Chapter 5, Problem 5.5.25P

A beam with a T-section is supported and loaded as shown in the figure. The cross section has width b = 2 1/2 in., height c = 3 in., and thickness t = 3/8 in.

1. Determine the maximum tensile and compressive stresses in the beam.

If the allowable stresses in tension and compression are 18 ksi and 12 ksi, respectively, what is the required depth h of the beam? Assume that thickness t remains at 3/8 in. and that flange width/) = 2.5 in.

Find the new values of loads P and q so that the allowable tension (18 ksi) and compression (12 ksi) stresses are reached simultaneously for the beam. Use the beam cross section in part (a) (see figure) and assume that L_h and L₃are unchanged.

  

To determine

The maximum tensile stress.

The maximum compressive stress.

Explanation of Solution

Given information:

The uniform load is $110\text{lb}/\text{ft}$ , the width of the beam is $2\frac{1}{2}\text{in}$the height of the beam is $3\text{in}$and the thickness of the beam is $\frac{3}{8}\text{in}$ .

The following figure shows the free body diagram of the beam.

  

Figure-(1)

Write the expression for the distance of the neutral axis from the bottom layer.

  $c_2=\frac{b\times t\times \left(\frac{t}{2}\right)+t\times \left(h-t\right)\left(\frac{h-t}{2}+t\right)}{\left(b\times t\right)+t\times \left(h-t\right)}$  ......(I)

Here, the width of the beam is $b$ , the height of the t section is $h$ , the thickness of the beam is $b$ , and the distance of neutral axis is $c_2$ .

Write the expression for the distance of the neutral axis from the top layer.

  $c_1=h-c_2$   ......(II)

Here, the distance of neutral axis from the top layer.

  

Figure-(2)

Write the expression for the moment of inertia.

  $I=\frac{b{t}^3}{12}+bt{\left(c_2+\frac{t}{2}\right)}^2+\frac{{\left(h-t\right)}^3}{12}t+t\left(h-t\right){\left(c_1-\left(\frac{h-t}{2}\right)\right)}^2$  ......(III)

Here, the moment of inertia is $I$ .

Write the expression for the moment equilibrium about $A$ .

  $\begin{array}{l}{\displaystyle \sum M_A}=0\\ R_B\times L_2=P\times L_1+q{L}_3\left(L_2+\frac{L_3}{2}\right)\end{array}$   ......(IV)

Here, the reaction at point $B$is $R_B$ , the length of section $AB$is $L_2$and the length of section $BC$is $L_3$and the concentrated load is $P$ , and the distance at which concentrated load is acting is $L_1$ .

Write the expression for the force equilibrium in vertical direction.

  $\begin{array}{l}{\displaystyle \sum F_y}=0\\ R_A=P+q{L}_3-R_B\end{array}$  ......(V)

Here, the reaction at point $A$is $R_A$ .

Write the expression for the maximum sagging moment at load $P$ .

  $M_p=R_A\times L_1$  ......(VI)

Here, the maximum sagging moment is $M_P$ .

Write the expression for the maximum hogging moment at point $B$ .

  $M_B=q\times L_3\times \frac{L_3}{2}$  ......(VII)

Here, the maximum hogging moment is $M_B$ .

Write the expression for the maximum tensile stress at point $B$ .

  ${\sigma }_t=\frac{M_B{c}_1}{I}$   ......(VIII)

Here, the maximum tensile stress is ${\sigma }_t$ .

Write the expression for the maximum compressive stress at point $B$ .

  ${\sigma }_c=\frac{M_B{c}_2}{I}$   ......(IX)

Here the maximum compressive stress is ${\sigma }_c$ .

Calculation:

Substitute $2.5\text{in}$for $b$ , $\frac{3}{8}\text{in}$for $t$ , and $3\text{in}$for $h$in Equation (I).

  $\begin{array}{c}{c}_2=\frac{2.5\text{in}\times \frac{3}{8}\text{in}\times \left(\frac{\frac{3}{8}\text{in}}{2}\right)+\frac{3}{8}\text{in}\times \left(3\text{in}-\frac{3}{8}\text{in}\right)\left(\frac{3\text{in}-\frac{3}{8}\text{in}}{2}+\frac{3}{8}\text{in}\right)}{\left(2.5\text{in}\times \frac{3}{8}\text{in}\right)+\frac{3}{8}\text{in}\times \left(3\text{in}-\frac{3}{8}\text{in}\right)}\\ =\frac{0.1757\text{in}+1.6611\text{in}}{1.9219\text{in}}\\ =0.9558\text{in}\end{array}$

Substitute $0.9558\text{in}$for $c_2$ , and $3\text{in}$for $h$in Equation (II).

