Steel design william t segui 5th edition solution manual

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An Instructor’s Solutions Manual to Accompany

WILLIAM T SEGUI

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ISBN-13: 978-1-111-57601-1 ISBN-10: 1-111-57601-7

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I NSTRUCTOR ' S S OLUTIONS M ANUAL

Contents

Chapter 1 Introduction 1-1 Chapter 2 Concepts in Structural Steel Design 2-1

Chapter 3 Tension Members 3-1 Chapter 4 Compression Members 4-1

Chapter 6 Beam-Columns 6-1 Chapter 7 Simple Connections 7-1 Chapter 8 Eccentric Connections 8-1 Chapter 9 Composite Construction 9-1 Chapter 10 Plate Girders 10-1

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PREFACE

This instructor's manual contains solutions to the problems in Chapters 1–10 of Steel

Design, 5 th Edition Solutions are given for all problems in the Answers to Selected

Problems section of the textbook, as well as most of the others

In general, intermediate results to be used in subsequent calculations were recorded to four significant figures, and final results were rounded to three significant figures Students following these guidelines should be able to reproduce the numerical results given However, the precision of the results could depend on the grouping of the computations and on whether intermediate values are retained in the calculator between steps

In many cases, there will be more than one acceptable solution to a design problem; therefore, the solutions given for design problems should be used only as a guide in grading homework

I would appreciate learning of any errors in the textbook or solutions manual that you may discover You can contact me at wsegui@memphis.edu A list of errors and corrections in the textbook will be maintained at http://www.ce.memphis.edu/segui/errata.html

William T Segui

August 15, 2011

1.5-5 (Note: These results are very approximate and depend on how the curves are

drawn.)

(a)

0 10

Load Elongation Stress

(kips) (in.) (ksi) Strain

1.0 0.0010 5.094 0.00052.0 0.0014 10.19 0.00072.5 0.0020 12.74 0.00103.5 0.0024 17.83 0.00125.0 0.0036 25.47 0.00186.0 0.0044 30.57 0.00227.0 0.0050 35.66 0.00258.0 0.0060 40.75 0.00309.0 0.0070 45.85 0.003510.0 0.0080 50.94 0.004011.5 0.0120 58.58 0.006012.0 0.0180 61.13 0.0090

Load Elongation x 10 3 S tress St rain x 10 3

(kip s) ( in.) ( ksi) (in./in )

0 10

CHAPTER 2 - CONCEPTS IN STRUCTURAL STEEL DESIGN

(b) Combination 3 controls: P a  D  S  0 2  0 14  0 34 kips/ft

 93 34 kipsCombination 4: 1 2D  1 0W  0 5L  0 5L r

 1 213 3  1 0150 6  0 56 9  0 51 3  170 7 kipsCombination 5: 1 2D  1 0E  0 5L  0 2S

CHAPTER 3 - TENSION MEMBERS

The allowable service load is the smaller value: P n/t  56 6 kips

Alternate solution using allowable stress: For yielding,

For fracture,

F t  0 5F u  0 558  29 0 ksi

and the allowable load is F t A e  29 02 203  63 89  63 89 kips

The allowable service load is the smaller value 56 7 kips

b) The allowable strength based on yielding is

P n

t  201 3

2 00  101 kips

Alternate solution using allowable stress: For yielding,

Gross section: P n  F y A g  508 81  440 5 kips

Net section: Hole diameter 1  1

b) The allowable strength based on yielding is

The allowable service load is the smaller value: P n/t  48 5 kips

Alternate solution using allowable stress: For yielding,

F t  0 6F y  0 636  21 6 ksi

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and the allowable load is

P u  1 295  1 69  128 kips (133 kips controls)

For two angles, P n  248 38  96 76 kips

a) The design strength based on yielding is

 t P n  0 9086 4  77 76 kips

The design strength based on fracture is

 t P n  0 7596 76  72 57 kips

The design strength is the smaller value:  t P n  72 6 kips

P u  1 2D  1 6L  1 212  1 636  72 0 kips  72.6 kips (OK)

The member has enough strength

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b) For the gross section, The allowable strength is

and the allowable strength is F t A g  21 62  1 20  51 84 kips

For the net section, the allowable strength is

and the allowable strength is F t A e  292  0 834 1  48 38 kips

The net setion strength controls; the allowable strength is 48 4 kips When the onlyloads are dead load and live load, ASD load combination 2 will always control:

P a  D  L  12  36  48 kips  48.4 kips (OK)

The member has enough strength

Rupture:  t P n  0 75201 8  151 4 kips

(b) Yielding: Pn t  144 0

1 67  86 23 kipsRupture: Pn t  201 8

3 ∴ this shape does not qualify for U  0 9.

Since there are more than 3 bolts per line, U  0 85

and the allowable strength is F t A g  306 08  182 4 kips

For the net section, the allowable strength is Pn t  308 6

2 00  154 3 kips

Alternately, the allowable stress is F t  0 5F u  0 565  32 5 ksi

and the allowable strength is F t A e  32 54 747  154 3 kips

The net setion strength controls; the allowable strength is 154 kips When the onlyloads are dead load and live load, ASD load combination 2 will always control:

(a) Gross:  t P n  0 90121 7  110 kips

Net:  t P n  0 75176 4  132 kips Gross section controls:

 t P n  110 kips

t  121 7

1 67  72 9 kipsNet: Pn t  176 4

2 00  88 2 kips Gross section controls:

P n/t  72 9 kips

3.3-8

For A242 steel, F y  50 ksi and F u  70 ksi (based on flange thickness)

