Chapter 5, Problem 5.6.11P

A beam ABC with an overhang from B to C is constructed of a C 10 × 30 channel section with flanges facing upward (sec figure). The beam supports its own weight (30 lb/ft) plus a triangular load of maximum intensity g₀ acting on the overhang. The allowable stresses in tension and compression arc IS ksi and 12 ksi, respectively.

1. Determine the allowable triangular load intensity allow if tne distance L equals 4 ft.

What is the allowable triangular load intensity fallow ^tnc Dcam i^s rotated 180^e about its longitudinal centroidal axis so that the flanges are downward?

  

To determine

The allowable triangular load intensity.

Answer to Problem 5.6.11P

The allowable triangular load intensity is $419.21\text{lb}/\text{ft}$ .

Explanation of Solution

Given information:

The weight of the beam is $30\text{lb}/\text{ft}$ , the allowable stress in tension is $18\text{kpsi}$ , allowable stress in compression is $12\text{ksi}$ , the distance is $4\text{ft}$ .

The following figure shows the free body diagram.

  

Figure-(1)

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

  $\begin{array}{l}{B}_y\times L=\left(q\times 2L\right)\left(L\right)+\left(\frac{q_0\times L}{2}\right)\left(L+\frac{2L}{3}\right)\\ B_y=\left(q\times 2L\right)+\left(\frac{q_0\times L}{2}\right)\left(\frac{5}{3}\right)\end{array}$.....(I)

Here, the vertical reaction on point $B$ is $B_y$ , the length of the beam is $2L$ , self weight of the beam per unit length is $q$ ,and triangular load intensity is $q_0$ .

Write the expression for the equilibrium forces at point $A$ and $B$ .

  $A_y+B_y=\left(q\times 2L\right)+\left(\frac{q_0\times L}{2}\right)$.....(II)

Here, the vertical reaction on point $A$ is $A_y$ .

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

  $M_{\max }=A_y\times L+\left(q\times L\right)\left(\frac{L}{2}\right)$.....(III)

Here, the maximum moment is $M_{\max }$ .

Write the expression for the allowable bending moment based on tension.

  $M_t=\frac{{\sigma }_t\times I}{c_1}$.....(IV)

Here, maximum tensile stress is ${\sigma }_t$ , the moment of inertia is $I$ , and the distance of neutral axis is $c_1$ .

Write the expression for the allowable bending moment based on compression.

  $M_c=\frac{{\sigma }_c\times I}{c_c}$.....(V)

Here, maximum compressible stress is ${\sigma }_c$ , the moment of inertia is $I$ , and the distance of neutral axis is $c_2$ .

Write the expression for the load intensity of triangular load considering tension.

  $M_{\max }=L^2\left(\frac{q_0}{3}+\frac{q}{2}\right)$ ….. (VI)

Write the expression for the load intensity of triangular load considering compression.

  $M_{\max }=L^2\left(\frac{q_0}{3}+\frac{q}{2}\right)$ ….. (VII)

Calculation:

Substitute $B_y=\left(q\times 2L\right)+\left(\frac{q_0\times L}{2}\right)\left(\frac{5}{3}\right)$ for $B_y$ in Equation (II).

  $\begin{array}{l}{A}_y+\left(q\times 2L\right)+\left(\frac{q_0\times L}{2}\right)\left(\frac{5}{3}\right)=\left(q\times 2L\right)+\left(\frac{q_0\times L}{2}\right)\\ A_y=\left(q\times 2L\right)+\left(\frac{q_0\times L}{2}\right)\left(\frac{5}{3}\right)-\left(q\times 2L\right)+\left(\frac{q_0\times L}{2}\right)\\ A_y=-\left(\frac{q_0\times L}{3}\right)\end{array}$

Substitute $-\left(\frac{q_0\times L}{3}\right)$ for $A_y$ in Equation (III).

  $\begin{array}{c}{M}_{\max }=-\left(\frac{q_0\times L}{3}\right)L+\frac{\left(q\times L\right)L}{2}\\ =-\left(\frac{\left(q_{0}L\right)\times \left(L^2\right)}{3}\right)+\frac{\left(q\times L\right)L}{2}\\ =L^2\left(\frac{q_0}{3}+\frac{q}{2}\right)\end{array}$

Substitute $3.93{\text{in}}^4$ for $I$ , $18\text{kpsi}$ for ${\sigma }_t$ ,and $2.381\text{in}\text{.}$ for $c_1$ in Equation (IV).

