Question

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5.11-7 Select an A992 W-shape for beam AB of the floor system shown in Fig ure P5.11-7. In addition to the weight of the beam, the dead load consists of a 5-inch-thick reinforced concrete slab (normal-weight concrete). The live load is 80 psf, and there is a 20-psf partition load. The total deflection must not exceed L/240. a. Use LRFD b. Use ASD Problems 291 A 30 -H 4@5'=20 FIGURE P5.11-7

Answer 1

Tributary width of beam=5 ft

Dead weight of floor=(5/12)*150=62.5 psf

Umiform dead load on beam=62.5*5=312.5 plf=0.3125 kip/ft

Uniform live load on beam=(80+20)*5=500 plf=0.5 kip/ft

Total service level load on beam=0.3125+0.5=0.8125 kip/ft

Span of beam=30 ft

a)

Factored load on beam=1.2*0.3125+1.6*0.5=1.175 kip/ft

Factored momemt in beam=1.175*30²/8=132.2 kip-ft

Lightest section with this strength=W12x26

Moment of inertia=204 in⁴

Total service load=0.8125 kip/ft=0.0677 kip/in

Total deflection=(5/384)*0.0677*(30*12)⁴/(29000*204)=2.5 in

Allowable deflection=30*12/240=1.5 in

Minimum moment of inertia required=204*(2.5/1.5)=340 in⁴

Lightet section with moment of inertia greater than 340 is W16x26

Therefore section required is W16x26

It has LRFD strength of 166 kip-ft

b)

Deflection check is same as in LRFD

ASD strength of W16x26=110 kip-ft

ASD demand on beam=0.8125*30²/8=91.4 kip-ft

Therefore,W16x26 is required