Chapter 7.2, Problem 7.35P

7.35 and 7.36 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

Fig. P7.35

To determine

The shear and bending-moment diagrams.

Answer to Problem 7.35P

The shear diagram is drawn in figure 6 and bending momentum is drawn in figure 7.

Explanation of Solution

Refer Figure 1.

Refer Figure 2.

Write an expression to calculate the net counter clockwise moment at point C along AC.

$\Sigma M_C=0$ (I)

Here, $M_C$ is the moment at point C.

Write an expression to calculate the net vertical force along AC.

$\Sigma F_y=0$ (II)

Here, $F_y$ is the vertical force.

Refer Figure 3.

Write an expression to calculate the net counter clockwise moment at point D along CD.

$\Sigma M_D=0$ (III)

Here, $M_D$ is the moment at point D.

Write an expression to calculate the net vertical force along CD.

$\Sigma F_y=0$ (IV)

Refer Figure 4.

Write an expression to calculate the net counter clockwise moment at point E along DE.

$\Sigma M_E=0$ (V)

Here, $M_E$ is the moment at point E.

Write an expression to calculate the net vertical force along EB.

$\Sigma F_y=0$ (VI)

Refer Figure 5.

Write an expression to calculate the net counter clockwise moment at point B along EB.

$\Sigma M_B=0$ (VII)

Here, $M_B$ is the moment at point B.

Write an expression to calculate the net vertical force along EB.

$\Sigma F_y=0$ (VIII)

Conclusion:

Refer Figure 2 and equation (II). Calculate the net vertical force.

$-15\text{kN}-V=0$

Here, $V$ is the shear force.

Rearrange the equation to calculate $V$.

$V=-15\text{kN}$ (III)

Refer Figure 2 and equation (I). Calculate the net counter clockwise moment at point C.

$M+x\left(15\text{kN}\right)=0$

Here, $M$ is the moment and $x$ is the length of portion.

Rearrange the equation to calculate $M$.

$\begin{array}{c}M=\left(15\text{kN}\right)x\\ =\left(15\text{kN}\right)\left(1\text{m}\right)\\ =15\text{kN}\cdot \text{m}\end{array}$      

Refer Figure 3 and equation (IV). Calculate the net vertical force.

$-15\text{kN}-25\text{kN}-V=0$

Rearrange the equation to calculate $V$.

$\begin{array}{c}V=-15\text{kN}-25\text{kN}\\ =-\text{40}\text{kN}\end{array}$ 

Refer Figure 3 and equation (III). Calculate the net counter clockwise moment at point D.

$M+\left(x-1\text{m}\right)\left(25\text{kN}\right)+x\left(15\text{kN}\right)=0$

Rearrange the equation to calculate $M$.

$\begin{array}{c}M=-\left(x-1\text{m}\right)\left(25\text{kN}\right)-x\left(15\text{kN}\right)\\ =-\left(1.3\text{m}-1\text{m}\right)\left(25\text{kN}\right)-\left(1.3\text{m}\right)\left(15\text{kN}\right)\\ =-27\text{kN}\cdot \text{m}\end{array}$      

Refer Figure 4 and equation (VI). Calculate the net vertical force.

$-15\text{kN}-25\text{kN}+30\text{kN}-V=0$

Rearrange the equation to calculate $V$.

$\begin{array}{c}V=-15\text{kN}-25\text{kN}+30\text{kN}\\ =-1\text{0}\text{kN}\end{array}$ 

Refer Figure 4 and equation (V). Calculate the net counter clockwise moment at point E.

$M-x\left(30\text{kN}\right)+\left(0.3\text{m}+x\right)\left(25\text{kN}\right)+\left(1\text{m}+0.3\text{m}+x\right)\left(15\text{kN}\right)=0$

Rearrange the equation to calculate $M$.

$\begin{array}{c}M=x\left(30\text{kN}\right)-\left(0.3\text{m}+x\right)\left(25\text{kN}\right)-\left(1\text{m}+0.3\text{m}+x\right)\left(15\text{kN}\right)\\ =\left(0.4\text{m}\right)\left(30\text{kN}\right)-\left(0.3\text{m}+0.4\text{m}\right)\left(25\text{kN}\right)-\left(1\text{m}+0.3\text{m}+0.4\text{m}\right)\left(15\text{kN}\right)\\ =-31.0\text{kN}\cdot \text{m}\end{array}$      

Refer Figure 5 and equation (VIII). Calculate the net vertical force.

$-15\text{kN}-25\text{kN}+30\text{kN}-\text{20}\text{kN}-V=0$

Rearrange the equation to calculate $V$.

$\begin{array}{c}V=-15\text{kN}-25\text{kN}+30\text{kN}-20\text{kN}\\ =-3\text{0}\text{kN}\end{array}$ 

Refer Figure 5 and equation (VII). Calculate the net counter clockwise moment at point E.

$M+x\left(20\text{kN}\right)+\left(0.3\text{m}+0.4\text{m}+x\right)\left(25\text{kN}\right)+\left(1\text{m}+0.3\text{m}+0.4\text{m}+x\right)\left(15\text{kN}\right)-\left(0.4\text{m}+x\right)\left(30\text{kN}\right)=0$

Rearrange the equation to calculate $M$.

$\begin{array}{c}M=\left(\begin{array}{c}-x\left(20\text{kN}\right)-\left(0.3\text{m}+0.4\text{m}+x\right)\left(25\text{kN}\right)\\ -\left(1\text{m}+0.3\text{m}+0.4\text{m}+x\right)\left(15\text{kN}\right)+\left(0.4\text{m}+x\right)\left(30\text{kN}\right)\end{array}\right)\\ =\left(\begin{array}{c}-\left(0.8\text{m}\right)\left(20\text{kN}\right)-\left(0.3\text{m}+0.4\text{m}+0.8\text{m}\right)\left(25\text{kN}\right)\\ -\left(1\text{m}+0.3\text{m}+0.4\text{m}+0.8\text{m}\right)\left(15\text{kN}\right)+\left(0.4\text{m}+0.8\text{m}\right)\left(30\text{kN}\right)\end{array}\right)\\ =-55.0\text{kN}\cdot \text{m}\end{array}$      

Thus, draw the shear diagram.

Thus, draw the bending-moment.

To determine

The maximum absolute value of the shear and bending moment.

Answer to Problem 7.35P

The maximum absolute value of the shear force is $40.0\text{kN}$ on CD and bending moment is $55.0\text{kN}\text{.m}$ at B.

Explanation of Solution

Determine the maximum absolute shear from diagram 2. The maximum absolute value of bending moment is at maximum shear force.

Conclusion:

Refer figure 6. Determine the maximum absolute value of shear force.

$\left|V_{\max }\right|=40.0\text{kN}$

Here, $V_{\max }$ is the maximum value of shear force.

Refer figure 7. Determine the maximum value of the bending moment at position B.

$\left|M_{\max }\right|=55.0\text{kN}\text{.m}$

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

Thus, the maximum absolute value of the shear force is $40.0\text{kN}$ on CD and bending moment is $55.0\text{kN}\text{.m}$ at B.