  $\begin{array}{c}{M}_t=\frac{\left(18\text{kpsi}\right)\times \left(3.93{\text{in}}^4\right)}{\left(2.381\text{in}\right)}\\ =\frac{\left(18\text{kpsi}\right)\left(\frac{1000\text{psi}}{1\text{kpsi}}\right)\times \left(3.93{\text{in}}^4\right)}{\left(2.381\text{in}\right)}\\ =\frac{\left(18000\text{psi}\right)\left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)\times \left(3.93{\text{in}}^4\right)}{\left(2.381\text{in}\right)}\\ =29710\text{lb}\cdot \text{in}\text{.}\end{array}$

Substitute $3.93{\text{in}}^4$ for $I$ , $12\text{ksi}$ for ${\sigma }_c$ ,and $0.649\text{in}\text{.}$ for $c_2$ in Equation (V).

  $\begin{array}{c}{M}_c=\frac{\left(12\text{kpsi}\right)\times \left(3.93{\text{in}}^4\right)}{\left(0.649\text{in}\right)}\\ =\frac{\left(12\text{kpsi}\right)\left(\frac{1000\text{psi}}{1\text{kpsi}}\right)\times \left(3.93{\text{in}}^4\right)}{\left(0.649\text{in}\right)}\\ =\frac{\left(12000\text{psi}\right)\left(\frac{1\text{lb}/{\text{in}}^2}{1\text{psi}}\right)\times \left(3.93{\text{in}}^4\right)}{\left(0.649\text{in}\right)}\\ =29710\text{lb}\cdot \text{in}\end{array}$

Substitute $29710\text{lb}\cdot \text{in}$ for $M_{\max }$ , $4\text{ft}$ for $L$ ,and $30\text{lb}/\text{ft}$ for $q$ in Equation (VI).

  $\begin{array}{l}\left(29710\text{lb}\cdot \text{in}\right)={\left(4\text{ft}\right)}^2\left(\frac{q_0}{3}+\frac{30\text{lb}/\text{ft}}{2}\right)\\ \left(\left(29710\text{lb}\cdot \text{in}\right)\times \frac{1\text{ft}}{12\text{in}}\right)=\left({\left(4\text{ft}\right)}^2\left(\frac{q_0}{3}+\frac{30\text{lb}/\text{ft}}{2}\right)\right)\\ q_0=\left(\left(\frac{2475.83\text{lb}\cdot \text{ft}}{{\left(4\text{ft}\right)}^2}-\frac{30\text{lb}/\text{ft}}{2}\right)\times 3\right)\\ q_0=419.21\text{lb}/\text{ft}\end{array}$

Substitute $72666\text{lb}\cdot \text{in}$ for $M_{\max }$ , $4\text{ft}$ for $L$ ,and $30\text{lb}/\text{ft}$ for $q$ in Equation (VII).

  $\begin{array}{l}\left(72666\text{lb}\cdot \text{in}\right)={\left(4\text{ft}\right)}^2\left(\frac{q_0}{3}+\frac{30\text{lb}/\text{ft}}{2}\right)\\ \left(\left(72666\text{lb}\cdot \text{in}\right)\times \frac{1\text{ft}}{12\text{in}}\right)=\left({\left(4\text{ft}\right)}^2\left(\frac{q_0}{3}+\frac{30\text{lb}/\text{ft}}{2}\right)\right)\\ q_0=\left(\left(\frac{6055.5\text{lb}\cdot \text{ft}}{{\left(4\text{ft}\right)}^2}-\frac{30\text{lb}/\text{ft}}{2}\right)\times 3\right)\\ q_0=1090.41\text{lb}/\text{ft}\end{array}$

Conclusion:

The allowable triangular load intensity is $419.21\text{lb}/\text{ft}$ .

To determine

The allowable triangular load intensity when the beam is rotated $180°$ .

Answer to Problem 5.6.11P

The allowable triangular load intensity when the beam is rotated $180°$ is $264.48\text{lb}/\text{ft}$ .