  $\begin{array}{c}{c}_1=3\text{in}-0.9558\text{in}\\ \text{=2}\text{.0442}\text{in}\end{array}$

Substitute $2.5\text{in}$for $b$ , $\frac{3}{8}\text{in}$for $t$ , $0.9558\text{in}$for $c_2$ , $2.442\text{in}$for $c_1$and $3\text{in}$for $h$in Equation (III).

  $\begin{array}{c}I=\left[\begin{array}{l}\frac{\left(2.5\text{in}\right){\left(\frac{3}{8}\text{in}\right)}^3}{12}+\left(2.5\text{in}\times \frac{3}{8}\text{in}\right){\left(\left(0.9558\text{in}\right)+\frac{\left(\frac{3}{8}\text{in}\right)}{2}\right)}^2+\frac{{\left(\left(3\text{in}\right)-\left(\frac{3}{8}\text{in}\right)\right)}^3}{12}\left(\frac{3}{8}\text{in}\right)\\ +\left(\frac{3}{8}\text{in}\right)\left(\left(3\text{in}\right)-\left(\frac{3}{8}\text{in}\right)\right){\left(\left(2.0442\text{in}\right)-\left(\frac{3\text{in}-\frac{3}{8}\text{in}}{2}\right)\right)}^2\end{array}\right]\\ =1.2411{\text{in}}^4+0.619{\text{in}}^4\\ =1.8602{\text{in}}^4\end{array}$

Substitute $8\text{ft}$for $L_2$ , $750\text{lb}$for $P$ , $5\text{ft}$for $L_3$ , $3\text{ft}$for $L_1$and $110\text{lb}/\text{ft}$for $q$ , in Equation (IV).

  $\begin{array}{l}\left(R_B\times 8\text{ft}\right)=\left(750\text{lb}\times 3\text{ft}\right)+\left(110\text{lb}/\text{ft}\right)\left(5\text{ft}\right)\left(8\text{ft}+\frac{3\text{ft}}{2}\right)\\ R_B=\frac{\left(750\text{lb}\times 3\text{ft}\right)+\left(110\text{lb}/\text{ft}\right)\left(5\text{ft}\right)\left(8\text{ft}+\frac{3\text{ft}}{2}\right)}{8\text{ft}}\\ R_B=1003.125\text{lb}\end{array}$

Substitute $8\text{ft}$for $L_2$ , $750\text{lb}$for $P$ , $5\text{ft}$for $L_3$ , $1003.125\text{lb}$for $R_B$and $110\text{lb}/\text{ft}$for $q$ , in Equation (V).

  $\begin{array}{c}{R}_A=\left(750\text{lb}\right)+\left(110\text{lb}/\text{ft}\right)\left(5\text{ft}\right)-\left(1003.125\text{lb}\right)\\ =\left(750\text{lb}\right)+\left(550\text{lb}\right)-\left(1003.125\text{lb}\right)\\ =296.875\text{lb}\end{array}$

Substitute $296.875\text{lb}$for $R_A$ , $3\text{ft}$for $L_1$in Equation (VI).

  $\begin{array}{c}{M}_P=\left(296.875\text{lb}\right)\times \left(3\text{ft}\right)\\ =890.625\text{lb}\cdot \text{ft}\end{array}$

Substitute $5\text{ft}$for $L_3$ , $110\text{lb}/\text{ft}$for $q$in Equation (VII).

  $\begin{array}{c}{M}_B=\left(110\text{lb}/\text{ft}\right)\times 5\text{ft}\times \frac{5\text{ft}}{2}\\ =\left(110\text{lb}/\text{ft}\right)\times 12.5{\text{ft}}^2\\ =1375\text{lb}\cdot \text{ft}\end{array}$

Substitute $1375\text{lb}\cdot \text{ft}$for $M_B$ , $1.8602{\text{in}}^4$for $I$and $2.0442\text{in}$for $c_1$ , in Equation (VIII).

