Chapter 13, Problem 54P

To determine

Calculate the support reactions for the given beam using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given beam.

Answer to Problem 54P

The horizontal reaction at A is $\underset{\_}{A_x=0}$.

The vertical reaction at A is $\underset{\_}{A_y=45.88\text{kN}(\uparrow )}$.

The vertical reaction at C is $\underset{\_}{C_y=100.48\text{kN}(\uparrow )}$.

The vertical reaction at E is $\underset{\_}{E_y=198.23\text{kN}(\uparrow )}$.

The vertical reaction at G is $\underset{\_}{G_y=45.41\text{kN}(\uparrow )}$.

Explanation of Solution

Given information:

The settlement at A $\left({\Delta }_A\right)$ is $10\text{mm}=0.01\text{m}$.

The settlement at C $\left({\Delta }_C\right)$ is $65\text{mm}=0.065\text{m}$.

The settlement at E $\left({\Delta }_E\right)$ is $40\text{mm}=0.04\text{m}$.

The settlement at G $\left({\Delta }_G\right)$ is $25\text{mm}=0.025\text{m}$.

The moment of inertia $\left(I\right)$ is $500\times {10}^6{\text{mm}}^4$.

The modulus of elasticity $\left(E\right)$ is 200 GPa.

The structure is given in the Figure.

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:

Sketch the free body diagram of the structure as shown in Figure 1.

Calculate the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 5 support reactions and the number of equilibrium equations is 3.

Therefore, the degree of indeterminacy of the beam is $i=2$.

Select the bending moments at the supports E and C as redundant.

Let ${\theta }_{CL}$ and ${\theta }_{CR}$ are the slopes at the ends C of the left and right spans of the primary beam due to external loading.

Let ${\theta }_{EL}$ and ${\theta }_{ER}$ are the slopes at the ends E of the left and right spans of the primary beam due to external loading.

Let $f_{CCL}$ and $f_{CCR}$ are the flexibility coefficient representing the slopes at C of the left and right spans due to unit value of redundant $M_C$.

Let $f_{EEL}$ and $f_{EER}$ are the flexibility coefficient representing the slopes at E of the left and right spans due to unit value of redundant $M_E$.

Let $f_{CE}$ be the flexibility coefficient representing the slope at point C of the primary beam due to unit value at point E.

Use beam deflection formulas:

Calculate the value of ${\theta }_{CL}$ using the relation:

${\theta }_{CL}=\frac{Pa\left(L^2-a^2\right)}{6EIL}$

Substitute $120\text{kN}$ for $P$, $6\text{m}$ for a, and $10\text{m}$ for L.

$\begin{array}{c}{\theta }_{CL}=\frac{120\times 6\left({10}^2-6^2\right)}{6EI\left(10\right)}\\ =\frac{768\text{kN}\cdot {\text{m}}^2}{EI}\end{array}$

Substitute $500\times {10}^6{\text{mm}}^4$ for E and 200 GPa for E.

$\begin{array}{c}{\theta }_{CL}=\frac{768\text{kN}\cdot {\text{m}}^2}{200\text{GPa}\times \frac{{10}^6\text{kN}/{\text{m}}^2}{1\text{GPa}}\times 500\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =0.00768\text{rad}\end{array}$

Calculate the value of ${\theta }_{CR}$ using the relation:

${\theta }_{CR}=\frac{Pb\left(L^2-b^2\right)}{6EIL}$

Substitute $120\text{kN}$ for $P$, $4\text{m}$ for b, $2I$ for $I$, and $10\text{m}$ for L.

$\begin{array}{c}{\theta }_{CR}=\frac{120\times 4\left({10}^2-4^2\right)}{6E\left(2I\right)\left(10\right)}\\ =\frac{336\text{kN}\cdot {\text{m}}^2}{EI}\end{array}$

Substitute $500\times {10}^6{\text{mm}}^4$ for E and 200 GPa for E.

$\begin{array}{c}{\theta }_{CR}=\frac{336\text{kN}\cdot {\text{m}}^2}{200\text{GPa}\times \frac{{10}^6\text{kN}/{\text{m}}^2}{1\text{GPa}}\times 500\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =0.00336\text{rad}\end{array}$

Calculate the value of ${\theta }_{EL}$ using the relation:

${\theta }_{EL}=\frac{Pa\left(L^2-a^2\right)}{6EIL}$

Substitute $120\text{kN}$ for $P$, $6\text{m}$ for a, $2I$ for $I$, and $10\text{m}$ for L.

$\begin{array}{c}{\theta }_{EL}=\frac{120\times 6\left({10}^2-6^2\right)}{6E\left(2I\right)\left(10\right)}\\ =\frac{384\text{kN}\cdot {\text{m}}^2}{EI}\end{array}$

Substitute $500\times {10}^6{\text{mm}}^4$ for E and 200 GPa for E.

