Chapter 6, Problem 1P

A column has both ends pinned and has a length of 32 in. It is made of SAE 1040 HR steel and has a circular shape with a diameter of 0.75 in. Determine the critical load.

To determine

The critical load that can be applied to a column of circular cross section whose both ends are pinned.

Answer to Problem 1P

The critical load of the column is given by 4490 lb

Explanation of Solution

Given Information:

Cross section of the column: Circular

Diameter of the column, D = 0.75 in

Length of the column, L = 32 in

End conditions: Both ends pinned

Material = SAE 1040 HR steel

So, Young’s Modulus, E = $30\times {10}^6\text{ psi}$

Yield strength, S_y= 42000 psi

Concept used:

For long Column Euler’s Formula used

  $P_{cr}=\frac{{\pi }^{2}E{I}_{\min }}{L_e^2}\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (1)}$

For short Column Euler’s Formula used

  $P_{cr}=A{s}_y\left[1-\frac{s_y{S}_r^2}{4{\pi }^{2}E}\right]\text{. }\text{. }\text{. }\text{. }\text{. (2)}$

Here,

P_cr is Critical load

E is Young’s Modulus

I_min is Minimum Moment of Inertia

L_e is effective length (Based on the end conditions of Column)

For both ends pinned condition, L_e = L

S_y is yield stress

S_r Slenderness ratio = $\frac{L_e}{r}$

r is radius of gyration

A cross section area

Calculation:

For solid round section,

r = D/4

= 0.75/4

= 0.188 in

Slenderness ratio,

  $\begin{array}{l}{S}_r=\frac{32}{0.188}\\ =171\end{array}$

Compute Column constant for the above material

  $\begin{array}{c}{C}_c=\sqrt{\frac{2{\pi }^{2}E}{S_y}}\\ =\sqrt{\frac{2{\pi }^2\left(30\times {10}^6\right)}{42000}}\\ =119\end{array}$

S_r>C_c, so Column is long, Use Euler’s Formula

Moment of Inertia,

  $\begin{array}{c}I=\frac{\pi D^4}{64}\\ =\frac{\pi \left({0.75}^4\right)}{64}\\ =0.01553m{m}^4\end{array}$

From Equation (1)

  $\begin{array}{c}{P}_{cr}=\frac{{\pi }^2\left(30\times {10}^6\right)\left(0.01553\right)}{{32}^2}\\ =4490lb\end{array}$

Thus, critical load is 4490 lb.