  $\begin{array}{c}{\sigma }_t=\frac{\left(1375\text{lb}\cdot \text{ft}\right)\times \left(2.0442\text{in}\right)}{1.8602{\text{in}}^4}\\ =\frac{\left(1375\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{lb}\cdot \text{in}}{1\text{lb}\cdot \text{ft}}\right)\times \left(2.0442\text{in}\right)}{1.8602{\text{in}}^4}\\ =\left(18132.08\text{lb}/{\text{in}}^2\right)\left(\frac{1\text{psi}}{1\text{lb}/{\text{in}}^2}\right)\\ =18132.08\text{psi}\end{array}$

Substitute $1375\text{lb}\cdot \text{ft}$for $M_B$ , $1.8602{\text{in}}^4$for $I$and $0.9558\text{in}$for $c_2$ , in Equation (IX).

  $\begin{array}{c}{\sigma }_c=\frac{\left(1375\text{lb}\cdot \text{ft}\right)\times \left(0.9558\text{in}\right)}{1.8602{\text{in}}^4}\\ =\frac{\left(1375\text{lb}\cdot \text{ft}\right)\left(\frac{12\text{lb}\cdot \text{in}}{1\text{lb}\cdot \text{ft}}\right)\times \left(0.9558\text{in}\right)}{1.8602{\text{in}}^4}\\ =\left(8477.95\text{lb}/{\text{in}}^2\right)\left(\frac{1\text{psi}}{1\text{lb}/{\text{in}}^2}\right)\\ =8477.95\text{psi}\end{array}$

Conclusion:

The maximum tensile stress is $7810\text{psi}$ .

The maximum compressive stress is $13885\text{psi}$ .

To determine

The required depth of the beam.

Answer to Problem 5.5.25P

The required depth of the beam is $3.26\text{in}$ .

Explanation of Solution

Given Information:

The allowable stress in tension is $18\text{ksi}$ , the allowable stress in compression is $12\text{ksi}$ .

Write the expression for the maximum tensile stress at point $B$ .

  ${\sigma }_t=\frac{M_B{c}_1}{I}$   ......(X)

Write the expression for the distance of the neutral axis from the bottom layer.

  $c_1=\frac{b\times t\times \left(\frac{t}{2}\right)+t\times \left(h-t\right)\left(\frac{h-t}{2}+t\right)}{\left(b\times t\right)+t\times \left(h-t\right)}$  ......(XI)

Write the expression for the moment of inertia.

  $I=\frac{b{t}^3}{12}+bt{\left(c_2+\frac{t}{2}\right)}^2+\frac{{\left(h-t\right)}^3}{12}t+t\left(h-t\right){\left(c_1-\left(\frac{h-t}{2}\right)\right)}^2$  ......(XII)

Calculation:

Substitute $2.5\text{in}$for $b$ , and $\frac{3}{8}\text{in}$for $t$in Equation (XI).

  $\begin{array}{l}{c}_1=\frac{2.5\text{in}\times \frac{3}{8}\text{in}\times \left(\frac{\frac{3}{8}\text{in}}{2}\right)+\frac{3}{8}\text{in}\times \left(h-\frac{3}{8}\text{in}\right)\left(\frac{h-\frac{3}{8}\text{in}}{2}+\frac{3}{8}\text{in}\right)}{\left(2.5\text{in}\times \frac{3}{8}\text{in}\right)+\frac{3}{8}\text{in}\times \left(h-\frac{3}{8}\text{in}\right)}\\ c_1=\frac{\left(\frac{3}{16}{h}^2+\frac{51}{64}h+\frac{153}{1024}\right)}{\left(\frac{51}{64}+\frac{3}{89}h\right)}\end{array}$

Substitute $2.5\text{in}$for $b$ , $\frac{3}{8}\text{in}$for $t$ , $0.9558\text{in}$for $c_2$ , $2.442\text{in}$for $c_1$and $3\text{in}$for $h$in Equation (XII).

