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S te e l S tr u c tu r es b y V i n n ak o ta

C h ap te r 4

p age 4 -1

CHAPTER 4

P4.1.

Determine the wind pressure on the the windward wall, leeward wall, and roof of a 40 ft by 80 ft by 20 ft bu ild ing wit h a flat r oo f. Th e b uil din g is loc ated on lev el g rou nd in a su bu rba n ar ea o f Ch icag o. Ste el ro of deck connected to open-web steel joists and girders acts as roof diaphragms. Solution

Height of building, h = 20 ft Plan dimensions of the building : Parallel to wind direction, L direction, L = 40 ft No rma l to win d d irec tio n, B n, B = 80 ft a.

Velocity pressures, q From Fig. 6.1 of the ASCES, the basic wind speed, V , for Chicago Chicago is 90 mph . The buildin g is located in a suburban area. Therefore, use Exposure B (See ASCES Section 6.5.6). The building function is not considered an essential facility or likely to be occupied by 300 people in a single area at one time. Therefore, the building is of Category II, from Table 1-1 of the ASCES. The importance factor I factor I w equals 1.0 for Category II structures (see ASCES Table 6 -1). The topographic factor K factor K t is taken to be 1.0 because the building is on a level ground. The wind directionality factor K factor K d equals 0.85 for buildings, from Table 6-6 of the ASCES. Velocity pressure at height z height z above ground level (Eq. 6-13 of the ASCES): q z = 0.00256 K 0.00256 K zt K d V 2 I w K z = 0.00256×1.0×0. 0.00256×1.0×0.85×90 85×902 ×1.0 K ×1.0 K z = 17.6 K 17.6 K z Velocity pressure exposure coefficients K coefficients K z for the main wind force resisting system (Case 2) and Exposu re B may be obtained from ASCES Table 6-5. Values of K of K and z and the resulting velocity pre ssu res q z at floor and roof levels are given in Table SP4 .1.1. TABLE SP4.1.1

(ft) z (ft)

K z

q z (psf)

p ez (psf)

15 20

0.57 0.62

10.0 10.9

6.8 7.4

The velocity pressure at mean roof height, q h, is 10.9 psf. b.

External wind pressures, p pressures, p e, for the main wind force resisting system

PROPRIETARY MATER IAL. © 2006 T he McGraw-Hill Companies, Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

Chapter 4

page 4-2

From Eq. 6.15 of the ASCES, p e = q z G C p for windward wall = q h G C p for leeward wall and roof Least width of building L min = 40 ft , indicating that it is a rigid structure. So, from ASCES Section 6.5.8.1, gust factor G = 0.85. The values for the external pressu re coefficients C p are obtained from Fig. 6-3 o f the ASCES : –

The windward wall pressure coefficient is 0.8. So, the external wind pressures, p ez = q z G C p , for the windward wall are as given in Table X4.7.1. (Ans.)

–

Th e le eward wall pr essure coe fficie nt is a fu nc tion of th e L / B ratio. For wind normal to the 80 ft face, . Therefore, the leeward wall pressure co efficient is - 0.5. The leeward wall pressure = q h G C p = 10.9×0 .85×(- 0.5) = - 4.6 psf

–

Fo r th e gi ven st ruc ture with pre ssu res:

Roof wind pressure: 0 - 20 ft = 20 - 40 ft = c.

(Ans.)

, two roo f zo ne s are specified for roo f wi nd

10.9×0 .85×(- 0.9) = - 8.3 psf 10.9×0 .85×(- 0.5) = - 4.6 psf

(Ans.) (Ans.)

Internal wind pressures p i for the main wind force resisting system. The build ing is assumed (con servatively) to be a partially enclosed bu ilding. From Table 6-7 of the ASCES, GC pi = ± 0.55, for partially enclosed buildings. Internal wind pressures, p i = q h (GC pi) = ± 10.9×0 .55 = ± 6.0 psf (Ans.)

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Steel Structures by Vinnakota

P4.2.

Chapter 4

page 4-3

Determine the external pressure acting over the windw ard wall, leeward wall, and roof of a gabled building, located on open flat terrain in Boston, for the wind load perpendicular to the ridge of the building. The fully enclosed agricultural building has plan dimensions of 60 ft by 120 ft, height to eave of 20 ft, and a roof slope of 35o. Also, determine the internal pressure in the building. Solution

Gabled building. Height to eave, h o = 20 ft Wind perpendicular to the ridge of the building. Plan dimensions of the building : Parallel to wind direction, L = 60 ft No rmal to win d d irec tion, B = 120 ft Roof slope, = 35 o; rise = 30 tan = 21 ft Mean roof height, h = 30.5 ft a.

Velocity pressures, q From Fig. 6.1 of the ASCES, the basic wind speed, V , for Chicago is 110 mp h. The buildin g is located in a suburban area. Therefore, use Exposure B (See ASCES Section 6.5.6). The building is an agricultural building that presents a low hazard to human life in the event of failure. Therefore, the building is of Category I, from Table 1-1 o f the ASCES. For Category I structures in hurricane prone regions with V > 100 mph the importance factor I w equals 0.77 (see ASCES Table 6-1). The topographic factor K t is taken to be 1.0 becau se the building is o n a flat terrain. The wind directionality factor K d equals 0.85 for buildings, from Table 6-6 of the ASCES. Velocity pressure at height z above ground level (Eq. 6-13 of the ASCES): q z = 0.00256 K zt K d V 2 I w K z = 0.00256×1.0×0.85×1102 ×0.77 K z = 20.3 K z Velocity pressure exposure coefficients K z for the main wind force resisting system (Case 2) and Exposu re B may be obtained from ASCES Table 6-5. Values of K and the resulting velocity z pre ssu res q z at floor and roof levels are given in Table SP4 .2.1. TABLE SP4.2.1

z (ft)

K z

q z (psf)

p ez (psf)

15 20 30.5

0.57 0.62 0.70

11.6 12.6 14.2

7.9 8.6

The velocity pressure at mean roof height, q h, is 14.2 psf. b.

