Beam bending under a uniformly distributed load

[dave319]

Homework Statement

I have a bumper and I am trying to determine whether or not the rectangular tubing I am using to build it is strong enough to withstand a given load (rear end collision).

The horizontal member is 4x3x.1875 tubing (4" base, 3" high, 95" long). Two 3x3x.1975 tubes are used as supports with the center of these supports being 32" from each side. If it wasn't for the horizontal edge pieces hanging past the supports than it would fit the model for a beam that is fixed at both ends. This bumper must be designed to withstand 66,000 lbs force using a safety factor of 4 based on the tensile strength of the material used (SA 36: 58,000 psi).

Homework Equations

σ = 58000 psi/4 = 14,500 psi
M = wl^2/12 (maximum bending moment for a fixed-fixed beam under uniformly distributed load)
w = distributed load per longitudinal unit
l = 28 inches (distance between supports)
S = M/σ (elastic section modulus)

The Attempt at a Solution

Because the horizontal member stretches past the supports can I distribute the total load across the entire member? Example:

w = 66000 lb/95 in = 694.74 lb/in (load distributed across entire length)
M = 694.74 lb/in * 28^2 in^2 / 12 = 45389.7 in-lb (bending moment only concerned about length between the two supports)
S = 45389.7 in-lb/14500 lb = 3.13 in^3

Now calculate the elastic section modulus based upon the shape of the tubing where:
b = 3 in
d = 4 in
b1 = 2.625 in
d1 = 3.625 in

S = (bd^3-b1d1^3)/(6d) = 2.79 in^3

Since the actual S is less than the required S it seems that this member is not strong enough. Is this the correct procedure?
Thanks for your help!

 

[PhanthomJay]

if the 66000 pound force is uniformly distributed along the full 95 inch length, the controlling case for max moment is the cantilever moment of the overhanging part at the support, wl^2/2, where l is the length of the overhang (30.5"). But I don't want to get into bumper design for safety reasons, and other unknown factors.