Chapter 13, Problem 1P

To determine

Find the reactions and sketch the shear and bending moment diagrams for the given beam using method of consistent deformation.

Answer to Problem 1P

The reactions acting on the beam are $\underset{\_}{A_x=0}$, $\underset{\_}{A_y=99.26\text{kN}(\uparrow )}$, $\underset{\_}{D_y=60.74\text{kN}(\uparrow )}$, and $\underset{\_}{M_A=233.34\text{kN}\cdot \text{m}}$.

Explanation of Solution

Given information:

Apply the sign conventions for calculating reactions, forces, and moments 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:

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

Refer Figure 1.

Consider the horizontal and vertical reaction at A are denoted by $A_x$ and $A_y$.

Consider the moment at A is denoted by $M_A$.

Consider the vertical reaction at D is denoted by $D_y$.

The reactions acting in the beam is 4.

The number of Equilibrium reaction is 3. Then,

The degree of indeterminacy of the beam is 1.

Take the vertical reaction at D as the redundant.

Modify the Figure 1 as shown in Figure 2.

Refer Figure 2.

The deflection at D due to the external moment is denoted by ${\Delta }_D$.

The deflection coefficient representing the deflection at D due to unit value of redundant $D_y$ is $f_{DD}$.

The deflection at D due to unknown redundant $D_y$ is $f_{DD}{D}_y$.

Show the deflection at the free end of the cantilever beam with a point load P acting at a distance from the fixed end as follows:

$\Delta =-\frac{P{a}^2}{6EI}\left(3l-a\right)$ (1)

Here, l is the length of the beam, E is modulus of Elasticity of the beam, I is the moment of inertia.

Find the deflection at the free end of the beam due to the load of $60\text{kN}$ as follows:

Substitute $3\text{m}$ for a, $9\text{m}$ for l, and $60\text{kN}$ for P in Equation (1).

$\begin{array}{c}{\Delta }_1=-\frac{60\times 3^2}{6EI}\left(3\times 9-3\right)\\ =-\frac{2160}{EI}\end{array}$

Find the deflection at the free end of the beam due to the load of $100\text{kN}$ as follows:

Substitute $6\text{m}$ for a, $9\text{m}$ for l, and $100\text{kN}$ for P in Equation (1).

$\begin{array}{c}{\Delta }_2=-\frac{100\times 6^2}{6EI}\left(3\times 9-6\right)\\ =-\frac{12600}{EI}\end{array}$

Calculate the deflection at D due to the loading on the beam as follows:

$\begin{array}{c}{\Delta }_D={\Delta }_1+{\Delta }_2\\ =-\frac{2160}{EI}-\frac{12600}{EI}\\ =-\frac{14760\text{kN}\cdot {\text{m}}^3}{EI}\end{array}$ (2)

Calculate the value of $f_{DD}$ as follows:

Substitute $9\text{m}$ for a, $9\text{m}$ for l, and $1\text{kN}$ for P in Equation (1).

$\begin{array}{c}{f}_{DD}=\frac{1\times 9^2}{6EI}\left(3\times 9-9\right)\\ =\frac{243\text{kN}\cdot {\text{m}}^3/\text{kN}}{EI}\end{array}$ (3)

Show the compatibility Equation as follows:

${\Delta }_D+f_{DD}\times D_y=0$

Modify the above Equation using Equation (2) and (3).

$\begin{array}{l}{\Delta }_D+f_{DD}\times D_y=0\\ -\frac{14760\text{kN}\cdot {\text{m}}^3}{EI}+\frac{243\text{kN}\cdot {\text{m}}^3/\text{kN}}{EI}\times D_y=0\\ D_y=\left(\frac{EI}{243\text{kN}\cdot {\text{m}}^3/\text{kN}}\times \frac{14760\text{kN}\cdot {\text{m}}^3}{EI}\right)\\ D_y=60.74\text{kN}(\uparrow )\end{array}$

Refer Figure 1.

Take the sum of the forces in the vertical direction as zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ A_y+D_y-60-100=0\\ A_y+60.74-160=0\\ A_y=99.26\text{kN}\\ A_y=99.26\text{kN}(\uparrow )\end{array}$

Take the sum of the forces in the horizontal direction as zero.

$\begin{array}{l}{\displaystyle \sum F_x}=0\\ A_x=0\end{array}$

Calculate the moment at A as follows:

$\begin{array}{c}{M}_A=D_y\times 9-\left(100\times 6\right)-\left(60\times 3\right)\\ =\left(60.74\times 9\right)-\left(100\times 6\right)-\left(60\times 3\right)\\ =-233.34\text{kN}\end{array}$

Take the moment at D as zero.

$\begin{array}{l}{\displaystyle \sum M_D}=0\\ M_A-A_y\times 9+60\times 6+100\times 3=0\end{array}$

Substitute $99.26\text{kN}$ for $A_y$.

$\begin{array}{l}{M}_A-99.26\times 9+60\times 6+100\times 3=0\\ M_A-233.34=0\\ M_A=233.34\text{kN}\cdot \text{m}\end{array}$

Thus, the reactions acting on the beam are $\underset{\_}{A_x=0}$, $\underset{\_}{A_y=99.26\text{kN}(\uparrow )}$, $\underset{\_}{D_y=60.74\text{kN}(\uparrow )}$, and $\underset{\_}{M_A=233.34\text{kN}\cdot \text{m}}$.

Show the loading on the beam as shown in the Figure 3.

Refer Figure 3.

Find the shear force at support A as follows:

$\begin{array}{c}{V}_A=A_y\\ =99.26\text{kN}\end{array}$

Find the shear force at support B as follows:

$\begin{array}{c}{V}_B=A_y-60\\ =99.26-60\\ =39.26\text{kN}\end{array}$

Find the shear force at support C as follows:

$\begin{array}{c}{V}_C=A_y-60-100\\ =99.26-60-100\\ =-60.74\text{kN}\end{array}$

Find the shear force at support D as follows:

$\begin{array}{c}{V}_D=D_y\\ =60.74\text{kN}\end{array}$

Calculate the bending moment at A as follows:

$M_A=-233.34\text{kN}\cdot \text{m}$

Calculate the bending moment at B as follows:

$\begin{array}{c}{M}_B=M_A-A_y\times 3\\ =233.34-99.26\times 3\\ =64.44\text{kN}\cdot \text{m}\end{array}$

Calculate the bending moment at C as follows:

$\begin{array}{c}{M}_C=M_A-A_y\times 6+60\times 3\\ =233.34-99.26\times 6+180\\ =182.22\text{kN}\cdot \text{m}\end{array}$

Calculate the bending moment at D as follows:

$\begin{array}{c}{M}_D=M_A-A_y\times 9+60\times 6+100\times 3\\ =233.34-99.26\times 9+360+300\\ =0\text{kN}\cdot \text{m}\end{array}$

Sketch the shear force and bending moment diagram as shown in Figure 4.