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Steel Structures by S. Vinnakota Chapter 11 page 11-1

PROPRIETARY MATER IAL. © 2006 T he McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be

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CHAPTER 11

P11.1. A W12 ×40 o f A992 steel is used as a 13-ft column in a braced frame to carry an axial factored load of 160kips. Will it be adequate to carry mome nts about the strong axis of 100 ft-kips at each end, bendin g thememb er in single curvature? For buckling about the minor axis consider the column to be pinned at bothends and provided with a brace at mid height.

Solution

uFactored axial load on the column, P = 160 kipsLength of column, L = 13.0 ftColumn p art of braced frame. Member pinn ed at both ends, with a brace at mid height.

x y L = 13.0 ft; L = 6.50 ft

x x y y K L = 1.0 × 13.0 = 13.0 ft; K L = 1.0 × 6.50 = 6.50 ft

W12×40 of A992 steel.From LRFDM Table 4-2, for a W12×40. So,

d exAlso from this table, for a W12×40 with KL = 6.50 ft, P = 442 kips and P ( KL ) = 8 79 0 × 1 0 .2 4

Axial load ratio, . Use LRFDS Eq. H1-1 a .

u yBending is about major axis only 6 M = 0*

Structure is braced in yy plane

Column is subjected to symmetric single curvature bending moments of 100 ft-kips.

bUnbraced length, L = 6.5 ft.

p bFrom beam selection tables (LRFDM T able 5-3), corresponding to a W 12×40, L = 6.85 ft > L ,

and The LHS of the interaction equation gives






So, the W12× 40 of A992 steel is adequate. (Ans.)






P11.2. A W14 ×61 is used as a 15 ft long column in a braced frame to carry an axial factored load of 200 kips.Determine the maximum moment that may be applied about the strong axis on the up per end wh en the lower

end is hinged.

Solution

uFactored axial load, P = 200 kips.Length of column, L = 15 ft. W14×61 of A992 steel.

x yColum n part of braced frame. Lower end hinged. Assume (conservatively), K = 1.0 and K = 1.0.

x x y y K L = 1.0 × 15.0 = 15.0 ft; K L = 1.0 × 15.0 = 15.0 ft

y y d From LRFDM Table 4-2, corresponding to W14×61 and KL = K L = 15.0 ft, P = 513 kips.

exAlso, for this section, P ( KL ) = 18300 × 10 .2 4

Axial load ratio, . So, use LRFDS Eq. H1.1 a .

There are no minor axis moments, and the structure is braced in yy plane. So,

As, from Table 10.4.1.

bFrom beam selection plots (LRFDM Table 5-5), corresponding to a W1 4×61 with L = 15 ft,

b pxAlso, N M = 383 ft-kips for this shape. So, the design bending strength

of the member,

We have

So, the maximum moment that may be applied about the strong axis on the upper end = 263 ft-kips.(Ans.)






P11.3. Figure 11-3 shows a 15-ft long W12×96 column with pinned end s. Two girders and a beam bring in 200kips of axial load. A column from above d elivers 90 kips at an eccentricity of 18-in. through a bracket

connected to the web of the column. The structure is braced in both xx and yy planes. The loads arefactored loads. Is the column adequate?

See Fig. P11.3 of the text book.

Solution

Length of column, L = 15 ft

x yColumn part of braced frame in both xx and yy planes. So, K # 1.0 and K # 1.0

x yPinned ends. So, K = 1.0 and K = 1.0.

x x y y K L = 1.0 × 15.0 = 15.0 ft; K L = 1.0 × 15.0 = 15.0 ft

y y d From LRFDM Table 4-2, corresponding to W12×96 and KL = K L = 15.0 ft, P = 935 kips.

eyAlso, for this section, P ( KL ) = 7730 × 10 .2 4

uFactored axial load, P = 200 + 90 = 290 kips.

Axial load ratio, . So, use LRFDS Eq. H1.1 a .

lty ltxStructure is braced in xx and yy planes. So M and M moments are zero. The 90 kips eccentricload produces moment about minor axis only. So

Eccentricity, e = 18 in.

nty 2 y M = M = 135 ft-kips

b pyFrom LRFDM Table 5-3 for a W12×96, N M = 250 ft-kipsWe have

So, W12×96 of A992 steel is adequate. (Ans.)






P11.4. An interior column of a building structure has floor girders framing into it at top and bottom with momentresisting connection s. It must carry a factored axial load of 800 kips, including the girder and beam

reactions, and the self-weight of the column. Live load imbalance in checkerboard loading causes a potentialmaximum moment at the top and bottom of 18 0 ft-kips, as shown in Figure P11-4. K for the weak axis is1.0, and K for the strong axis is estimated as 0.9. The column is 12 ft 6 in. long. Is a W12 ×106 of A992steel adequate to carry the load?

See Fig. P11.4 of the text book .

Solution

uFactored axial load on the column, P = 800 kipsLength of column, L = 12.5 ft

x y K = 0.9; K = 1.0

x x y y K L = 0.9 × 12.5 = 11.3 ft; K L = 1.0 × 12.5 = 12.5 ft.W12×106 of A992 steel.

y y d From LRFDM T able 4-2, corresponding to KL = K L = 12.5 ft and a W12×106, P = 1115 kips.exAlso, P ( KL ) = 2 67 00 × 1 02 4

Axial load ratio, . Use LRFDS Eq. H1-1 a .

Structure is braced in xx and yy planes

ntyBending is about major axis only 6 M = 0Column is subjected to symmetric single curvature bending moments of 180 ft-kips.

bUnbraced length, L = 12.5 ft.

b bFrom beam selection plots (LRFDM Table 5-4), corresponding to a W12 ×106, L = 12.5 ft, and C = 1.0,

The LHS of the interaction equation gives

So, the W12× 106 of A992 steel is adequate. (Ans.)






uP11.5. A W14×22 tension member of A572 Grade 42 steel is subjected to a factored tensile load T = 60 kips and

ux uy bfactored bending moments M of 28 ft-kips and M of 6 ft-kips. Is the member satisfactory if L = 8.0 ft

band C = 1.67?

Solution

From LRFDM T able 1-1, for a W14×22, we obtain:

y y A = 6.49 in ; Z = 4.39 in ; S = 2.80 in.2 3 3

x x y Z = 33.2 in. ; S = 29.0 in. ; r = 1.04 in.3 3

1 2 X = 1600 ksi; X = 0.0278 (1/ksi) 2

For a W14×2 2 of A572 Grade 42 steel:

y L y r F = 42 ksi; F = F - F = 42 - 10 = 32 ksi

dy b py b yy M = min [ N M ; 1.5 N M ] = min [13.8 ; 1.5 × 8.82] = 13.2 ft-kips

b px b rx r pBF = ( N M - N M ) / ( L - L ) = (105 - 69 .6) /(11.0 - 4.01 ) = 5.06 kips

p b r As ( L = 4.01 ft) < ( L = 8 ft) < ( L = 11.0 ft)

Design tensile strength of the member, corresponding to the limit state of yield on gross area

d d 1 t g yT = T = N A F = 0.9 × 6.49 × 42 = 245 kips

. So, use Eq. 11.9.11 a (LRFDS Eq. H1-1 a )

= 0.24 + 0.64 = 0.88 < 1.0 safeSo, the W14× 22 of A572 Grade 42 is adequate. (Ans.)






d 1 Note that, for convenience, we hav e assumed tha t ten sion yield ing ( T ) controls over tension

d 2rupture ( T ). If this is not the case, the design tensile strength should b e based upon the tension

rupture design strength.






P11.6. A W10×68 of A992 steel is used as an exterior column of a building structure braced in both directions (seeFig. 5.2.5 c for a plan view). It must carry 310 kips from the girders con nected to the web, plus a 100 -kip

reaction from the floor beam that frames into the colum n flange with an eccentricity of 8 in. The colum n is15-ft long and is assumed to be p inned about both axes as both ends. The structure is braced in both xx and

yy planes of the column. Is the member adequate?

See Fig. 5.2.5c of the text book .