External wind pressures, p e, for the main wind force resisting system

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Chapter 4

From Eq. 6.15 of the ASCES, p e = q z G C p = q h G C p Least width of building L min = 60 ft

page 4-4

for windward wall for leeward wall and roof

, indicating that it is a rigid structure. So, from ASCES Section 6.5.8.1, gust factor G = 0.85. The values for the external pressu re coefficients C p are obtained from Fig. 6-3 o f the ASCES: –

The windward wall pressure coefficient is 0.8. So, the external wind pressures, p ez = q z G C p , for the windward wall are as given in Table X4.7.1. (Ans.)

–

Th e le eward wall pr essure coe fficie nt is a fu nc tion of th e L / B ratio. For wind normal to the 120 ft face, . Therefore, the leeward wall pressure co efficient is - 0.5. The leeward wall pressure = q h G C p = 14.2×0 .85×(- 0.5) = - 6.0 psf

–

For th e gi ven stru ctu re w ith , two roo f pr essu re c oefficie nts are specified for windward roof slope (namely, C p = -0.2 and 0.3) and one roof pressure coefficient for leeward roof slope (namely, -0.6). Roof wind pressures: Windward roof slope Leeward roof slope

c.

(Ans.)

= 14.2×0.85×(- 0.2) = - 2.4 psf (suction) = 14.2×0.85×( 0.3) = 3.6 psf (pressure) = 14.2×0 .85×(- 0.6) = - 7.2 psf (suction)

Internal wind pressures p i for the main wind force resisting system. The buildin g is a fully enclosed building. From Table 6-7 of the ASCES, GC pi = ± 0.18. Internal wind pressures, p i = q h (GC pi) = ± 14.2×0 .18 = ± 2.6 psf

(Ans.) (Ans.) (Ans.)

(Ans.)

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Steel Structures by Vinnakota

P4.3.

Chapter 4

page 4-5

Determine the design snow load acting on the building given in Problem P4.1. The building is in an industrial park with no trees or other structures offering shelter. Solution a.

From Fig. 7.1 of the ASCES the ground snow load p g for Chicago is 25 p sf. Suburban area Exposure B. The exposure factor C e can be taken equal to 0.9 from Table 7.2 of the ASCES because there are no trees or other structures offering shelter (fully exposed) and Exposure B.

Thermal factor C t is 1.0 from Table 7-3 of the ASCES because the building is a heated structure. The building function is not considered an essential facility or likely to be occupied by 300 people in a single area at one time. Therefore, the building is of Category II, from Table 1-1 of the ASCES. The importance factor I equals 1.0 for Category II structures (see ASCES Table 7 -4). From Eq. 7.1 of the ASCES, the flat roof snow load is obtained as p f = 0.7 C e C I t p g = 0.7×0.9×1.0×1.0×25 = 15.8 psf Also, from Section 7.3 of ASCES, for p g > 20 psf, minimum vale for low slope roofs, p f, min = 20 ( I ) = 20 (1.00) = 20.0 psf So, roof snow load to be used is 20.0 psf.

(Ans.)

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Chapter 4

P4.4.

page 4-6

Determine the balanced and unbalanced design snow loads for the roof of the gabled frame of Problem P4.2. Solution Span, L = 60 ft a. Roof slope, =

35 o

From Fig. 7.1 of the ASCES the ground snow load p g for Boston is 35 psf. Suburban area Exposure B. The exposure factor C e can be taken equal to 0.9 from Table 7.2 of the ASCES because there are minimal obstructions (fully exposed) and Exposure B. Thermal factor C t is 1.0 from Table 7-3 of the ASCES because the building is a heated structure. The building is an agricultural building that presents a low hazard to human life in the event of failure. Therefore, the building is of Category I, from Table 1-1 o f the ASCES. For Category I structures the importance factor I equals 0.8 (see ASCES Table 7-4). From Eq. 7.1 of the ASCES, the flat roof snow load is obtained as p f = 0.7 C e C I t p g = 0.7×0.9×1.0×0.8×35 = 17.6 psf From Fig. 7-2 of the ASCES, the roof slope factor C is 0.88 for warm roofs with roo f slope, = s 35 o. From Eq. 7.2 of the ASCES, the sloped-roof snow load is obtained as (Ans.) This load is called the balanced snow load and is applied to the entire roof of the structure. b.

Unbalanced snow load. Roof length p arallel to the ridge line L = 120 ft Horizontal distance from eave to ridge W = 30 ft

So, from Eq. 7.3 o f the ASCES, the gable roo f drift parameter equals 1.0 From Eq. 7.4 of the ASCES, snow density, = 0.13 p g + 14 = 0.13×35.0 + 14 = 18.6 pcf < 30 pcf O.K. Minimum slope, As W > 20 ft and > min, the structure must be designed to resist an unbalanced uniform snow load on the leeward side (see ASCES Table 7-5). Intensity of unbalanced snow load on the windward side, (Ans.)

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Steel Structures by Vinnakota

Chapter 4

page 4-7

Intensity of unbalanced snow load on the leeward side,

(Ans.)

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Chapter 4

P4.5.

page 4-8

The axial compressive force on a building column from the code-spec ified loads have been determined as: 180 kips dead load, 30 kips from the roof snow load, 140 kips (reduced) floor live load, ± 120 kips from wind, and ± 40 kips from earthquake. Determine the factored axial load, P u , for which the column is to be designed . If the resistance factor is 0.85, what is the required nominal strength? Solution Given: Building Column D = 180 kips L (reduced) = 140 kips; S = 30 kips; W = ±120 kips; E = ± 40 kips

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

0.5 L = 70 kips L rSR = max ( L r, S , R) = 30 kips 0.8W = 96 kips

Factored Axial Force (kips)

1.4 D = 1.4×180 1.2 D + 1.6 L + .5 L rSR = 1.2×180 + 1.6×140 + .5×30 1.2 D + 1.6 L rSR + .8W = 1.2×180 + 1.6×30 + .8×120 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×180 + 1.6×120 + .5×140 + .5×30 1.2 D + 1.0 E + .5 L + .2S = 1.2×180 + 1.0× 40 + .5×140 + .2×30 .9 D + 1.6W = .9×180 – 1.6×120 .9 D + 1.0 E = .9×180 – 1.0×40

= = = = = = =

252 455 360 493 341 30.0 122

No te: C om pre ssiv e fo rce s are conside red po sitive. The controlling load combination is LC-4 (Wind). Factored axial compressive load, P u = 493 kips Design axial compressive strength, P d P req = P u = 493 kips No min al ax ial c om pre ssiv e st ren gth , P n = Pd / 493 /.85 = 580 kips

(Ans.) (Ans.)

