Question

I have a solid square aluminum beam. It is 12 inches long with cross sectional area .25" x .25". A force is applied to the center of the rod and supported at the ends. I need to know the maximum force before the aluminum fails. Equations and work done would be great so I can change the length and cross section if needed bigger or smaller max load.

Answer 1

Assuming the solid aluminum square section beam is simply supported at ends and carries a point load of P (lb). Also let fy ( psi) be maximum permissible tensile /compressive stress in bending which would also be equal to yeild strength of aluminum and depends upon grade of aluminum. Also let Ty (psi) be maximum permissible shear stress.

Now due to loading , maximum shear force will occur at the supports and would be = P/2 ( because of symmetry the vertical support reactions would be equal )

However maximum shear stress at supports will occur at neutral axis and = 1.5*(P/2) /( b*d) where b and d is the cross sectional dimensions of width and depth respectively.

Thus maximum shear stress = 1.5*(P/2)/(0.25*0.25) = 12P (psi)

Maximum bending moment will occur at mid span and = PL/4 = 3P lb-in

For square cross section , section modulus =Z =( b^3)/6 = 0.25*0.25*0.25 /6

Thus maximum stress due to moment = M/Z = 3P *6 / (0.25*0.25*0.25) = 1152 P (psi)

Now there are two possible failures , by shearing or by elastic bending thus maximum force P for shear failure = Ty/12 (lb )

Similarly maximum force P for elastic bending failure = fy/1152

Thus maximum force P that can be applied before failure of beam = minimum of Ty/12 and fy/1152. Subsituting values of Ty and fy ( parameters as explained above) we can get maximum force P .