Question

For a 2" by 2" beam simply supported at the ends and loaded with a concentrated load at mid-span, determine the length so that failure will occur by shear and compression at the same load. Assume an allowable compressive stress of 12,000 psi and an allowable shear stress of 1,200 psi.

Answer 1

Length of beam = 10in

Step-by-step explanation:

We first calculate the maximum shear force.

`Max shear force =(V*A*y')/(Ib)`

`1200=(V(2*1)0.5)/((2*2^3)/(12)*2)`

Calculating for V, we have:

V = 3200Ib

P = 3200 *2

P = 6.4k

Let's now use the equation for the load for a length.

`M =(PL)/(4)`

`M =(6.4*L)/(4)`

Using the stress equation,

`(M)/(I)=(o)/(y)`

Inputting figures in the equation, we have:

`(6400*L)/(4*(2*2^3)/(12))=(12000)/(1)`

Solving for L, we have:

L = 10in

Length of beam will = 10in