As there is a reversal of sign in load combination LC-6, the member (and the end connections) should also be designed for an axial tensile load, T u = 30 kips. (Ans.)

This problem provides introduction to the use of the load factors and how the load Comments: combin ations work. The main design consideration is the compression loading. While the tension is small, it should still be noted.

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Steel Structures by Vinnakota

P4.6.

Chapter 4

page 4-9

Repeat Problem 4.5 for a garage column . Solution

G iv en :

Hint:

G arag e C olu mn D = 180 kips; L (reduced) = 140 kips S = 30 kips L rSR = 30 kips W = 120 kips; E = 40 kips See Exception s in ASCES Section 2.3.2 Basic comb inations. The load factor on L in combin ations LC-3, LC-4, and LC-5 equals 1.0 for garages.

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

Factored Axial Force (kips)

1.4 D = 1.4×180 1.2 D + 1.6 L + .5 L rSR = 1.2×180 + 1.6×140 + .5×30 1.2 D + 1.6 L rSR + 1.0 L = 1.2×180 + 1.6×30 + 140 1.2 D + 1.6W + 1.0 L + .5 L rSR = 1.2×180 + 1.6×120 + 140 + .5×30 1.2 D + 1.0 E + 1.0 L + .2S = 1.2×180 + 40 + 140 + .2×30 .9 D + 1.6W = .9×180 – 1.6×120 .9 D + 1.0 E = .9×180 – 40

= 252 = 455 = 404 = 563 = 402 = -30.0 = 122

No te: C om pre ssiv e fo rce s are conside red po sitive. The controlling load combination is LC-4 (Wind). Factored axial compressive load, Pu = 563 kips Design axial compressive strength, P d P req = P u = 563 kips No min al ax ial c om pre ssiv e st ren gth , P n = Pd / 563/.85 = 662 kips

(Ans.) (Ans.)

As there is a reversal of sign in load combination, LC-6, the member (and the end connections) should be designed for an axial tensile load, Tu = 30 kips. (Ans.)

Like the first problem, this provides a good idea of how to use the load factors and load Comments: combin ations, and also raises the awareness of special conditions for garage structures. The main design consideration is the compressive loading. While the tension is small, it should still be noted.

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Chapter 4

P4.7.

page 4-10

Loads acting on a roof deck include a dead load of 30 psf, a roof live load of 20 psf, a snow load of 35 psf, and a wind pressure of 18 psf (upward or downward). Determine the governing loading on the decking. Solution G iv en : R oo f D ec kin g D = 30 psf Roof Live Load, L r = 20 psf ; W = ±18 psf

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

S = 35 psf

Factored Loading (psf)

1.4 D = 1.4×30 1.2 D + 1.6 L + .5 L rSR = 1.2×30 + 1.6×0 + .5×35 1.2 D + 1.6 L rSR + .8W = 1.2×30 + 1.6×35 + .8×18 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×30 + 1.6×18 + .5×0 + .5×35 1.2 D + 1.0 E + .5 L + .2S = 1.2×30 + 0 + .5×0 + .2×35 .9 D + 1.6W = .9×30 – 1.6×18 .9 D + 1.0 E = .9×30 – 0

= =

42.0 53.5 = 106 = 82.3 = 43.0 = - 1.80 = 27.0

No te: M inu s sig n sign ifies suctio n ( up lift) on the decking . The controlling load combination is LC-3 (Roof live load). Factored load (pressure) on the decking, p u = 106 psf As there is a reversal of sign, in load com bination LC -6, the decking (and the welds or nails connec ting the decking to the suppo rts), should be designed for a suction of 1.8 psf.

(Ans) (Ans)

This problem makes one to consider different directions of wind forces. The main design Comments: consideration is the LC-3. While uplift from LC-6 is small, it should still be noted.

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Steel Structures by Vinnakota

P4.8.

Chapter 4

page 4-11

The axial forces in a diagonal memb er of a roof truss are as follows: 50 kips dead load (T), 40 kips sn ow load (T), 20 kips roof live load (T), 15 kips rain load (T), and a wind load of 18 kips (C). Here, T represents tension and C, compression. Determine the required design strength of the diagonal member. Solution Given: Roof truss diagonal member D = 50 kips (T); Roof Live Load, L r = 20 kips (T) S = 40 kips (T); R = 15 kips (T) W = 18 kips (C) Here, T represents tension and C compression.

Combination

Factored Axial Force (kips)

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

1.4 D = 1.4×50 1.2 D + 1.6 L + .5 L rSR = 1.2×50 + 1.6×0 + .5×40 1.2 D + 1.6 L rSR + .8W = 1.2×50 + 1.6×40 + .8×0 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×50 + 1.6×0 + .5×0 + .5×40 1.2 D + 1.0 E + .5 L + .2S = 1.2×50 + 0 + .5×0 + .2×40 .9 D + 1.6W = .9×50 – 1.6×18 .9 D + 1.0 E = .9×50 – 0

= =

70.0 80.0 = 124 = 80.0 = 68.0 = 16.2 = 45.0

No te: Te nsi le fo rce s are considered po sitive. The controlling load combination is LC-3 (Roof loading). Factored axial tensile load, T u = 124 kips (Ans) There is no reversal of sign in load combinations LC-6 and LC-7, indicating that the member is not subjected to any compression under factored loads.

Comments: It should be noted that the compressive force contributed by wind loading was not included in LC-3 or LC-4, since it did not contribute to an increase in the member axial load.

