Chapter 7, Problem 1FP

Determine the normal force, shear force, and moment at point C.

Prob. F7-1

To determine

The normal force, shear force, and moment at point C.

Answer to Problem 1FP

The normal force at point C, $N_C$ is $\underset{\_}{\text{zero}}$.

The shear force at point C, $V_C$ is $\underset{\_}{1.25\text{kN}}$.

The moment at point C, $M_C$ is $\underset{\_}{18.75\text{kN-m}}$.

Explanation of Solution

Assumption:

• Consider the state of the member as tension, where the force is pulling the member, and as compression, where the force is pushing the member.
• The normal force is positive if it creates tension; a positive shear force acting on the segment causes it to rotate clockwise, and a positive bending moment acting on the segment will cause it to bend and concave upwards. Loadings that are opposite to these are considered negative.
• Consider the force indicating the right side as positive and the left side as negative, in the horizontal components of forces.
• Consider the force indicating upside as positive and downside as negative, in the vertical components of forces.
• Consider the clockwise movement as negative and the anti-clockwise movement as positive, wherever applicable.
• The method of sections can be used to determine the internal loadings.

Determine the reactions:

Entire beam:

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

Using Figure (1),

Moment at point A:

Determine the vertical reaction at point B by taking the moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ B_y\left(6\right)-15\left(4.5\right)-10\left(1.5\right)=0\end{array}$ (I)

Solve Equation (I).

$\begin{array}{l}6B_y-67.5-15=0\\ 6B_y=82.5\\ B_y=\frac{82.5}{6}\\ B_y=13.75\text{kN}\end{array}$

Determine the forces and moment:

Segment CB:

Show the free body diagram of the segment CB as in Figure (2).

Using Figure (2),

Along the vertical direction:

Determine the shear force at point C by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ V_C+B_y-15=0\end{array}$ (II)

Along the horizontal direction:

Determine the normal force at point C by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ N_C=0\end{array}$ (III)

Moment about point C:

Determine the moment at point C by taking the moment about point C.

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ B_y\left(3\right)-15\left(1.5\right)-M_C=0\end{array}$ (IV)

Conclusion:

Substitute 13.75 kN for $B_y$ in Equation (II).

$\begin{array}{l}{V}_C+13.75-15=0\\ V_C=1.25\text{kN}\end{array}$

Thus, the shear force at point C, $V_C$ is $\underset{\_}{1.25\text{kN}}$.

Refer to Equation (III).

$N_C=0$

Thus, the normal force at point C, $N_C$ is $\underset{\_}{\text{zero}}$.

Substitute 13.75 kN for $B_y$ in Equation (IV).

$\begin{array}{l}13.75\left(3\right)-15\left(1.5\right)-M_C=0\\ 41.25-22.5-M_C=0\\ M_C=18.75\text{kN-m}\end{array}$

Thus, the moment at point C, $M_C$ is $\underset{\_}{18.75\text{kN-m}}$.