Solucionario vinnakota Capitulo 8.pdf

Steel Structures by S. Vinnakota Chapter 8 page 8-1

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CHAPTER 8: COLUMNS

cr y cP8.1. Plot the LRFDS column curve NF / F vs. 8 . Show all the salient points.

crP8.2. Use the LRFD Specification and plot the design axial compressive stress NF vs. effective slenderness ratio

KL/r for steel columns. Include the following steels.

(a) A36 (b) A572 Grade 50 (c) A572 Grade 65

(d) A514 Grade 90 (e) A514 Grade 100

P8.3. Determine the elastic flexural buckling stress and elastic flexural buckling load for the pin-ended columns

using the Euler equation. Assume E = 29,000 ksi.

A solid square bar 2.0 in. by 2.0 in. (a) L = 6.0 ft (b) L = 12 ft

Solution

Pin ended column

Section: Square

b = 2.0 in.; d = 2.0 in.

A = bd = 2.0 × 2.0 = 4.00 in. 2

I = b d / 12 = 2.0 × 2.0 ÷ 12 = 1.33 in. 3 3 4

a. Column length, L = 6.0 ft = 72.0 in.

Elastic flexural buckling load, (Ans.)

E E Elastic flexural buckling stress, f = P / A = 73.4 ÷ 4.00 = 18.4 ksi (Ans.)

b. Column length, L = 12.0 ft = 144 in.

Elastic flexural buckling load, (Ans.)

E E Elastic flexural buckling stress, f = P / A = 18.4 ÷ 4.00 = 4.60 ksi (Ans.)






P8.4. Determine the elastic flexural buckling stress and elastic flexural buckling load for the pin-ended columns

using the Euler equation. Assume E = 29,000 ksi. A W 12×96

(a) L = 50 ft (b) L = 25 ft

Solution

Pin ended column

x ySection: W 12×96 6 A = 28.2 in. ; I = 833 in. ; I = 270 in.2 4 4

a. Column length, L = 50 ft = 600 in.

From Eq. 8.4.19:

From Eq. 8.4.20, elastic flexural buckling load,

(Ans.)

E E Elastic flexural buckling stress, f = P / A = 215 ÷ 28.2 = 7.62 ksi (Ans.)

b. Column length, L = 25 ft = 300 in.

From Eq. 8.4.19:

From Eq. 8.4.20, elastic flexural buckling load,

(Ans.)

E E Elastic flexural buckling stress, f = P / A = 859 ÷ 28.2 = 30.5 ksi (Ans.)






P8.5. Determine the effective lengths of each of the columns of the frame shown in Fig. P8.5. The members are

oriented so that the webs are in the plane of the frame. The structure is unbraced in the plane of the frame.

All connections are rigid, unless indicated otherwise. In the direction perpendicular to the frame, the frame

is braced at the joints. The connections at these points of bracing are simple connections (no rotational

restraint).

(a) Use the alignment charts.

(b) Use the equations given in Section 8.5.

See Fig. P8.5 of the text book.

Solution

Unbraced frame.

x xColumn sections: W10×39 6 I = 209 in. ; W10×45 6 I = 248 in.4 4

xW10×49 6 I = 272 in. 4

x xGirder sections: W16×40 6 I = 518 in. ; W16×50 6 I = 650 in.4 4

x xW14×30 6 I = 291 in. ; W14×34 6 I = 340 in. 4 4

Column lengths: c1, c2, c3, c4 = 14 ft; c5, c6 = 12 ft

Girder lengths: g1 = 24 ft; g2, g4 = 16 ft; g3 = 32 ft

a. Using alignment charts

Column c1

g1" = 0.5 (far end pinned)

AG = 10.0 (recommended value for hinged base)

xFrom alignment chart for sidesway uninhited columns (Fig. 8.56b), K . 1.98 (Ans.)

Column c4




Column c2






g1 g2" = 0.0 (near end pinned); " = 1.0 (both ends rigidly jointed)

AG = 1.0 (recommended value for fixed base)


Column c3




Column c5

A BG = 1.98 (same as G value calculated for column c2)


Column c6


xFrom alignment chart for sidesway uninhited columns (Fig. 8.56b), K . 1.30 (Ans)






b. Using the equations given in Section 8.5

From Eq. 8.5.14, the effective length of a column in an unbraced frame is,

A B xColumn c1: G = 10.0; G = 1.38; K . 1.99 (Ans.)











P8.6. Repeat Problem P8.5, assuming the structure is braced in the plane of the frame also, by diagonal members in

the middle bay.

Solution

Braced frame.

x xColumn sections: W10×39 6 I = 209 in. ; W10×45 6 I = 248 in.4 4

xW10×49 6 I = 272 in. 4

x xGirder sections: W16×40 6 I = 518 in. ; W16×50 6 I = 650 in.4 4

x xW14×30 6 I = 291 in. ; W14×34 6 I = 340 in. 4 4

Column lengths: c1, c2, c3, c4 = 14 ft; c5, c6 = 12 ft

Girder lengths: g1 = 24 ft; g2, g4 = 16 ft; g3 = 32 ft

a. Using alignment charts

Column c1



xFrom alignment chart for sidesway inhited columns (Fig. 8.56a), K . 0.81 (Ans.)

Column c4




Column c2









Column c3




Column c5



Column c6








b. Using the equations given in Section 8.5

From Eq. 8.5.18, the effective length of a column in an unbraced frame is,












P8.7. Determine the factored axial compressive strength of an 18 ft long W8×48 column fixed about both axes at its

ends. Assume A242 Grade 50 steel. Solve using specification equations only. Check calculations by using

various tables in the LRFD Manual.

Solution

Given

x yMember: W8 × 48 6 A = 14.1 in. ; r = 3.61 in. ; r = 2.08 in.2 2

yMaterial: A242 Gr 50 6 F = 50 ksi

Length, L = 18ft

x yAs there are no intermediate supports, L = L = 18 ft

x yEnd conditions: Fixed at both ends about both axes. From Fig. 8.5.4a: K = K = 0.65

As the end conditions are the same about both the axes, and as there are no intermediate braces, minor

axis buckling controls. So,

a. Using LRFDS Equations in Section E2

From Eq. E2-4 of the LRFDS, slenderness parameter,

cAs 8 < 1.5, buckling occurs in the inelastic domain. Use LRFDS Equation E2-2,

cr y crF = F 6 F = 35.8 ksi

d c cr g� P = N F A = 0.85 × 35.8 × 14.1 = 429 kips (Ans.)

b. Using LRFDS Table 4

cEnter LRFDS Table 4 with 8 = 0.892 and read,

d c cr g� P = N F A = 0.609 × 50 × 14.1 = 429 kips (Ans.)

c. Using LRFDS Table 3-50

c crEnter LRFDS Table 3-50, with KL /r = 67.5, and obtain N F = 30.5 ksi by interpolation.

d c cr g� P = N F A = 30.5 × 14.1 = 430 kips (Ans.)