For yielding of the gross section,

Alternate computation of allowable load using allowable stress: For yielding,

F t  0 6F y  0 650  30 0 ksi

and the allowable load is F t A g  30 04 79  143 7 kips

For fracture, F t  0 5F u  0 570  35 ksi

and the allowable load is

F t A e  353 482  121 9 kips ∴ P  122 kips

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a Gross section: P n  F y A g  365/8  12  270 kips P n  270 kips

A e  A n  tw n  5/810 25  6 406 in.2

3.4-2

Gross section: P n  F y A g  365/8  10  225 0 kips

Net section: Hole diameter 7

w n  10 − 31.0  74  12 25 in

The effective net area is

A e  A n  tw n  5/88 5  5 313 in.2

P n  F u A e  585 313  308 2 kips

a) The design strength based on yielding is  t P n  0 90225 0  202 5 kips

The design strength based on fracture is  t P n  0 75308 2  231 2 kips

b) The allowable strength based on yielding is

Alternate computation of allowable load using allowable stress: For yielding,

Gross section: P n  F y A g  507 02  351 0 kips

Net section: Hole diameter 3

8 − 2 5242 5  76  7 256 in.2

Use A n  6 320 in.2

U  1 − x̄ℓ  1 − 0 981

5 5  0 821 6The effective net area is

(b) Net section: Hole diameter 3

− 0 3057/8 − 0 305 7

8 − 3243  11 19 in.2

Gross section: P n  F y A g  362 89  104 0 kips

Net section: Hole diameter 3

P n  F y A g  505 87  293 5 kips for one channel.

Net section: Hole diameter 1

or 5 87− 0 3795/8 − 0 379 5

8 − 4245  5 699 in.2

Use A n  5 633 in.2

U  1 − x̄ℓ  1 − 4  40 606  0 962 1

The effective net area is

A e  A n U  5 6330 9621  5 420 in.2

P n  F u A e  655 420  352 3 kips for one channel

(a) Gross:  t P n  0 90293 5  264 2 kips

Check upper limit:

Check upper limit:

The shear areas are

The tension area is

A nt  3/84 − 1.0  1 125 in.2

F y  36 ksi, F u  58 ksi

R n  0 750.6F u A nv  U bs F u A nt  0 750 6584 875  1 0581 125

 176 kipsCheck upper limit:

0 750 6Fy A gv  U bs F u A nt  0 750 6366 75  1 0581 125

 158 kips  176 kipsThe upper limit controls; R n  158 kips

A e  A n U  6 3440 80  5 075 in.2  5 06 in.2 (OK) Use L8 6  9

16

(b) P a  D  L  50  100  150 kips

or D  0 75L  0 750 6W  50  0 75100  0 750 645  145 3 kips Use P a  150 kips

A g  7 84 in.2  6 94 in.2 (OK)

rmin  r z  1 59 in  0.8 in (OK)

A g  2 07  2  4 14 in.2  3 70 in.2 (OK)

r x  0 853 in., r y  1 43 in., ∴ rmin  0 853 in  0.6 in (OK)

A n  4 14 − 1 01/4  3 89 in.2

From Case 8 in AISC Table D3.1, use U  0 80

A e  A n U  3 890 80  3 11 in.2  2 76 in.2 (OK)

4, long legs back-to-back:

A g  2 07  2  4 14 in.2  3 70 in.2 (OK)

r x  0 853 in., r y  1 43 in., ∴ rmin  0 853 in  0.6 in (OK)

A n  4 14 − 1 01/4  3 89 in.2

From Case 8 in AISC Table D3.1, use U  0 80

A e  A n U  3 890 80  3 11 in.2  2 76 in.2 (OK)

Use U  0 70 (Case 7, AISC Table D3.1)

A e  A n U  10 980 70  7 69 in.2  6 95 in.2 (OK) Use S12 40.8

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Use U  0 70 (Case 7, AISC Table D3.1)

A e  A n U  13 50 70  9 45 in.2  8 31 in.2 (OK) Use S15 50

A g  2 07 in.2  1 65 in.2 (OK) rmin  1 17 in  0.32 in (OK)

A g  2 07 in.2  1 67 in.2 (OK) rmin  1 17 in  0.32 in (OK)

A g  5 87 in.2  4 28 in.2 (OK)

rmin  r y  0 690 in  0 7 in (say OK)

A e  A g U  5 870 85  4 99 in.2 3 96 in.2 (OK)

Compute U with Equation 3.1.

U  1 − x̄ℓ  1 − 0 6069  0 932 7

The next lighter shape that meets slenderness requirements is a C10 15.3 with

A g  4 48 in.2, rmin  0 711 in., and x̄  0.634 in.

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Required d  1 53 in Use 15/8in.

(b) ASD: Load combination 2 controls: P a  D  L  43  4  47 kips

(a) Dead load beam weight  0.036 kips/ft

Because of symmetry, the tension is the same in both rods

Load at each support 50 2712  603 2 lb, Load on rod  603 2

2  301 6 lb

[3-36]

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T T

603.2 lb

lb6.301

02.6032

A e  A g U  3 240 9027  2 92 in.2  2 34 in.2 (OK)

This shape can be used for all of the web tension members Although each membercould be a different size, this would not usually be practical.

3.8-3

Use load combination 3: 1 2D  1 6S.

Tributary surface area per joint18 92  9/62  164 2 ft2

Dead load per truss 4  12  340 79  230  58030  5 85  104 lb

Snow load per truss 208030  48, 000 lb

D  58500/8  7313 lb, S  48, 000/8  6000 lb

Load combination 3 controls:

Factored joint load 1 2D  1 6S  1 27 313  1 66  18 38 kips

Bottom chord: Member FE (member adjacent to the support) has the largest tension

force

[3-41]

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