Chapter 3, Problem 1P

To determine

Calculate the support reactions of the beam.

Answer to Problem 1P

The horizontal reaction at A is $\underset{\_}{A_x=6\text{kips}(\to )}$.

The vertical reaction at A is $\underset{\_}{A_y=8.62\text{kips}(\uparrow )}$.

The vertical reaction at B is $\underset{\_}{B_y=19.38\text{kips}(\uparrow )}$.

Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions 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 $A_x$ and $A_y$ be the horizontal and vertical reactions at the hinged support A and $B_y$ be the vertical reaction at the roller support B.

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

Refer to Figure 1.

Use equilibrium equations:

Find the vertical reaction at B:

Summation of moment about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -B_y\times 13+5\times 4\times \left(\frac{4}{2}+9\right)+10\left(\frac{4}{5}\right)\times 4=0\\ B_y=19.38\text{kips}(\uparrow )\end{array}$

Therefore, the vertical reaction at B is $\underset{\_}{B_y=19.38\text{kips}(\uparrow )}$.

Find the horizontal reaction at A:

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x-10\left(\frac{3}{5}\right)=0\\ A_x=6\text{kips}(\to )\end{array}$

Therefore, the horizontal reaction at A is $\underset{\_}{A_x=6\text{kips}(\to )}$.

Find vertical reaction at A:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y-10\left(\frac{4}{5}\right)-5\times 4+19.38=0\\ A_y=8.62\text{kips}(\uparrow )\end{array}$

Therefore, the vertical reaction at A is $\underset{\_}{A_y=8.62\text{kips}(\uparrow )}$.