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Chapter 4

P4.9.

page 4-12

A beam is part of the framing system for the floor of a residential building. The end moments caused by the service loads are: 160 ft-kips (clockwise) from dead load, 23 0 ft-kips (clockwise) from live load, and 15 0 ftkips (clockwise or anticlockwise) from wind load. Determine the maximum factored bending mom ent. What is the controlling ASCES load combin ation? Also, if the resistance factor is 0.9, what is the required nominal bending strength in ft-kips? Solution Given: Beam in a residential floor framing system D = 160 ft-kips (CW) L = 230 ft-kips (CW) ; 0.5 L = 115 ft-kips W = 150 ft-kips (CW or CCW) ; 0.8W = 120 ft-kips

Combination

Factored Bending Moment (ft-kips)

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

1.4 D = 1.4×160 1.2 D + 1.6 L + .5 L rSR = 1.2×160 + 1.6×230 + .5×0 1.2 D + 1.6 L rSR + 0.8W = 1.2×160 + 1.6×0 + .8×150 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×160 + 1.6×150 + .5×230 + .5×0 1.2 D + 1.0 E + .5 L + .2S = 1.2×160 + 0 + .5×230 + .2×0 .9 D + 1.6W = .9×160 – 1.6×150 .9 D + 1.0 E = .9×160 – 0

= 224 = 560 = 312 = 547 = 307 = - 96.0 = 144

No te: Clo ckw ise mo ments are c on sid ere d p osi tive . The controlling load combination is LC-2 (Live load). Factored b ending moment, M u = 560 ft-kips (clockwise) Design bending strength, M d M req = M u = 560 ft-kips No min al b endin g st ren gth , M n = M d / 560 /.9 = 622 ft-kips

(Ans) (Ans)

As there is a reversal of sign in load combination LC-6, the member (and the end connections) should also be designed for a factored anticlockwise bending moment, M u = 96 ft-kips. (Ans)

The main design consideration is the clockwise bending. While the counterclockwise Comments: loading is small, it should still be noted .

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Steel Structures by Vinnakota

P4.10.

Chapter 4

page 4-13

The maximum mom ents caused by the service loads on the roof beam of an office building are as follows: dead load, 58 ft-kips; snow load, 75 ft-kips; roof live load, 50 ft-kips; rain load, 25 ft-kips; and win d load 30 ft-kips. All these momen ts occur at the same location on the beam (center) and can therefore be comb ined. Determine the required bending strength of the roof beam. Solution Given: Office building roof beam D = 58 ft-kips S = 75 ft-kips; L r = 50 ft-kips; R = 25 ft-kips W = - 30 ft-kips

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

Factored Bending Moment (ft-kips)

1.4 D = 1.4×58 1.2 D + 1.6 L + .5 L rSR = 1.2×58 + 1.6×0 + .5×75 1.2 D + 1.6 L rSR + .8W = 1.2×58 + 1 .6×75 + .8×0 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×58 + 1.6×0 + .5×0 + .5×75 1.2D + 1.0 E + .5 L + .2S = 1.2×58 + 0 + .5×0 + .2×75 .9 D + 1.6W = .9×58 – 1.6×30 .9 D + 1.0 E = .9×58 – 0

= = = = = = =

81.2 107 190 107 84.6 4.20 52.2

Factored bending moment, M u = 190 ft-kips

The main design consideration is the positive bending moment. There is no reversal of Comments: moment to be considered, under the given loading condition.

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Chapter 4

P4.11.

page 4-14

A beam-colum n is subjected to the following forces by the service loads indicated. Axial compressio n, P : dead load, 120 kips; live load, 220 kips. Bendin g moments, M : dead load 180 ft-kips, live load 24 0 ft-kips. Determine the factored axial load, Pu , and bending moment, M u , for which the beam-column is to be designed. Solution G iv en : B eam -c olu mn Axial compressive loads D = 120 kips L = 220 kips

Bending moments D = 180 ft-kips L = 240 ft-kips

a. Combination

LC-1 LC-2

Factored Axial Force (kips)

1.4 D = 1.4×120 1.2 D + 1.6 L + .5 L rSR = 1.2×120 + 1.6×220 + .5×0

= 168 = 496

b. Combination

LC-1 LC-2

Factored Bending Moment (ft-kips)

1.4 D = 1.4×180 1.2 D + 1.6 L + .5 L rSR = 1.2×180 + 1.6×240 + .5×0

= 252 = 600

c. Combination

Factored Axial force (kips)

Factored Bending Moment (ft-kips)

LC-1 LC-2

16 8 496

25 2 600

Factored axial load, P u = 496 kips Factored b ending moment, M u = 600 ft-kips No te th at b oth max imu m axial forc e an d m axim um bend ing mo me nt o ccu r for load c om bination LC-2, in this example. Comments:

This problem is good because it shows the analysis of members subject to more than one

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Steel Structures by Vinnakota

Chapter 4

page 4-15

type of load. In this example, the same combination prod uces maximum axial force and maximum momen t. This is not necessarily true in all cases. The main design con sideration is the axial compression and the positive bendin g. Also there are no load reversals to consider und er the given loading condition.

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Chapter 4

page 4-16

Additional problem not in Text Book P4.12.

Repeat Problem 4.7 if the wind load is always a suction (- 18 psf). Solution G iv en : R oo f d ec kin g D = 30 psf Roof Live Load, L r = 20 psf ; W = 18 psf

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

S = 35 psf

Factored Loading (psf)

1.4 D = 1.4×30 1.2 D + 1.6 L + .5 L rSR = 1.2×30 + 1.6×0 + .5×35 1.2 D + 1.6 L rSR + .8W = 1.2×30 + 1.6×35 + .8×0 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×30 + 1.6×0 + .5×0 + .5×35 1.2 D + 1.0 E + .5 L + .2S = 1.2×30 + 0 + .5×0 + .2×35 .9 D + 1.6W = .9×30 – 1.6×18 .9 D + 1.0 E = .9×30 – 0

= = = = = = =

42.0 53.5 92.0 53.5 43.0 - 1.8 27.0

No te: M inu s sig n sign ifies suctio n ( up lift) on the decking . The controlling load combination is again LC-3 (Roof load). Factored load (pressure) on the decking, p u = 92 psf As there is a reversal of sign, in load com bination LC -6, the decking (and the welds or nails connec ting the decking to the suppo rts), should be designed for a suction of 1.8 psf.

(Ans) (Ans)

When wind load is always a suction (negative), the wind load is not included in load Comments: combin ation LC-3 (taken as 0). The main design consideration is the LC-3 while uplift from LC-6 is small, it should still be noted.

