Textbook Problem

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13.9 through 13.12 Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P13.9–P13.12 using the method of consistent deformations. Select the reaction at the interior support to be the redundant.

FIG. P13.9, P13.30, P13.50

To determine

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

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 beam as shown in the 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 C is denoted by $M_C$.

Consider the vertical reaction at C and E is denoted by $C_y$ and $E_y$.

Find the degree of indeterminacy (DI).

The reactions (R) acting in the beam is 4.

The number of Equilibrium equation (E) is 3.

$\begin{array}{c}DI=R-E\\ =4-3\\ =1\end{array}$

Take the vertical reaction at C as the redundant.

Modify the Figure 1 as shown in Figure 2.

Refer Figure 2.

The deflection at C due to the external force is denoted by ${\Delta }_{CO}$.

The deflection coefficient representing the deflection at C due to unit value of redundant $C_y$ is $f_{CC}$.

The deflection at C due to unknown redundant $C_y$ is $f_{CC}{C}_y$.

Show the deflection at the mid of the beam due to the action of two point loads P, each acting at a distance of a from the supports as follows:

$\Delta =-\frac{Pab}{3lEI}\left(l^2-b^2-a^2\right)$        (1)

Here, l is the length of the beam, E is modulus of Elasticity of the beam, b is the distance between the loads, and I is the moment of inertia.

Find the deflection at the free end of the beam due to the load of 50 k as follows:

Substitute $12\text{ft}$ for a, $24\text{ft}$ for b, $48\text{ft}$ for l, and $50\text{k}$ for P in Equation (1).

$\begin{array}{c}{\Delta }_{CO}=-\frac{50\times 12\times 24}{3\times 48\times EI}\left({48}^2-{24}^2-{12}^2\right)\\ =-\frac{158400\text{k}\cdot {\text{ft}}^3}{EI}\end{array}$

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

Show the deflection at the mid of the beam due to the action of central point load P as follows:

${\Delta }_2=\frac{P{l}^3}{48EI}$        (2)

Substitute $1\text{k}$ for P and $48\text{ft}$ for l in Equation (2).

$\begin{array}{c}{f}_{CC}=\frac{{48}^3}{48EI}\\ =\frac{2304\text{k}\cdot {\text{ft}}^3/\text{k}}{EI}\end{array}$        (3)

Show the compatibility Equation as follows:

${\Delta }_{OC}+f_{CC}\times C_y=0$

Modify the above Equation using Equation (1) and (3).

$\begin{array}{l}{\Delta }_{CO}+f_{CC}\times C_y=0\\ -\frac{158400\text{k}\cdot {\text{ft}}^3}{EI}+\frac{2304\text{k}\cdot {\text{ft}}^3/\text{k}}{EI}\times C_y=0\\ C_y=\left(\frac{EI}{2304\text{k}\cdot {\text{ft}}^3/\text{k}}\times \frac{158400\text{k}\cdot {\text{ft}}^3}{EI}\right)\\ C_y=68.75\text{k}(\uparrow )\end{array}$

Refer Figure 1.

Take the sum of the moment at A as zero.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ -\left(50\times 12\right)+C_y\times 24+E_y\times 48-\left(50\times 36\right)=0\\ -2400+C_y\times 24+E_y\times 48=0\end{array}$

Substitute $68$