Solution

Length of column, L = 15 ftColumn part of a braced frame in both xx and yy planes 6

In the absence of rational analysis assume (LRFDS Section C2.1),

bUnbraced length, L = 15 ft.

u xFactored axial load, P = 310 + 100 = 410 kips. Eccentricity of 100 kip load, e = 8 in.Factored major axis moment,

lt As the structure is braced in both xx and yy planes, there are no M moments. Also, as there are no bend ing mo ments abo ut min or axis ,

y y d exFrom LRFDM T able 4-2, for a W10×68 with KL = K L = 15.0 ft, P = 597 kips, and P ( KL ) = 11300 × 102 4

Axial load ratio, . Use Eq. 11.9.11 a (LRFDS Eq. H1-1 a ).

6

M From Table 10.4.1, corresponding to a linear variation of moments over the unbraced length and r

b= 0, C = 1.67. From beam selection plots (LRFDM Table 5-5), corresponding to a W10×68

b b pxwith L = 15 ft, Also, N M = 320 ft-kips for this shape. So, the design

bend ing strength of the mem ber,






So, the W10×68 of A992 steel is adequate. (Ans.)






P11.7. Figure P11.7 sho ws a 15-ft-long hinged b ase column pinne d at both ends. It must carry a factored axial loadof 400 kips and a uniformly applied wind load of 1.5 klf that causes bending abou t the major axis. The

structure is braced in both xx and yy planes. Check if a W10× 60 of A992 steel will be adequate.See Fig. P11.7 of the text book.

Solution


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

x x y y No inte rmediate b racing. Colum n p inn ed at both ends. So, K L = 15 ft, K L = 15 ft.

Section W10×60 of A992 steel

d From LFRDM Table 4-2, for a W10×60 with KL = 15 ft, P = 523 kips

exAlso, P ( KL ) = 9 76 0 × 1 0 .2 4

Axial load ratio, . So, Eq. 11.9.11 a (LRFDS Eq. H1-1 a ) governs.

uUniformly distributed load, q = 1.5 klf

Maximum first-order moment,

b yUnbraced length, L = L = 15.0 ft.

bFrom beam selection plots (LRFDM Table 5-5), for a W10×60 with L = 15 ft,

From Fig. 10.4.3 a , for a simple beam with uniformly distributed transverse load and braced at the

bends only, C = 1.14. We have

From LRFDC Table C.C2.1, for a beam-column with pinned ends and uniformly distributed

mtransverse load, C = 1.0.

;

Substituting in the interaction equation, we get:

The W10×60 of A992 steel selected is adequate. (Ans.)






P11.8. A 18-ft-long hinged base column carries a factored axial load of 160 kips as shown in Fig. P11.8. Thecolumn h as its weak axis braced at third-points by channel girts. The wind load on the wall is picked up by

the girts and transferred to the column as point loads (8 kips each, under factored wind). The structure is braced in b oth xx and yy planes. Check if a W10× 33 of A992 is adequate.

See Fig. P11.8 of the text book.

Solution


x y1 y2 y3 y L = 18 ft; L = L = L = L = 6 ftStructure braced in both xx and yy planes of the column section. Member pinned at both ends.

x x y y bSo, K L = 18 ft; K L = 6 ft. Unbraced length, L = 6 ft.

Column section: W10×33 o f A992 steel.From LRFDM Table 4-2, for a W10×33. So,

Also from LRFDM Table 4-2, for a W10×33 with KL = 8.33 ft,

d ex P = 340 kips, and P ( KL ) = 4890 × 102 4

Axial load ratio,

The pin ended member un der two equal concentrated loads at third points is subjected to uniformmoment over the middle bay.

p b pxFrom LRFDM T able 5-3, for a W10×33, L = 6.85 ft and N M = 146 ft-kips

b p dx b pxAs L < L , M = N M = 146 ft-kips

ltx ltyThe column is part of a braced structure in both xx and yy planes 6 M = 0.0, M = 0.0We have

As per LRFDS Section C1.2, for compression members subject to transverse loading between their

msupports, C = 1.00 when member ends are unrestrained.






The interaction equation gives

So, the W10×33 of A992 steel provided is adequate. (Ans.)






P11.9. Investigate the adequacy of a W8×24 beam-column used with its web vertical, as the top chord member of atruss. Under factored loads of a load combination , the member is subjected to an axial force of 110 kips and

a uniformly distributed lateral load of 0.32 klf. The memb er is 12 ft long and could be assumed to be pinnedabout both axes at both ends. There is no intermediate bracing present. Assume A242 steel. Neglect self-weight.

See Fig. P11.9 of the text book .

Solution

a. Da taFrom LRFDM T able 1-1, we have for a W8×24:

y A = 7.08 in. ; S = 5.63in.2 3

y y y Z = 8.57 in. ; I = 18.3 in. ; r = 1.61 in.3 4

From LRFDM Tables 2-4 and 2-1, we observe that the W8×24, a Group 1 shape, is available in

yA242 steel in Grade 50 only. That is, F = 50 ksi.

uRequired axial compressive strength, P = 110 kips x yColumn length, L = 12 ft; K = K = 1.0

b. Des ign axial streng th

c cr From LRFDS Table 3-50, for KL /r = 89.4, N F = 23.7 ksi, resulting in:

d P = 23.7 × 7.08 = 168 kips

Therefore, Eq. 11.9.11 a (LRFDS Eq. H1-1 a ) governs.

c. Des ign ben ding streng thAs the bending is about the minor axis, there is no concern for lateral-torsional buckling.

We have:

d. Req uired ben din g s tren gth

uxThere is no bendin g about the major axis. Hence, M = 0*

ltAs the member is part of a truss, there is no relative translation of the ends; i.e., M momen ts are

zero. Hence,






Maximum factored, first-order, no-translation moment,

From Table 11 .13.1, for

mAs the beam-column is pinned at the ends, and subjected to a uniformly distributed lateral load, C = 1.0 from LRFDC Table C-C1.1. So,

The factored, second-order bending moment,

e. Check interaction

So, the W8×2 4 of A242 Grade 50 steel is O.K. (Ans.)






P11.10. A W12×50 of A992 steel is used as the bottom chord member of a welded roof truss. The 12-ft longmember, between panel points, is subjected to an axial tensile force of 200 kips under factored loads.

Determine the maximum factored concentrated load that could be suspend ed at midspan. Assume that theload causes bending about the major axis and that lateral supports are provided at the panel points only.

Solution

Truss bottom chord member.Section: W12×50 of A992 steel

bMember length, L = 12 ft; unbraced length, L = 12 ft

uAxial tension under factored loads, T = 200 kips

uFactored concentrated load acting at midspan, QTransverse load causes bending about major axis. Assume pinned end s (neglect restraint provided

by the end con nectio ns) . Max imu m b ending moment occurs at mid span. We have:

ux u u M = ( Q L) ÷ 4 = 3 Q ft-kips

From LRFDM T able 1-1, for a W12×50, we obtain, A = 14.6 in.2

From beam selection tables for W-Shapes (LRFDM Table 5-3, for example), for a W12×50 shape

yof F = 50 ksi steel, we have:

p b px L = 6.92 ft; N M = 270 ft-kips

r L = 21.5 ft; BF = 5.30 kips

From Fig. 10.4.3 e, a simply supported beam under a concentrated load at midspan and laterally

b braced at th e en ds on ly has a C vallue of 1.32.

p b r As, ( L = 6.92 ft) < ( L = 12 ft) < ( L = 21.5 ft),

dxAlternatively, the value of 243 ft-kips for M could be read from LRFDM Table 5-5.o

Design tensile strength of the member, corresponding to the limit state of yield on gross area

d d 1 t g yT = T = N A F = 0.90 × 14.6 × 50.0 = 657 kips

Axial load ratio,

So, use Eq. 11.9.11 a (LRFDS Eq. H1-1 a )






So, the maximum factored concentrated load that could be suspended at midspan is 70.5 kips.

(Ans.) Note:For lack of information about the end connection details, we have assumed that tension yielding

d 1 d 2(T ) controls over tension rupture ( T ). If this is not the case, the design tensile strength should be based upon the ten sion rup ture design strength.






P11.11. A W14×8 2 is used as a column in an unbraced portal frame. It is 13 ft long and pinned at the base. Under

nt lt factored loads of a load comb ination, it is subjected to an axial load of 191 kip s, M = 156 ft-kips and M =

x y104 ft-kips at the top. Assume K = 1.7 and K = 1.0. Check the adequacy of the member.

