Chapter 1: Q30E (page 360)

A vertical, solid steel post 25 cm in diameter and 2.50 m long is required to support a load of 8000kg. 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?

Short Answer

Expert verified

(a) $- 1.60 \times 10^{6} Pa$

(b) $- 8 \times 10^{- 6}$

Step by step solution

Given information

Length $l_{o} = 2.50 m$ , diameter $= 2.5 \times 10^{- 2} m$ , Load $m = 8000 k g$

Concept/Formula used

$Y = \frac{l_{0} F}{A ∆ l}$

Where, Y is Young’s modulus, l₀ is length of muscle, F is muscle force, A is cross-sectional area and $∆ l$ is elongation.

Cross –sectional area of the post

$A = ττ (\frac{d^{2}}{4}) = ττ (\frac{{(2.50 \times 10^{- 2} m)}^{2}}{4}) = 0.05 m^{2}$

Force application on the post

$\begin{array}{l} F = m g \\ = (8000 kg) (9.80 m I s^{2}) \\ = 7.84 \times 10^{4} N \end{array}$

Young’s Modulus for steel is $Y = 2.0 \times 10^{11} Pa$

Stress in the post

(a)

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

Note: negative sign shows that the stress is compressive.

Strain in the post

(b)

$stress = \frac{stress}{Y} = - \frac{1.60 \times 10^{6} Pa}{2 \times 10^{11} Pa} = - 8 \times 10^{6}$

Note: negative sign shows that the length decreases.

Change in the post length

(C)

$∆ l = l_{0} (strain) = (2.5 m) (- 8 \times 10^{- 6}) = - 2.0 \times 10^{- 5} m$

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A vertical, solid steel post 25 cm in diameter and 2.50 m long is required to support a load of 8000kg. 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?

Short Answer

Step by step solution

Given information

Concept/Formula used

Cross –sectional area of the post

Force application on the post

Stress in the post

Strain in the post

Change in the post length

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