Student ID: December 13, 2014 Problem 3: (20 Points) Given: The beam shown in th
ID: 1714647 • Letter: S
Question
Student ID: December 13, 2014 Problem 3: (20 Points) Given: The beam shown in the figure supports uniformly distributed loads and is braced for lateral displacement and torsion at the supports and at mid-span only. Service loads are shown in the figure, however, dead load does not include the weight of the beam. The beam is ASTM A992 steel. 12.5 ft. 25 ft. Find: Answers: 1. The lightest W shape considering both flexure and shear strength using ASD 2. Maximum live load beam deflection (AL) 3. Maximum total beam deflection (AD+L) 4. Does the beam satisfy the maximum deflection requirements of 1 1/360 and do in in. 1240? Yes/No Method/Solution:Explanation / Answer
1)Unbraced length of beam = 12.5 ft
Maximum moment in beam due to dead load = 0.4*252/8=31.25 kip-ft
Maximum moment in beam due to live load = 1*252/8=78.1 kip-ft
Total moment in beam = 31.25+78.1= 109.4 kip-ft
The lightest beam that has ASD moment capapcity greater than 109.4 kip-ft for unbraced length of 12.5 ft is
Let us consder w14x30
moment capacity of w14x30 for unbraced length of 12.5 ft = 177-(177-110)/(14.9-5.26)*(12.5-5.26)=126.7 kip-ft
total dead load on beam = 0.4+0.03=0.43 klf
total live load on beam = 1 klf
2)moment of inertia of w14x30 = 299 in4
live load = 1 klf = (1/12)=0.0833 kli
span of beam = 25 ft=300 in
live load deflection = (5/384)*(0.0833*3004)/(29000*299)=1.01 in
3) Total load on beam = 0.43+1=1.41 klf=0.1175 kli
total deflection of beam = (5/384)*(0.1175*3004)/(29000*299)=1.42 in
4)live load deflection limit = L/360 = 300/360=0.833 in
total deflection limit = L/240=300/240=1.25 in
since live load deflection is greater than L/360 and total deflection is greater than L/240, the beam does not satisfy deflection limits