Solution

a. Da taColumn length, L = 13 ft

x y x x y y K = 1.7; K = 1.0 6 K L = 22.1 ft; K L = 13.0 ft

uFactored axial load, P = 191 kips

nt lt Major axis moments at the top: M = 156 ft-kips; M = 104 ft-kipsMajor axis momen ts at the base: 0.0 (pinned base)

uyThere are no minor axis moments. 6 M = 0.0*

yA 992 steel: F = 50 ksi

W14×82 section: x I = 881 in. ; A = 24.0 in.4 2

x yr = 6.05 in.; r = 2.48 in.

b. Des ign axial streng th

;

c cr From LRFDS Table 3-50, design axial compressive stress, N F = 31.8 ksi

d c cr Axial strength of the column, P = N F A = 31.8 × 24.0 = 763 kips

Axial load ratio,

So, we have to use Eq. 11.9.11 a (or, LRFDS Eq. H1-1 a ) for verification of memb er strength.

c. Des ign flexura l strength

bFor major axis bending, the laterally unbraced length, L is 13.0 ft

yFrom LRFDM Table 5-3, for a W14×82 section of F = 50 ksi steel, read

yFrom beam selection plots for W-Shapes (LRFDM Table 5-5), for a W14×82 of F = 50 ksi steel,

b b dxC = 1.0, and L = 13 ft, M = 490 ft-kips.o

From Table 10.4.1, a beam segment that has a linear distribution of bending moment over its

b bunbraced length, with a zero value at one end, has a C value of 1.67. As the unbraced length, L , is

pgreater than L , the design major axis bending strength of the member is:

dx b dx b px M = min[ C M ; N M ] = min[ 1.67 × 490 ; 521] = 521 ft-kipso






d. Second-order moments

For bending about the major axis, the maximum 1 order moment,st

The linear variation of major axis mome nt over the heigh t of the column, w ith a zero value at the

mx base, re sul ts in C = 0.60 (see Table 10.4.1).

6

ux 1 x ntx 2 x ltx M = B M + B M = 1.00 × 156 + 1.06 × 104 = 266 ft-kips*

e. Check interaction formula

Thus, the W14×82 of A992 steel is acceptable as a beam-column, to carry the given factored loads,according to LRFDS . (Ans.)






P11.12. An unbraced frame contains a 14-ft column which is subjected to an axial load of 428 kips, a no-translation

nt lt moment M = 205 ft-kips, and a translation-permitted moment M = 123 ft-kips, at the top-end with half of

these momen ts at the bottom-end cau sing reverse curvature. The forces given are under factored loads of aload combin ation, and the bending is about the major axis. Check if a W12 ×96 will be able to carry this

x yloading. Assume that the effective length factor in the plane is K = 1.8 and K = 1.0.

Solution

a. Da taLength of the column, L = 14 ft


uy No bend ing moments abo ut the y-axis. 6 M = 0.0*

Maximum 1 order moments about the x-axisst

ntx Top: no-translation moment, M = 205 ft-kipsltxtranslation moment, M = 123 ft-kips

B otto m: n o-tran slatio n m om en t = - 1 03 ft-kip s

translation mome nt = - 61.5 ft-kips

b. Check the section

y y yEnter column selection table for W12-shapes (LRFDM Table 4-2) with KL = K L = 14 ft; F =

d y x50 ksi, to find that for a W12×96, P = 966 kips, and I = 833 in. 4

d indicating that major axis buckling will control. Reenter the table with KL = 14.3 ft and get P =d x P = 957 kips.

So, use interaction Eq. 11.9.11 a .

bThe laterally unbraced length L is 14 ft.

yFrom beam selection plots for W-Shapes (LRFDM Table 5-5), we have for a W12×9 6 of F = 5 0

b px p b b dxksi steel N M = 551 ft-kips, L = 10.9 ft and, for L = 14 ft and C = 1.0, M = 534 ft-kips.o

The ratio of (total) end-moments

bSo, from Table 10.4.1, C = 2.17.

b pAs L > L , we have :

dx b dx b px M = min [ C M ; N M ] = min [2.17{534}; 551] = 551 ft-kipso






1 x 2 x Nex t B and B are calculated. We have:


Substituting in the interaction Eq. 11.9.11 a :

Close. Accept.

Hence the W12 ×96 of A992 steel is acceptable. (Ans.)






P11.13. Check the adequacy of a W14×48 as a 12-ft column in an unbraced frame. The member is subjected to a

nt factored axial load of 200 kips, a no-translation moment M = 100 ft-kips, and a translation-permitted

lt moment M = 108 ft-kips at the top of the member. One-half of these momen ts are applied at the other end xof the member, bending in single-curvature. All momen ts are applied about the strong axis. Assume K =

y1.6 and K = 0.9.

Solution



uy No bend ing moments abo ut the y-axis. 6 M = 0.0*

Maximum 1 order moments about the x-axisst

ntx Top: no-translation moment, M = 100 ft-kipsltxtranslation moment, M = 108 ft-kips

B otto m: n o-tran slatio n m om en t = 5 0.0 ft-k ip stranslation mome nt = 54.0 ft-kips

b. Check the section

y y yEnter column selection table for W14-shapes (LRFDM Table 4-2) with KL = K L = 10.8 ft; F =

d y x50 ksi, to find that for a W14×48, P = 428 kips, and I = 484 in. .4

d d yindicating that minor axis buckling will control. So, P = P = 428 kips



yFrom beam selection plots for W-Shapes (LRFDM Table 5-5), we have for a W14×4 8 of F = 5 0

b px p b b dxksi steel N M = 294 ft-kips, L = 6.75 ft and, for L = 12 ft and C = 1.0, M = 258 ft-kips.o


bSo, from Table 10.4.1, C = 1.25.

b pAs L > L , we have :

dx b dx b px M = min [ C M ; N M ] = min [1.25{258}; 294] = 294 ft-kipso

1 x 2 x Nex t B and B are calculated. We have:







Substituting in the interaction Eq. 11.9.11 a :

Hence the W14 ×48 of A992 steel is not adequate. (Ans.)






P11.15. Select a W12-shape for a pin-ended column with a length of 16 ft to carry a factored load of 360 kips andend mom ents of 480 ft-kips producing symmetric single curvature bending about the major axis. The

memb er is part of a braced frame in its xx and yy planes.

Solution

a. Da ta

uFactored axial load, P = 360 kipsFor a column in braced frame, K = 1.0 for design (LFRDS Section C2.1)

x x y y K L = K L = 1.0 × 16 ft = 16.0 ftThere are no minor axis moments.

ltxBecause the colu mn is part of a braced frame, M = 0, and Eq. 11.9.3 (LRFDS Eq . C1-2) reduces

to

ntxMaximum 1 order major axis moment, M = 480 ft-kipsst

b. Preliminary selectio n

1 xAssume B = 1.05 6

Method 1 :As bending is abo ut major axis only, we select a trail shape using,

From Table 11.14.1, for a W12 with KL = 16 ft, m = 1.5, which when substituted in the aboverelation results in:

reqEntering the column load table for W12-shapes (LRFDM Table 4-2) with KL = 16 ft and P =

d x1120 kips, we observe that a W12×120 h as a design strength P = 1140 kips. Also, I = 1070 in. 4

Check the W12×120.

c. Column effect For the W12×120 section selected, axial load ratio,

uySo, use the interaction equation 11.9.11a ( LRFDS Eq. H1-1a), which for M = 0 reduces to

d. Bea m effec t

b yThe laterally unbraced length, L = L = 16 ft.

bFor a beam segment under u niform moment C = 1.0 .

p b p bFrom LRFDM Table 5-5, for a W12×120 , L = 11.1 ft. Also, for L = 16 ft > L and C = 1.0, we

have,






The mom ent magnification factor is calculated next using Eq. 11.9.4. For a pin-ended m ember

munder uniform moment C = 1.0 .

e. Check the interaction formula

O.K.

So, the W12× 120 of A992 steel selected is O.K. (Ans.)






P11.16. Select a W14-shape for a pin-ended column with a length of 24 ft to carry a factored load of 640 kips andend mom ents of 240 ft-kips producing symmetric single curvature bending about the major axis. The

memb er is part of a braced frame in its xx and yy planes.

