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Civil PESampleExaminationThird Edition

Michael R. Lindeburg, PE

Professional Publications, Inc. • Belmont, California

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Afternoon SessionStructural

121. A two-lane highway bridge is constructed usingprecast concrete girders. The girders are simply sup-ported and span 60 ft. The weight of girders and deck issuch that the dead load bending moment at the criticallocation for bending moment is 500 ft-kips. If thebridge is designed for AASHTO HL-93 loading usingLRFD strength criteria, and only one lane is loaded ata time, and if girder load distribution is not considered,the design bending moment per lane at the criticallocation is most nearly

(A) 2500 ft-kips

(B) 3000 ft-kips

(C) 3500 ft-kips

(D) 4000 ft-kips

122. A rigid diaphragm transfers a lateral wind forceof 0.4 kip/ft into a system of shear walls whose relativerigidities, in multiples of R, against forces in the northdirection are shown in the plan. The force in wall A ofthe system is most nearly

wall A(4R)

wall D(3R)

wall B(R)

wall C(R)

wall E(3R)

40 ft120 ft

60 ft

0.4 kip/ft

N

(A) 15 kips

(B) 22 kips

(C) 27 kips

(D) 33 kips

123. The roof framing of a single story commercialbuilding consists of wood joists supported by timberbeams and sheathed with a properly nailed and blockedplywood diaphragm. Seismic lateral forces for NS groundmotion are shown. Assume sufficiently rigid plywoodshear walls 14 ft high and 24 ft long are constructed at

lines 1, 2, and 3. Disregard accidental torsion that may berequired by code. The axial compression and tensionforces in the shear wall boundary members at line 2 underthe given loadings are most nearly

240 lbf/ft

100 ft1 2 3

24 ft

54 ft

300 lbf/ft

60 ft

plywood diaphragm

N

(A) 10 kips

(B) 12 kips

(C) 16 kips

(D) 20 kips

124. For the truss shown, the modulus of elasticity forall members is 29,000 ksi. The cross-sectional area of themembers is 8 in2. The horizontal deflection at joint D ofthe truss is most nearly

20 ft

10 ft

15 ft

D

A B

C

30 kips

(A) 0.01 in

(B) 0.02 in

(C) 0.04 in

(D) 0.08 in

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125. A two-story building is 14 ft from ground tosecond floor and 12 ft from second floor to roof. Theexterior wall projects 3 ft above the roof level to createa parapet. The exterior wall weighs 15 psf, the secondfloor dead load is 30 psf, and the roof dead load is20 psf. The building is wood framed with plywooddiaphragms and shear walls resisting all lateral forces.The building is situated in seismic performance cate-gory D, where the design spectral response accelera-tion at short periods is 0.6, the design spectralresponse acceleration at one second period is 0.2, andthe importance factor for seismic response is 1.0.Assume the building qualifies as a building frame sys-tem with light-frame walls with shear panels. Theseismic base shear for north-south (NS) ground motionby the IBC static force procedure, on a working loadbasis, is most nearly

120 ft1 2

60 ft

plywood diaphragm

N

(A) 25 kips

(B) 45 kips

(C) 65 kips

(D) 80 kips

126. The compound beam shown has an internalhinge (M=0) at point B and is simply supported onhinges or rollers at points A, C, and E. The ordinate ofthe influence line for the bending moment at point D,which is 12 ft to the right of support C, is most nearly

8 ft 8 ft 12 ft 8 ft 5 ft

A B C D E

(A) 1 ft-kip/kip

(B) 3 ft-kips/kip

(C) 5 ft-kips/kip

(D) 7 ft-kips/kip

127. A continuous 8 in thick bridge deck is made ofreinforced normal weight concrete. It is supported bysteel girders spaced 8.5 ft on center, with flangewidths of 1 ft. The positive bending moment, per footof width, for dead weight of the slab is 1.0 ft-kip/ft,and is 0.3 ft-kip/ft for a future wearing course. Thedeck is continuous over three or more spans and is tobe designed by the traditional approach using theAASHTO LRFD Bridge Design Specifications. The fac-tored positive bending moment per foot of deck widththat controls deck strength is most nearly

8.5 ft1 ft (typ.)

8 in

(A) 8 ft-kips/ft

(B) 10 ft-kips/ft

(C) 12 ft-kips/ft

(D) 14 ft-kips/ft

128. The circular shaft shown is subjected to an axialtension force P at its free end and a compressive force of50 kips at point B. Note that the shaft is hollow betweenpoints A and B. The allowable normal tension stress is22 ksi, the modulus of elasticity is 29,000 ksi, and themaximum allowable elongation is 0.04 in. The maximumallowable value of P is most nearly

3 in

0.75 in

50 kips

20 in 30 inP

A BC

(A) 111 kips

(B) 117 kips

(C) 155 kips

(D) 171 kips

129. A beam is simply supported over a 22 ft span andoverhangs the left support 8 ft. Uniformly distributeddead loading of 2 kips/ft and live loading of 3 kips/ftare applied. The live load is positioned to produce

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34 C I V I L P E S A M P L E E X A M I N A T I O N

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SolutionsAfternoon SessionStructural

121. The HL-93 consists of an HS20-44 design truckcombined with a 640 lbf/ft lane load. The resultant ofthe three wheel loads for an HS20-44 loading is a 72 kipsforce located 4.67 ft from the 32 kips center force.

x ¼ åPixiR

¼ ð32 kipsÞð14 ftÞ þ ð8 kipsÞð28 ftÞ72 kips

¼ 9:33 ft

Maximum wheel-load bending moment occurs when themidspan lies halfway between the resultant and thecentral 32 kips force. Thus, the position for maximumwheel-load bending moment is

A B

8 kips32 kips

(0.5)(4.67 ft)

14 ft 14 ft

30 ft

9.33 ft

R

RA RB

32 kips

60 ft

Maximum wheel-load bending moment from the truckoccurs under the 32 kips load to the right of midspan.

