Chapter 11, Problem 11.3.25P

An S6 × 12.5 steel cantilever beam AB is supported by a steel tic rod at B as shown. The tie rod is just taut when a roller support is added at Cat a distance s to the left of £, then the distributed load q is applied to beam segment AC, Assume E = 30 × 10⁶ psi and neglect the self-weight of the beam and tie rod. Sec Table F-2(a) in Appendix F for the properties of the S-shape beam.

(a)

What value of uniform load q will, if exceeded, result in buckling of the tie rod if L₁, =6 ft, s = 2 ft, H = 3 ft, and d = 0.25 in.?

(b)

What minimum beam moment of inertia i_bis required to prevent buckling of the tie rod if q = 200 lb/ft, L₁, = 6 ft, H = 3 ft, d = 0.25 in., and s = 2 ft?

(c)

For what distance s will the tic rod be just on the verge of buckling if q = 200 lb/ft, L₁= 6 ft, M = 3 ft, and d = 0.25 in.?

  

To determine

The value of uniform load $q$ .

Answer to Problem 11.3.25P

The value of uniform load $q$ is $\underset{\_}{1.483\text{lb}/\text{in}}$ .

Explanation of Solution

Given information:

The young’s modulus of beam and tie rod is $30\times {10}^6\text{psi}$ , the length of the beam is $6\text{ft}$ , the distance between point $C$ and $B$ is $2\text{ft}$ , the length of tie rod is $3\text{ft,}$ and the diameter of tie rod is $0.25\text{in}$ .

Write the expression for the deflection in beam at point $B$ due to uniform load only.

  ${\left({\delta }_B\right)}_1=\frac{q{\left(L_1-s\right)}^4}{24E{I}_b}$

Here, the uniformly distributed load on beam is $q$ , the length of the beam is $L_1$ , the distance between point $C$ and $B$ is $s$ , the young’s modulus of beam is $E,$ and the moment of inertia of beam is $I_b$ .

Write the expression for the force generated in the tie rod.

  $Q=\frac{{\pi }^{2}EI}{4H^2}$   ......(I)

Here, the length of the tie rod is $H$ and moment of inertia of tie rod is $I$ .

Write the expression for the deflection in beam at point $B$ due to force generated in tie rod.

  ${\left({\delta }_B\right)}_2=\frac{Q{s}^3}{3E{I}_b}+\frac{Qs\left(L_1-s\right)s}{3E{I}_b}$   ......(II)

Write the expression for the compression of length of the tie rod.

  ${\delta }_{rod}=\frac{QH}{E{A}_r}$   ......(III)

Here, the cross section area of tie rod is $A_r$ .

Write the expression for compatibility equation.

  ${\left({\delta }_B\right)}_1-{\left({\delta }_B\right)}_2={\delta }_{rod}$   ......(IV)

Substitute $\frac{q{\left(L_1-s\right)}^3}{24E{I}_b}$ for ${\left({\delta }_B\right)}_1$ in equation (IV).

  $\begin{array}{l}\frac{q{\left(L_1-s\right)}^4}{24E{I}_b}-{\left({\delta }_B\right)}_2={\delta }_{rod}\\ \frac{q{\left(L_1-s\right)}^4}{24E{I}_b}={\delta }_{rod}+{\left({\delta }_B\right)}_2\\ q=\frac{24E{I}_b}{{\left(L_1-s\right)}^4}\left({\delta }_{rod}+{\left({\delta }_B\right)}_2\right)\end{array}$   ......(V)

Write the expression for the moment of inertia of tie rod.

  $I=\frac{\pi }{64}{d}^4$   ......(VI)

Write the expression for the area of tie rod.

  $A_r=\frac{\pi }{4}{d}^2$   ......(VII)

Calculation:

Substitute $0.25\text{in}$ for $d$ in equation (VI).

  $\begin{array}{c}I=\frac{\pi }{64}{\left(0.25\text{in}\right)}^4\\ =\frac{\pi }{64}\left(3.90625\times {10}^{-3}\right){\text{in}}^4\\ =1.917475\times {10}^{-4}{\text{in}}^4\end{array}$

Substitute $0.25\text{in}$ for $d$ in equation (VII).

  $\begin{array}{l}{A}_r=\frac{\pi }{4}{\left(0.25\text{in}\right)}^2\\ =0.04908{\text{in}}^2\end{array}$

Substitute $30\times {10}^6\text{psi}$ for $E$ , $1.917475\times {10}^{-4}{\text{in}}^4$ for $I$ and $3\text{ft}$ for $H$ in equation (I).