Solution

a. Da ta


x x y y K L = K L = 1.0 × 24 ft = 24.0 ftThere are no minor axis moments.

ltxBecause the colu mn is part of a braced frame, M = 0, and Eq. 11.9.3 (LRFDS Eq . C1-2) reduces

to




Method 1 :As bending is abo ut major axis only, we select a trail shape using,

From Table 11.14.1, for a W12 with KL = 24 ft, m = 1.2, which when substituted in the aboverelation results in:

reqEntering the column load table for W12-shapes (LRFDM Table 4-2) with KL = 24 ft and P = 971

d x pkips, we observe that a W14×120 has a design strength P = 972 kips. Also, I = 1380 in. and L4

= 13.2 ft.Check the W14×120.

c. Column effect For the W14×120 section selected, axial load ratio,

uySo, use the interaction equation 11.9.11a ( LRFDS Eq. H1-1a), which for M = 0 reduces to

d. Bea m effec t

b yThe laterally unbraced length, L = L = 16 ft.

bFor a beam segment under u niform moment C = 1.0 .

b p bFrom LRFDM Table 5-5, for a W14×120 , for L = 24 ft > L and C = 1.0, we have,









.







P11.17. Repeat Problem P11.16, if the member is provided with a brace at mid-length.[ P11 .16 . Select a W14-shape fo r a pin -en ded colum n with a length of 24 ft to carry a fa cto red loa d o f 64 0

kips and end moments of 240 ft-kips producing symmetric single curvature bending about the major axis.The member is part of a braced frame in its xx and yy planes. ]

Solution

Column length, L = 24 ftPinned at both ends; brace at mid-height. Colum n part of braced frame in both planes.

x x y y b K L = 24.0 ft; K L = 12.0 ft; L = 12.0 ft


uxEnd moments produce, symmetric single curvature bending about x-axis. M = 240 ft-kips

a. Select a preliminary section

x y x x yAssume r /r . 1.66 6 ( K L ) = 24.0 ÷ 1.66 . 14 ft. From Table 11.14.1, read m = 1.4.

1 x ux 1 x uxAssume B = 1.2 6 M = B M = 1.2 × 240 = 288 ft-kips*

u eq u uxFrom Eq. 11.14.5 a , P = P + m M = 640 + 1.4 × 288 = 1040 kips*

reqEnter LRFDM T able 4-2 for W14-shapes, with KL = 14 ft and P = 1040 kips, and select a

d W14×99 with P = 1060 kips.

b. Check the selected sectionFrom LRFDM Table 4-2, for a W14×99:

x y x x y y y r /r = 1.66 6 ( K L ) = 24.0 ÷ 1.66 = 14.5 ft > K L = 12.0 ft

d u d For KL = 14 .5 ft, P = 1050 kips 6 P / P = 0.610 > 0.2 6 Use Eq. 11.9.11 a

mxFor symmetric, single curvature no translation moments, C = 1.0

ex ntxFrom LRFDM Table 4-2, for a W14×99: P ( KL ) = 3 1,8 00 × 1 0 . As, ( KL ) = 2 4.0 ft,2 - 4

;

ux 1 x ntxWe obtain, M = B M = 1.20 × 240 = 288 ft-kips*

y b px pFrom LRFDM Table 5-3, for a W14×99 and F = 50 ksi: N M = 646 ft-kips; L = 13.5 ft

b p dx b pxAs L = 12.0 ft < L = 13.5 ft, M = N M = 646 ft-kips

We therefore have:

So, select a W14 ×99 of A992 steel. (Ans.)






P11.18. Select a W14 section of A992 steel for a 14-ft-long beam-column in a framed structure braced in both

udirections. Factored axial load, P = 840 kips. The first order, symmetric, single curvature end-moments

ntx nty x yare: M = 280 ft-kips and M = 40 ft-kips. Assume K = K = 1.0.

See Fig. 11.18 of the text book .

Solution

a. Da taA572 Gr 50 steel

x y x x y y L = 14 ft; K = K = 1.0; K L = K L = 1.0 × 14.0 = 14.0 ft

u ntx nty P = 840 kips; M = 280 ft-kips; M = 40.0 ft-kips

ltx lty ux 1 x ntx uy 1 y ntyBraced frame, so M = M = 0 and M = B M ; M = B M * *


1 x 1 yAssume B = 1.05; B = 1.20

ux uy M = 1.05 × 280.0 = 294.0 ft-kips; M = 1.20 × 40.0 = 48.0 ft-kips* *

Method 2 :Since the axial load is relatively large, we get the med ian value of m from LRFDM Table 6-1. For

bW14-shapes with L = 14 ft, we obtain m = 0.826×10 . Also, the median value of n for W14--3

shapes is 1.64×10 . Equation 11.14.10 a therefore yields:-3

From LRFDM Table 6-2, a W14×120 with KL = 14 ft has b = 0.773 × 10 < 0.808 × 10 . Also,-3 -3

bm = 1.13×10 for a member with L = 14 ft. Addition ally, n = 2.34×10 . Substituting these-3 -3

values of b , m, and n in Eq. 11.14.10 a , we obtain:

By repeating the calculations for the next heavier section, namely, W14×132, we obtain LHS =0.991 < 1.0. O.K.Select W14×132.

c. Check W14×132 for streng th

As use Eq. 11.9.11 a (LRFDS Eq. H1-1 a ).

x y A = 38.8 in. ; I = 1530 in. ; I = 548 in.2 4 4






mx myFor symmetric, single curvature moments given, C = C = 1.0.

ux M = 1.06 × 280 = 297 ft-kips*

uy M = 1.18 × 40 = 47.2 ft-kips*

yFrom beam selection plots for W-Shapes (LRFDM Table 5-5), for a W14×132 of F = 50 ksi steel,

b b dx L = 14 ft, and uniform bending ( C = 1.0), M = 872 ft-kips.

b pyAlso, from LRFDM Table 5-3, N M = 419 ft-kips, for this shape.

dy b py M = N M = 419 ft-kips

Using Eq. 11.13.11 a (or, LRFDS Eq. H1-1 a ):

6 0.990 < 1.0 O.K.


P11.19. A pin-ende d colum n in a braced frame must carry a factored axial load of 222 kip s along with a factored

uniformly distributed transverse load of 1.2 klf. The column is 14-ft long and is braced at mid length. Thetransverse load is applied to put bendin g about the strong axis. Select the lightest W10 -shape of A992 steel.






P11.20. Solve Example 11.15.1, if the solution is limited to W12-shapes.

[EXAMPLE 11.15.1: Select a W14-shape. for a column to support, under factored loads, an axial load of 1600 kips and a major axis moment at the top and bottom of 210 ft-kips as determined by a first-order analy sis of the structure. The moments are in the opposite direction resulting in single curvature. Theinterior column considered is part of a braced frame in b oth directions. Use A992 steel. ]

See Fig. X11.15.1 of the text book .

Solution

a. Da ta


x x y y K L = K L = 1.0 × 11 ft = 11 ft

uyThere are no minor axis moments. 6 M = 0*

ltxColum n is part of a braced frame 6 M = 0 6


b. Preliminary selection

1 xAssume B = 1.1 6

Method 1 :As bendin g is about major axis only, we select a trail shape using

For a W12 with KL = 11 ft, m = 1.65 from Table 11.14.1. So,


d 1980 kips, we observe that a W12×190 h as a design strength P = 2100 kips.Check the W12×190.

c. Column effect

d For the W12×1 90 section selected, with P = 2100 kips,

Axial load ratio,

uySo, use the interaction equation 11.9.11 a ( LRFDS Eq. H1-1 a ), which for M = 0 reduces to

d. Bea m effec t

b yThe laterally unbraced length, L = L = 11 ft

p b px xFrom LRFDM T able 5-3, L = 11.5 ft, N M = 1170 ft-kips and I = 1890 in. , for a W12×190.4






b pAs L < L , we have,




O.K.







P11.21. Solve Example 11.15.1, if the member is 16-ft long.

[EXAMPLE 11.15.1: Select a W14-shape. for a column to support, under factored loads, an axial load of 1600 kips and a major axis moment at the top and bottom of 210 ft-kips as determined by a first-order analy sis of the structure. The moments are in the opposite direction resulting in single curvature. Theinterior column considered is part of a braced frame in b oth directions. Use A992 steel. ]

See Fig. X11.15.1 of the text book.