Explanation of Solution

Write the expression for the allowable bending moment based on tension.

  $M_t=\frac{{\sigma }_t\times I}{c_1}$.....(VIII)

Write the expression for the allowable bending moment based on compression.

  $M_c=\frac{{\sigma }_c\times I}{c_2}$.....(IX)

Write the expression for the load intensity of triangular load considering tension.

  $M_{\max }=L^2\left(\frac{q_0}{3}+\frac{q}{2}\right)$ ….. (X)

Write the expression for the load intensity of triangular load considering compression.

  $M_{\max }=L^2\left(\frac{q_0}{3}+\frac{q}{2}\right)$….. (XI)

Calculation:

Substitute $3.93{\text{in}}^4$ for $I$ , $18\text{ksi}$ for ${\sigma }_t$ ,and $0.649\text{in}\text{.}$ for $c_1$ in Equation (VIII).

  $\begin{array}{c}{M}_t=\left(\frac{18\text{ksi}\times 3.93{\text{in}}^4}{0.649\text{in}}\right)\\ =\left(\frac{18\text{ksi}\times 3.93{\text{in}}^4}{0.649\text{in}}\right)\left(\frac{{10}^3\text{lb}/{\text{in}}^2}{1\text{ksi}}\right)\\ =108988.45\text{lb}\cdot \text{in}\end{array}$

Substitute $3.93{\text{in}}^4$ for $I$ , $12\text{ksi}$ for ${\sigma }_c$ ,and $2.381\text{in}$ for $c_2$ in Equation (IX).

  $\begin{array}{c}{M}_c=\left(\frac{12\text{ksi}\times 3.93{\text{in}}^4}{2.381\text{in}}\right)\\ =\left(\frac{12\text{ksi}\times 3.93{\text{in}}^4}{2.381\text{in}}\right)\left(\frac{{10}^3\text{lb}/{\text{in}}^2}{1\text{ksi}}\right)\\ =19806.8\text{lb}\cdot \text{in}\end{array}$

Substitute $108998.45\text{lb}\cdot \text{in}$ for $M_{\max }$ , $4\text{ft}$ for $L$ ,and $30\text{lb}/\text{ft}$ for $q$ in Equation (VI).

  $\begin{array}{l}\left(108998.45\text{lb}\cdot \text{in}\right)={\left(4\text{ft}\right)}^2\left(\frac{q_0}{3}+\frac{30\text{lb}/\text{ft}}{2}\right)\\ \left(\left(108998.45\text{lb}\cdot \text{in}\right)\times \frac{1\text{ft}}{12\text{in}}\right)=\left({\left(4\text{ft}\right)}^2\left(\frac{q_0}{3}+\frac{30\text{lb}/\text{ft}}{2}\right)\right)\\ q_0=\left(\left(\frac{9083.20\text{lb}\cdot \text{ft}}{{\left(4\text{ft}\right)}^2}-\frac{30\text{lb}/\text{ft}}{2}\right)\times 3\right)\\ q_0=1658.10\text{lb}/\text{ft}\end{array}$

Substitute $19806.8\text{lb}\cdot \text{in}$ for $M_{\max }$ , $4\text{ft}$ for $L$ ,and $30\text{lb}/\text{ft}$ for $q$ in Equation (VII).

  $\begin{array}{l}\left(19806.8\text{lb}\cdot \text{in}\right)={\left(4\text{ft}\right)}^2\left(\frac{q_0}{3}+\frac{30\text{lb}/\text{ft}}{2}\right)\\ \left(\left(19806.8\text{lb}\cdot \text{in}\right)\times \frac{1\text{ft}}{12\text{in}}\right)=\left({\left(4\text{ft}\right)}^2\left(\frac{q_0}{3}+\frac{30\text{lb}/\text{ft}}{2}\right)\right)\\ q_0=\left(\left(\frac{1650.57\text{lb}\cdot \text{ft}}{{\left(4\text{ft}\right)}^2}-\frac{30\text{lb}/\text{ft}}{2}\right)\times 3\right)\\ q_0=264.48\text{lb}/\text{ft}\end{array}$

Conclusion:

The allowable triangular load intensity when the beam is rotated $180°$ is $264.48\text{lb}/\text{ft}$ .