  $\begin{array}{l}Q=\frac{{\pi }^2\times 30\times {10}^6\text{psi}\times 1.917475\times {10}^{-4}{\text{in}}^4}{4{\left(3\text{ft}\right)}^2}\\ =\frac{{\pi }^2\times 30\times {10}^6\text{psi}\times \left(\frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{1\text{psi}}\right)\times 1.917475\times {10}^{-4}{\text{in}}^4}{4{\left(3\text{ft}\times \left(\frac{12\text{in}}{1\text{ft}}\right)\right)}^2}\\ =\frac{{\pi }^{2}30\times {10}^6\text{lb}/{\text{in}}^{\text{2}}\times 1.917475\times {10}^{-4}{\text{in}}^4}{4\times 1296{\text{in}}^2}\\ =\frac{56774.1591}{5184}\text{lb}\end{array}$

  $=10.951\text{lb}$

Refer to table $\text{F-2}\left(\text{a}\right)$ “properties of $\text{S}$ -shape beam” in Appendix $\text{F}$ to obtain the value of moment of inertia $I_b$ of beam as $22{\text{in}}^4$

Substitute $10.951\text{lb}$ for $Q$ , $2\text{ft}$ for $s$ , $30\times {10}^6\text{psi}$ for $E$ , $22{\text{in}}^4$ for $I_b$ and $6\text{ft}$ for $L_1$ in equation (II).

  $\begin{array}{c}{\left({\delta }_B\right)}_2=\frac{10.951\text{lb}\times {\left(2\text{ft}\right)}^3}{3\left(30\times {10}^6\text{psi}\right)\left(22{\text{in}}^4\right)}+\frac{10.951\text{lb}\times 2\text{ft}\left(6\text{ft}-2\text{ft}\right)\times 2\text{ft}}{3\left(30\times {10}^6\text{psi}\right)\left(22{\text{in}}^4\right)}\\ =\left[\begin{array}{l}\frac{10.951\text{lb}\times {\left(2\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\right)}^3}{3\left(30\times {10}^6\text{psi}\left(\frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{\text{1}\text{psi}}\right)\right)\left(22{\text{in}}^4\right)}+\\ \frac{10.951\text{lb}\times 2\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\left(6\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)-2\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\right)\times 2\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)}{3\left(30\times {10}^6\text{psi}\left(\frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{\text{1}\text{psi}}\right)\right)\left(22{\text{in}}^4\right)}\end{array}\right]\end{array}$

  $\begin{array}{l}=\frac{\left(10.951\text{lb}\right)\times \left(\text{13824}{\text{in}}^3\right)}{3\times \left(30\times {10}^6\text{lb}/{\text{in}}^2\right)\times \left(22{\text{in}}^4\right)}+\frac{\left(10.951\text{lb}\right)\times \left(24\text{in}\right)\times \left(48\text{in}\right)\times \left(24\text{in}\right)}{3\times \left(30\times {10}^6\text{lb}/{\text{in}}^2\right)\times 22{\text{in}}^4}\\ =7.64578\times {10}^{-5}\text{in}+1.529157\times {10}^{-4}\text{in}\\ =\text{2}\text{.29373}\times {\text{10}}^{-4}\text{in}\end{array}$

Substitute $10.951\text{lb}$ for $Q$ , $3\text{ft}$ for $H$ , $30\times {10}^6\text{psi}$ for $E$ and $0.04908{\text{in}}^2$ for $A_r$ in equation (III).

  $\begin{array}{c}{\delta }_{rod}=\frac{10.951\text{lb}\times 3\text{ft}}{30\times {10}^6\text{psi}\times 0.04908{\text{in}}^2}\\ =\frac{10.951\text{lb}\times 3\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)}{30\times {10}^6\text{psi}\left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)\times 0.04908{\text{in}}^2}\\ =\frac{10.951\text{lb}\times 36\text{in}}{30\times {10}^6\text{lb}/{\text{in}}^2\times 0.04908{\text{in}}^2}\end{array}$

  $\begin{array}{c}{\delta }_{rod}=\frac{394.236\text{lb}\cdot \text{in}}{1472400\text{lb}}\\ =2.677506\times {10}^{-4}\text{in}\end{array}$

Substitute $30\times {10}^6\text{psi}$ for $E$ , $22{\text{in}}^4$ for $I_b$ , $6\text{ft}$ for $L_1$ , $2\text{ft}$ for $s$ , $2.677506\times {10}^{-4}\text{in}$ for ${\delta }_{rod}$ , $\text{2}\text{.29373}\times {\text{10}}^{-4}\text{in}$ for ${\left({\delta }_B\right)}_2$ in equation (V).

  $\begin{array}{l}q=\frac{24\times 30\times {10}^6\text{psi}\times 22{\text{in}}^4}{{\left(6\text{ft}-2\text{ft}\right)}^4}\left(2.677506\times {10}^{-4}\text{in}+\text{2}\text{.29373}\times {\text{10}}^{-4}\text{in}\right)\\ =\frac{720\times {10}^6\text{psi}\left(\frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{\text{1}\text{psi}}\right)\times 22{\text{in}}^4}{{\left(4\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\right)}^4}\left(4.971236\times {10}^{-4}\text{in}\right)\\ =\frac{720\times {10}^6\text{lb}/{\text{in}}^{\text{2}}\times 22{\text{in}}^4}{{\left(48\text{in}\right)}^4}\left(4.971236\times {10}^{-4}\text{in}\right)\\ =\frac{7874437.824\text{lb}\cdot {\text{in}}^3}{5308416{\text{in}}^4}\end{array}$

  $=1.483\text{lb}/\text{in}$

Conclusion:

The value of uniform load $q$ is $\underset{\_}{1.483\text{lb}/\text{in}}$ .

