978-1337094740 Chapter 4 Part 1

Type Homework Help

Pages 14

Words 3589

Textbook Steel Design 6th Edition

Authors William T. Segui

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CHAPTER 4 -COMPRESSION MEMBERS

4.3-1

2. 65 45. 28

4. 71 E

4. 71 29000

50 113. 4

2. 65 135. 8

4. 71 E

4. 71 29000

50 113. 4

4.3-2

rKL

r1. 025 12

3. 68 81. 52

4. 71 E

4. 71 29000

35 135. 6

4.3-3

4.3-4

(a) Lc

rKL

r0. 6515 12

4.3-5

(a) Lc

rKL

r0. 813 12

2. 48 50. 32

Since Lc/r50. 32 113. 4, use AISC Eq. E3-2:

c

1. 67 597. 1 kips Pn

c

(b) From Manual Table 4-14, for Lc/r50. 32 and Fy50 ksi,

c

4.3-6

Since Lc/r42. 83 113. 4, use AISC Eq. E3-2:

[4-3]

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c

1. 67 919. 2 kips Pn

c

(b) From Manual Table 4-14, for Lc/r42. 83 and Fy50 ksi,

c

4.3-7

(a) Let PucPn

4.3-8

[4-4]

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4.4-1

Compute the overall, or flexural, buckling strength.

46 118

Since 87.8 118, use AISC Equation E3-2.

The nominal strength is

Check width-thickness ratios. From the dimensions and properties table in the Manual,

the width-thickness ratio for the larger overall dimension is

The ratio for the smaller dimension is

From AISC Table B4.1, case 12 (and Figure 4.9 in this book), the upper limit for

nonslender elements is

tand b

tare 1.40 E

strength must be computed. (Although the limiting width-thickness ratio is labeled b/t

in the table, that is a generic notation, and it applies to h/tas well.)

For the larger dimension,

Since 54. 5 45. 55,

Feℓ

From AISC Table E7-1, Case b,

The elastic buckling stress is

Feℓc2r



Fy1. 38 35. 15

54. 5

24636. 44 ksi

The unreduced length of the longer side can be taken as

and the effective width is

Feℓ

27. 39

27. 39 8. 41 in.

For the smaller dimension,

Theeffectivecross-sectionalareais

The nominal compressive strength is

4.4-2

Compute the overall, or flexural, buckling strength.

Since 87.20 113.4, use AISC Equation E3-2.

The nominal strength is

Check width-thickness ratios. From the dimensions and properties table in the Manual,

the width-thickness ratio of the web is

From AISC Table B4.1 (and Figure 4.9 in this book), the upper limit for nonslender

Since h

1.49 E

, the web is slender.

For the flange,

For the web,

[4-7]

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Since 47. 37, beb, andthereisnoreductioninarea. Pn855 kips

4.6-1

From the column load tables for Lc18 ft, a W14 74 has a design strength of 563

kips. UseaW1474

From the column load tables for Lc18 ft, a W14 82 has an allowable strength of

413 kips. UseaW1482

Try W16 77 (a nonslender shape), Ag22. 6 in.2,ry2.47 in.

∴Fcr 0. 658Fy/FeFy0. 65860/37.436030. 67 ksi

Compute the overall, or flexural, buckling strength.

[4-8]

or posted to a publicly accessible website, in whole or in part.

Since 87.8 103.5, use AISC Equation E3-2.

The nominal flexural buckling strength is

Check width-thickness ratios. From the dimensions and properties table in the Manual,

the width-thickness ratio of the web is

From AISC Table B4.1 (and Figure 4.9 in this book), the upper limit for nonslender

elements is

For the flange,

For the web,

UseaW1667

(b-2) Assume Fcr 25 ksi

[4-9]

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0. 6Fcr

0. 62526. 3 in.2

2. 49 86. 75 200 (OK)

2. 47 87. 45 200 (OK)

4. 71 E

4. 71 29000

60 103. 5 87. 45

4.6-2

LcKL 2. 02040 ft

(a-1) Pu1. 2D1. 6L1. 21101. 6110308 kips

From the column load tables for Lc40 ft, a W12 120 has a design strength of 339

1. 69 284. 0 200

2. 63 182. 5 200

4. 71 E

4. 71 29000

55 108. 2 182. 5

∴Fcr 0. 877Fe0. 8778. 5947. 537 ksi

2. 74 175. 2 200

Fe2E

2. 72 176. 5 200

[4-11]

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Fe2E

0. 6Fcr

0. 6845. 8 in.2

2. 74 175. 2 200

4. 71 E

4. 71 29000

55 108. 2 175. 2

4.6-3

LcKL 2. 11225. 2 ft

(a) Pu1. 2D1. 6L1. 21001. 6300600. 0 kips

HSS 16 12 3

4.6-4

LcKL 0. 81512 ft

4.6-5

LcKL 2. 11225. 2 ft

4.6-6

LcKL 0. 81512 ft

4.6-7

Pu1. 2D1. 6L1. 2901. 6260524 kips

LcKL 0. 6515. 339. 965 say 10 ft

4.6-8

PaDL90 260 350 kips

4.6-9

c

4.7-1

[4-14]

or posted to a publicly accessible website, in whole or in part.

4.7-2

For an HSS 10 65/16, Ag8. 76 in.2,rx3. 66 in., ry2. 47 in., and there are

no slender elements (see Manual Table 1-12A).

c

1. 67 205. 0 kips Pn

c

4.7-3

[4-15]

or posted to a publicly accessible website, in whole or in part.

Since Lc/r62. 95 103. 5, use AISC Eq. E3-2:

4.7-4

From the column load tables for Lc12 ft, try a W14 74

Lcx

22

4.7-5

From the column load tables for Lc15ft,therearenoW8orW10shapeswith

enough strength. Try a W12 230 :

cPn2450 kips for Lc15 ft

Lcx

rx/ry

35

1. 80 19. 44 ft 15 ft

For Lc19 ft, cPn2150 kips 2280 kips (N.G.)

Try a W14 233:

KxL

rx/ry

35

1. 62 21. 60 ft 15 ft

c

Lcx

rx/ry

35

1. 80 19. 44 ft 15 ft

For Lc19 ft, Pn

c

1430 kips 1520 kips (N.G.)

Try a W12 252:

Lcx

35

c

Try a W14 233:

Lcx

rx/ry

35

1. 62 21. 60 ft 15 ft

For Lc21. 60 ft, Pn

c

1530 kips 1520 kips (OK)

4.7-6

The x-axis strength will control when rx/ryLcx

Lcy

2. 0. This is true for all

rectangular HSS.

4.7-7

Lcx KxL27 ft, Lcy KyL15 ft

3. 85 84. 16 Lcy

3. 25 55. 38

4. 71 E

4. 71 29000

46 118. 3

Since Lc/r84. 16 118. 3, use AISC Eq. E3-2:

Fcr 0. 658Fy/FeFy0. 65846/40.414628. 57 ksi

Fcr

39. 09 38. 13

Since 54. 5 38. 13,

Feℓ



54. 5

The unreduced length of the longer side can be taken as

Fcr

39. 09

39. 09 7. 246 in.

Theeffectivecross-sectionalareais

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978-1337094740 Chapter 4 Part 1

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978-1337094740 Chapter 4 Part 1

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