Chapter 3, Problem 15P

To determine

Find the expression for vertical reaction at the supports in terms of distance x.

Sketch the graph for reactions as a function of x.

Answer to Problem 15P

The vertical reaction at A for the interval $0\le x\le 20\text{m}$ is $\underset{\_}{A_y=45-2x\text{kN}\uparrow }$.

The vertical reaction at B for the interval $0\le x\le 20\text{m}$ is $\underset{\_}{B_y=5+2x\text{kN}\uparrow }$.

The horizontal reaction at A for the interval $0\le x\le 20\text{m}$ is $\underset{\_}{A_x=0}$.

The vertical reaction at A for the interval $20\text{m}\le x\le 25\text{m}$ is $\underset{\_}{A_y=\frac{{\left(25-x\right)}^2}{5}\text{}\text{kN}\uparrow }$.

The vertical reaction at B for the interval $20\text{m}\le x\le 25\text{m}$ is $\underset{\_}{B_y=\frac{\left(625-x^2\right)}{5}\text{kN}\uparrow }$.

Explanation of Solution

Given information:

The structure is given in the Figure.

Length of the trolley is 5 m.

Apply the sign conventions for calculating reaction forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.
• For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Let the roller at B exerts the vertical reaction $B_y$.

Let $A_x$ and $A_y$ be the horizontal and vertical reactions at the hinged support A.

Sketch the free body diagram of the beam bridge as shown in Figure 1.

Use equilibrium equations for the interval $0\le x\le 20\text{m}$:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ -A_y\left(25\right)+10\left(5\right)\left(25-\left(x+2.5\right)\right)=0\\ -25A_y=-50\left(25-x-2.5\right)\\ -25A_y=-50\left(22.5-x\right)\end{array}$

$\begin{array}{c}{A}_y=\frac{1,125-50x}{25}\\ =45-2x\text{kN}\uparrow \end{array}$

Therefore, the vertical reaction at A for the interval $0\le x\le 20\text{m}$ is $\underset{\_}{A_y=\left(45-2x\right)\text{kN}\uparrow }$.

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y-10\left(5\right)+B_y=0\\ B_y=-A_y+50\end{array}$

Substitute $\left(45-2x\right)\text{kN}$ for $A_y$.

$\begin{array}{c}{B}_y=-\left(45-2x\right)+50\\ =-45+2x+50\\ =\left(5+2x\right)\text{kN}\uparrow \end{array}$

Therefore, the vertical reaction at B for the interval $0\le x\le 20\text{m}$ is $\underset{\_}{B_y=\left(5+2x\right)\text{kN}\uparrow }$.

Summation of forces along x-direction is equal to 0.

$\begin{array}{l}+\to \sum F_x=0\\ A_x=0\end{array}$

Therefore, the horizontal reaction at A for the interval $0\le x\le 20\text{m}$ is $\underset{\_}{A_x=0}$.

Sketch the free body diagram of the beam bridge for the interval $20\text{m}\le x\le 25\text{m}$ as shown in Figure 2.

Use equilibrium equations for the interval $20\text{m}\le x\le 25\text{m}$:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ -A_y\left(25\right)+10\left(25-x\right)\left(\frac{25-x}{2}\right)=0\\ -25A_y=-5{\left(25-x\right)}^2\\ A_y=\frac{5{\left(25-x\right)}^2}{25}\end{array}$

$A_y=\frac{{\left(25-x\right)}^2}{5}\text{}\text{kN}\uparrow$

Therefore, the vertical reaction at A for the interval $20\text{m}\le x\le 25\text{m}$ is $\underset{\_}{A_y=\frac{{\left(25-x\right)}^2}{5}\text{}\text{kN}\uparrow }$.

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y-10\left(25-x\right)+B_y=0\\ B_y=-A_y+10\left(25-x\right)\end{array}$

Substitute $\frac{{\left(25-x\right)}^2}{5}\text{}\text{kN}$ for $A_y$.

$\begin{array}{c}{B}_y=-\frac{{\left(25-x\right)}^2}{5}+10\left(25-x\right)\\ =\frac{-{\left(25-x\right)}^2+50\left(25-x\right)}{5}\\ =\frac{-\left({25}^2-50x+x^2\right)+1,250-50x}{5}\\ =\frac{-625+50x-x^2+1,250-50x}{5}\end{array}$

$B_y=\frac{\left(625-x^2\right)}{5}\text{kN}\uparrow$

Therefore, the vertical reaction at B for the interval $20\text{m}\le x\le 25\text{m}$ is $\underset{\_}{B_y=\frac{\left(625-x^2\right)}{5}\text{kN}\uparrow }$.

Sketch the graph of the reaction $A_y$ vs. x as shown in Figure 3.

Sketch the graph of the reaction $B_y$ vs. x as shown in Figure 3.