Author Topic: 8/5/2014 Mechanics of Materials  (Read 2021 times)

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8/5/2014 Mechanics of Materials
« on: August 05, 2014, 07:39:49 PM »
Question:


A simply supported steel beam, with a span of 15 feet,  is loaded with a distributed force of 15,000 lb/ft .

Calculate the maximum beam deflection (inches): Assume I = 254 ft4

FastTrack Guru

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Re: 8/5/2014 Mechanics of Materials
« Reply #1 on: August 11, 2014, 08:44:51 AM »
Refer to page 81 of the FE Handbook: "Simply Supported Beam Deflections and Slopes:"

Middle column for deflection:

Vmax = 5wL^4/ 384EI

Givens:

L = 15 ft, so L^4 = (15ft)^4 = 50,625 ft^4
w = 15,000 lb/ft
I = 254 ft4
E = The problem statement says a steel beam, therefore, Refer to page 80 of the FE Handbook to find the value of E for steel:

"Typical Material Properties:"
For steel, E = 29 Mpsi, or  29,000,000 psi

Therefore:
Vmax = 5wL^4/ 384EI
Vmax = [5(15,000 lb/ft) (50,625 ft4)] / [384(29 Mpsi) (254 ft^4)]
Vmax = [3,796,875,000 lb ft3)] / [2.8 x 1012   (lb/in2) ft4)] (144 in2/ft2)    (note: watch your units! we need to convert psi into psf)
Vmax = [3,796,875,000 lb ft3)] / [4 x 1014 (lb ft2)]
 Vmax = 9.3 x 10-6 ft    (note: the problem statement wants the answer to be in inches, not feet)
Vmax =  9.3 x 10-6 ft (12 in / 1 ft)
Vmax = .00001 in

The answer is  .00001 in.

andrei

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Re: 8/5/2014 Mechanics of Materials
« Reply #2 on: February 24, 2015, 08:04:28 AM »
I tried to solve this problem in inches, and the result
comes up 10 times greater.
I wonder if yo could let me know where am I wrong?

L = 15*12     # in
w = 15000 /12 # lb/in
I = 254*12^4  # in4
E = 29e6      # psi
Vmax = 5*w*L^4/(384*E*I) # = 0.00011