Chapter 5, Problem 1P

To determine

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

Answer to Problem 1P

The axial force at point A is $\underset{\_}{-40\text{kN}}$.

The shear force at point A is $\underset{\_}{32.14\text{kN}}$.

The bending moment at point A is $\underset{\_}{524.98\text{kN}\cdot \text{m}}$.

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

The shear force at point B is $\underset{\_}{-87.14\text{kN}}$.

The bending moment at point B is $\underset{\_}{261.42\text{kN}\cdot \text{m}}$.

Explanation of Solution

Given information:

The point load acting at the distance of 5 m from the left support $\left(P_1\right)$ is 60 kN.

The inclined point load acting at the distance of 10 m from the left support $\left(P_2\right)$ is 80 kN with angle of $60°$.

The point load acting at the distance of 15 m from the left support $\left(P_3\right)$ is 50 kN.

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:

Draw the free body diagram of the entire beam as in Figure (1).

Determine the horizontal force at point C using equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ C_x-P_2\cos 60°=0\end{array}$

Substitute 80 kN for $P_2$.

$\begin{array}{l}{C}_x-80\cos 60°=0\\ C_x=40\text{kN}(\to )\end{array}$

Consider clockwise moment as positive and counter clockwise moment as negative.

Determine the vertical force at the left support using equilibrium condition.

Taking moment about point D.

$\begin{array}{l}{\displaystyle \sum M_D}=0\\ C_y\times L-P_1\times \left(L-l_1\right)-P_2\sin 60°\left(l_4+l_5+l_6\right)-P_3\left(l_5+l_6\right)=0\end{array}$

Substitute 20 m for L, 60 kN for $P_1$, 5 m for $l_1$, 80 kN for $P_2$, 5 m for $l_4$, 2 m for $l_5$, 3 m for $l_6$, and 50 kN for $P_3$.

$\begin{array}{l}{C}_y\times 20-60\times \left(20-5\right)-80\sin 60°\left(5+2+3\right)-50\left(2+3\right)=0\\ 20C_y=1,842.8\\ C_y=\frac{1,842.8}{20}\\ C_y=92.14\text{kN}\end{array}$

Determine the vertical force at the right support using equilibrium condition.

$\begin{array}{l}{\displaystyle \sum F_Y}=0\\ C_y+D_y-P_1-P_2\sin 60°-P_3=0\end{array}$

Substitute 92.14 kN for $C_y$, 60 kN for $P_1$, 80 kN for $P_2$, and 50 kN for $P_3$.

$\begin{array}{l}92.14+D_y-60-80\sin 60°-50=0\\ D_y=87.14\text{kN}\end{array}$

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

Draw the sections aa and bb as in Figure (2).

Consider section aa.

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

Axial force:

Consider the external forces acting to the left as positive.

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

$\begin{array}{l}+\leftarrow {\displaystyle \sum F_x=0}\\ 40+Q_A=0\\ Q_A=-40\text{kN}\end{array}$

Therefore, the axial force at point A is $\underset{\_}{-40\text{kN}}$.

Shear force:

Consider the external forces acting upward as positive.

Determine the shear at point A using the relation.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ C_y-P_1-S_A=0\end{array}$

Substitute 92.14 kN for $C_y$ and 60 kN for $P_1$,

$\begin{array}{l}92.14-60-S_A=0\\ S_A=32.14\text{kN}\end{array}$

Therefore, the shear force at point A is $\underset{\_}{32.14\text{kN}}$.

Bending moment:

Consider the clockwise moments of the external forces about A as positive.

Determine the moment at point A equilibrium equations.

Taking moment about point A.

$\begin{array}{l}+\circlearrowright {\displaystyle \sum M=0}\\ -M_A+C_y\times \left(l_1+l_2\right)-P_1\times l_2=0\end{array}$

Substitute 92.14 kN for $C_y$, 5 m for $l_1$, 2 m for $l_2$, and 60 kN for $P_1$.

$\begin{array}{l}-M_A+92.14\times \left(5+2\right)-60\times 2=0\\ M_A=524.98\text{kN}\cdot \text{m}\end{array}$

Therefore, the bending moment at point A is $\underset{\_}{524.98\text{kN}\cdot \text{m}}$.

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.

Axial force:

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

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

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

Shear force:

Determine the shear at point B using the relation.

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

Substitute 87.14 kN for $D_y$.

$\begin{array}{l}-S_B-87.14=0\\ S_B=-87.14\text{kN}\end{array}$

Therefore, the shear force at point B is $\underset{\_}{-87.14\text{kN}}$.

Bending moment:

Consider the counter clockwise moments of the external forces about B as positive.

Determine the moment at point B equilibrium equations.

Taking moment about point B.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M=0}\\ -M_B+D_y\times l_6=0\end{array}$

Substitute 87.14 kN for $D_y$ and 3 m for $l_6$.

$\begin{array}{l}-M_B+87.14\times 3\\ M_B=261.42\text{kN}\cdot \text{m}\end{array}$

Therefore, the bending moment at point B is $\underset{\_}{261.42\text{kN}\cdot \text{m}}$.