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C h ap te r 4

p age 4 -1

CHAPTER 4

P4.1.

Determine the wind pressure on the the windward wall, leeward wall, and roof of a 40 ft by 80 ft by 20 ft bu ild ing wit h a flat r oo f. Th e b uil din g is loc ated on lev el g rou nd in a su bu rba n ar ea o f Ch icag o. Ste el ro of deck connected to open-web steel joists and girders acts as roof diaphragms. Solution

Height of building, h = 20 ft Plan dimensions of the building : Parallel to wind direction, L direction, L = 40 ft No rma l to win d d irec tio n, B n, B = 80 ft a.

Velocity pressures, q From Fig. 6.1 of the ASCES, the basic wind speed, V , for Chicago Chicago is 90 mph . The buildin g is located in a suburban area. Therefore, use Exposure B (See ASCES Section 6.5.6). The building function is not considered an essential facility or likely to be occupied by 300 people in a single area at one time. Therefore, the building is of Category II, from Table 1-1 of the ASCES. The importance factor I factor I w equals 1.0 for Category II structures (see ASCES Table 6 -1). The topographic factor K factor K t is taken to be 1.0 because the building is on a level ground. The wind directionality factor K factor K d equals 0.85 for buildings, from Table 6-6 of the ASCES. Velocity pressure at height z height z above ground level (Eq. 6-13 of the ASCES): q z = 0.00256 K 0.00256 K zt K d V 2 I w K z = 0.00256×1.0×0. 0.00256×1.0×0.85×90 85×902 ×1.0 K ×1.0 K z = 17.6 K 17.6 K z Velocity pressure exposure coefficients K coefficients K z for the main wind force resisting system (Case 2) and Exposu re B may be obtained from ASCES Table 6-5. Values of K of K and z and the resulting velocity pre ssu res q z at floor and roof levels are given in Table SP4 .1.1. TABLE SP4.1.1

(ft) z (ft)

K z

q z (psf)

p ez (psf)

15 20

0.57 0.62

10.0 10.9

6.8 7.4

The velocity pressure at mean roof height, q h, is 10.9 psf. b.

External wind pressures, p pressures, p e, for the main wind force resisting system

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Chapter 4

page 4-2

From Eq. 6.15 of the ASCES, p e = q z G C p for windward wall = q h G C p for leeward wall and roof Least width of building L min = 40 ft , indicating that it is a rigid structure. So, from ASCES Section 6.5.8.1, gust factor G = 0.85. The values for the external pressu re coefficients C p are obtained from Fig. 6-3 o f the ASCES : –

The windward wall pressure coefficient is 0.8. So, the external wind pressures, p ez = q z G C p , for the windward wall are as given in Table X4.7.1. (Ans.)

–

Th e le eward wall pr essure coe fficie nt is a fu nc tion of th e L / B ratio. For wind normal to the 80 ft face, . Therefore, the leeward wall pressure co efficient is - 0.5. The leeward wall pressure = q h G C p = 10.9×0 .85×(- 0.5) = - 4.6 psf

–

Fo r th e gi ven st ruc ture with pre ssu res:

Roof wind pressure: 0 - 20 ft = 20 - 40 ft = c.

(Ans.)

, two roo f zo ne s are specified for roo f wi nd

10.9×0 .85×(- 0.9) = - 8.3 psf 10.9×0 .85×(- 0.5) = - 4.6 psf

(Ans.) (Ans.)

Internal wind pressures p i for the main wind force resisting system. The build ing is assumed (con servatively) to be a partially enclosed bu ilding. From Table 6-7 of the ASCES, GC pi = ± 0.55, for partially enclosed buildings. Internal wind pressures, p i = q h (GC pi) = ± 10.9×0 .55 = ± 6.0 psf (Ans.)

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Steel Structures by Vinnakota

P4.2.

Chapter 4

page 4-3

Determine the external pressure acting over the windw ard wall, leeward wall, and roof of a gabled building, located on open flat terrain in Boston, for the wind load perpendicular to the ridge of the building. The fully enclosed agricultural building has plan dimensions of 60 ft by 120 ft, height to eave of 20 ft, and a roof slope of 35o. Also, determine the internal pressure in the building. Solution

Gabled building. Height to eave, h o = 20 ft Wind perpendicular to the ridge of the building. Plan dimensions of the building : Parallel to wind direction, L = 60 ft No rmal to win d d irec tion, B = 120 ft Roof slope, = 35 o; rise = 30 tan = 21 ft Mean roof height, h = 30.5 ft a.

Velocity pressures, q From Fig. 6.1 of the ASCES, the basic wind speed, V , for Chicago is 110 mp h. The buildin g is located in a suburban area. Therefore, use Exposure B (See ASCES Section 6.5.6). The building is an agricultural building that presents a low hazard to human life in the event of failure. Therefore, the building is of Category I, from Table 1-1 o f the ASCES. For Category I structures in hurricane prone regions with V > 100 mph the importance factor I w equals 0.77 (see ASCES Table 6-1). The topographic factor K t is taken to be 1.0 becau se the building is o n a flat terrain. The wind directionality factor K d equals 0.85 for buildings, from Table 6-6 of the ASCES. Velocity pressure at height z above ground level (Eq. 6-13 of the ASCES): q z = 0.00256 K zt K d V 2 I w K z = 0.00256×1.0×0.85×1102 ×0.77 K z = 20.3 K z Velocity pressure exposure coefficients K z for the main wind force resisting system (Case 2) and Exposu re B may be obtained from ASCES Table 6-5. Values of K and the resulting velocity z pre ssu res q z at floor and roof levels are given in Table SP4 .2.1. TABLE SP4.2.1

z (ft)

K z

q z (psf)

p ez (psf)

15 20 30.5

0.57 0.62 0.70

11.6 12.6 14.2

7.9 8.6

The velocity pressure at mean roof height, q h, is 14.2 psf. b.

External wind pressures, p e, for the main wind force resisting system

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Chapter 4

From Eq. 6.15 of the ASCES, p e = q z G C p = q h G C p Least width of building L min = 60 ft

page 4-4

for windward wall for leeward wall and roof

, indicating that it is a rigid structure. So, from ASCES Section 6.5.8.1, gust factor G = 0.85. The values for the external pressu re coefficients C p are obtained from Fig. 6-3 o f the ASCES: –

The windward wall pressure coefficient is 0.8. So, the external wind pressures, p ez = q z G C p , for the windward wall are as given in Table X4.7.1. (Ans.)