P8.8. A W12×65 column 30 ft long is fixed at both ends about both axes. In addition it is braced in the weak direction

at the one-third points. A242 Grade 42 steel is used. Determine the design axial compressive strength of the

column.

Solution

g x yMember: W12×65 6 A = 19.1 in. ; r = 5.28 in.; r = 3.02 in.2


Column length, L = 30ft

For major axis buckling:

x xL = 30.0 ft; fixed - fixed 6 K = 0.65

For minor axis buckling:

y1 yL = 10.0 ft; fixed - pinned 6 K = 0.80

y2 y L = 10.0 ft; pinned - pinned 6 K = 1.00

y3 y L = 10.0 ft; pinned - fixed 6 K = 0.80

x xK L = 0.65 × 30.0 = 19.5 ft

y y 1 y y 3 y y 2(K L ) = 0.65 × 10.0 = 6.5 ft = (K L ) ; (K L ) = 1.0 × 10.0 = 10.0 ft

y yK L = 10.0 ft

;

Slenderness parameter,

cEnter LRFDS Table 4 with 8 = 0.537 and obtain,

c cr 6 N F = 0.753 × 42 = 31.6 ksi







P8.9. A S10×35 standard I-shape is used as a column in an industrial building. The column considered is 21 ft long.

It can be assumed to be hinged about both axes at the top and fixed about both axes at the base. In addition,

transverse braces (girts) are connected to the web of the column. What factored axial load is permitted by the

LRFDS, if the bracing is positioned at:

(a) points 7 ft apart (b) points 5 ft 3 in. apart?

Solution

x yMember: S10 × 35 6 A = 10.3 in. ; r = 3 .78 in.; r = 0.898 in.2

yMaterial: A36 steel (Preferred material specification, for S-shapes) 6 F = 36 ksi

Length, L = 21 ft

x x x xL = 21 ft; fixed-pinned 6 K = 0.80 6 K L = 0.8 × 21 = 16.8 ft

a. Bracing at 7 ft intervals

3 segments.

y1 y1 y y 1L = 7.0 ft; fixed-pinned 6 K = 0.8 6 (K L ) = 0.8 × 7.0 = 5.60 ft

y2 y2 y y 2L = 7.0 ft; pinned-pinned 6 K = 1.0 6 (K L ) = 1.0 × 7.0 = 7.00 ft


y yK L = 7.00 ft

Enter LRFDS Table 3-36 and read,


b. Bracing at 5 ft 3 in. apart

4 segments.

y1 y1 y y 1L = 5.25 ft; fixed-pinned 6 K = 0.8 6 (K L ) = 0.8 × 5.25 = 4.20 ft


y yK L = 5.25 ft

Enter LRFDS Table 3-36, and read,







P8.10. A W 12×72 of A588 Grade 50 steel is used as a column 16 feet long. The column ends are pinned and the weak

axis is braced 10 ft from the lower end. The structure is braced in both xx and yy planes of the column.

Determine the design axial compressive strength of the column.

Solution

x ySection: W 12×72 6 A = 21.1 in. ; r = 5.31 in.; r = 3.04 in.2

yMaterial: A588 Grade 50 steel 6 F = 50 ksi

Column length, L = 16 ft

x yStructure braced in both xx and yy planes of the column 6 K # 1.0; K # 1.0

x xFor buckling about the major axis: 6 pinned at both ends 6 K = 1.0, L = 16 ft

For buckling about the minor axis: 6 pinned at both ends and brace at 10 ft from base

y y 1 y y 26 (K L ) = 1.0 × 10.0 = 10.0 ft; (K L ) = 1.0 × 6.0 = 6.00 ft

y y6 K L = max (10.0; 6.00) = 10.0 ft

;

c crFrom LRFDM Table 3-50, for K L/r = 39.5, N F = 37.9 ksi

Design axial compressive strength,

d c cr gP = = N F A = 37.9 × 21.1 = 800 kips (Ans.)

Alternatively

x x y yFor the W12x72 column given, K L = 16.0 ft; K L = 10 ft

From LRFDM Table 4-2, for a W12×72,

dFrom LRFDM Table 4-2, for a W12×72 with KL = 10 ft, P = 800 kips. (Ans.)






P8.11. A W12×58 steel shape, strengthened by welding two d× 8 in plates to the outside face of the flanges is used

as a column. The member has a length of 15 ft and is assumed to be pin connected at both ends. It is braced

at mid-length to prevent movement in the x direction only. Determine the factored axial compressive strength

of the column, if A572 Grade 60 steel is used.

Solution

Section: W12×58 with two d×8 plates

W xW yW W12×58: A = 17.0 in. ; I = 475 in. ; I = 107 in.2 4 4

pl Plates : A = 2 × 8 × 3/8 = 6.0 in.2

yMaterial: A572 Gr 60; F = 60 ksi


x xL = 15.0 ft; end conditions: pinned-pinned 6 K = 1.00

y1 y2 yL = L = 7.5 ft; end conditions: pinned-pinned 6 K = 1.00

Properties of built-up section

A = 17.0 + 6.0 = 23.0 in. 2

x I = [475 + 0] + 2 [0 + 8.0× 3/8 × (6.29) ] = 712 in.2 4

y I = [107 + 0] + 2[1/12 × (3/8) (8.0) ] = 139 in.3 4









P8.12. A 20 ft long, A572 Grade 42 steel column is obtained by welding a ½ × 12 in. plate to two MC 13 × 50 channel

shapes, to form a doubly symmetric section as shown in Fig. P8.12. Determine the factored axial compressive

x ystrength of the column as per the LRFDS. Assume K = 1.0 and K = 1.4.

See Figure P8.12 of the text book.

Solution

Column length = 20ft

x yGiven: K = 1.00, K = 1.4

Dubly symmetric section built-up from:

C xC yCMC 13×50: A = 14.7 in. ; I = 314 in. ; I = 16.4 in.2 4 4

wCt = 0.787 in.; x = 0.974 in.

plPlate ½×12: A = 6.0 in.2

For the built-up section:

A = 14.7×2 + 12× ½ = 35.4 in.2

xI = = 628 + 0 = 628 in.4

1d = 6 + 0.787 - 0.974 = 5.813 in.

yI = = 1100 in.4

;









P8.13. The cross-section of a structural steel column is a built-up section obtained by welding the web of a W12×26

shape to the flange of a W8×48 to form a mono-symmetric section shown in Fig. P8.13. The 20-ft long column

is fixed at the base and pinned at the top about both axes. The column is part of a braced frame in both the xx

and yy planes. Assume A572 Grade 65 steel. Determine the design axial compressive strength of the column.


