2
and
the horizontal one to
H
, thus
T
max =
p
V
2
+
H
2
=
s
wL
2
2
+
H
2
=
H
s
1 +
wL/
2
H
2
(31.11)
Combining this with Eq.
we obtain
.
T
max =
H
p
1 + 16
r
2
≈
H
(1 + 8
r
2
)
(31.12)
8
Had we assumed a uniform load
w
per length of cable
(rather than horizontal projection),
the equation would have been one of a catenary
v
=
H
w
cosh
w
H
L
2
−
x
+
h
(31.13)
The cable between transmission towers is a good example of a catenary.
1
Recalling that (
a
+
b
)
n
=
a
n
+
na
n
−
1
b
+
n
(
n
−
1)
2!
a
n
−
2
b
2
+
·
or (1 +
b
)
n
= 1 +
nb
+
n
(
n
−
1)
b
2
2!
+
n
(
n
−
1)(
n
−
2)
b
3
3!
+
· · ·
;
Thus for
b
2
<<
1,
√
1 +
b
= (1 +
b
)
1
2
≈
1 +
b
2
2
Derivation of this equation is beyond the scope of this course.
Victor Saouma
Structural Engineering
Draft
Chapter 32
Case Study III: MAGAZINI
GENERALI
Adapted from (Billington and Mark 1983)
32.1
Geometry
1
This sotrage house, built by Maillart in Chiasso in 1924, provides a good example of the
mariage between aesthetic and engineering.
2
The most strking feature of the Magazini Generali is not the structure itself, but rather the
shape of its internal supporting frames, Fig.
3
The frame can be idealized as a simply supported beam hung from two cantilever column
supports. Whereas the beam itself is a simple structural idealization, the overhang is designed
in such a way as to minimize the net moment to be transmitted to the supports (foundations),
Fig.
32.2
Loads
4
The load applied on the frame is from the weights of the roof slab, and the frame itself. Given
the space between adjacent frames is 14.7 ft, and that the roof load is 98
psf
, and that the
total frame weight is 13.6 kips, the total uniform load becomes, Fig.
q
roof
= (98)
psf
(14
.
7)
ft
= 1
.
4
k/ft
(32.1-a)
q
f rame
=
(13
.
6)
k
(63
.
6)
ft
= 0
.
2
k/ft
(32.1-b)
q
total
= 1
.
4 + 0
.
2 = 1.6
k/ft
(32.1-c)
Draft
32.3 Reactions
623
ROOF
q = 1.4 k/ft
FRAME
+ q = 0.2 k/ft
ROOF
q = 1.4 k/ft
FRAME
q = 1.6 k/ft
TOTAL
+ q = 0.2 k/ft
Figure 32.3: Magazzini Generali; Loads (Billington and Mark 1983)
32.3
Reactions
5
Reactions for the beam are determined first taking advantage of symmetry, Fig.
W
= (1
.
6)
k/ft
(63
.
6)
ft
= 102
k
(32.2-a)
R
=
W
2
=
102
2
= 51
k
(32.2-b)
We note that these reactions are provided by the internal shear forces.
63.6 ft
51 k
51 k
TOTAL
q = 1.6 k/ft
Figure 32.4: Magazzini Generali; Beam Reactions, (Billington and Mark 1983)
32.4
Forces
6
The internal forces are pimarily the shear and moments. Those can be easily determined for
a simply supported uniformly loaded beam. The shear varies linearly from 51 kip to -51 kip
with zero at the center, and the moment diagram is parabolic with the maximum moment at
the center, Fig.
, equal to:
M
max
=
qL
2
8
=
(1
.
6)
k/ft
(63
.
6)
ft
2
8
= 808
k.ft
(32.3)
7
The externally induced moment at midspan must be resisted by an equal and opposite internal
moment. This can be achieved through a combination of compressive force on the upper fibers,
and tensile ones on the lower. Thus the net axial force is zero, however there is a net internal
couple, Fig.