–

Th e le eward wall pr essure coe fficie nt is a fu nc tion of th e L / B ratio. For wind normal to the 120 ft face, . Therefore, the leeward wall pressure co efficient is - 0.5. The leeward wall pressure = q h G C p = 14.2×0 .85×(- 0.5) = - 6.0 psf

–

For th e gi ven stru ctu re w ith , two roo f pr essu re c oefficie nts are specified for windward roof slope (namely, C p = -0.2 and 0.3) and one roof pressure coefficient for leeward roof slope (namely, -0.6). Roof wind pressures: Windward roof slope Leeward roof slope

c.

(Ans.)

= 14.2×0.85×(- 0.2) = - 2.4 psf (suction) = 14.2×0.85×( 0.3) = 3.6 psf (pressure) = 14.2×0 .85×(- 0.6) = - 7.2 psf (suction)

Internal wind pressures p i for the main wind force resisting system. The buildin g is a fully enclosed building. From Table 6-7 of the ASCES, GC pi = ± 0.18. Internal wind pressures, p i = q h (GC pi) = ± 14.2×0 .18 = ± 2.6 psf

(Ans.) (Ans.) (Ans.)

(Ans.)

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Steel Structures by Vinnakota

P4.3.

Chapter 4

page 4-5

Determine the design snow load acting on the building given in Problem P4.1. The building is in an industrial park with no trees or other structures offering shelter. Solution a.

From Fig. 7.1 of the ASCES the ground snow load p g for Chicago is 25 p sf. Suburban area Exposure B. The exposure factor C e can be taken equal to 0.9 from Table 7.2 of the ASCES because there are no trees or other structures offering shelter (fully exposed) and Exposure B.

Thermal factor C t is 1.0 from Table 7-3 of the ASCES because the building is a heated structure. The building function is not considered an essential facility or likely to be occupied by 300 people in a single area at one time. Therefore, the building is of Category II, from Table 1-1 of the ASCES. The importance factor I equals 1.0 for Category II structures (see ASCES Table 7 -4). From Eq. 7.1 of the ASCES, the flat roof snow load is obtained as p f = 0.7 C e C I t p g = 0.7×0.9×1.0×1.0×25 = 15.8 psf Also, from Section 7.3 of ASCES, for p g > 20 psf, minimum vale for low slope roofs, p f, min = 20 ( I ) = 20 (1.00) = 20.0 psf So, roof snow load to be used is 20.0 psf.

(Ans.)

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Chapter 4

P4.4.

page 4-6

Determine the balanced and unbalanced design snow loads for the roof of the gabled frame of Problem P4.2. Solution Span, L = 60 ft a. Roof slope, =

35 o

From Fig. 7.1 of the ASCES the ground snow load p g for Boston is 35 psf. Suburban area Exposure B. The exposure factor C e can be taken equal to 0.9 from Table 7.2 of the ASCES because there are minimal obstructions (fully exposed) and Exposure B. Thermal factor C t is 1.0 from Table 7-3 of the ASCES because the building is a heated structure. The building is an agricultural building that presents a low hazard to human life in the event of failure. Therefore, the building is of Category I, from Table 1-1 o f the ASCES. For Category I structures the importance factor I equals 0.8 (see ASCES Table 7-4). From Eq. 7.1 of the ASCES, the flat roof snow load is obtained as p f = 0.7 C e C I t p g = 0.7×0.9×1.0×0.8×35 = 17.6 psf From Fig. 7-2 of the ASCES, the roof slope factor C is 0.88 for warm roofs with roo f slope, = s 35 o. From Eq. 7.2 of the ASCES, the sloped-roof snow load is obtained as (Ans.) This load is called the balanced snow load and is applied to the entire roof of the structure. b.

Unbalanced snow load. Roof length p arallel to the ridge line L = 120 ft Horizontal distance from eave to ridge W = 30 ft

So, from Eq. 7.3 o f the ASCES, the gable roo f drift parameter equals 1.0 From Eq. 7.4 of the ASCES, snow density, = 0.13 p g + 14 = 0.13×35.0 + 14 = 18.6 pcf < 30 pcf O.K. Minimum slope, As W > 20 ft and > min, the structure must be designed to resist an unbalanced uniform snow load on the leeward side (see ASCES Table 7-5). Intensity of unbalanced snow load on the windward side, (Ans.)

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Steel Structures by Vinnakota

Chapter 4

page 4-7

Intensity of unbalanced snow load on the leeward side,

(Ans.)

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Chapter 4

P4.5.

page 4-8

The axial compressive force on a building column from the code-spec ified loads have been determined as: 180 kips dead load, 30 kips from the roof snow load, 140 kips (reduced) floor live load, ± 120 kips from wind, and ± 40 kips from earthquake. Determine the factored axial load, P u , for which the column is to be designed . If the resistance factor is 0.85, what is the required nominal strength? Solution Given: Building Column D = 180 kips L (reduced) = 140 kips; S = 30 kips; W = ±120 kips; E = ± 40 kips

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

0.5 L = 70 kips L rSR = max ( L r, S , R) = 30 kips 0.8W = 96 kips

Factored Axial Force (kips)

1.4 D = 1.4×180 1.2 D + 1.6 L + .5 L rSR = 1.2×180 + 1.6×140 + .5×30 1.2 D + 1.6 L rSR + .8W = 1.2×180 + 1.6×30 + .8×120 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×180 + 1.6×120 + .5×140 + .5×30 1.2 D + 1.0 E + .5 L + .2S = 1.2×180 + 1.0× 40 + .5×140 + .2×30 .9 D + 1.6W = .9×180 – 1.6×120 .9 D + 1.0 E = .9×180 – 1.0×40

= = = = = = =

252 455 360 493 341 30.0 122

No te: C om pre ssiv e fo rce s are conside red po sitive. The controlling load combination is LC-4 (Wind). Factored axial compressive load, P u = 493 kips Design axial compressive strength, P d P req = P u = 493 kips No min al ax ial c om pre ssiv e st ren gth , P n = Pd / 493 /.85 = 580 kips

(Ans.) (Ans.)

