Chapter 2, Problem 2.6.21P

-21 Plastic bar AB of rectangular cross section (6 = 0.75 in. and h = 1.5 in.) and length L = 2 Ft is Fixed at A and has a spring support (Ar = 18 kips/in.) at C (see figure). Initially, the bar and spring have no stress. When the temperature of the bar is raised hy foot. the compressive stress on an inclined plane pq at Lq = 1.5 Ft becomes 950 psi. Assume the spring is massless and is unaffected by the temperature change. Let a = 55 × l0-⁶p and E = 400 ksi.

(a) What is the shear stresst₉ on plane pq? What is angle 07 =1 Draw a stress element oriented to plane pq, and show the stresses acting on all laces of this element.

(c) If the allowable normal stress is ± 1000 psi and the allowable shear stress is ±560 psi, what is the maximum permissible value of spring constant k if the allowable stress values in the bar are not to be exceeded?

(d) What is the maximum permissible length L of the bar if the allowable stress values in the bar are not be exceeded? (Assume £ = IB kips/in.)

(e) What is the maximum permissible temperature increase (A7") in the bar if the allowable stress values in the bar are not to be exceeded? (Assume L = 2 ft and k = L& kips/in

To determine

The angle $\theta$at which bar carries maximum load is = $30.96°$.

Answer to Problem 2.6.21P

The angle $\theta$at which bar carries maximum load is = $30.96°$.

Explanation of Solution

The following figure shows the reaction force on the bar:

  

       Figure-(1)

Write the area of rectangular cross-section of the bar.

  $A=bh$....... (I)

Here, the width of the rectangle section is $b$, the height is $h$.

Write the expression for the elongation of the bar.

  $\delta =\alpha \text{Δ}TL$....... (II)

Here, the thermal coefficient is $\alpha$, change in temperature is $\text{Δ}T$, and length of the bar is $L$.

Write the expression for the elongation of the bar.

  $\delta =\frac{R_{C}L}{EA}+\frac{1}{k}$....... (III)

Here, the reaction at support C is $R_C$, area of the cross-section of the bar is $A$, and the stiffness of the spring is $k$.

Substitute $\alpha \text{Δ}TL$for $\delta$in Equation (III).

  $\begin{array}{l}\alpha \text{Δ}TL=-\left(R_C\right)\left(\frac{L}{EA}+\frac{1}{k}\right)\\ R_C=\frac{-\alpha \text{Δ}TL}{\left(\frac{L}{EA}+\frac{1}{k}\right)}\end{array}$....... (IV)

Write the expression for the axial stress acting on the element at angle $\theta$.

  ${\sigma }_{x^{\prime }}=\frac{R_C}{A}{\cos }^2\theta$....... (V)

Write the expression for the shear stress acting on the element at angle $\theta$.

  ${\tau }_{x^{\prime }{y}^{\prime }}=-\frac{R_C}{A}\sin \theta \cos \theta$....... (VI)

Write the expression for the normal stress in y-direction.

  ${\sigma }_{y^{\prime }}=\frac{R_C}{A}\left[{\cos }^2\left(\theta +90°\right)\right]$....... (V)

Calculation:

Substitute $0.75\text{in}$for $b$, $1.5\text{in}$for $h$in Equation (I).

  $\begin{array}{c}A=\left(0.75\text{in}\right)\left(1.5\text{in}\right)\\ =1.125{\text{in}}^2\end{array}$

Substitute $55\times {10}^{-6}/°\text{F}$for $\alpha$, $100°\text{F}$for $\text{Δ}T$, $2\text{ft}$for $L$, $1.125{\text{in}}^2$for $A$, $18000\text{lbf}/\text{in}$for $k$, and $400\text{ksi}$for $E$in Equation (IV).

  $\begin{array}{c}{R}_C=\frac{-\left(55\times {10}^{-6}/°\text{F}\right)\left(100°\text{F}\right)\left(2\text{ft}\right)}{\left(\frac{2\text{ft}}{\left(400\text{ksi}\right)\left(1.125{\text{in}}^2\right)}+\frac{1}{\left(18000\text{lbf}/\text{in}\right)}\right)}\\ =\frac{-\left(55\times {10}^{-4}\right)\left(2\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{\left(\frac{\left(2\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{\left(400\text{ksi}\right)\left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)\left(1.125{\text{in}}^2\right)}+\frac{1}{\left(18000\text{lbf}/\text{in}\right)}\right)}\\ =-1212.25\text{lb}\end{array}$

Substitute $-950\text{psi}$for ${\sigma }_{x^{\prime }}$, $-1212.25\text{lb}$for $R_C$, $1.125{\text{in}}^2$for $A$in Equation (V).

  $\begin{array}{l}-950\text{psi}=\frac{-1212.25\text{lb}}{1.125{\text{in}}^2}{\cos }^2\theta \\ 950\text{psi}=\left(1077.55\text{psi}\right){\cos }^2\theta \\ \theta =20.12°\end{array}$

Substitute $20.12°$for $\theta$, $-1212.25\text{lb}$for $R_C$, $1.125{\text{in}}^2$for $A$in Equation (VI).

  $\begin{array}{c}{\tau }_{x^{\prime }{y}^{\prime }}=-\frac{-1212.25\text{lb}}{1.125{\text{in}}^2}\sin \left(20.12°\right)\cos \left(20.12°\right)\\ =\frac{1212.25\text{lb}}{1.125{\text{in}}^2}\left(0.323\right)\\ =348.1\text{psi}\end{array}$

Substitute $20.12°$for $\theta$, $-1212.25\text{lb}$for $R_C$, $1.125{\text{in}}^2$for $A$in Equation (VI).

