Chapter 6, Problem 6.31P

For the beam shown, derive the expressions for V and M, and draw the shear force and bending moment diagrams. Neglect the weight of the beam.

To determine

Derive equation for $V$ and $M$, draw the shear force and bending moment diagram.

Answer to Problem 6.31P

The equation for $V$ can be expressed as $V=\{\begin{array}{l}900-100x^{2}0\le x\le 3\text{ ft}\\ 100x^2-900x+18003\text{ ft}\le x\le 6\text{ ft}\end{array}$ and the equation for $M$ can be expressed as $M=\{\begin{array}{l}900x-33.33x^{3}0\le x\le 3\text{ft}\\ 33.33x^3-600x^2+2700x-18003\text{ ft}\le x\le 6\text{ ft}\end{array}$, the shear force diagram is illustrated in figure (a) and Bending moment diagram is illustrated in figure (b).

Explanation of Solution

Draw the Free body diagram for the beam as shown below.

  

Write the expression for the equilibrium of forces in $y$ -direction.

  ${\displaystyle \sum F_y=0}$

Here, ${\displaystyle \sum F_y}$ is the summation of all the forces in $y$ -direction.

Take the summation of all the forces in $y$ -direction.

  $A_y+N_c-\frac{1}{2}\left(6\right)\left(600\right)=0$

Simplify the above expression.

  $A_y+N_c=1800$ ...... (1)

Here, $A_y$ is the vertical reaction at point $A$ and $N_c$ is the reaction provided by support at $C$.

Take moment about point $A$.

  $\begin{array}{c}{\displaystyle \sum M_A}=N_c\left(6\right)-\frac{1}{2}\left(600\right)\left(6\right)\left(3\right)\\ =6N_c-5400\end{array}$

Substitute $0$ for ${\displaystyle \sum M_A}$ in above expression and simplify for $N_C$.

  $\begin{array}{c}{N}_C=\frac{5400}{6}\\ =900\text{}\text{lb}\end{array}$

Substitute $900\text{ lb}$ for $N_C$ in equation (1).

  $A_y+900=1800$

Simplify the above for $A_y$.

  $A_y=900\text{ lb}$

Take section $x-x$ in between points $A$ and $B$.

  

From the above diagram, by the property of similar triangles magnitude of UVL at distance $x$ is.

  $\frac{600}{3}=\frac{P}{x}$

Simplify the above expression for $P$.

  $P=200x$

Here, $P$ is the magnitude of load at section $x-x$.

Take shear force on left side of the section $x-x$.

  $\begin{array}{l}{V}_x=A_y-\frac{1}{2}\left(200x\right)\left(x\right)\\ V_x=A_y-100x^2\end{array}$

Here, $V_x$ is the shear force at section $x-x$.

Substitute $900\text{ lb}$ for $A_y$ in above expression.

  $V_x=900-100x^2$ ...... (2)

Take moment about section $x-x$.

  $\begin{array}{c}{M}_x=-\frac{1}{2}\left(200x\right)\left(x\right)\left(\frac{x}{3}\right)+A_y\left(x\right)\\ =-\frac{100x^3}{3}+A_{y}x\end{array}$

Here, $M_x$ is the moment about section $x-x$.

Substitute $900\text{ lb}$ for $A_y$ in above expression.

  $M_x=900x-33.33x^3$ ...... (3)

Take section $x-x$ in between points $B$ and $C$.

  

From the above diagram, by the property of similar triangles magnitude of UVL at distance $x$ is.

  $\frac{600}{3}=\frac{P}{\left(6-x\right)}$

Simplify the above expression for $P$.

  $P=1200-200x$

Take shear force on left side of the section $x-x$.

  $\begin{array}{c}{V}_x=A_y-\frac{1}{2}\left(600\right)\left(3\right)-\frac{1}{2}\left(1200-200x\right)\left(x-3\right)\\ =A_y-900-\left(600x-1800-100x^2+300x\right)\\ =A_y+900-900x+100x^2\end{array}$

Here, $V_x$ is the shear force at section $x-x$.

Substitute $900\text{ lb}$ for $A_y$ in above expression.

  $V_x=100x^2-900x+1800$ ...... (4)

Take moment about section $x-x$.

  $\begin{array}{c}{M}_x=N_c\left(6-x\right)-\frac{1}{2}\left(1200-200x\right)\left(6-x\right)\left(\frac{6-x}{3}\right)\\ =N_c\left(6-x\right)-\left(600-100x\right)\left(\frac{{\left(6-x\right)}^2}{3}\right)\\ =N_c\left(6-x\right)-\left(7200+200x^2-2400x-1200x-33.33x^3+400x^2\right)\\ =6N_c-x{N}_c-7200-600x^2+3600x+33.33x^3\end{array}$

Substitute $900\text{}\text{lb}$ for $N_c$ in above expression.

  $M_x=6\left(900\right)-x\left(900\right)-7200-600x^2+3600x+33.33x^3$

Simplify the above expression.

  $M_x=33.33x^3-600x^2+2700x-1800$ ...... (5)

Thus, from equation (2) and (4) the equation for $V$ can be expressed as $V=\{\begin{array}{l}900-100x^{2}0\le x\le 3\text{ ft}\\ 100x^2-900x+18003\text{ ft}\le x\le 6\text{ ft}\end{array}$ and from equation (3) and (5) the equation for $M$ can be expressed as $M=\{\begin{array}{l}900x-33.33x^{3}0\le x\le 3\text{ft}\\ 33.33x^3-600x^2+2700x-18003\text{ ft}\le x\le 6\text{ ft}\end{array}$.

Draw the shear force diagram for the beam as shown below.

  

Draw the Bending moment diagram for the beam as shown below.