As there is a reversal of sign in load combination LC-6, the member (and the end connections) should also be designed for an axial tensile load, T u = 30 kips. (Ans.)

This problem provides introduction to the use of the load factors and how the load Comments: combin ations work. The main design consideration is the compression loading. While the tension is small, it should still be noted.

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Steel Structures by Vinnakota

P4.6.

Chapter 4

page 4-9

Repeat Problem 4.5 for a garage column . Solution

G iv en :

Hint:

G arag e C olu mn D = 180 kips; L (reduced) = 140 kips S = 30 kips L rSR = 30 kips W = 120 kips; E = 40 kips See Exception s in ASCES Section 2.3.2 Basic comb inations. The load factor on L in combin ations LC-3, LC-4, and LC-5 equals 1.0 for garages.

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

Factored Axial Force (kips)

1.4 D = 1.4×180 1.2 D + 1.6 L + .5 L rSR = 1.2×180 + 1.6×140 + .5×30 1.2 D + 1.6 L rSR + 1.0 L = 1.2×180 + 1.6×30 + 140 1.2 D + 1.6W + 1.0 L + .5 L rSR = 1.2×180 + 1.6×120 + 140 + .5×30 1.2 D + 1.0 E + 1.0 L + .2S = 1.2×180 + 40 + 140 + .2×30 .9 D + 1.6W = .9×180 – 1.6×120 .9 D + 1.0 E = .9×180 – 40

= 252 = 455 = 404 = 563 = 402 = -30.0 = 122

No te: C om pre ssiv e fo rce s are conside red po sitive. The controlling load combination is LC-4 (Wind). Factored axial compressive load, Pu = 563 kips Design axial compressive strength, P d P req = P u = 563 kips No min al ax ial c om pre ssiv e st ren gth , P n = Pd / 563/.85 = 662 kips

(Ans.) (Ans.)

As there is a reversal of sign in load combination, LC-6, the member (and the end connections) should be designed for an axial tensile load, Tu = 30 kips. (Ans.)

Like the first problem, this provides a good idea of how to use the load factors and load Comments: combin ations, and also raises the awareness of special conditions for garage structures. The main design consideration is the compressive loading. While the tension is small, it should still be noted.

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Chapter 4

P4.7.

page 4-10

Loads acting on a roof deck include a dead load of 30 psf, a roof live load of 20 psf, a snow load of 35 psf, and a wind pressure of 18 psf (upward or downward). Determine the governing loading on the decking. Solution G iv en : R oo f D ec kin g D = 30 psf Roof Live Load, L r = 20 psf ; W = ±18 psf

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

S = 35 psf

Factored Loading (psf)

1.4 D = 1.4×30 1.2 D + 1.6 L + .5 L rSR = 1.2×30 + 1.6×0 + .5×35 1.2 D + 1.6 L rSR + .8W = 1.2×30 + 1.6×35 + .8×18 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×30 + 1.6×18 + .5×0 + .5×35 1.2 D + 1.0 E + .5 L + .2S = 1.2×30 + 0 + .5×0 + .2×35 .9 D + 1.6W = .9×30 – 1.6×18 .9 D + 1.0 E = .9×30 – 0

= =

42.0 53.5 = 106 = 82.3 = 43.0 = - 1.80 = 27.0

No te: M inu s sig n sign ifies suctio n ( up lift) on the decking . The controlling load combination is LC-3 (Roof live load). Factored load (pressure) on the decking, p u = 106 psf As there is a reversal of sign, in load com bination LC -6, the decking (and the welds or nails connec ting the decking to the suppo rts), should be designed for a suction of 1.8 psf.

(Ans) (Ans)

This problem makes one to consider different directions of wind forces. The main design Comments: consideration is the LC-3. While uplift from LC-6 is small, it should still be noted.

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Steel Structures by Vinnakota

P4.8.

Chapter 4

page 4-11

The axial forces in a diagonal memb er of a roof truss are as follows: 50 kips dead load (T), 40 kips sn ow load (T), 20 kips roof live load (T), 15 kips rain load (T), and a wind load of 18 kips (C). Here, T represents tension and C, compression. Determine the required design strength of the diagonal member. Solution Given: Roof truss diagonal member D = 50 kips (T); Roof Live Load, L r = 20 kips (T) S = 40 kips (T); R = 15 kips (T) W = 18 kips (C) Here, T represents tension and C compression.

Combination

Factored Axial Force (kips)

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

1.4 D = 1.4×50 1.2 D + 1.6 L + .5 L rSR = 1.2×50 + 1.6×0 + .5×40 1.2 D + 1.6 L rSR + .8W = 1.2×50 + 1.6×40 + .8×0 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×50 + 1.6×0 + .5×0 + .5×40 1.2 D + 1.0 E + .5 L + .2S = 1.2×50 + 0 + .5×0 + .2×40 .9 D + 1.6W = .9×50 – 1.6×18 .9 D + 1.0 E = .9×50 – 0

= =

70.0 80.0 = 124 = 80.0 = 68.0 = 16.2 = 45.0

No te: Te nsi le fo rce s are considered po sitive. The controlling load combination is LC-3 (Roof loading). Factored axial tensile load, T u = 124 kips (Ans) There is no reversal of sign in load combinations LC-6 and LC-7, indicating that the member is not subjected to any compression under factored loads.

Comments: It should be noted that the compressive force contributed by wind loading was not included in LC-3 or LC-4, since it did not contribute to an increase in the member axial load.