Solution


x yNo intermediate braces 6 L = L = 20.0 ft

x yEnd conditions: fixed at the base and pinned at the top 6 K = 0.80; K = 0.80


Mono-symmetric section built-up from two W-shapes:

1 x1 y1W8×48: A = 14.1 in. ; I = 184 in. ; I = 60.9 in.2 4 4

1 d = 8.50 in.

2 x2 y2W12×26: A = 7.65 in. ; I = 204 in. ; I = 17.3 in.2 4 4

w2 t = 0.230 in.


A = 14.1 + 7.65 = 21.8 in.2

from bottom fibre

1 2d = 5.77 - 4.25 = 1.52 in.; d = 8.615 - 5.77 = 2.845 in.

xI = [184 + 14.1×1.52 ] + [17.3 + 7.65×2.845 ] = 296 in.2 2 4

yI = [ 60.9 + 0] + [204 + 0] = 265 in.4

x yr = 3.68 in.; r = 3.49 in.









P8.14. A cantilever column (fixed at the base and free at the top, about both axes) is obtained by welding two WT 9

× 53's to the web of a W24 × 76 section, to form a doubly symmetric compound section shown in Fig. P8.14.

The column is 20 ft long. Assume A572 Grade 60 steel and determine the design axial compressive strength of

the column as per the LRFD Specification.

See Figure. P8.14 of the text book.

Solution


x yNo intermediate braces 6 L = L = 20.0 ft

End conditions: fixed at the base and free at the top

The recommended value for the effective length factor of a cantilever column is 2.1, from Fig. 8.5.4.

x y Hence, K = 2.10; K = 2.10.


Doubly symmetric section built-up from:

W wW xW yW W24×76 6 A = 22.4 in. ; t = 0.44 in.; I = 2100 in. ; I = 82.5 in.2 4 4

T T xT yT WT 9×53 6 A = 15.6 in. ; d = 9.365 in.; I = 104 in. , I = 110 in. 2 4 4

T y = 1.97 in.


A = 22.4 + 2 × 15.6 = 53.6 in.2

xI = [2100 + 0] + [2×110 + 0] = 2320 in.4

yI = [82.5 + 0] + 2 [104 + 15.6 (0.22 + 9.365 - 1.97) ] = 2100 in.2 4

;



So, the design strength of the column is

d c cr gP = N F A = 0.479 × 60 × 53.6 = 1540 kips (Ans.)






P8.15. A heavy column used in the ground floor of a high-rise building is formed by welding two 4×24 in. plates to the

flanges of a W14×730 section as shown in Fig. P8.15. Assume A572 Grade 42 steel and determine the design

x x y yaxial compressive strength of the column, if K L = 32 ft and K L = 28 ft. Use the LRFD Specification.


Solution

x x y yGiven: K L = 32 ft; K L = 28 ft


Doubly symmetric section built-up from:

W fW xW yW W14×730: A = 215 in. ; b = 17.9 in.; I = 14,300 in. ; I = 4,720 in.2 4 4

pl PL 24 × 4 : A = 24 × 4 = 96 in. ; 2


A = 215 + 2 (96) = 407 in.2

xI = [14300 + 0] + 2 × [4608 + 0] = 23,520 in.4

yI = [4720 + 0] + 2 × [128 + 96 (8.95 + 2.0) ] = 28,000 in.2 4

;

For the column:

controls;



So, the design strength of the column is

d c cr gP = N F A = 0.727 × 60 × 407 = 12,430 kips (Ans.)






P8.16. Determine the design axial compressive strength of the columns considered in the Problem P8.5, assuming A992

steel is used.

P8.17. Determine the design axial compressive strength of the columns considered in the Problem P8.5, assuming A992

steel is used.






P8.18. An eight-story office building has a bay size of 40 feet by 36 feet. The structure is designed for a roof dead load

of 80 psf, a roof live load of 20 psf, a floor dead load of 80 psf, and a floor live load of 50 psf. Make a

preliminary design, and select a suitable W14-shape of A992 steel for an interior column at ground level.

x y x yAssume that K = K = 1.0 and L = L = 14 ft.

Solution

x x y yGiven: K L = 14.0 ft; K L = 14.0 ft

Tributary area at each level = 40 × 36 = 1440 ft2

The column under consideration receives loads from roof and seven floors.

Roof dead load = 80 psf

rRoof live load, L = 20 psf

Floor dead load = 80 psf

Floor live load, L = 50 psf

With only gravity loads acting on the column, load combination LC-4 controls the required axial

strength of the column

r1.2D + 1.6L + 0.5L

Contribution to axial load, from roof,

Contribution to axial load, from the seven floors,

So, factored load on the column,

uP = 153 + 1770 = 1920 kips

y y reqEnter LRFDM Table 4-2 for W14 shapes with KL = K L = 14 ft, P = 1920 kips and select a

dW14×176 with P = 1940 kips.

Provide a W14×176 of A992 steel. (Ans.)






P8.19. Select the lightest W section for a column supporting a factored axial load of 620 kips. Support conditions,

x yapplicable for both principal axes, are full fixity at the bottom and a pinned connection at the top. L = L = 17

ft 6 in. Assume A992 steel is used.

Solution

yMaterial: A 992 steel 6 F = 50 ksi

uFactored load, P = 620 kips

From Fig. 8.5.4c, for a column fixed at one end and pinned at the other end, the recommended value

of K = 0.80. Hence, we have:

x x y yK L = K L = 0.80 × 17.5 = 14.0 ft

As KL is same about both axes, buckling in the weak direction will control the strength of the column.

So, we enter the column selection tables in Part 4 of the LRFDM with a value of KL = 14 ft :

dW10 series 6 W10 × 68 6 P = 625 kips > 620 kips O.K.



Use W12 × 65, the lightest section.






P8.20. The interior column of an industrial building supports a roof dead load of 55 psf and a snow load of 35 psf. The

column is 25 feet long. The bottom end is fixed, and the top end pinned. In addition, its weak axis is braced by

struts at a point 6 feet down from the column top as shown in Figure P8.20. The column has a tributary area

of 1800 ft . The structure is braced in both the xx and yy planes of the column. Select a suitable W shape of2

A588 Grade 50 steel.


Solution

Tributary area = 1800 ft2

Dead load, D = 55 psf

Snow load, S = 35 psf

Assuming 50 plf for self weight of column, dead load of column = say 1.5

Factored axial load on the column,

x x y yAs the structure is braced in both the xx and yy planes of the column, K L # 25 ft, and K L # 25 ft

x xFor buckling about major axis 6 Fixed at base and pinned at top 6 K L = 0.80 × 25 = 20 ft

For buckling about the minor axis 6 Fixed at base, pinned at top and braced 19 ft from the base:

y y 16 (K L ) = 0.8 × 19 = 15.2 ft

y y 2(K L ) = 1.0 × 6 = 6.00 ft

y y6 K L = max [15.2; 6.00] = 15.2 ft

y yEnter LRFDM Table 4-2 for W10 shapes with KL = K L = 15.2 and observe that a W10×39 has a

dyP = 263 kips > 221 kips. Also, resulting in:

So, major axis buckling will not control the design strength.