Solution

a. Da ta


x x y y K L = K L = 1.0 × 11 ft = 11 ft

uyThere are no minor axis moments. 6 M = 0*

ltxColum n is part of a braced frame 6 M = 0 6




Method 1 :As bendin g is about major axis only, we select a trail shape using

For a W14 with KL = 11 ft, m = 1.3 from Table 11.14.1. So,


d 1915 kips, we observe that a W14×193 h as a design strength P = 2050 kips.Check the W14×193.

c. Column effect

d For the W14×193 section selected, with P = 2050 kips,

Axial load ratio,

uySo, use the interaction equation 11.9.11 a ( LRFDS Eq. H1-1 a ), which for M = 0 reduces to

d. Bea m effec t

b yThe laterally unbraced length, L = L = 16 ftW14×193 of A992 steel. From LRFDM Table 5-3:






p r L = 14.3 ft; L = 70.1 ft

b pxN M = 1 33 0 ft-kip s; B F = 7 .1 9 kip s

bWith uniform moment over the unbraced length, C = 1.0 p b r As L = 1 4.3 ft < L = 16.0 ft < L = 70.1 ft

dx dx b px b pM = M = N M - BF ( L - L )o




O.K.








[EXAMPLE 11.15.2: Select the lightest W12 section of A992 steel for an interior column in a framed stru ctu re b raced in bo th xx a nd yy d irectio ns. The colu mn is 1 4 ft lon g. Under facto red gra vity loa ds thecolumn is subjected to an axial load of 600 kips and the major and minor axis moments shown in Fig.

x y X11.1 5.2 . Assume K = K = 1.0 and n o l ateral brac ing b etween the floor leve ls. ]


Solution

a. Data

x x y y Column length, L = 14 ft; K L = 14 ft; K L = 14 ft

ltx ltyBraced frame : M = 0.0; M = 0.0

x2 x1 M = 240 ft-kips; M = 120 ft-kips;

M b mFrom Table 10 .4.1, corresponding to r = - 0.5, C = 1.25; C = 0.8

y2 my M = 6 C = 1.0

b. Selection of trial section

1 x 1 yAssume: B = 1.0; B = 1.1So ,

yFor a W14 column ( F = 50 ksi steel) KL = 14 ft, m = 1.4 from Table 11.14.1. Assume u = 2.0.So :

y yFrom column selection tables (LRFDM Table 4-2), for KL = K L = 14 ft select:

d reqW14×109 with P = 1,270 kips > P = 1,120 kips

So, the W14×1 09 is a potential trial shape. From Table 11.14 .2, it is seen that this shape has a uvalue of 1.97 resulting in a revised value:

Reentering the column selection tables, it is seen that the W14×109 with an axial load capacity of 1270 kips appears to be an acceptable shape. So, let us try W14 ×109 -shape.

c. Des ign stre ng thsFrom Table 1-1 of the LRFDM, for a W14×109 -shape:






x y A = 32.0 in. ; I = 1240 in. ; I = 447 in.2 4 4

d Also, from column load tables, P = 1,270 kips.

Axial load ratio, . Therefore, use Eq. 11.9.11 a .

bUnbraced length, L = L = 14 ft

yFrom beam selection plots for W-shapes (LRFDM Table 5-5), for a W14×109-shape of F = 50 ksi

b b dxsteel, L = 14 ft and C = 1.0, M = 715 ft-kipso

p b r As L < L < L we have

dx b dx b px M = min [ C M ; N M ] = min [l.25{715}; 720 ] = 720 ft-kipso


d. Second-order moments

1 x 2 xThe magnification factors B and B are calculated from Eq. 11.9.4.

;

The second-order moments are therefore

1 x ntx = B M = 1.00 × 240 = 240 ft-kips

1 y nty = B M = 1.15 × 60.0 = 69.0 ft-kips

The LHS of the interaction equation 11.9.11 a ( LRFDS Eq. H1-1 a ) is:







P11.23. Solve Problem P11.18, if the solution is limited to W12-shapes.

[ P11 .18 . Select a W14 section o f A9 92 stee l fo r a 14-ft- lon g b eam-column in a framed s tructu re b raced inuboth directions. Factored axial load, P = 840 kips. The first order, symmetric, single curvature end-

ntx nty x ymoments are: M = 280 ft-kips and M = 40 ft-kips. Assume K = K = 1.0. ]

See Fig. 11.18 of the text book .

Solution

a. Da ta

x y x x y y L = 14 ft; K = K = 1.0; K L = K L = 1.0 × 14.0 = 14.0 ft

u ntx nty P = 840 kips; M = 280 ft-kips; M = 40.0 ft-kips

ltx lty ux 1 x ntx uy 1 y ntyBraced frame, so M = M = 0 and M = B M ; M = B M * *


1 x 1 yAssume B = 1.05; B = 1.20ux uy M = 1.05 × 280 = 294 ft-kips; M = 1.20 × 40.0 = 48.0 ft-kips* *

yFor a W12 column ( F = 50 ksi steel) KL = 14 ft, m = 1.5 from Table 11.14.1. Assume u = 2.0.So :

y yFrom column selection tables (LRFDM Table 4-2), for KL = K L = 14 ft select:

d reqW12×152 with P = 1,550 kips > P = 1,425 kips

So, the W12×1 52 is a potential trial shape. From Table 11.14 .2, it is seen that this shape has a uvalue of 2.11 resulting in a revised value:

So, the W12×1 52 with an axial load capacity of 1550 kip s appears to be an acceptable shape. For

x ythe W12 ×152-shape: I = 1430 in. ; I = 454 in.4 4

c. Check W12×152 for streng th

As use Eq. 11.9.11 a (LRFDS Eq. H1-1 a ).






mx myFor symmetric, single curvature moments given, C = C = 1.0.

ux M = 1.06 × 280 = 297 ft-kips*

uy M = 1.22 × 40 = 48.8 ft-kips*

b yThe laterally unbraced length, L = L = 14 ftW12×152 of A992 steel. From LRFDM Table 5-3:

p r L = 11.3 ft; L = 62.1 ft

b pxN M = 9 11 ft-kips; BF = 5.5 9 kips

bWith uniform moment over the unbraced length, C = 1.0

p b r As L = 1 1.3 ft < L = 14.0 ft < L = 62.1 ft

dx dx b px b pM = M = N M - BF ( L - L )o

b pyAlso, from LRFDM Table 5-3, N M = 410 ft-kips, for this shape.

dy b pySo, M = N M = 410 ft-kips

Using Eq. 11.9.11 a (or, LRFDS Eq. H1-1 a ):








[EXAMPLE 11.15.4: A 3 0 f t long chord mem ber of a truss is sub jected , un der fa cto red loads, to an axial compressive force of 690 kips and a uniformly distributed factored load of 0.3 klf causing bending abou t itsweak axis. Use A992 steel and select the appro priate W14 section. Include the influence of the self weight of the member which causes bending about the major a xis. ]


Solution

a. Da ta

u P = 690 kips

Assume self-weight of the beam = 190 plf The factored distributed load due to self weight, . The

maximum major-axis bending moment at the center of the beam,

uxFactored uniformly distributed transverse load, q = 0.30 klf The maximum, minor-axis bending moment at the center of the beam is:


1 x 1 yAs the membe r is quite flexible, we will assume B = 1.2 and B = 2.5, resulting in

ux lx ntx M = B M = 1.20 × 25.7 = 30.8 ft-kips*

uy l y nty M = B M = 2.50 × 33.8 = 84.5 ft-kips*

For the biaxially bent beam-column:

yFor a W12 with KL = 30 ft and F = 50 ksi, m = 1.2 from Table 11.14.1 . Also, for W12 -sectionsheavier than a W12 ×152, u . 2.11 from Table 11.14.2. Hence, we have

y d In the column load tables for W-Shapes (LRFDM Table 4-2) for F = 50 ksi and KL = 30 ft, P =

967 kips ( > = 941 kips) for a W12×190 selected. So, try a W12×190.

c. Check the section

d Axial strength of the column, P = 967 kips

The axial load ratio, . So, use Eq. 11.9.11 a .