To determine

The minimum moment of inertia of beam to prevent buckling in tie rod.

Answer to Problem 11.3.25P

The minimum moment of inertia of beam to prevent buckling in tie rod is $\underset{\_}{247.182{\text{in}}^4}$ .

Explanation of Solution

Given information:

Intensity of uniformly distributed load on beam is $200\text{lb}/\text{ft}$ .

Calculation:

Substitute $30\times {10}^6\text{psi}$ for $E$ , $200\text{lb}/\text{ft}$ for $q$ , $6\text{ft}$ for $L_1$ , $2\text{ft}$ for $s$ , $2.677506\times {10}^{-4}\text{in}$ for ${\delta }_{rod}$ , $\text{2}\text{.29373}\times {\text{10}}^{-4}\text{in}$ for ${\left({\delta }_B\right)}_2$ in equation (V).

  $\begin{array}{c}200\text{lb}/\text{ft}=\frac{24\times \left(30\times {10}^6\text{psi}\right)\times I_b}{{\left(6\text{ft}-2\text{ft}\right)}^4}\left(2.677506\times {10}^{-4}\text{in}+\text{2}\text{.29373}\times {\text{10}}^{-4}\text{in}\right)\\ \frac{1}{I_b}=\frac{24\times \left(30\times {10}^6\text{psi}\left(\frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{\text{1}\text{psi}}\right)\right)}{200\text{lb}/\text{ft}\left(\frac{1\text{lb}/\text{in}}{12\text{lb}/\text{ft}}\right)\times {\left(\text{4}\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)\right)}^4}\left(2.677506\times {10}^{-4}\text{in}+\text{2}\text{.29373}\times {\text{10}}^{-4}\text{in}\right)\\ \frac{1}{I_b}=\frac{720\times {10}^6\text{lb}/{\text{in}}^{\text{2}}}{16.66667\text{lb}/\text{in}\times {\left(48\text{in}\right)}^4}\left(4.971236\times {10}^{-4}\right)\\ \frac{1}{I_b}=4.0455\times {10}^{-3}{\text{in}}^4\end{array}$

  $I_b=247.182{\text{in}}^4$

Conclusion:

The minimum moment of inertia of beam to prevent buckling in tie rod is $\underset{\_}{247.182{\text{in}}^4}$ .

To determine

The distance between point $B$ and $C$ .

Answer to Problem 11.3.25P

The distance between point $B$ and $C$ is $\underset{\_}{21.782\text{in}}$ .

Explanation of Solution

Given information:

Intensity of uniformly distributed load on beam is $200\text{lb}/\text{ft}$ .

Calculation:

Substitute $30\times {10}^6\text{psi}$ for $E$ , $200\text{lb}/\text{ft}$ for $q$ , $6\text{ft}$ for $L_1$ , $22{\text{in}}^4$ for $I_b$ , $2.677506\times {10}^{-4}\text{in}$ for ${\delta }_{rod}$ , $\text{2}\text{.29373}\times {\text{10}}^{-4}\text{in}$ for ${\left({\delta }_B\right)}_2$ in equation (V).

  $\begin{array}{c}200\text{lb}/\text{ft}=\left[\frac{24\times \left(30\times {10}^6\text{psi}\right)\times 22{\text{in}}^4}{{\left(6\text{ft}-s\right)}^4}\times \left(\begin{array}{l}2.677506\times {10}^{-4}\text{in}+\\ \text{2}\text{.29373}\times {\text{10}}^{-4}\text{in}\end{array}\right)\right]\\ 200\text{lb}/\text{ft}\times \left(\frac{1\text{lb}/\text{in}}{12\text{lb}/\text{ft}}\right)=\left[\begin{array}{l}\frac{24\times \left(30\times {10}^6\text{psi}\left(\frac{\text{1}\text{lb}/{\text{in}}^{\text{2}}}{\text{1}\text{psi}}\right)\right)\times 22{\text{in}}^4}{{\left(\text{4}\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)-s\right)}^4}\times \\ \left(2.677506\times {10}^{-4}\text{in}+\text{2}\text{.29373}\times {\text{10}}^{-4}\text{in}\right)\end{array}\right]\\ {\left(\text{4}\text{ft}\left(\frac{12\text{in}}{1\text{ft}}\right)-s\right)}^4=\frac{720\times {10}^6\text{lb}/{\text{in}}^{\text{2}}\times 22{\text{in}}^4}{16.6667\text{lb}/\text{in}}\times \left(4.971236\times {10}^{-4}\text{in}\right)\\ {\left(48\text{in}-s\right)}^4=\left(950398099.2{\text{in}}^3\right)\times \left(4.971236\times {10}^{-4}\text{in}\right)\end{array}$$\begin{array}{c}48\text{in}-s=26.2175\text{in}\\ s=21.782\text{in}\end{array}$

Conclusion:

The distance between point $B$ and $C$ is $\underset{\_}{21.782\text{in}}$ .