Victor Saouma
Structural Engineering
Draft
Chapter 33
BUILDING STRUCTURES
33.1
Introduction
33.1.1
Beam Column Connections
1
The connection between the beam and the column can be, Fig.
θ
b
θ
c
θ
b
θ
c
θ
b
θ
c
=
θ
b
θ
c
=
θ
b
θ
c
=
θ
θ
θ
θ
b
b
c
c
M=K(
-
)
s
s
Flexible
Rigid
Semi-Flexible
Figure 33.1: Flexible, Rigid, and Semi-Flexible Joints
Flexible
that is a hinge which can transfer forces only. In this case we really have cantiliver
action only. In a flexible connection the column and beam end moments are both equal
to zero,
M
col
=
M
beam
= 0. The end rotation are not equal,
θ
col
6
=
θ
beam
.
Rigid:
The connection is such that
θ
beam
=
θ
col
and moment can be transmitted through the
connection. In a rigid connection, the end moments and rotations are equal (unless there
is an externally applied moment at the node),
M
col
=
M
beam
6
= 0,
θ
col
=
θ
beam
.
Semi-Rigid:
The end moments are equal and not equal to zero, but the rotation are different.
θ
beam
6
=
θ
col
,
M
col
=
M
beam
6
= 0. Furthermore, the difference in rotation is resisted by
the spring
M
spring
=
K
spring
(
θ
col
−
θ
beam
).
33.1.2
Behavior of Simple Frames
2
For vertical load across the beam rigid connection will reduce the maximum moment in the
beam (at the expense of a negative moment at the ends which will in turn be transferred to
Draft
33.1 Introduction
629
w
P
w/2
-w/2
w/2
-w/2
w/2
-w/2
w/2
M’
M’
M’/L
-M’/L
M’/L
-M’/L
a
b
h
L
c
d
e
f
g
h
i
j
k
l
p
M
w/2
w/2
M’
-M’/L
p
-M’/L
p
w/2
-w/2
M/h
-M/h
p/2
p/2
-M/L
w/2
-w/2
0.36M/h
-0.36M/h
-M/L
p/2
p/2
-w/2
0.68M/h
-0.5M’/L
p/2
p/2
M
M
w/2
w/2
w/2
w/2
M
M
M/h
w/2
w/2
M’/2
M’/2
p/2
M/L
-M/L
0.4M
0.4M
0.64M
0.4M/h
w/2
w/2
M’/2
p/2
-M’/L
M’/L
0.55M
0.45M
M’/4
M’/4
M’/4
0.68M/h
p/2
w/2
w/2
M’/2L
-M’/2L
POST AND BEAM STRUCTURE
SIMPLE BENT FRAME
THREE-HINGE PORTAL
THREE-HINGE PORTAL
TWO-HINGE FRAME
RIGID FRAME
-0.68M/h
M’/4
0.45M
M’/2
Deformation
Shear
Moment
Axial
Frame Type
W=wL, M=wL/8, M’=Ph
2
Figure 33.3: Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal
Frames Subjected to Vertical and Horizontal Loads
Victor Saouma
Structural Engineering
Draft
33.2 Buildings Structures
631
33.2
Buildings Structures
11
There are three primary types of building systems:
Wall Subsytem:
in which very rigid walls made up of solid masonry, paneled or braced timber,
or steel trusses constitute a rigid subsystem. This is only adequate for small rise buildings.
Vertical Shafts:
made up of four solid or trussed walls forming a tubular space structure. The
tubular structure may be interior (housing elevators, staircases) and/or exterior. Most
efficient for very high rise buildings.
Rigid Frame:
which consists of linear vertical components (columns) rigidly connected to stiff
horizontal ones (beams and girders). This is not a very efficient structural form to resist
lateral (wind/earthquake) loads.
33.2.1
Wall Subsystems
12
Whereas exterior wall provide enclosure and interior ones separation, both of them can also
have a structural role in trnsfering vertical and horizontal loads.
