Chapter 13, Problem 24P

a.

To determine

To Find: The stress if the temperature is decreased by given amount.

Answer to Problem 24P

The stress if the temperature is decreased by $-{30}^0\text{C}$ is $1.4\times {10}^8{\text{N/m}}^{\text{2}}$

Explanation of Solution

Given:

Coefficient of linear expansion= $12\times {10}^{-6}{/}^0\text{C}$

Change in temperature= ${60}^0\text{C}$

Modulus of elasticity= $200\times {10}^9\text{}\text{N/m}$

Formula used:

  $\text{Stress}=\mathrm{\alpha }\mathrm{E}\mathrm{\Delta }\mathrm{T}$

Here,

Coefficient of linear expansion= $\mathrm{\alpha }$

Change in temperature= $\mathrm{\Delta }\mathrm{T}$

Modulus of elasticity= $\mathrm{E}$

Calculation:

Substituting the values,

  $\begin{array}{c}\mathrm{S}=\mathrm{\alpha }\mathrm{E}\mathrm{\Delta }\mathrm{T}\\ =(12\times {10}^{-6}/{}^0\text{C})(200\times 10{\text{N/m}}^2)(60{}^0\text{C})\\ =1.4\times {10}^8{\text{N/m}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the stress is $1.4\times {10}^8{\text{N/m}}^{\text{2}}$ .

b.

To determine

To identify: Whether the ultimate strength of the steel exceeds or not.

Answer to Problem 24P

The ultimate strength of steel will not exceed.

Explanation of Solution

Given:

Coefficient of linear expansion= $12\times {10}^{-6}{/}^0\text{C}$

Change in temperature= ${60}^0\text{C}$

Modulus of elasticity= $200\times {10}^9\text{}\text{N/m}$

The ultimate tensile strength of the steel is constant and its value is $5\times {10}^8{\text{N/m}}^{\text{2}}$ is fixed for steel.

Hence the ultimate steel is not exceeded.

c.

To determine

To Calculate: The stress of the concrete for the given cross-sectional area.

To identify: Whether it will fracture or not.

Answer to Problem 24P

  $2\times {10}^6{\text{N/m}}^{\text{2}}$

Yes, it will fracture.

Explanation of Solution

Given:

A= 0.13m²$\mathrm{\Delta }\mathrm{T}={60}^0\text{C}$

  $\mathrm{\alpha }$ = $12\times {10}^6{/}^0\text{C}$

E = $20\times {10}^9\text{N/m}$ .

Formula used:

  $\text{Stress}=\mathrm{\alpha }\mathrm{E}\mathrm{\Delta }\mathrm{T}$

Here,

Coefficient of linear expansion= $\mathrm{\alpha }$

Change in temperature= $\mathrm{\Delta }\mathrm{T}$

Modulus of elasticity= $\mathrm{E}$

Calculation:

Substituting the values,

  $\begin{array}{c}\mathrm{S}=\mathrm{\alpha }\mathrm{E}\mathrm{\Delta }\mathrm{T}\\ =(12\times {10}^6{/}^0\text{C})(20\times {10}^9{\text{N/m}}^2)(60{}^0\text{C})\\ =1.4\times {10}^7{\text{N/m}}^{\text{2}}\end{array}$

The ultimate tensile strength of the concrete is constant and its value is $2\times {10}^6{\text{N/m}}^{\text{2}}$ . And the calculated value for the concrete is $1.4\times {10}^7{\text{N/m}}^{\text{2}}$ . So there is fracture in concrete.