An axial load P is applied to the 32-mm-square aluminum

Step 1 of 3

Given data:

The side of a square aluminum bar is given as, \(b = h = 32\;{\rm{mm}} = 32 \times {10^{ - 3}}\;{\rm{m}}\).

Horizontal deflection is given as, \({y_{\max }} = 4\;{\rm{mm}} = 4 \times {10^{ - 3}}\;{\rm{m}}\).

Young's modulus is given as, \(E = 70\;{\rm{GPa}}\).

Allowable load is given as, \(P = 24\;{\rm{kN}}\)

The length is given as, \(L = 0.65\;{\rm{m}}\).

Step 2 of 3

(a)

Horizontal deflection can be determined as,

\({y_{\max }} = e\left( {\sec \left( {\frac{\pi }{2}\sqrt {\frac{{{{\left( {2L} \right)}^2}P}}{{{\pi ^2}E\left( {\frac{{b{h^3}}}{{12}}} \right)}}} } \right) - 1} \right) \)

Here, e is the eccentricity.

Substitute the values in the above formu