Chapter 9, Problem 1P

To determine

Find the reactions and sketch the shear and moment diagram.

Locate the point of maximum deflection for the beam.

Answer to Problem 1P

The reaction at A and C are $\underset{\_}{R_A=20.45\text{k}}$ and $\underset{\_}{R_C=15.55\text{k}}$.

The bending moment at A is $\underset{\_}{M_A=-90.75\text{kips}\cdot \text{ft}\left(\text{Anticlockwise}\right)}$.

Explanation of Solution

Given information:

EI is constant.

Calculation:

Consider the vertical reaction at C is denoted by $R_C$.

Sketch the beam as shown in Figure 1.

Refer Figure 1.

The horizontal reaction at A is $A_x=0$.

The vertical reaction at C is the constraint in the structure.

Select the vertical reaction at C as the redundant.

Remove the redundant from the beam to get the released structure.

Calculate the gap $\left({\Delta }_{CO}\right)$ produced by the applied load on the released structure using the relation;

$\begin{array}{c}{\Delta }_{CO}=\frac{w{L}_1^3}{3EI}+\left(\frac{w{L}_1^2}{2EI}\times L_2\right)\\ =\frac{36\times 9^3}{3EI}+\left(\frac{36\times 9^2}{2EI}\times 6\right)\\ =\frac{8,748}{EI}+\frac{8,748}{EI}\\ =\frac{17,496}{EI}\end{array}$

Calculate the gap ${\delta }_{CC}$ produced by the unit load $w=1\text{kip}$ applied on the released structure using the relation:

$\begin{array}{c}{\delta }_{CC}=\frac{w{L}^3}{3EI}\\ {\delta }_{CC}=\frac{\left(1\right)\times {15}^3}{3EI}\\ =\frac{1,125}{EI}\end{array}$

The deflection ${\Delta }_C$ of the beam at the hinge support is zero.

Refer Figure 1.

Take the upward deflection as positive and downward deflection as negative.

Apply the compatibility Equation as follows:

$\begin{array}{l}{\Delta }_C={\Delta }_{CO}+{\Delta }_{CC}\\ {\Delta }_C={\Delta }_{CO}+{\delta }_{CC}\times R_C\\ 0=\frac{-17,496}{EI}+\frac{1,125}{EI}{R}_C\\ R_C=\frac{17,496}{EI}\times \frac{EI}{1,125}\\ R_C=15.55\text{k}\end{array}$

Thus, the reaction at C is $\underset{\_}{R_C=15.55\text{k}}$.

Find the reaction $R_A$ at A as follows:

$\begin{array}{l}{R}_A+R_C=36\\ R_A+15.55=36\\ R_A=20.45\text{k}\end{array}$

Thus, the reaction at A is $\underset{\_}{R_A=20.45\text{k}}$.

Find the moment at A as follows:

$\begin{array}{c}{\displaystyle \sum M_A}=0\\ M_A=R_C\times 15-36\times 9\\ =15.55\times 15-36\times 9\\ =-90.75\text{kip}\cdot \text{ft}\end{array}$

Thus, the bending moment at A is $\underset{\_}{M_A=90.75\text{kip}\cdot \text{ft}\left(\text{Anticlockwise}\right)}$.

Find the moment at B as follows:

$\begin{array}{c}{\displaystyle \sum M_B}=0\\ M_B=R_C\times 6\\ =15.55\times 6\\ =93.3\text{kips}\cdot \text{ft}\end{array}$

The bending moment $M_x$ at the point of contra flexure is zero.

Consider the point of contra flexure is at a distance x from the fixed support A as follows:

$\begin{array}{l}{M}_x-M_A=R_A\left(x\right)\\ 0-\left(-90.72\text{kip}\cdot \text{ft}\right)=\left(20.45\text{kips}\right)\times x\\ x=\frac{90.72}{20.45}\\ x=4.436\text{ft}\end{array}$

The bending moment $M_C$ at the free end C is zero.

Find the shear-force at A, B, and C as follows:

$\begin{array}{c}{V}_A=R_A\\ =20.45\text{k}\end{array}$

$\begin{array}{c}{V}_{B,left}=R_A\\ =20.45\text{k}\end{array}$

$\begin{array}{c}{V}_{B,right}=R_A-36\\ =20.45-36\\ =-15.55\text{k}\end{array}$

$\begin{array}{c}{V}_C=R_A-36+15.55\\ =20.45-36+15.55\\ =-15.55\text{+15}\text{.55}\\ =0\end{array}$

Show the shear force, bending moment diagram as shown in Figure 2.

Refer Figure 2.

The point of maximum deflection occurs at 9 ft from the left support.