  $\begin{array}{c}{\sigma }_{y^{\prime }}=\left(\frac{-1212.25\text{lb}}{1.125{\text{in}}^2}\right)\left[{\cos }^2\left(90°+20.12°\right)\right]\\ =\left(\frac{-1212.25\text{lb}}{1.125{\text{in}}^2}\right)\left[0.118\right]\\ \approx -127.5\text{psi}\end{array}$

Conclusion:

The angle $\theta$at which bar carries maximum load is = $30.96°$.

To determine

The sketch of stresses acting on the element.

Explanation of Solution

The sketch of the stresses acting on the element is shown below:

  

       Figure-(2)

To determine

The maximum permissible value of spring constant.

Answer to Problem 2.6.21P

The maximum permissible value of spring constant is = $-15625\text{lbf}/\text{in}$.

Explanation of Solution

Write the expression for the maximum axial force.

  $P_n={\sigma }_{1}A$....... (VIII)

Calculation:

Substitute $1.125{\text{in}}^2$for $A$, and $1000\text{psi}$for ${\sigma }_1$in Equation (VIII).

  $\begin{array}{c}{P}_n=\left(1000\text{psi}\right)\left(1.125{\text{in}}^2\right)\\ =1125\text{lb}\end{array}$

The maximum axial force will be equal to the reaction at C.

Substitute $55\times {10}^{-6}/°\text{F}$for $\alpha$, $100°\text{F}$for $\text{Δ}T$, $2\text{ft}$for $L$, $1.125{\text{in}}^2$for $A$, $1125\text{lb}$for $R_C$, and $400\text{ksi}$for $E$in Equation (IV).

  $\begin{array}{l}1125\text{lb}=\frac{-\left(55\times {10}^{-6}/°\text{F}\right)\left(100°\text{F}\right)\left(2\text{ft}\right)}{\left(\frac{2\text{ft}}{\left(400\text{ksi}\right)\left(1.125{\text{in}}^2\right)}+\frac{1}{k}\right)}\\ 1125\text{lb}=\frac{-\left(55\times {10}^{-4}\right)\left(2\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{\left(\frac{\left(2\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{\left(400\text{ksi}\right)\left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)\left(1.125{\text{in}}^2\right)}+\frac{1}{k}\right)}\\ k=-15625\text{lbf}/\text{in}\end{array}$

Conclusion:

The maximum permissible value of spring constant is = $-15625\text{lbf}/\text{in}$.

To determine

The maximum permissible length of the bar.

Answer to Problem 2.6.21P

The maximum permissible length of the bar is = $20.835\text{in}$.

Explanation of Solution

Calculation:

Substitute $55\times {10}^{-6}/°\text{F}$for $\alpha$, $100°\text{F}$for $\text{Δ}T$, $18000\text{lbf}/\text{in}$for $k$, $1.125{\text{in}}^2$for $A$, $1125\text{lb}$for $R_C$, and $400\text{ksi}$for $E$in Equation (IV).

  $\begin{array}{l}1125\text{lb}=\frac{-\left(55\times {10}^{-6}/°\text{F}\right)\left(100°\text{F}\right)\left(L\right)}{\left(\frac{L}{\left(400\text{ksi}\right)\left(1.125{\text{in}}^2\right)}+\frac{1}{\left(18000\text{lbf}/\text{in}\right)}\right)}\\ 1125\text{lb}=\frac{-\left(55\times {10}^{-4}\right)L}{\left(\frac{L}{\left(400\text{ksi}\right)\left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)\left(1.125{\text{in}}^2\right)}+\frac{1}{18000\text{lbf}/\text{in}}\right)}\\ L=20.835\text{in}\end{array}$

Conclusion:

The maximum permissible length of the bar is = $20.835\text{in}$.

To determine

The maximum permissible temperature change.

Answer to Problem 2.6.21P

The maximum permissible temperature change is $-92.80°\text{F}$.

Explanation of Solution

Calculation:

Substitute $55\times {10}^{-6}/°\text{F}$for $\alpha$, $2\text{ft}$for $L$, $18000\text{lbf}/\text{in}$for $k$, $1.125{\text{in}}^2$for $A$, $1125\text{lb}$for $R_C$, and $400\text{ksi}$for $E$in Equation (IV).

  $\begin{array}{l}1125\text{lb}=\frac{-\left(55\times {10}^{-6}/°\text{F}\right)\left(\text{Δ}T\right)\left(2\text{ft}\right)}{\left(\frac{2\text{ft}}{\left(400\text{ksi}\right)\left(1.125{\text{in}}^2\right)}+\frac{1}{\left(18000\text{lbf}/\text{in}\right)}\right)}\\ 1125\text{lb}=\frac{-\left(55\times {10}^{-6}/°\text{F}\right)\left(\text{Δ}T\right)\left(2\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{\left(\frac{\left(2\text{ft}\right)\left(\frac{12\text{in}}{1\text{ft}}\right)}{\left(400\text{ksi}\right)\left(\frac{{10}^3\text{psi}}{1\text{ksi}}\right)\left(1.125{\text{in}}^2\right)}+\frac{1}{18000\text{lbf}/\text{in}}\right)}\\ \text{Δ}T=-92.80°\text{F}\end{array}$

Conclusion:

The maximum permissible temperature change is = $-92.80°\text{F}$.