So, select a W10×39 of A588 Grade 50 steel. (Ans.)






P8.21. Select the lightest W12 of A588 Grade 50 steel to carry an axial compressive load of 500 kips dead load and 800

kips live load. The 30 ft long truss member is assumed to be pinned at both ends about both axes. Make all

checks.

Repeat the problem if, in addition, the column is provided with weak direction support at mid-height.

Solution

Service loads: D = 500 kips, L = 800 kips

uUse load combination, LC-2 6 P = 1.2×500 + 1.6×800 = 1880 kips


End conditions: Pinned at both ends about both axes.

yMaterial: A588 Grade 50 6 F = 50 ksi

a. No intermediate bracing

x x y yK L = K L = 1.00 × 30 ft = 30.0 ft. So, buckling about the minor axis will control the design.

y yEnter Column Tables for W12 shapes, in LRFDM Table 4-2, with KL = K L = 30 ft and select a

d dy uW12×336 that provides P = P = 1910 kips > P = 1880 kips.

Check:

x yFor a W12×336: A = 98.8 in. ; r = 6.41 in.; r = 3.47 in.2

;

c crEnter LRFDS Table 3-50 and read N F = 19.3 ksi

d c cr g� P = N F A = 19.3 × 98.8 = 1907 kips > 1880 kips O.K.


b. Intermediate bracing provided at midheight

x x y yK L = 1.00 × 30 ft = 30.0 ft; K L = 1.00 × 15 ft = 15.0 ft

y yEnter Column tables for W12 shapes in LRFDM Part 4, with KL = K L = 15.0 ft and tentatively

dy u x yselect a W12×190 with P = 1900 kips > P = 1880 kips. Also r /r = 1.79 for this section.

x x y dxReenter Column tables for W12 shapes, with KL = (K L ) = 16.8 ft and select a W12×210 with P

u= 1992 kips > P = 1880 kips. For this section :

x yA = 61.8 in. ; r = 5.89 in.; r = 3.28 in.2






;

c crFrom LRFDS Table 3-50, N F = 32.4 ksi

d c cr g P = N F A = 32.4 × 61.8 = 2,000 kips > 1,880 kips O.K.







P8.22. Select the lightest W14 of A572 Grade 50 steel to carry an axial compression load of 100 kips dead load and

360 kips live load. The 28 ft long column is pinned at both ends about both axes and in addition has weak

direction support at mid height. Make all checks.

Solution


yMaterial: A572 Grade 50 steel 6 F = 50 ksi

Service loads: D = 100 kips; L = 360 kips

Factored axial load on the column (from LC 4-2),

uP = 1.2×100 + 1.6×360 = 696 kips

Column pinned at both ends about both axes, and has support at midheight for buckling about the

minor axis, resulting in:

x x y yK L = 1.0×28.0 = 28.0 ft; K L = 1.0×14.0 = 14.0 ft

y yEnter Column Load Tables for W14 shapes, in LRFDM Part 4, with KL = K L = 14.0 ft and observe

dy u x ythat for a W14×82, P = 729 kips > P = 696 kips. Also, for this section r /r = 2.44

d� W14×82 with P = 729 kips is the lightest W14 shape.

Checks

1. Strength

x y For a W14 × 82, LRFDM Table 1-1 gives: A = 24.0 in. ; r = 6.05 in; r = 2.48 in.2

;

From LRFDS Table 3-50, for

d c cr g� P = N F A = 30.4 × 24.0 = 730 kips > 696 kips O.K.

2. Local buckling of plates

With the help of Table 8.7.1 and LRFDM Table 1-1, we obtain for

Flanges: � O.K.

Web: � O.K.

So, plate local buckling will not reduce the member design strength calculated above.

So, select W14×82 of A572 Grade 50 steel. (Ans.)






P8.23. Design the lightest W-shape of A992 steel to support a factored axial load of 960 kips. The effective length with

respect to its minor axis is 14 ft and the effective length with respect to its major axis is 30 ft.

Solution



uFactored axial compressive load, P = 960 kips

x x y yGiven: K L = 30ft; K L = l4 ft

a. Member selection

y y reqEnter LRFDM Table 4-2 for W10 shapes, with KL = K L = 14.0 ft, P = 960 kips and observe that

dy x ya W10×112 gives P = 1050 kips. Also, for this section, r /r = 1.74 resulting in

x x y (K /L ) = 17.2 > 14.0 ft

x x yReentering the table with KL = (K L ) = 17.2 ft, we observe that the heaviest W10 section (W10

×112) does not provide adequate axial strength.

y y reqNext, enter LRFDM Table 4-2 for W12 shapes, with KL = K L = 14.0 ft, P = 960 kips and note that

dy x ya W12×96 gives P = 966 kips and has r /r = 1.76.

x x y� (K L ) = 17.0 > 14.0 ft

x x y dReentering Table 4-2 with KL = (K /L ) = 17.0 ft, we observe that a W12×106 provides, P = 968

ukips > P = 960 kips.

dy x yProceeding in a similar manner, we observe that for a W14×90: P = 969 kips; r /r = 1.66.

x x y� (K /L ) = 18.1 > 14.0 ft

x x y dReentering Table 4-2 with KL = (K /L ) = 18.1 ft, we find that a W14×99 provides, P = 962 kips

u> P = 960 kips.

From W12×106 and W14×99, W14×99 is the lighter, so we select W14×99.

b. Checks

1. Strength

x y W14 × 99 : A = 29.1 in. , r = 6.17 in., r = 3.71 in.2

;

c cr From LRFDS Table 3-50, for , N F = 33.1 ksi

d c cr g � P = N F A = 963 kips > 960 kips








Flanges: � O.K.

Web: � O.K.








P8.24. A column is built into a wall so that it may be considered as continuously braced in the weak direction. The

column is 18 ft long and may be considered hinged at both ends with respect to strong axis buckling. The

factored axial load to be carried is 700 kips. Use A992 steel. Select the lightest W-shape used as columns.

See Fig. X8.5.1 f of the text book for guidance.

Solution



Material: A992 steel

y yBecause column is continuously braced in the weak direction, K L = 0.

The calculation is shown in a tabular form below. The loads and other information are from LRFDM

Table 4-2. The first trial shape is obtained by entering the table with KL = 0.

Section W10 × 60 W12 × 58 W14 × 61

dyP (kips) 748 723 761

x yr /r 1.71 2.10 2.44

x x y(K L ) (ft) 10.5 8.57 7.38

Revised

section

W10×68 W12×65 W14×68

dxP (kips) 715 745 773

The lightest section is a W12 × 65.