RA ¼ årF

L

¼ð72 kipsÞ

�30 ftþ ð0:5Þð4:67 ftÞ

�60 ft

¼ 38:8 kips

M truck ¼årF ¼ ð38:8 kipsÞ�30 ftþ ð0:5Þð4:67 ftÞ

�

� ð32 kipsÞð14 ftÞ¼ 807 ft-kips

The maximum moment due to the lane loading is

MLL ¼ wL2

8¼

0:640kip

ft

� �ð60 ftÞ2

8

¼ 288 ft-kips

The AASHTO specification requires an increase in thewheel-load bending moment to account for dynamicloading, but this is not applied to the dead or laneloading. A multiple presence factor, MPF, is requiredwhen only one lane is loaded.

IM ðdynamic load allowanceÞ ¼ 0:33

MPF ðmultiple presence factorÞ ¼ 1:2

DC ðcomponent dead load factorÞ ¼ 1:25

LL ðlive load factorÞ ¼ 1:75

Mu ¼ ðDCÞMD þ ðLLÞðMPFÞ��MLL þ ð1þ IMÞM truck

�

¼ ð1:25Þð500 ft-kipsÞ þ ð1:75Þð1:2Þ� �288 ft-kipsþ ð1þ 0:33Þð807 ft-kipsÞ�

¼ 3483:8 ft-kips ð3500 ft-kipsÞ

The answer is (C).

122. The resultant lateral force is

V ¼ wL ¼ 0:4kip

ft

� �ð160 ftÞ

¼ 64 kips

This resultant force acts 80 ft from the west wall. Thecenter of rigidity of the wall group is

x ¼ åRixi

åRi

¼ 4Rð0 ftÞ þ 3Rð120 ftÞ þ 3Rð160 ftÞ4Rþ 3Rþ 3R

¼ 84 ft ½from the west side of wall A�

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105

(This disregards the accidental torsion of 5% that maybe required by code or ASCE7.)

From symmetry,

y ¼ 30 ft ½from the south wall�

The wall system is subjected to a torsional moment of

Mt ¼ V x � L2

� �

¼ ð64 kipsÞ 84 ft� 160 ft2

� �

¼ 256 ft-kips ½clockwise�

The polar moment of inertia for the walls resisting thetorsional moment is

J ¼åðRyix2i þ Rxiy

2i Þ

¼ 4Rð�84 ftÞ2 þ 3Rð120 ft� 84 ftÞ2

þ 3Rð160 ft� 84 ftÞ2

þ Rð30 ftÞ2 þ Rð30 ftÞ2

¼ 51;240R ft2

The maximum lateral force resisted by wall A is thecombined direct force plus the force caused by thetorsional moment, both acting in the same sense.

VA ¼ 4R

åRyiV þMtRixi

J

¼ 4R10R

� �ð64 kipsÞ þ ð256 ft-kipsÞ4Rð84 ftÞ

51;240R ft2

¼ 27:3 kips ð27 kipsÞ

The answer is (C).

123. The plywood diaphragm is considered flexible,and the lateral forces transfer to the shear wall on thebasis of their tributary width. Thus, the lateral forceacting on the shear wall at line 2 is

V ¼åwB

¼ 240lbfft

� �100 ft

2

� �þ 300

lbfft

� �60 ft2

� �

¼ 21;000 lbf ð21 kipsÞ

T

C

V = 21 kips

h = 14 ft

L = 24 ft

elevation of wall on line B

The overturning moment on the wall is

MOT ¼ Vh

¼ ð21 kipsÞð14 ftÞ¼ 294 ft-kips

The axial force in the shear wall boundary members is

T ¼ C ¼ MOT

L

¼ 294 ft-kips

24 ft

¼ 12:3 kips ð12 kipsÞThe answer is (B).

124. Using the dummy load method, the unit virtualforce is applied at D in the direction of the requireddeflection.

20 ft

10 ft

15 ft 15 ft

20 ft

D D

A B A B

C C

10 ft

30 kips

1 lbf

load system P load system Q

The member forces for the real loads, load system P, andfor the dummy load system, system Q, are found usingbasic statics.

memberNP

(kips)NQ

(lbf)L(in)

NPNQL(kips-lbf-in)

AB 15.0 1.0 180 2700AC 25.0 0 150 0AD 0 1.33 240 0BC �25.0 –1.67 150 6263CD 0 –1.67 150 0

8963

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106 C I V I L P E S A M P L E E X A M I N A T I O N