13
Walls are constructed out of masonry, timber concrete or steel.
14
If the wall is braced by floors, then it can provide an excellent resitance to horizontal load
in the plane of the wall (but not orthogonal to it).
15
When shear-walls subsytems are used, it is best if the center of orthogonal shear resistance
is close to the centroid of lateral loads as applied. If this is not the case, then there will be
torsional design problems.
33.2.1.1
Example: Concrete Shear Wall
From (Lin and Stotesbury 1981)
16
We consider a reinforced concrete wall 20 ft wide, 1 ft thick, and 120 ft high with a vertical
load of 400 k acting on it at the base. As a result of wind, we assume a uniform horizontal
force of 0.8 kip/ft of vertical height acting on the wall. It is required to compute the flexural
stresses and the shearing stresses in the wall to resist the wind load, Fig.
1. Maximum shear force and bending moment at the base
V
max
=
wL
= (0
.
8)
k.ft
(120)
ft
= 96
k
(33.8-a)
M
max
=
wL
2
2
=
(0
.
8)
k.ft
(120)
2
ft
2
2
= 5
,
760
k.ft
(33.8-b)
2. The resulting eccentricity is
e
Actual
=
M
P
=
(5
,
760)
k.ft
(400)
k
= 14
.
4
ft
(33.9)
3. The critical eccentricity is
e
cr
=
L
6
=
(20)
ft
6
= 3
.
3
ft
< e
Actual
N.G.
(33.10)
thus there will be tension at the base.
Victor Saouma
Structural Engineering
Draft
33.2 Buildings Structures
633
9. The compressive stress of 740 psi can easily be sustained by concrete, as to the tensile
stress of 460 psi, it would have to be resisted by some steel reinforcement.
10. Given that those stresses are
service
stresses and not factored ones, we adopt the WSD
approach, and use an allowable stress of 20 ksi, which in turn will be increased by 4
/
3 for
seismic and wind load,
σ
all
=
4
3
(20) = 26
.
7
ksi
(33.16)
11. The stress distribution is linear, compression at one end, and tension at the other. The
length of the tension area is given by (similar triangles)
x
460
=
20
460 + 740
⇒
x
=
460
460 + 740
(20) = 7
.
7
ft
(33.17)
12. The total tensile force inside this triangular stress block is
T
=
1
2
(460)
ksi
(7
.
7
×
12)
in
(12)
in
| {z }
width
= 250
k
(33.18)
13. The total amount of steel reinforcement needed is
A
s
=
(250)
k
(26
.
7)
ksi
= 9
.
4
in
2
(33.19)
This amount of reinforcement should be provided at both ends of the wall since the wind
or eartquake can act in any direction. In addition, the foundations should be designed to
resist tensile uplift forces (possibly using piles).
33.2.1.2
Example: Trussed Shear Wall
From (Lin and Stotesbury 1981)
17
We consider the same problem previously analysed, but use a trussed shear wall instead of
a concrete one, Fig.
1. Using the maximum moment of 5
,
760 kip-ft (Eq.
), we can compute the compression
and tension in the columns for a lever arm of 20 ft.
F
=
±
(5
,
760)
k.ft
(20)
ft
=
±
288
k
(33.20)
2. If we now add the effect of the 400 kip vertical load, the forces would be
C
=
−
(400)
k
2
−
288 =
−
488
k
(33.21-a)
T
=
−
(400)
k
2
+ 288 = 88
k
(33.21-b)
3. The force in the diagonal which must resist a base shear of 96 kip is (similar triangles)
F
96
=
p
(20)
2
+ (24)
2
20
⇒
F
=
p
(20)
2
+ (24)
2
20
(96) = 154
k
(33.22)
4. The design could be modified to have no tensile forces in the columns by increasing the
width of the base (currently at 20 ft).
Victor Saouma
Structural Engineering