1. Strength

x yW12×65: A = 19.1 in. ; r = 5.28 in.; r = 3.02 in.2


d c cr gP = N F A = 37.6 × 19.1 = 718 > 700 kips � O.K.



Flanges: � O.K.

Web: � O.K.


So, select a W12×65 of A992 steel. (Ans.)











P8.25. Repeat Problem P8.24, if the selection is extended to W-shapes up to 18-in. nominal depth.

Solution

u yGiven: L = 18ft; P = 700 kips; F = 50 ksi

y y K L = 0

Aim is to see, if a lighter section than W12x65 (obtained in Problem 8.24) could be found by selecting

a deeper section. The design calculation is given below in a tabular form.

Section W16 × 67 W16 × 57 W18 × 60 W18 × 65

xr (in) 6.96 6.72 7.47 7.49

x x x(K L )/r 31 32 28.9 28.8

c crN F (ksi) 39.62 39.43 39.99 40.001

A (in. ) 19.7 16.8 17.6 19.12

dP (kips) 781 662 (NG) 704 764

f fb /2t 7.7 5.0 5.4 5.1

13.5 13.5 13.5 13.5

wh /t 35.9 (NG) 33.0 38.7 (NG) 35.7

35.8 35.8 35.8 35.8

Note 1: Values are from LRFDS Table 3-50.

The lightest acceptable W-shape is a W18×65; as the web of the W18×60 shape does not satisfy the

web local buckling criteria.

Checks

1. Strength

x yW18×65: A = 19.1 in. ; r = 7.49 in.; r = 1.69 in.2



2. Local buckling of flange plates and local buckling of web plate are O.K. as shown in the table.







P8.26. A steel column is built into a wall so that it may be considered as continuously braced for buckling about its y-

x xaxis. If the factored load on the column is 1000 kips and K L = 22 ft, select the lightest W section for the

column. The depth of the section provided is to be limited to 18 in. (nominal).

See Fig. X8.5.1f of the text book.

Solution

y y x x uGiven: K L = 0; K L = 22 ft; P = 1000 kips


a. Member selection

y y reqEnter Column Tables for W10 shapes in LRFDM Table 4-2, with K L = 0 and P = 1000 kips, and

dy x ytentatively select a W10×88 with P = 1100 kips and r /r = 1.73.

x x y(K L ) = 22/1.73 = 12.71

dxRe-enter the table and select a W10×112 with P = 1102 kips. O.K.

Use similar procedure to select a W12×96 and a W14×99, as the lightest W12 and W14 shapes.

Deeper shapes are not tabulated in column tables. But, select sections weighing close to 99 plf and

lighter, and verify if they provide adequate strength. Observe that:

Lighest W16 ÷ W16×89

Lighest W18 ÷ W18×97

The lightest of all is a W16×89

b. Checks

1. Strength

xW16×89: A = 26.4 in. ; r = 7.05 in.2





Flanges: � O.K.

Web: � O.K.


Select a W16×89 of A572 Grade 50 steel. (Ans.)






See the design calculation shown in the following table

From LRFDM Table 4-2 From LRFDM Table 1-1

Section W10×112 W12×96 W14×99 Section W16×89 W18×97

dyP

(kips)

1,400 1,200 1,240 A

(in. )2

26.4 28.5

x y xr /r 1.74 1.76 1.66 r

(in.)

7.05 7.82

x x y(K L )

(ft)

12.6 12.5 13.3 37.4 33.5

dxP

(kips)

1,106 1,008 1,081 1,014 1,114

f fb /2t 4.17 6.76 9.34 5.92 6.41

13.5 13.5 13.5 13.5 13.5

wh/t 10.4 17.7 23.5 25.9 30.0

35.8 35.8 35.8 35.8 35.8






P8.27. Select the lightest W12 shape of A588 Grade 50 steel to carry an axial compressive force of 600 kips under

factored loads. Assume the member to be part of a braced frame in both planes. The idealized support

conditions are that the member is hinged in both principal directions at the top of a 30 ft height; supported in

the weak direction at 12 ft and 22 ft from the bottom; and fixed in both directions at the base.

Solution

uFactored axial compressive load, P = 600 kips



Column part of a braced frame in both planes 6 No relative translation of the top end with respect

to the bottom end.

For buckling about x-axis: fixed at base - hinged at top

x xK L = 0.8 × 30 = 24.0 ft

For buckling about y-axis: fixed at base -hinged at top and intermediate braces at 12 ft and 22 ft from

base

y y 1 y y 2 y y 3(K L ) = 0.8 × 12 = 9.60 ft; (K L ) = 1.0 × 10 = 10.0 ft; (K L ) = 1.0 × 8 = 8.00 ft

y yK L = 10.0 ft

a. Member selection

y y dyEnter LRFDM Table 4-2 with KL = K L = 10.0 ft and tentatively select a W12 × 58 with P = 611

x ykips and r /r = 2.10. So,

x x y(K L ) = 24.0 ÷ 2.10 = 11.4 > 10.0 ft

x x y dxRe-entering Table 4-2 with KL = (K L ) = 11.4 ft, we observe that the W12×58 gives P = 581 kips

x y < 600. We also observe that for sections heavier than W12×58, r /r . 1.75. So,

x x y(K L ) = 13.7 > 10.0 ft

x x y dxRe-entering LRFDM Table 4-2 with KL = (K L ) = 13.7 ft, we select a W12×65 with P = 653

kips.

b. Checks

1. Strength

x yW12 × 65: A = 19.1 in. ; r = 5.28 in.; r = 3.02 in.2

;

c crEnter LRFDS Table 3-50 and read, N F = 34.2 ksi

d c cr g� P = N F A = 653 kips > 600 kips O.K.


Verify and show O.K.












P8.28. The roof structure of an industrial building is supported by A992 steel columns, located between the roof beam

and floor girder as shown in Fig. P8.28. The columns are connected in such a manner that the ends may be

assumed pinned against rotation about the strong (xx) axis, and fixed with respect to the weak (yy) axis. If the

length of the column is 24 ft and the factored axial load is 242 kips, find the lightest W10 shape.


Solution


Material: A572 Grade 50

Factored Load: 242 kips

x x xK = 1.00; K L = 1.00 × 24 = 24.0 ft

y y yK = 0.65; K L = 0.65 × 24 = 15.6 ft

a. Member selection

y y dy reqEnter LRFDM Table 4-2 with KL = K L = 15.6 ft, and note that for a W10 × 39, P = 254 > P

x yand r /r = 2.16. So,

x x y (K L ) = 24.0 ÷ 2.16 = 11.1 < 15.6

� W10 × 39 selected is O.K.

b. Checks

1. Strength

x yW10 × 39: A = 11.5 in. ; r = 4.27 in.; r = 1.98 in.2

;


d c cr g� P = N F A = 254 kips > 242 kips � O.K.


Show O.K.

Select a W10×39 of A992 steel. (Ans.)