  $\begin{array}{c}I=\left[\begin{array}{l}\frac{\left(2.5\text{in}\right){\left(\frac{3}{8}\text{in}\right)}^3}{12}+\left(2.5\text{in}\times \frac{3}{8}\text{in}\right){\left(\left(0.9558\text{in}\right)+\frac{\left(\frac{3}{8}\text{in}\right)}{2}\right)}^2+\frac{{\left(h-\left(\frac{3}{8}\text{in}\right)\right)}^3}{12}\left(\frac{3}{8}\text{in}\right)\\ +\left(\frac{3}{8}\text{in}\right)\left(\left(h\right)-\left(\frac{3}{8}\text{in}\right)\right){\left(\left(2.0442\text{in}\right)-\left(\frac{h-\frac{3}{8}\text{in}}{2}\right)\right)}^2\end{array}\right]\\ =\left[\frac{45}{4906}+\frac{15}{16}{\left(h-c_1-\frac{3}{16}\right)}^2+\frac{3}{96}{\left(h-\frac{3}{8}\right)}^3+\frac{3}{8}\left(h-\frac{3}{8}\right){\left(c_1-\left(\frac{h}{2}-\frac{3}{16}\right)\right)}^2\right]\end{array}$

Substitute $18\text{ksi}$for ${\sigma }_t$ , $1375\text{lb}\cdot \text{ft}$for $M_B$ , $\frac{\left(\frac{3}{16}{h}^2+\frac{51}{64}h+\frac{153}{1024}\right)}{\left(\frac{51}{64}+\frac{3}{89}h\right)}$for $c_1$and $\left[\frac{45}{4906}+\frac{15}{16}{\left(h-c_1-\frac{3}{16}\right)}^2+\frac{3}{96}{\left(h-\frac{3}{8}\right)}^3+\frac{3}{8}\left(h-\frac{3}{8}\right){\left(c_1-\left(\frac{h}{2}-\frac{3}{16}\right)\right)}^2\right]$for $I$in Equation (X).

.   ......(XIII)

Substitute $3\text{in}$for $h$in Equation (XIII).

  $\begin{array}{l}\left(\frac{12\text{in}}{11}\right)=\frac{\left(\frac{\left(\left(\frac{3}{16}\text{in}\right){\left(3\text{in}\right)}^2+\left(\frac{51}{64}\text{in}\right)\left(3\text{in}\right)+\left(\frac{153}{1024}\text{in}\right)\right)}{\left(\frac{51}{64}\text{in}+\left(\frac{3}{89}\text{in}\right)\left(3\text{in}\right)\right)}\right)}{\left[\begin{array}{l}\frac{45}{4906}{\text{in}}^2+\frac{15}{16}\text{in}{\left(\left(3\text{in}\right)-\frac{3}{16}\right)}^2+\frac{3}{96}\text{in}{\left(\left(3\text{in}\right)-\frac{3}{8}\right)}^3\\ +\frac{3}{8}\text{in}\left(\left(3\text{in}\right)-\frac{3}{8}\text{in}\right){\left(c_1-\left(\frac{\left(3\text{in}\right)}{2}-\frac{3}{16}\text{in}\right)\right)}^2\end{array}\right]}\\ \frac{12\text{in}}{11}=\frac{2.1997{\text{in}}^2}{1.7031{\text{in}}^2}\\ 1.091=1.2915\end{array}$

Substitute $3.26\text{in}$for $h$in Equation (XIII).

  $\begin{array}{l}\left(\frac{12\text{in}}{11}\right)=\frac{\left(\frac{\left(\left(\frac{3}{16}\text{in}\right){\left(3.26\text{in}\right)}^2+\left(\frac{51}{64}\text{in}\right)\left(3.26\text{in}\right)+\left(\frac{153}{1024}\text{in}\right)\right)}{\left(\frac{51}{64}\text{in}+\left(\frac{3}{89}\text{in}\right)\left(3.26\text{in}\right)\right)}\right)}{\left[\begin{array}{l}\frac{45}{4906}{\text{in}}^2+\frac{15}{16}\text{in}{\left(\left(3.26\text{in}\right)-\frac{3}{16}\text{in}\right)}^2+\frac{3}{96}\text{in}{\left(\left(3.26\text{in}\right)-\frac{3}{8}\right)}^3\\ +\frac{3}{8}\text{in}\left(\left(3.26\text{in}\right)-\frac{3}{8}\text{in}\right){\left(\left(\frac{\left(3.26\text{in}\right)}{2}-\frac{3}{16}\text{in}\right)\right)}^2\end{array}\right]}\\ \frac{12\text{in}}{11}=\frac{2.3472{\text{in}}^2}{2.1400\text{in}}\\ 1.091=1.091\end{array}$

Conclusion:

The required depth of the beam is $3.26\text{in}$ .