Ma jor axis b endin g

yFrom LRFDM Table 5-3, for a W12×190 and F = 50 ksi:

b px x yN M = 1170 ft-kips; I = 1890 in. ; I = 589 in.4 4






b py p r N M = 523 ft -kips L = 11.5 ft; L = 76.6 ft; BF = 5.79

p b r As L = 1 1.5 ft < L = 14.0 ft < L = 76.6 ftdx b px b p M = N M - BF ( L - L )o

A simply supported beam under uniformly distributed load and laterally braced at the ends only, has

b b p ra C value of 1.14 from Table 5-1 of the LRFDM. As the unbraced length, L , is between L and L ,the design major axis bending strength of the member is:


For a beam-column simply supported at the ends and subjected to uniformly distributed lateral load,

mC = 1.0. Also,

ux l x ntx M = B M = 1.20 × 25.7 = 30.8 ft-kips*

Minor axis bend ing


mFor a beam-column with uniformly distributed transverse load, C = 1.0. We have:

d. Check for limit state of strength

The W12 ×190 of A992 steel is therefore acceptable. (Ans.)







[EXAMPLE 11.15.5: A W10-shap e. is used a s a continuou s top chord mem ber of the 12 8 f t long roof trussof an industrial building. The eight-panel Pratt truss is 12 ft deep at the center and 8 ft deep at the

sup portin g colu mns. The to p chord sup ports purlins at the panel poin ts and mid way between th e pan el po ints. Under facto red gra vity loa ds, each purlin transmit s a loa d o f 35 kips (end p urlin 17 .5 kips). Theseloads include pro vision for the dead weight of the truss too. Use A992 steel and select the lightest section,in the center panel. ]


Solution

a. General

4 5From Example 11.15.5, member U U is the critical element of the truss top-chord. It acts as a beam-colum n and is to be designed fo r:

As there is no relative translation of the ends of a truss member, K = 1.0. We assume that lateralsupport is provided for the top-chord member at its ends (panel points) and at the center by the

purlin s. Th us,

b. Trial sectionWe will assume a KL value of 10 ft (as major axis buckling may control).For a W12-shape of A992 steel, KL = 10 ft, we read a m value of 1.7 from Table 11.14.1. Assume

1 x B = 1.1.

y yEntering column selection table for W12 series (LRFDM Table 4-2) with KL = K L = 8 ft,

dy x y 6 W12×79 has P = 917 kips > 878 kips required; r /r = 1.75

As the design strength about both axes is greater than the required value of 878 kips, we willconsider the W12×7 9-shape.

c. Axial an d b endin g s treng ths

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

From LRFDS Table 3-50, for






So, use Eq. 11.9.11 a .

From beam selection tables for W-Shapes (LRFDM Table 5-3, for example), for a W12×79,

b pAs L = 8 ft < L = 10.8 ft,

d. Second-order moment ,

For bending about the x-axis, we have from Table 11.9.1 , for:

From Table C-C1.1 of the LRFDC, we have for a fixed ended beam with a central transverse load:

e. Check for limit state of strengthFor the beam-column under major axis bending, we have:

O.K.








[EXAMPLE 11.15.6: A W14 colum n, 14 ft lo ng , is pa rt o f an unbraced frame in the pla ne of the web an d x y pa rt o f a braced frame in the perpe nd icu lar direct ion such tha t K = 1.5 an d K = 1.0 . Firs t-order

fac tored loa d a na lysis g ives an axial loa d o f 650 kips und er (1.2D + 0.5 L + 1.6 W), the no n-sway gravi ty

nt Rt moments M under (1.2D + 0.5L) and the sway moments M under 1.6W, as shown in Fig. X11.15.6. For

uthe story und er consideration, 3 P = 20 ,00 0 k ips . The a llowab le s tory d rift ind ex is 0 .00 2 d ue to total horizontal (unfactored) wind forces 3 H = 500 k ips . Select the lightest sect ion using A992 stee l. ]


Solution


ntxMaximum 1 order no-translation moment, M = 100 ft-kipsst

ltxMaximum 1 order translation moment, M = 200 ft-kipsst

oStory drift index, ) /h = 0.002


1 x 2 xAssume B = 1.0, and B = 1.1, giving

y yFor a W12 section of Grade 50 steel and KL = K L = 14 ft, value of m is 1.5 from Table 11.14.1.Hence

reqEnter column selection table for W12-shapes (LRFDM Table 4-2) with P = 1130 kips; KL = 14

yft; F = 50 ksi, to find that for a W12 ×120 ,

d y req P = 1220 kips > P = 1130 kips and

d dyindicating that major axis buckling will not control, and P = P = 1,220 kips

c. Check the section

xFor a W12×120 section, A = 35.3 in. ; I = 1070 in.2 4



yFrom beam selection plots for W-Shapes (LRFDM Table 5-5), we have for a W12×1 20 of F = 50






b px b b dxksi steel, N M = 698 ft-kips and, for L = 14 ft and C = 1.0, M = 692 ft-kips.o


b b pSo, from Table 10.4.1, C = 2.16. As L > L , we have :

dx b dx b px M = min [ C M ; N M ] = min [2.16{692}; 698 ] = 698 ft-kipso

1 x Nex t B is calculated from Eq. 11.9.4. We have:

Factored, second-order moment is

Substituting in the interaction Eq. 11.9.11a :

OK

Hence adopt a W12×120 of A572 Grade 50 steel. (Ans.)







[ EXAM PLE 11 .15 .7: Select a W14 section o f A9 92 steel fo r a 14 ft lo ng beam-co lum n, p art of an x yunbraced frame in y-plane and part of a braced frame in the x-plane. K = 1.4 and K = 1.0. Factored

u ntxaxial load on column P is 840 kips. First order, single curvature moments under gravity loads are: M =

nty280 ft-kips, M = 40 ft-kips. The first order reverse-curvature moments under factored wind loads are

Rtx1 Rtx2 M = 200 f t-kips and M = 50 ft-kips. The d rift ind ex i s 1/40 0 u nder E H = 100 k ips . Total fac tor ed gravi ty load a bo ve this story is 6 ,80 0 k ips . ]


Solution

a. Da taW12-shape. of A992 steel

x y L = 14.0 ft; K = 1.4; K = 1.0

x x y y b K L = 1.4×14 = 19 .6 ft; K L = 1.0×14.0 = 14.0 ft = Lu ntx nty P = 840 kips; M = 280 ft-kips; M = 40.0 ft-kips

ltx1 ltx2 ltx M = 200 ft-kips; M = 50.0 ft-kips; M = 200 ft-kips

lty M = 0 (column part of braced frame in xx plane)

o u) /h = 1/400; E H = 100 kips; E P = 6,800 kips


1 x 1 y 2 xAssume B = 1.05; B = B = 1.2


uy 1 y nty M = B M = 1.2 × 40.0 = 48.0 ft-kips*

From Table 11.14.1, for a W12-shape with KL = 14 ft, get m = 1.. Assum e u = 2.0.

= 840 + 1.6 × 534 + 1.6 × 2.0 × 48.0 = 1850 kips

y y dyFrom LRFDM Table 4-2, for a W12×190 and KL = K L = 14 ft, P = 1950 kips > 1850 kips.

x y x x y y yAlso r /r = 1.79 for this section, resulting in ( K L ) = 19.6 /1.79 = 11.0 < K L = 14 ft.

y ySo, K L controls the design. For the W12×190 section, u = 2.11 from Table 11.14.2 , resulting ina revised value,

u eq P = 840 + 1.6 × 534.0 + 1.6 × 2.11 × 48.0 = 1856 kipsThe W 12×190 is still valid.

c. Check the selected section

x x y y y d For the W12×190 of A992 steel column with ( K L ) = 11.0 ft and K L = 14 ft, P = 1,950 kips.

so use Eq. 11 .9.11 a .

mx myFor symmetric, single curvature no translation moments, C = C = 1.0. From LRFDM Table 4-2, for a W12×190:

ex ey P ( KL ) = 54,100 × 10 ; P ( KL ) = 16,900 × 102 - 4 2 - 4






ntx ntyConservatively take ( KL ) = ( KL ) = 1.0 × 14 = 14.0 ft. Then,

;

Also,

We obtain


uy 1 y nty M = B M = 1.16 × 40.0 = 46.4 ft-kips*

yFrom LRFDM Table 5-3, for a W12×190 and F = 50 ksi:

b px pN M = 1170 ft-kips; L = 11.5 ft

b py r N M = 523 ft-kips L = 76.6 ft; BF = 5.79

p b r As L = 1 1.5 ft < L = 14.0 ft < L = 76.6 ft

dx b px b p M = N M - BF ( L - L )o


b b pSo, from Table 10.4.1, C = 1.27. As L > L , the design major axis bending strength of themember is:


Also,

We therefore have:








