Chapter 13, Problem 13.1P

through 13.6 Calculate the reactions at points A and BFor the beams shown.

To determine

The reaction forces at points A and B

Answer to Problem 13.1P

  $R_A=72kip$upward

  $R_B=72kip$upward

Explanation of Solution

Given:

The beam diagram and loading are given as shown below:

  

Concept Used:

Replacing the uniformly distribute load by a concentrated external load $W$ acting through the midpoint of the beam length.

  

Calculation:

  $W=6\text{ k/ft}\text{.}\times 24\text{ ft}\text{.=144 kip}$

Taking anticlockwise moment positive, the summation of all moment of forces at point A equals to zero

  $\begin{array}{l}{R}_B\times 24-W\times 12=0\\ R_B\times 24-144\times 12=0\\ R_B=\frac{144\times 12}{24}kip=72kip\end{array}$

For vertical equilibrium, the summation of forces in vertical direction must be zero

  $\begin{array}{l}{\displaystyle \sum F_V=0}\\ R_A+R_B-W=0\\ R_A=W-R_B\\ R_A=144-72=72kip\end{array}$

Conclusion:

Therefore, the reaction force at points A, $R_A=72kip\uparrow$ upward and the reaction force at points B, $R_B=72kip\uparrow$ upward.

To determine

The reaction forces at points A and B

Answer to Problem 13.1P

  $R_A=13.125k\uparrow$

  $R_B=6.875k\uparrow$

Explanation of Solution

Given:

The beam diagram and loading are given as shown below

  

Concept Used:

The reaction forces at point A and B are shown below in the figure and assumed to act in the upward (positive) direction

  

Calculation:

Taking anticlockwise moment positive, the summation of all moment of forces at point A equals to zero

  $\begin{array}{l}{R}_B\times 16-10\times 3-10\times 8=0\\ R_B\times 16=30+80\\ R_B=\frac{110}{16}k=6.875k\end{array}$

For vertical equilibrium, the summation of forces in vertical direction must be zero

  $\begin{array}{l}{\displaystyle \sum F_V=0}\\ R_A+R_B=10k+10k\\ R_A+6.875=10k+10k\\ R_A=(20-6.875)k=13.125k\end{array}$

Conclusion:

Therefore, the forces are $R_A=13.125k$ and $R_B=6.875k$ .