Chapter 5, Problem 1P

To determine

Find the axial force, shear, and bending moment at points A and B of the beam.

Answer to Problem 1P

The axial force at point A is $\underset{\_}{50\text{kN}}$.

The shear at point A is $\underset{\_}{91.7\text{kN}}$.

The moment at point A is $\underset{\_}{1,333.6\text{kN-m}}$.

The axial force at point B is $\underset{\_}{0}$.

The shear at point B is $\underset{\_}{-194.9\text{kN}}$.

The moment at point B is $\underset{\_}{584.7\text{kN-m}}$.

Explanation of Solution

Sign conversion:

Apply the sign convention for calculating the equations of equilibrium as below.

• For the horizontal forces equilibrium condition, take the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.
• For the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.
• For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive.

Apply the following sign convention for calculating the axial forces, shear and bending moments.

• When the portion of the beam considered is left of the section, then the external force acting to the left are considered as positive.
• When the portion of the beam considered is right of the section, then the external force acting to the right are considered as positive.
• When the portion of the beam considered is left of the section, then the external force acting upward are considered as positive.
• When the portion of the beam considered is right of the section, then the external force acting downward are considered as positive.
• When the portion of the beam considered is left of the section, then the clockwise moments are considered as positive.
• When the portion of the beam considered is right of the section, then the counterclockwise moments are considered as positive.

Calculation:

Show the free-body diagram of the entire beam as in Figure 1.

Find the horizontal reaction at point C by resolving the horizontal equilibrium.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -C_x+100\cos 60°=0\\ C_x=50\text{kN}\leftarrow \end{array}$

Find the vertical reaction at point C by taking moment about the point D.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -C_y\left(20\right)+120\left(15\right)+200\left(10\right)+100\sin 60°\left(5\right)=0\\ -20C_y+1800+2000+433.01=0\\ C_y=211.7\text{kN}\uparrow \end{array}$

Find the vertical reaction at point D by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ C_y-120-200-100\sin 60°+D_y=0\\ 211.7-320-86.60+D_y=0\\ D_y=194.9\text{kN}\uparrow \end{array}$

Pass the sections aa and bb at points A and B respectively.

Show the sections aa and bb as in Figure 2.

Consider section aa:

Consider the left side of the section aa for calculation of internal forces.

Show the free-body diagram of the left side of the section aa as in Figure 3.

Find the axial force at point A by resolving the horizontal equilibrium.

$\begin{array}{l}+\leftarrow {\displaystyle \sum F_x=0}\\ 50-Q_A=0\\ Q_A=50\text{kN}\end{array}$

Find the shear at point A by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ 211.7-120-S_A=0\\ S_A=91.7\text{kN}\end{array}$

Find the moment at point A by taking moment about the section aa.

$\begin{array}{l}+\circlearrowright {\displaystyle \sum M=0}\\ 211.7\left(8\right)-120\left(3\right)-M_A=0\\ 1693.6-360-M_A=0\\ M_A=1,333.6\text{kN-m}\end{array}$

Thus, the axial force at point A is $\underset{\_}{50\text{kN}}$.

Thus, the shear at point A is $\underset{\_}{91.7\text{kN}}$.

Thus, the moment at point A is $\underset{\_}{1,333.6\text{kN-m}}$.

Consider section bb:

Consider the right side of the section bb for calculation of internal forces.

Show the free-body diagram of the right side of the section bb as in Figure 4.

Find the axial force at point B by resolving the horizontal equilibrium.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ -Q_B+0=0\\ Q_B=0\end{array}$

Find the shear at point B by resolving the vertical equilibrium.

$\begin{array}{l}+\downarrow {\displaystyle \sum F_y=0}\\ -S_B-194.9=0\\ S_B=-194.9\text{kN}\end{array}$

Find the moment at point B by taking moment about the section bb.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M=0}\\ -M_B+194.9\left(3\right)=0\\ M_B=584.7\text{kN-m}\end{array}$

Thus, the axial force at point B is $\underset{\_}{0}$.

Thus, the shear at point B is $\underset{\_}{-194.9\text{kN}}$.

Thus, the moment at point B is $\underset{\_}{584.7\text{kN-m}}$.