Q11-30E - University Physics with Modern Physics

Question

A vertical, solid steel post 25 cm in diameter and 2.50 m long is required to support a load of 8000 Kg. You can ignore the weight of the post. What are (a) the stress in the post; (b) the strain in the post; and (c) the change in the post’s length when the load is applied?

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right) -1.60\times {10}^{6}Pa \\ \left(b\right) -8\times {10}^{-6} \\ \left(c\right) -2.0\times {10}^{-5}m \end{gathered}$$

Step 1: Given information

Length I₀ = 2.50m, diameter d = 2.5x10^-2m, Load m = 8000kg

Step 2: Concept/Formula used

$\mathit{Y}\mathbf{=}\frac{{\mathbf{l}}_{\mathbf{0}}\mathbf{F}}{\mathbf{A\Delta l}}$

Where, Y is Young’s modulus, I₀ is length of muscle, F is muscle force, A is cross-sectional area and $\mathbf{∆}\mathit{l}$ is elongation.

Step 3: Cross –sectional area of the post

$\begin{array}{rcl}A& =& \pi \left(\frac{d^2}{4}\right)\\ & =& \pi \left(\frac{{\left(2.50\times {10}^{-2}m\right)}^2}{4}\right)\\ & =& 0.05m^2\end{array}$

Step 4: Force application on the post

$$\begin{gathered} F=mg \\ =\left(8000kg\right)\left(9.80m/s^2\right) \\ =7.84\times {10}^{4}N \\ Young’s Modulus for steel is Y=2.0\times {10}^{11}Pa \end{gathered}$$

Step 5: Stress in the post

(a)

$$\begin{gathered} stress=\frac{F}{A} \\ =-\frac{7.84\times {10}^{4}N}{0.05m^2} \\ =-1.60\times {10}^{6}Pa \end{gathered}$$

Note: negative sign shows that the stress is compressive.

Step 6: Strain in the post

(b)

$\begin{array}{rcl}strain& =& \frac{stress}{Y}\\ & =& -\frac{1.60\times {10}^{6}Pa}{2\times {10}^{11}Pa}\\ & =& -8\times {10}^{-6}\end{array}$

Note: negative sign shows that the length decreases.

Step 7: Change in the post length

(C) 

$\begin{array}{rcl}\Delta l& =& l_0\left(strain\right)\\ & =& \left(2.5m\right)\left(-8\times {10}^{-6}\right)\\ & =& -2.0\times {10}^{-5}m\end{array}$