[EXAMPLE 11.15.8: Select a W12 section of A992 steel for a 16 ft long beam-column. The member isu ntx nty pa rt o f a sym metric un bra ced frame abou t both axes. Under fa cto red loads, P = 600 k ips, M = M =

Rtx Rty x y0, M = 240 ft-kips and M = 100 ft-kips. Also K = 1.5 and K = 1.3. For all columns in the story

ui e2x e2yunder consideration, E P = 10 ,80 0 k ips, E P = 120,0 00 kips and E P = 82 ,00 0 k ips . ]


Solution

a. Da ta

b L = 16.0 ft; L = 16.0 ft

x x x K = 1.5; K L = 1.5 × 16.0 = 24.0 ft

y y yK = 1.3; K L = 1.3 × 16.0 = 20.8 ft

u ntx nty P = 600 kips; M = M = 0

ltx lty M = 240 ft-kips; M = 100 ft-kipsui e2 x e2yE P = 10,800 kips; E P = 120,000 kips; EP = 82,000 kips

From Eq. 11.9.9 ( LRFDS Eq. C1-5) :

ux 1 x ntx 2 x ltx M = B M + B M = 0 + 1 .10 × 240 = 264 ft-kips*

uy 1 y nty 2 y lty M = B M + B M = 0 + 1.15 × 100 = 115 ft-kips*


y y yEntering Table 11.14.1 with KL = K L = 20.8 ft, F = 50 ksi and W14-shapes we obtain m =1.2. Assume u = 2.0. We obtain

= 600 + 1.2 × 264 + 1.2 × 2.0 × 115 = 1193 kips

y y yFrom LRFDM Table 4-2 for W-Shapes, for KL = K L = 20.8 ft and F = 50 ksi, a W14×132 has

dy P = 1192 kips . 1 193 kips. Also, u = 1.99 from Table 11.14.2, resulting in: = 600 + 1.2 × 264 + 1.2 × 1.99 × 115 = 1191 kips






So, try a W14×132.

c. Check the selected sectionFrom LRFDM T able 5-3, for a W14×132 of A992 steel:

p r L = 13.3 ft; L = 49.7 ft

b px b pyN M = 87 8 ft-kip s; B F = 6 .8 8 k ip s; N M = 419 ft-kips

bWith equal, reverse-curvature moments at the ends C = 2.27

p b r As L = 1 3.3 ft < L = 16.0 ft < L = 49.7 ft

dx b px b p M = N M - BF ( L - L )o

dxM = min [ ] = min [2.27 × 859; 878] = 878 ft-kips

The axial load ratio, so use Eq. 11.9.11 a (LRFDS Eq.H1-1 a ).

LHS

So, the W14× 132 of A992 steel selected is acceptable. (Ans.)







[EXAMPLE 11.15.9: Select a W14 section of A992 steel for a beam-column in an unbraced frame withu ntx nty fac tored loa ds: P = 400 kips ; symmetric sing le cu rva ture moments, M = 150 f t-kips, M = 50 ft-kips,

Rtx Rtyreverse curvature moments M = 116 ft-kips and M = 72 ft-kips. There are no transverse loads along

x ythe span. Story height is 15 ft. K = 1.4 and K = 1.2. The allowable story drift index is 1/500, due tounfactored horizontal wind forces in the yy plane of 120 kips (causing bending about xx axis) and 82 kips in

u x-x direction (causing ben din g a bout yy a xis) . Total factored gravi ty lo ad ab ove this lev el is E P = 7,200kips.]


Solution

a. Da ta

b L = 15 ft; L = L = 15.0 ft

x x x K = 1.4; K L = 1.4 × 15 = 21.0 ft y y y K = 1.2; K L = 1.2 × 15 = 18.0 ft

uP = 400 kips

ntx nty M = 150 ft-kips; M = 50 .0 ft-kips; symmetric single curvature

ltx lty M = 116 ft-kips; M = 72.0 ft-kips; reverse curvature

uE P = 7,200 kips

Drift index, , correspon ding to:

– a horizontal force H = 120 kips for bending about x-axis of the column, and – a horizontal force H = 82 kips for bending about y-axis.


1 x 1 yAssume B = 1.05; B = 1.10. From LRFDS Eq. C1-1:


uy 1 y nty 2 y lty M = B M + B M = 1.10 × 50.0 + 1.21 × 72.0 = 142 ft-kips*

y y yFrom Table 11.14.1, for a W12 column of F = 50 ksi steel and KL = K L = 18 ft, the coefficientm = 1 .4. Assum e u = 2.0.

ueq P = 400 + 1.4 × 290 + 1.4 × 2.0 × 142 = 1200 kips






y y yBy interpolation in the KRFDM Table 4-2, corresponding to KL = K L = 18 ft and F = 50 ksi,

d req x ywe observe that a W12×136 has P = 1210 kips ( > P = 1200 kips). This selection has r /r =

1.77, resulting in:

indicating that x-axis buckling will not control the design. Also, u = 2.09 for a W12×1 36, fromTable 11.14.2. The revised value of

ueq P = 400 + 1.4×290 + 1.4×2.09×142 = 1220 kips

Still try a W12×136.

c. Check selected section

y y yFrom LRFDM Table 4-2, for a W12×136 column with F = 50 ksi, KL = K L = 18 ft,

d ex ey P = 1210 kips; P ( KL ) /10 = 35,500; P ( KL ) /10 = 11,4002 4 2 4

1 x 1 y ntxThe magnification factors B and B are determined, using co nservatively, ( KL ) = L = 15 ft and

nty( KL ) = L = 15 ft. Thus

mx myFor the symmetric, single-curvature, first-order, no-translational moments given , C = 1.0 and C = 1.0. Thus

The second-order factored moments are

ux M = 1.04 × 150 + 1.14 × 116 = 288 kips*

uy M = 1.13 × 50.0 + 1.21 × 72.0 = 144 kips*

From LRFDM T able 5-3, for a W12×136 of A992 steel:

p r L = 11.2 ft; L = 55.7 ft

b px b pyN M = 80 3 ft-kip s; B F = 5 .4 9 k ip s; N M = 361 ft-kips

bWe will conservatively assume that C = 1.0.

p b r bAs L = 11.2 ft < L = 1 5.0 ft < L = 55.7 ft and C = 1.0






dx b px b p M = N M - BF ( L - L )


As the axial load ratio, , we use Eq. 11.9.11 a .

Substituting the various terms in the interaction formula , we obtain:

Select a W12 ×136 of A992 steel. (Ans.)







[EXAMPLE 11 .15.10: It is req uired to design a n inte rior crane colu mn for a m ill b uildin g. The colu mn of A992 stee l an d 2 8 f t high carries two cra ne girders in add ition to the roo f load . The cran e ru nwa y on each side o f the colu mn is to be suppo rted by means of a bracket welded to the flange o f the colu mn at a d istance12 ft below the top. The roof covering including dec king, insulation, joists, truss, and piping is 30 psf of the roof surface and the snow load is assumed to be 30 psf. The contributory area for the column under conside ration is 1800 sq. ft. Maximu m vertical load from wheels, including impact of wheels is given as 40kips. The horizonta l load from the crane wheels is 6 kips. The weight of the crane girder and rails is 3.2kips and the weight of each bracket is estimated as 1.8 kips. The eccentricity of the crane loads is 16 in.Select a suitable W12-shape.. Assume the column is pinned at both ends about both axes and laterally

supported at bra cket level by the crane g irders. ]


Solutiona. Fac tored loa ds

u ltxThe column will be designed for an axial load P of 228 kips and a bending mom ent, M = 114.0ft-kips (for details see Example 11.15.10).Also, we have

b. Preliminary m ember se lect ionThe equivalent axial load on the beam-column is:

yAssume minor axis buckling controls the design axial strength. For a W14 colu mn with KL = K

y 1 x L = 16 ft, m = 1.3 from Table 11.14.1. Assume B = 1.2. So

d Using the column load table for W-Shapes, select a W14×613, which for KL = 16 ft has P = 486kips ( > = 406 kips) and . We have:

as assumed. So, try W14 ×61

c. Column action

Axial load ratio,

So, the interaction formula to be checked is Eq. 11.9.11 a .






d. Beam ac tionFrom LRFDM T able 5-3, for a W14×61 beam of A992 steel

p b px L = 8.65 ft; N M = 3 83 ft-kips; B F = 6 .5 0 kip sr x L = 25.0 ft; I = 640 in. 4

The lower segment which is longer, subjected to heavier axial load, and higher maximum bending

p b r momen t, is more critical. As L < L = 16 ft < L

The variation of bending moment over the the segment is linear, and asfrom Table 10.4.1, and

ux 1 x ntx M = B M *

with

mxThe moment reduction factor C is conservatively obtained from LRFDC Table C-C1.1corresponding to a pin ended column with a central concentrated load. As the bending is about themajor axis

e. Limit st ate of streng th


Note: T he next ligh ter W1 4 (a W1 4×53) will n ot work (LHS = 1.0 4).






