Chapter 4, Problem 99RP

A T-shaped bracket supports the four loads shown. Determine the reactions at A and B (a) if $a=10$ in., (b) if $a=7$ in.

To determine

(a)

The reaction at A and B for the T shaped bracket.

Answer to Problem 99RP

The reaction at A is $-20\widehat{j}\text{ lb}$ and reaction at B is $150\widehat{j}\text{ lb}$.

Explanation of Solution

Given:

The value of $a$ is $10\text{ in}$

Concept used:

Write the expression for net force in $x$ direction.

$\sum F_x=0$

Write the expression for reaction at B in $x$ direction.

$R_x^B=0$

Here, $R_x^B$ is the reaction force applied at $x$ direction.

Write the expression for moment about A.

${\overrightarrow{M}}_A=\left[\begin{array}{l}\left(6\widehat{i}\times -40\widehat{j}\right)+\left(12\widehat{i}\times -50\widehat{j}\right)+\left(\left(12+a\right)\widehat{i}\times -30\widehat{j}\right)\\ +\left(\left(20+a\right)\widehat{i}\times -10\widehat{j}\right)+\left(12\widehat{i}\times R_B\widehat{j}\right)\end{array}\right]$ ....... (1)

Write the expression for moment about B.

${\overrightarrow{M}}_B=\left[\left(6\widehat{i}\times 40\widehat{j}\right)+\left(\left(a\right)\widehat{i}\times -30\widehat{j}\right)+\left(\left(8+a\right)\widehat{i}\times -10\widehat{j}\right)+\left(6\widehat{i}\times R_A\widehat{j}\right)\right]$ ....... (2)

Calculation:

The moment about A is zero due to bracket is in equilibrium condition.

Substitute $0$ for ${\overrightarrow{M}}_A$ in equation (1)

$\begin{array}{c}0=\left[\begin{array}{l}\left(6\widehat{i}\times -40\widehat{j}\right)+\left(12\widehat{i}\times -50\widehat{j}\right)+\left(\left(12+a\right)\widehat{i}\times -30\widehat{j}\right)\\ +\left(\left(20+a\right)\widehat{i}\times -10\widehat{j}\right)+\left(12\widehat{i}\times R_B\widehat{j}\right)\end{array}\right]\\ 0=-240\widehat{k}-600\widehat{k}-360a\widehat{k}-30a\widehat{k}-200\widehat{k}-10a\widehat{k}-12R_B\widehat{k}\\ 12R_B\widehat{k}=1400\widehat{k}+40a\widehat{k}\end{array}$

The expression for the resultant at B is,

$R_B=\frac{1400+40a}{12}$ ....... (3)

The moment about B is zero due to bracket is in equilibrium condition.

Substitute $0$ for ${\overrightarrow{M}}_B$ in equation (1)

$\begin{array}{c}0=\left[\left(6\widehat{i}\times 40\widehat{j}\right)+\left(\left(a\right)\widehat{i}\times -30\widehat{j}\right)+\left(\left(8+a\right)\widehat{i}\times -10\widehat{j}\right)+\left(6\widehat{i}\times R_A\widehat{j}\right)\right]\\ 0=240\widehat{k}-30a\widehat{k}-80\widehat{k}-10a\widehat{k}+12R_A\widehat{k}\\ 12R_A\widehat{k}=40a\widehat{k}-160\widehat{k}\\ R_A=\frac{40a-160}{12}\end{array}$

The expression for the resultant at A is,

$R_A=\frac{40a-160}{12}$ ....... (4)

Substitute $10$ for $a$ in equation (3)

$\begin{array}{c}{R}_B=\frac{1400+40\left(10\right)}{12}\\ =150\text{ lb}\end{array}$

Substitute $10$ for $a$ in equation (4)

$\begin{array}{c}{R}_A=\frac{40\left(10\right)-160}{12}\\ =20\text{ lb}\end{array}$

Conclusion:

Thus, the reaction at A is $-20\widehat{j}\text{ lb}$ and reaction at B is $150\widehat{j}\text{ lb}$.

To determine

(b)

The reaction at A and B for the T shaped bracket.

Answer to Problem 99RP

The reaction at A is $-10\widehat{j}\text{ lb}$ and reaction at B is $140\widehat{j}\text{ lb}$.

Explanation of Solution

Given:

The value of $a$ is $7\text{ in}$

Calculation:

Substitute $7$ for $a$ in equation (3)

$\begin{array}{c}{R}_B=\frac{1400+40\left(7\right)}{12}\\ =140\text{ lb}\end{array}$

Substitute $7$ for $a$ in equation (4)

$\begin{array}{c}{R}_A=\frac{40\left(7\right)-160}{12}\\ =10\text{ lb}\end{array}$

Conclusion:

Thus, the reaction at A is $-10\widehat{j}\text{ lb}$ and reaction at B is $140\widehat{j}\text{ lb}$.