P8.29. (a) Find the most economical W12 section to carry an 800 kip factored axial compressive load. The 22-ft

long column is pinned at both ends about both axes. In addition, it is braced at mid-height. Use A572

Grade 50 steel and the LRFD Specifications.

(b) After the column sections were ordered, the architect of the project decided not to provide lateral

bracing at mid-height of the column as assumed in (a). The fabricator strengthened the shapes

received, by welding two d ×14 in. plates of A572 Gr 50 steel in stock. What arrangement did he use?

Substantiate your results.


Solution

Length, L = 22 ft


Factored load: 800 kips

a. With bracing at mid-height

x xK L = 1.0 × 22 = 22.0 ft

y yK L = 1.0 × 11 = 11.0 ft

dyEnter LRFDM Table 4-2 with KL = 11 ft and tentatively select a W12×79 which provides a P of 860

x ykips. Also, r / r = 1.75.

x x y� (K L ) = 22/1.75 = 12.6 ft > 11.0 ft

x x y dxRe-enter the table and observe that for KL = (K L ) = 12.6 ft, W12×79 provides P = 824 kips. So,

the W12×79 section is still O.K.

b. Checks

1. Strength check

x yW12×79: A = 23.2 in. ; r = 5.34 in.; r = 3.05 in.2

;


d c cr g P = N F A = 826 kips > 800 kips � O.K.



Flanges: � O.K.

Web: � O.K.







b. With no intermediate bracing

x x y yK L = 1.0 × 22 = 22.0 ft; K L = 1.0 × 22 = 22.0 ft

Alternative 1: Plates parallel to the xx axis, welded to the flanges

1. Strength of revised member

W xW yWW12×79: A = 23.2 in.; I = 662 in. ; I = 216 in. 4 4

W fWd = 12.38 in.; b = 12.1 in.


A = 23.2 + (14 × 3/8 × 2) = 33.7 in.2


d c cr g� P = N F A = 27.2 × 33.7 = 917 kips > 800 kips � O.K.

2. Local buckling of cover plates. Plate between welds is to be checked as a stiffened element.

Distance between welds equals the width of the flange.

Alternative 2: The plates are welded to the flanges, parallel to the web plane.

1. Strength of revised member

f W 2b = 12.1 in.; d = 6.05 + 0.19 = 6.24 in






;

;


d c cr gP = N F A = 32.3 × 33.7 = 1089 kips > 800 kips

2. Local buckling of plates.

Distance between welds equals the depth of the section.

So second arrangement, with the plates welded to the flanges, parallel to the web, is better! (Ans.)






P8.30. An interior column of a building is to support a factored axial load of 640 kips. It is 22 ft long and is to be of

A992 steel. The column base is rigidly fixed to the footing, and the top of the column is rigidly connected to

stiff trusses. Assume that bracing is provided to the structure to prevent side sway in the xx (strong) plane of

the column, but the side sway in the yy (weak) plane of the column is not prevented. Select the lightest column

shape.

Solution

Column length = 22 ft

Material: A992


For buckling about:

xx-axis: Column fixed at base and top, and part of an unbraced structure K = 1.2

y y-axis: Column fixed at base and top, and part of a braced frame, K = 0.65

x xK L = 1.2 × 22 = 26.4 ft

y yK L = 0.65 × 22 = 14.3 ft

a. Member selection

y y uEnter LRFDM Table 4-2 for W10 shapes, with KL = K L = 14.3 ft, P = 640 kips and observe that


x x y� (K L ) = 15.3 > 14.3 ft

x x y dReentering Table 4-2 with KL = (K L ) = 15.3 ft, we observe that the W10×77 provides, P = 667

reqkips > P = 640 kips.

dy x y x x y dxSimilarly, for a W12 × 72, P = 710 kips, r / r = 1.75, (K L ) = 15.1 ft, and P = 692 kips.

� W12 x 72 is O.K.

dy x y x x y d dyFinally, for a W14 × 74, P = 652 kips, r / r = 2.44, (K L ) = 10.8 ft, and P = P = 652 kips.

� W14 × 74 is O.K.

Of these three sections, W12 × 72 is the lightest. � Select it

b. Checks

1. Strength check

x yW12×72: A = 21.1 in. ; r = 5.31 in.; r = 3.04 in.2

;


d c cr g u P = N F A = 692 kips > P = 640 kips

2) Local buckling of plates

Flange and web plates satisfy the requirements (show).












P8.31. Select the lightest W shape of A588 Grade 50 steel for a 18 ft long column to support an axial dead load of 290

kips and a live load of 280 kips. The column is assumed to be hinged at the base, while the top of the column

is rigidly framed to very stiff girders. Assume that bracing is provided to the structure to prevent side sway in

the xx (strong) plane of the column, but the side sway in the yy (weak) plane is not prevented. Select the lightest

column shape.

Solution


Loads: Dead load = 290 kips; Live load = 280 kips

uFactored load, P = 1.2D + 1.6L = 1.2 × 290 + 1.6 × 280 = 796 kips


Buckling about xx axis:

xSway permitted. Hinged base. Fixed top. 6 K = 2.0 from Fig. 8.5.4 f

Buckling about yy axis:

ySway prevented. Hinged base. Fixed top. 6 K = 0.8 from Fig. 8.5.4 b

x x y yK L = 2.0 × 18 = 36.0 ft; K L = 0.80 × 18 = 14.4 ft

a. Member selection


dy x ya W10×88 provides P = 803 kips. Also, for this section, r / r = 1.73 resulting in :

x x y (K L ) = 20.8 > 14.4 ft

x x yReentering the table with KL = (K L ) = 20.8 ft, we observe that the heaviest W10 section (W10

×112) does not provide adequate axial strength.

y y reqNext, enter LRFDM Table 4-2 for W12 shapes, with KL = K L = 14.4 ft, P = 796 kips and observe

dy x ythat a W12×87 gives P = 863 kips and has r /r = 1.75.

x x y� (K L ) = 20.6 > 14.4 ft

x x y dReentering Table 4-2 with KL = (K L ) = 20.6 ft, we observe that a W12×106 provides, P = 835


dy x yProceeding in a similar manner, we observe that for a W14×90: P = 960 kips; r /r = 1.66.

x x y� (K L ) = 21.7 > 14.4 ft

x x y dReentering Table 4-2 with KL = (K L ) = 21.7 ft, we find that a W14×99 provides, P = 862 kips

req > P = 796 kips.

From W12×106 and W14×99, W14×99 is the lighter, so choose W14×99.

b. Checks

x yW14×99: A = 29.1 in. ; r = 6.17 in; r = 3.71 in.2


d c cr g� P = N F A = 864 kips > 796 kips � O.K.







