Textbook Problem

Find support reactions at A and B and then calculate the axial force N. shear force J and bending moment M at mid-span of AB. Let L = 14 ft. q₀ = 12 lb/ft. P = 50 lb. and = 300 lb-ft.

  

To determine

You need to calculate support reactions at $A$and $B$; and then find axial force $N$, shear force $V$and bending moment $M$at mid span of the $AB$in the figure given below.

Answer to Problem 1.3.1P

The external reaction forces are $B_x=30$lb, $B_y=94.57$lb, $A_y=29.4$lb and internal reaction forces are axial force $N=0$, shear force $V=8.42$lb and moment $M=-143.06$lb-ft.

Explanation of Solution

Given Information:

You have a beam with different types of loading as shown in figure below.

Figure 1 It is given that, $L=14$ft, $q_0=12$lb/ft, $P=50$lb, and $M_0=300$lb-ft.

Calculation:

To find external support reaction forces, draw free body diagram as shown in figure below.

Figure 2: Free body diagram for external reactions Take equilibrium in $x-$direction (horizontal direction),

${\displaystyle \sum F_x=0\Rightarrow B_x=\frac{3}{5}P}=\frac{3}{5}\times 50\text{ lb}=30\text{ lb}$

Take equilibrium of moments about point $A$,

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ \Rightarrow M_0+B_{y}L=\frac{1}{2}{q}_{0}L\frac{2L}{3}+\frac{4}{5}P\left(L+\frac{L}{2}\right)\\ \Rightarrow B_y=\frac{1}{L}\left(-M_0+\frac{1}{2}{q}_{0}L\frac{2L}{3}+\frac{4}{5}P\left(L+\frac{L}{2}\right)\right)\\ \Rightarrow B_y=\frac{1}{14}\left(-300+\frac{1}{2}\times 12\times 14\times \frac{2\times 14}{3}+\frac{4}{5}\times 50\times \left(14+\frac{14}{2}\right)\right)\\ \Rightarrow B_y=94.57\text{ lb}\end{array}$

Take equilibrium in $y-$direction (vertical direction),

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y+B_y=\frac{4}{5}P+\frac{1}{2}{q}_{0}L\\ \Rightarrow A_y=-B_y+\frac{4}{5}P+\frac{1}{2}{q}_{0}L\\ \Rightarrow A_y=-94.57+\frac{4}{5}\times 50+\frac{1}{2}\times 12\times 14\\ \Rightarrow A_y=29.4\text{ lb}\end{array}$

To find internal forces at mid span of AB, draw the following free body diagram.

Figure 3 Free body diagram for internal forces Now take left hand side free body diagram.

Take equilibrium in $x-$direction (horizontal direction),

${\displaystyle \sum F_x=0\Rightarrow N=0}$

Take equilibrium in $y-$direction (vertical direction),

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \Rightarrow A_y=V+\frac{1}{2}\frac{q_0}{2}\frac{L}{2}\\ \Rightarrow V=A_y-\frac{1}{2}\frac{q_0}{2}\frac{L}{2}\\ \Rightarrow V=29.4-\frac{1}{2}\times \frac{12}{2}\times \frac{14}{2}\\ \Rightarrow V=8.42\text{ lb}\end{array}$

Take equilibrium of moments about point $A$,

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ \Rightarrow M_0+M=\frac{1}{2}\frac{q_{0}L}{4}\frac{2L}{3\times 2}+V\frac{L}{2}\\ \Rightarrow M=-M_0+\frac{1}{2}\frac{q_{0}L}{4}\frac{2L}{3\times 2}+V\frac{L}{2}\\ \Rightarrow M=-300+\frac{1}{2}\times \frac{12\times 14}{4}\times \frac{2\times 14}{3\times 2}+8.42\times \frac{14}{2}\\ \Rightarrow M=-143.06\text{ lb-ft}\end{array}$

Conclusion:

Thus the external reaction forces are $B_x=30$lb, $B_y=94.57$lb, $A_y=29.4$lb and internal reaction forces are axial force $N=0$, shear force $V=8.42$lb and moment $M=-143.06$lb-ft.