P11.31. Solve Example 11.1 5.11, if the portal frame is fixed at the base. Limit the selection of columns to W14-shapes.

[EXAMPLE 11 .15.11: Des ign the members of the hinged-ba se, r igid join ted po rtal frame ABCD shown in Fig . 5.1.2 , stu die d in Exa mple 5 .3.1 . The beam BC is sub jected to a u niformly dis trib ute d se rvic e load of 2.5 k R f (1 k R f D + 1 .5 k R f S). In addit ion , the frame is subje cted to a concen tra ted dead loa d o f 30 kips at the column tops B and C. The wind load on the frame consists of a 15 kip horizontal force acting at the

joint B. The dead loads given inc lude provi sion for self weig ht of members. Use A992 W-shapes with the ir webs in the plane of the frame. The columns are braced at the top and bottom against y-axis displacement and at mid-height against y-axis buckling. Lateral bracing for the beam is provided at the ends B and C,and at the q uarter points F, E, and G. Limit the drift to h/250 and the live load deflection to L/360. ]

P11.32. A 24-ft long W21×93 of A992 steel is used as a simply supported beam with regard to both principal axes.

Lateral braces are provided only at the suppo rts. The beam is subjected to a single concentrated load Q atmidspan. The load passes through the shear center of the cross section, and is inclined at an angle of 15 o

with the vertical axis (web axis). Neglect the self-weight of the bam and determine the maximum factored

uload Q as per LRFDS.






P11.33. Redesign th e crane runway beam o f Example 10.4.10 , if it has to carry a lateral force of 3 kips in addition tothe loads given there. The lateral force acts perpend icular to the beam, at each wheel, 4¼ in. above the top

flange.

[ EXAM PLE 10 .4.1 0: A 18 f t long , cra ne runwa y beam carries the two end wheels of a crane. The wheelsare spaced 12 ft on centers and each transfers a maximum vertical live load of 24 kips to the beam.

Neglect the latera l fo rce and longitud ina l fo rce on the member. The crane beam is without latera l supp ort except at the columns. The specified impact is 25% of the live load. Design the beam, using A242 Grade50 steel. The maximum service live load deflection is to be limited to L/500. ]


Solution

bSpan, L = 18 ft; Unbraced length, L = 18 ftWheel load, Q = 24 kips; Wheel spacing, a = 12 ft

Impact factor, I = 25%

a. Required st ren gth s

yLEquivalent statically applied concentrated wheel load, Q = 1.25×24.0 = 30.0 kipsThe maximum live load moment will occur at midspan, with one of the crane wheels at the center of the span (Fig. X10.4.11 b).Assume a 90-lb rail, weighing 90 lb/yard or 30 lb/ft. Also, assume the weight of the crane beam to

be 90 lb/ft.

Bending moment due to dead load,

Bending moment d ue to live load,

ux D LRequired bending strength about major axis, M = 1.2 M + 1.6 M = 1.2(4.86) + 1.6(135)= 222 ft-kips

xLHorizontal load on the beam = Lateral thrust from the moving crane = Q = 3.0 kips

f f Assume a W18×76 shape: d = 18.2 in.; b = 11.0 in.; t = 0.680 in.Height of rail = 4 ¼ in.From Eq. 11.16.4, equivalent flange force,

f xL f Q = 1.6 Q (½ d + 4¼ ) ÷ (½ d - ½ t ) = 7.32 kips

u f Factored bending moment about the y-axis of the flange = M = 7.32 × 18.0 ÷ 4 = 32.9 ft-kips

b. Strength limit statesbAs the bending moment is essentially due to the concentrated load at midspan, use C = 1.32 (value

from Fig. 10.4.1 e, given for a simple beam with lateral bracing at the supports on ly).






b yEntering the beam selection plots for W-Shapes (LRFDM Table 5-5) , with C = 1, F = 50 ksi and

b d b pxL = 18 ft, observe that a W18×76 has a bending strength M = 514 ft-kips and N M equals 612o

ft-kips, resulting in:d b d b px M = min [ C M ; N M ]= min [1.32×514 ; 612] = 612 ft-kipso

b py dyAlso from LRFDM Table 5-3, N M = 155 ft-kips = M Using Eq. 11.16.6:

As the weight of the beam selected (76 lbs) is less than the assumed value of 90 lbs used in the deadload calculations, no revision is necessary. The section satisfies the requirements for compactnessas there is no footnote in the beam selection tables stating otherwise.

Therefore, use a W18 ×76 of A242 Grade 50 steel. (Ans.)

P11.34. A crane runway girder 24 ft in length is to be designed to carry the two end wh eels of a 5-ton crane. Lateralsuppo rts are provided at the ends only. The whee ls of the crane are 8 ft on centers and each wheel transfersa maximum load of 12.5 kips to the top o f 60-lb rail. Assume 10 p ercent of the wheel load acting as a lateralload applied at 4¼ in. above the top of the compression flange. Select the lightest W14-shape of A992 steel.






P11.35. Select the lightest MC-shape of A36 steel to be used as purlins in an industrial building. The roof pitch is 5on 12 ; the rafters are 18 ft on centers; and the purlins are spaced at 8-ft intervals along the roof. The roo f

covering, including purlins, weighs 16 psf of the roof surface and the snow load is 24 psf of the horizontalsurface. The purlins should be designed without sag rods.

Solution

Inclination of roof, 2 = tan (5 ÷ 12) = 22.6-1 o

Spacing of trusses, L = 18 ftSpacing of purlins along roof = 8 ft

Horizontal Spacing of purlins =

Uniform service loads per foot of purlin:

Dead load, D =Snow load, S =

With only dead and snow loads acting on the purlin, load combination LC - 3, namely, [1.2D +1.6S] controls the design. Both these loads are vertical loads.

uV q = 1.2 D + 1.6 S = 1.2 × 0.128 + 1.6×0 .177 = 0.437 klf

The components of the factored loads acting in y- and x- directions at the mid-width of the purlinare:

uy uV q = q cos 2 = 0.437 cos 22.6 = 0.403 klf o

ux uV q = q sin 2 = 0.437 sin 22.6 = 0.168 klf o

uy ux Note that q pro du ces major axis moment, while q pro duces min or axis momen t. Th e purl in actsbas a simple beam of span L (= 18 ft) for major axis bendin g. It is not laterally suppo rted. So, L =

18 ft. With no sag rods placed, the purlin acts as a simple beam with respect to weak axis bending,with a span, L = 20 ft.

The required bend ing strengths are:

Or,

To select a trial shape, use the beam selection plots for C-shapes (LRFDM Table 5-11) and choose




a shape with a relatively large margin of strength with respect to major axis bending. For an

b y bunbraced length, L = 18 ft, F = 36 ksi and C = 1 an MC10×33.6 provides

Also, from LRFDM Table 5-10, for an MC10×33.6,

bFrom Table 1 0.4.1, C = 1.14 for a simply supported beam under uniform ly distributed load, withlateral supports at the ends on ly.Design flexural strength for major axis bending,

For C-shapes ben t about their minor axis, the shape factor is greater than 1.5. So, the designflexural strength for minor axis bending,

Substituting in the interaction Eq. 11.16.5 results in:

So, select a MC1 0×33 .6 (Ans.)

P11.36. Solve Problem P11.36, if one line of sag rods is used.

P11.37. Solve Problem P11.36, if W-shapes of A992 steel are used.

P11.38. Select the lightest W section o f A992 steel to carry 0.4 klf dead load, in addition to the weight of the beam ,and live load of 1.5 klf. The supe rimposed load is applied eccentrically 4 in. from the center of the web. The

beam is simply su pp orted and has a span of 2 8 ft. Assu me that late ral b racin g is pro vided at th e en d o nly.

solver vinnakota cap 11

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