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Chapter 4

P4.9.

page 4-12

A beam is part of the framing system for the floor of a residential building. The end moments caused by the service loads are: 160 ft-kips (clockwise) from dead load, 23 0 ft-kips (clockwise) from live load, and 15 0 ftkips (clockwise or anticlockwise) from wind load. Determine the maximum factored bending mom ent. What is the controlling ASCES load combin ation? Also, if the resistance factor is 0.9, what is the required nominal bending strength in ft-kips? Solution Given: Beam in a residential floor framing system D = 160 ft-kips (CW) L = 230 ft-kips (CW) ; 0.5 L = 115 ft-kips W = 150 ft-kips (CW or CCW) ; 0.8W = 120 ft-kips

Combination

Factored Bending Moment (ft-kips)

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

1.4 D = 1.4×160 1.2 D + 1.6 L + .5 L rSR = 1.2×160 + 1.6×230 + .5×0 1.2 D + 1.6 L rSR + 0.8W = 1.2×160 + 1.6×0 + .8×150 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×160 + 1.6×150 + .5×230 + .5×0 1.2 D + 1.0 E + .5 L + .2S = 1.2×160 + 0 + .5×230 + .2×0 .9 D + 1.6W = .9×160 – 1.6×150 .9 D + 1.0 E = .9×160 – 0

= 224 = 560 = 312 = 547 = 307 = - 96.0 = 144

No te: Clo ckw ise mo ments are c on sid ere d p osi tive . The controlling load combination is LC-2 (Live load). Factored b ending moment, M u = 560 ft-kips (clockwise) Design bending strength, M d M req = M u = 560 ft-kips No min al b endin g st ren gth , M n = M d / 560 /.9 = 622 ft-kips

(Ans) (Ans)

As there is a reversal of sign in load combination LC-6, the member (and the end connections) should also be designed for a factored anticlockwise bending moment, M u = 96 ft-kips. (Ans)

The main design consideration is the clockwise bending. While the counterclockwise Comments: loading is small, it should still be noted .

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Steel Structures by Vinnakota

P4.10.

Chapter 4

page 4-13

The maximum mom ents caused by the service loads on the roof beam of an office building are as follows: dead load, 58 ft-kips; snow load, 75 ft-kips; roof live load, 50 ft-kips; rain load, 25 ft-kips; and win d load 30 ft-kips. All these momen ts occur at the same location on the beam (center) and can therefore be comb ined. Determine the required bending strength of the roof beam. Solution Given: Office building roof beam D = 58 ft-kips S = 75 ft-kips; L r = 50 ft-kips; R = 25 ft-kips W = - 30 ft-kips

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

Factored Bending Moment (ft-kips)

1.4 D = 1.4×58 1.2 D + 1.6 L + .5 L rSR = 1.2×58 + 1.6×0 + .5×75 1.2 D + 1.6 L rSR + .8W = 1.2×58 + 1 .6×75 + .8×0 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×58 + 1.6×0 + .5×0 + .5×75 1.2D + 1.0 E + .5 L + .2S = 1.2×58 + 0 + .5×0 + .2×75 .9 D + 1.6W = .9×58 – 1.6×30 .9 D + 1.0 E = .9×58 – 0

= = = = = = =

81.2 107 190 107 84.6 4.20 52.2

Factored bending moment, M u = 190 ft-kips

The main design consideration is the positive bending moment. There is no reversal of Comments: moment to be considered, under the given loading condition.

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Chapter 4

P4.11.

page 4-14

A beam-colum n is subjected to the following forces by the service loads indicated. Axial compressio n, P : dead load, 120 kips; live load, 220 kips. Bendin g moments, M : dead load 180 ft-kips, live load 24 0 ft-kips. Determine the factored axial load, Pu , and bending moment, M u , for which the beam-column is to be designed. Solution G iv en : B eam -c olu mn Axial compressive loads D = 120 kips L = 220 kips

Bending moments D = 180 ft-kips L = 240 ft-kips

a. Combination

LC-1 LC-2

Factored Axial Force (kips)

1.4 D = 1.4×120 1.2 D + 1.6 L + .5 L rSR = 1.2×120 + 1.6×220 + .5×0

= 168 = 496

b. Combination

LC-1 LC-2

Factored Bending Moment (ft-kips)

1.4 D = 1.4×180 1.2 D + 1.6 L + .5 L rSR = 1.2×180 + 1.6×240 + .5×0

= 252 = 600

c. Combination

Factored Axial force (kips)

Factored Bending Moment (ft-kips)

LC-1 LC-2

16 8 496

25 2 600

Factored axial load, P u = 496 kips Factored b ending moment, M u = 600 ft-kips No te th at b oth max imu m axial forc e an d m axim um bend ing mo me nt o ccu r for load c om bination LC-2, in this example. Comments:

This problem is good because it shows the analysis of members subject to more than one

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Steel Structures by Vinnakota

Chapter 4

page 4-15

type of load. In this example, the same combination prod uces maximum axial force and maximum momen t. This is not necessarily true in all cases. The main design con sideration is the axial compression and the positive bendin g. Also there are no load reversals to consider und er the given loading condition.

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Chapter 4

page 4-16

Additional problem not in Text Book P4.12.

Repeat Problem 4.7 if the wind load is always a suction (- 18 psf). Solution G iv en : R oo f d ec kin g D = 30 psf Roof Live Load, L r = 20 psf ; W = 18 psf

Combination

LC-1 LC-2 LC-3 LC-4 LC-5 LC-6 LC-7

S = 35 psf

Factored Loading (psf)

1.4 D = 1.4×30 1.2 D + 1.6 L + .5 L rSR = 1.2×30 + 1.6×0 + .5×35 1.2 D + 1.6 L rSR + .8W = 1.2×30 + 1.6×35 + .8×0 1.2 D + 1.6W + .5 L + .5 L rSR = 1.2×30 + 1.6×0 + .5×0 + .5×35 1.2 D + 1.0 E + .5 L + .2S = 1.2×30 + 0 + .5×0 + .2×35 .9 D + 1.6W = .9×30 – 1.6×18 .9 D + 1.0 E = .9×30 – 0

= = = = = = =

42.0 53.5 92.0 53.5 43.0 - 1.8 27.0

No te: M inu s sig n sign ifies suctio n ( up lift) on the decking . The controlling load combination is again LC-3 (Roof load). Factored load (pressure) on the decking, p u = 92 psf As there is a reversal of sign, in load com bination LC -6, the decking (and the welds or nails connec ting the decking to the suppo rts), should be designed for a suction of 1.8 psf.

(Ans) (Ans)

When wind load is always a suction (negative), the wind load is not included in load Comments: combin ation LC-3 (taken as 0). The main design consideration is the LC-3 while uplift from LC-6 is small, it should still be noted.

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