P8.32. Design a 12 ft long W12 interior column of a building structure shown in Fig. P8.32, to support a factored axial

load of 820 kips. The column is part of an unbraced, rigid jointed frame in the plane of its web. W27×94

girders, 28 ft long, are rigidly connected to each flange, at the top and bottom of the column under consideration.

Assume same sections are used for columns just above and below. The column considered is braced normal to

its web at the top and bottom so that side sway is prevented in that plane. Assume A992 steel.


Solution



c gL = 12 ft; L = 28 ft

xBuckling about the xx axis: Column part of an unbraced frame. 6 K $ 1.0

y yBuckling about the yy axis: Column part of a braced frame. 6 K # 1.0. Assume K = 1.0.

Girders: W27×94

a. Member selection

xAssume I for the column as 600 in. From Eq. 8.5.7: 4

xEnter nomograph for sidesway uninhibited columns (Fig. 8.5.6b) and read, K = 1.15.

x xK L = 1.15 × 12 = 13.8 ft

y yK L = 1.0 × 12 = 12.0 ft

req y yEnter LRFDM Table 4-2 with P = 820 kips, KL = K L = 12.0 ft and tentatively select a W12 × 79

dy x y x x ywith P = 838 kips. For this section, r / r = 1.75 6 (K L ) = 7.89 ft < 12.0 ft.

� W12 × 79 is O.K.

b. Strength check

x x yW12×79: A = 23.2 in. ; I = 662 in. ; r = 5.34 in.; r = 3.05 in.2 4

x B A xUse the revised value of I and find G = G = 0.472, K = 1.16

� ;


d c cr g� P = N F A = 36.1 × 23.2 = 838 kips > 820 kips � O.K.

So, select a W12×79 (Ans.)






P8.33. Select a suitable W12 shape for the columns of a fixed-base, rigid-jointed, unbraced, portal frame shown in Fig.

P8.33, using A992 steel and LRFD Specification. The horizontal girder is a W18×76. The webs of these rolled

sections lie in the plane of the frame. In the plane perpendicular to the frame, bracing (girts) are provided at top,

and mid height of columns using simple flexible beam-to-column connections. The factored axial load on the

xcolumns is 400 kips. (Hint: As a starting point assume that I of column section lies between 300 and 500 in. ).4

See Figure P8.33 of the textbook.

Solution



x y1 y2Length, L = 16 ft; L = 16 ft; L = 8 ft; L = 8 ft

xBuckling about the x-axis: Column part of an unbraced frame. 6 K $ 1.0

Buckling about the y-axis: Column part of a braced frame. Fixed base. Girts at midheight.

y1 y2K = 0.8; K = 1.0

y y 1 y y 2 y y(K L ) = 0.8 × 8.0 = 6.40 ft; (K L ) = 1.0 × 8.0 = 8.00 ft; 6 K L = 8.00 ft

g cL = 32 ft; L = 16 ft

xW18×76 girder: I = 1330 in.4

xAssume I of column = 400 in.4

AG = 1 (Recommended value for a fixed base)

From Eq. 8.5.7:

xEnter the nomographs for the sway uninhibited case (Fig. 8.5.6b) and obtain, K = 1.21

x x� K L = 1.21 × 16 = 19.4 ft

u y yEnter LRFDM Table 4-2 with P = 400 kips, KL = K L = 8.00 ft and tentatively select a W12× 40

dy x y x x ywith P = 416 kips. For this section, r / r = 2.64 6 (K L ) = 7.5 ft < 8.00 ft.

� W12×40 is O.K.

b. Checks

x x yW12×40: A = 11.7 in. ; I = 307 in. ; r = 5.13 in.; r = 1.94 in.2 4

B x x xRevised value of G = 0.47. So, K changes to 1.10, and K L = 1.10 × 16 = 17.6 ft

;

c crFrom LRFDS Table 3-50, N F = 35.6 ksi, and

d c cr g P = N F A = 417 kips > 400 kips O.K.







P8.34. The column AB of a two bay, single story, rigid-jointed steel frame is subjected to a factored axial compressive

load of 1930 kips (Fig. P8.34). The W30×90 girders are 30 ft long, and the columns are 16 ft long. The bases

yare hinged. Side sway is possible in the plane of the frame. In the perpendicular direction K = 0.8. Assume

A242 Grade 46 steel and check the adequacy of the W14×193. Include inelastic behavior of the column.


Solution



x y yL = 16 ft; L = 16 ft; K = 0.8

g cL = 30 ft; L = 16 ft

x x yColumn: W14×193 6 I = 2400 in. ; A = 56.8 in. ; r = 6.50 in.; r = 4.05 in.4 2

xGirders: W30×90 6 I = 3610 in.4

AG = 10 (Recommended value for a pinned base)

From Eq. 8.5.7:

xEnter the nomographs for the sway uninhibited case (Fig. 8.5.6b) and obtain K = 1.80.

x x� K L = 1.80 × 16 = 28.8 ft

y yAlso, K L = 0.80 × 16.0 = 12.8 ft

;



d d e c cr g� P = P = N F A = 0.702 × 46 ×56.8 = 1835 kips < 1930 kips N.G.

Axial stress,

yFrom LRFDS Table 4-1, corresponding to F = 46 ksi, we obtain SRF = 0.331

Bi A� G = 0.623 × 0.331 = 0.205; G = 10

x cRe-enter nomograph and obtain, K = 1.75 6 8 = 0.655 6

d i c cr g� P = N F A = 0.710 × 46 × 56.8 = 1860 kips < 1930 kips N.G.






The W14×193 is therefore inadequate. (Ans.)






P8.35. Check the adequacy of a W12×152 for column AB of the rigid-jointed steel frame shown in Fig. P8.35. The

columns are 12 ft long, and the columns above and below AB may be assumed to be approximately of the same

size as AB. The adjacent girders are all of W18×50 and 32 ft long. The webs of all columns and girders shown

lie in the plane of the paper. For buckling in the perpendicular plane, take K = 0.9. All members are of A992

steel. Take inelastic behavior of columns into account. The factored axial load in the member is 1540 kips.


Solution



y y yK = 0.9; K L = 0.9 × 12.0 = 10.8 ft

c gL = 12 ft; L = 32 ft

x x yColumns: W12×152 6 I = 1430 in. ; A = 44.7 in. ; r = 5.66 in.; r = 3.19 in.4 2

xGirders: W18×50 6 I = 800 in. 4

From Eq. 8.5.7:

xEnter nomograph for sway uninhibited columns (Fig. 8.5.6 b) and get K = 2.23

;


d c cr g�P = N F A = 33.6 × 44.7 = 1500 kips < 1540 kips N.G.

Axial stress,

yEnter LRFDM Table 4-1 for F = 50 ksi and obtain, SRF = 0.461

iA iB e x� G = G = SRF G = 0.461 × 4.76 = 2.20 � K = 1.65

c crEnter LRFDS Table 3-50, and obtain N F = 37.4 ksi

d c cr gP = N F A = 37.4 × 44.7 = 1670 kips > 1540 kips � O.K.

