Lecture 40
Example Problem
Beam Deflection
Given:
the 24' long simply supported beam with a 2k/ft load distributed over it's entire length and a 16k concentrated load at the midspan. The distributed load includes the self weight of the W21x62 beam.
E = 29,000 ksi
I = 1330 in^4
Determine:
the actual deflection of the beam and compare it with an allowable deflection of 1/360th of the span (l/360).
Solution:
The loading is a combination of a distributed load and a concentrated load. The deflection for each of these loading cases should be found independently and then combined to give a total deflection. Be careful with units since not all given information has consistent unit designation.
First determine the deflection based solely on the distributed load.
delta = 5wL^4/384 EI
= (5)(2kip/ft)(24^4ft)(12in/ft)^3/(384)(29,000ksi)(1330in^4)
= .387 in
Second, determine the deflection based solely on the concentrated load:
delta = PL^3/48 EI
= (16kips)(24^3ft)(12in/ft)^3/(48)(29,000ksi)(1330in^4)
= .206 in
Add these two deflections to determine the total deflection:
.387in + .206in = 0.593in
compare this to the allowable deflection:
span/360 = (24ft)(12in/ft)/360 = 0.8 in
0.593in < 0.8in - OK
Copyright © 1995 by Chris Luebkeman and Donald Peting