� The W12×152 of A992 steel is adequate (Ans.)






P8.36. A W14×193 column is part of a rigid-jointed, unbraced frame in both xx and yy planes. Two W24×84 girders,

36 ft long are rigidly connected to the flanges, while two W18 × 60 girders, 18 ft long, are rigidly connected to

the web (Fig. P8.36). A992 steel is used for all members. The column in the tier above is also a W14×193,

while the column in the tier below is a W14×211. The story height is 12 ft 6 in. A particular load combination

results in a factored axial compressive load of 2000 kips in the column under consideration. Is the section safe

according to the LRFDS?


Solution

x yColumn length, L = 12.5 ft; L = L = 12.5 ft


A992 steel

xBuckling about the x- axis: Column part of an unbraced frame. 6 K $ 1.0

b cL = 36 ft; L = 12.5 ft

x xW14 × 193: I = 2400 in. ; r = 6.50 in.; A = 56.8 in.4 2

xW14 × 211: I = 2660 in. 4

xW24 × 84: I = 2370 in. 4

yBuckling about the y-axis: Column part of an unbraced frame. 6 K $ 1.0

b cL = 18 ft; L = 12.5 ft

y yW14 × 193: I = 931 in. ; r = 4.05 in.4

yW14 × 211: I = 1030 in.4

xW18 × 60: I = 984 in.4

For buckling about the major axis of the column section (Eq. 8.5.7):

;

xEnter nomograph for sway uninhibited columns (Fig. 8.5.6b) and get, K = 1.80

For buckling about the minor axis of the column section (Eq. 8.5.7):

;

yAgain, enter nomograph for sway uninhibited columns (Fig. 8.5.6b) and get, K = 1.40






;

c crFrom LRFDS Tables 3-50, N F = 34.9 ksi

d c cr gP = N F A = 34.9 × 56.8 = 1980 kips < 2000 kips N.G.

� The column is inadequate according to the LRFDS. (Ans.)






P8.37. Select suitable W12 shapes for columns AB and CD of the axially loaded, unbraced frame with leaning column

shown in Fig. P8.37. The girder BC is rigidly connected to the left column but has only a simple connection to

the right column. The columns are braced top and bottom against sidesway out of the plane of the frame. In

addition, the columns are braced at midheight by girts. Assume A992 steel. The loads given are factored loads.


Solution


Column height, L = 15 ft

xGirder: W18×60 6 I = 984 in.4

a. Design of the leaning column CD

x y y yK = 1.0; K = 0.50; K L = 0.5×15 = 7.50 ft

Required strength = 800 kips

y y yFrom LRFDM, Table 4-2, for F = 50 ksi and KL = K L = 7.50 ft, tentatively select a W14×74 for

which,

dyP = 841 kips > 800 kips O.K.

x y x x yr /r = 2.44 6 (K L ) = < 7.70 ft

d dySo, P = P = 841 kips > 800 kips OK

Use a W14×74 for column CD. (Ans.)

b. Design of Column AB (Yura’s Method )

uyRequired axial strength (out of plane), P = 800 kips

uxRequired axial strength (in-plane), P = 800 + 800 = 1,600 kips

c bL = 15 ft; L = 32 ft

x xTry W14×233: A = 68.5 in. ; I = 3,010 in. ; r = 6.63 in.2 4

xEnter nomograph for sway uninhibited columns (Fig. 8.5.6b), and get K = 3.20

dx uxP = 24.4×68.5 = 1,671 kips > P = 1,600 kips O.K.

y y dy uyK L = 0.5 × 15 = 7.50 ft Y P = 2,810 kips > P = 800 kips OK

Use W14×233






P8.38. Design the columns AB and CD of the axially loaded, unbraced frame with leaning columns shown in Fig. P8.38.

The girders are rigidly connected to the interior column while the other connections are simple. The columns

are braced top and bottom against sidesway out of the plane of the frame. In addition, the exterior columns are

braced at midheight by girts. Assume A992 steel. Use W12 shapes. The loads given are factored loads.


Solution

xW30×124 girders 6 I = 5,360 in.4

a. Design of the leaning column CD

uFactored load on the column, P = 2200 kips


x yL = L = 16.0 ft

x y y yK = K = 1.0 6 K L = 1.0×16.0 = 16.0 ft

dyFrom LRFDM, Table 4-2, try W14×211 having P = 2,240 kips > 2,200 kips O.K.

Use W14×211 for column CD. (Ans.)

b. Design of columns AB and EF (Yura’s Method)

c xColumn length, L = 16 ft = L = L

gGirder length, L = 36 ft

uyRequired axial strength (out of plane), P = 1,000 kips

uxRequired axial strength (in-plane), P = due to symmetry

x xTry W14×233: A = 68.5 in. ; I = 3010 in. ; r = 6.63 in.2 4

xEnter nomograph for sidesway uninhibited columns (Fig. 8.5.6b) and read K = 2.20

dx uxP = 31.6×68.5 � 2170 kips > P = 2100 kips O.K.

y y dy uyK L = 16 ft Y P = 2760 kips > P = 1000 kips O.K.

Use W14×233 for columns AB and EF. (Ans.)






P8.39. Design the columns AB and CD of the axially loaded, unbraced frame with leaning columns shown in Fig. P8.39.

The girders are rigidly connected to the interior columns while the connections to the exterior columns are

simple. The columns are braced top and bottom against sidesway out of the plane of the frame. In addition,

the exterior columns are braced at midheight by girts. Assume A992 steel. Use W12-shapes. The loads given

are factored loads.


Solution

xW30×124 girders 6 I = 5,360 in.4

a. Design of the leaning columns AB and EF

uFactored load on the column, P = 1000 kips


x y1 y2L = 16.0 ft; L = L = 8.0 ft

x y y yK L = 16.0; K L = 1.0×8.0 = 8.00 ft



x x y� (K L ) = 9.64 > 8.0 ft

x x y dReentering Table 4-2 with KL = (K L ) = 9.64 ft, we observe that a W14×90 provides, P = 1047


Use W14×90 for columns AB and EF. (Ans.)

b. Design of column CD (Yura’s Method)

c xColumn length, L = 16 ft = L = L

gGirder length, L = 36 ft

uyRequired axial strength (out of plane), P = 2200 kips

uxRequired axial strength (in-plane), P = 2200 + 1000 = 3200 kips due to symmetry

x xTry W14×342: A = 101 in. ; I = 4900 in. ; r = 6.98 in.2 4

xEnter nomograph for sidesway uninhibited columns (Fig. 8.5.6b) and read K = 2.14

dx uxP = 33.0× 101 � 3330 kips > P = 3200 kips O.K.

y y dy uyK L = 16 ft Y P = 3690 kips > P = 2200 kips O.K.

Use W14×342 (Ans.)

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