$\begin{array}{c}{\theta }_{EL}=\frac{384\text{kN}\cdot {\text{m}}^2}{200\text{GPa}\times \frac{{10}^6\text{kN}/{\text{m}}^2}{1\text{GPa}}\times 500\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =0.00384\text{rad}\end{array}$

Calculate the value of ${\theta }_{ER}$ using the relation:

${\theta }_{ER}=\frac{P{L}^2}{16EI}$

Substitute $150\text{kN}$ for $P$ and $8\text{m}$ for L.

$\begin{array}{c}{\theta }_{ER}=\frac{150{\left(8\right)}^2}{16EI}\\ =\frac{600\text{kN}\cdot {\text{m}}^2}{EI}\end{array}$

Substitute $500\times {10}^6{\text{mm}}^4$ for E and 200 GPa for E.

$\begin{array}{c}{\theta }_{ER}=\frac{600\text{kN}\cdot {\text{m}}^2}{200\text{GPa}\times \frac{{10}^6\text{kN}/{\text{m}}^2}{1\text{GPa}}\times 500\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =0.006\text{rad}\end{array}$

Calculate the value of $f_{CCL}$ using the relation:

$f_{CCL}=\frac{ML}{3EI}$

Substitute $1\text{kN}\cdot \text{m}$ for M and 10 m for L.

$\begin{array}{c}{f}_{CCL}=\frac{1\left(10\right)}{3EI}\\ =\frac{10\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{3EI}\end{array}$

Substitute $500\times {10}^6{\text{mm}}^4$ for E and 200 GPa for E.

$\begin{array}{c}{f}_{CCL}=\frac{10\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{3\times 200\text{GPa}\times \frac{{10}^6\text{kN}/{\text{m}}^2}{1\text{GPa}}\times 500\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =0.0000333\text{rad}/\text{kN}\cdot \text{m}\end{array}$

Calculate the value of $f_{CCR}$ using the relation:

$f_{CCR}=\frac{ML}{6EI}$

Substitute $1\text{kN}\cdot \text{m}$ for M and 10 m for L.

$\begin{array}{c}{f}_{CCL}=\frac{1\left(10\right)}{6EI}\\ =\frac{10\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{6EI}\end{array}$

Substitute $500\times {10}^6{\text{mm}}^4$ for E and 200 GPa for E.

$\begin{array}{c}{f}_{CCL}=\frac{10\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{6\times 200\text{GPa}\times \frac{{10}^6\text{kN}/{\text{m}}^2}{1\text{GPa}}\times 500\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =0.0000167\text{rad}/\text{kN}\cdot \text{m}\end{array}$

Calculate the value of $f_{CE}$ and $f_{EC}$ using the relation:

$f_{CE}=f_{EC}=\frac{ML}{6EI}$

Substitute $1\text{kN}\cdot \text{m}$ for M, $2I$ for $I$, and 10 m for L.

$\begin{array}{c}{f}_{CE}=f_{EC}=\frac{1\left(10\right)}{6E\left(2I\right)}\\ =\frac{10\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{12EI}\end{array}$

Substitute $500\times {10}^6{\text{mm}}^4$ for E and 200 GPa for E.

$\begin{array}{c}{f}_{CE}=f_{EC}=\frac{10\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{12\times 200\text{GPa}\times \frac{{10}^6\text{kN}/{\text{m}}^2}{1\text{GPa}}\times 500\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =0.00000833\text{rad}/\text{kN}\cdot \text{m}\end{array}$

Calculate the value of $f_{EEL}$ using the relation:

$f_{EEL}=\frac{ML}{6EI}$

Substitute $1\text{kN}\cdot \text{m}$ for M and 10 m for L.

$\begin{array}{c}{f}_{EEL}=\frac{1\left(10\right)}{6EI}\\ =\frac{10\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{6EI}\end{array}$

Substitute $500\times {10}^6{\text{mm}}^4$ for E and 200 GPa for E.

$\begin{array}{c}{f}_{EEL}=\frac{10\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{6\times 200\text{GPa}\times \frac{{10}^6\text{kN}/{\text{m}}^2}{1\text{GPa}}\times 500\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =0.0000166\text{rad}/\text{kN}\cdot \text{m}\end{array}$

Calculate the value of $f_{EER}$ using the relation:

$f_{EER}=\frac{ML}{3EI}$

Substitute $1\text{kN}\cdot \text{m}$ for M and 8 m for L.

$\begin{array}{c}{f}_{EER}=\frac{1\left(8\right)}{3EI}\\ =\frac{8\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{3EI}\end{array}$

Substitute $500\times {10}^6{\text{mm}}^4$ for E and 200 GPa for E.

$\begin{array}{c}{f}_{EER}=\frac{8\text{kN}\cdot {\text{m}}^2\text{/kN}\cdot \text{m}}{3\times 200\text{GPa}\times \frac{{10}^6\text{kN}/{\text{m}}^2}{1\text{GPa}}\times 500\times {10}^6{\text{mm}}^4\times {\left(\frac{1\text{m}}{1,000\text{mm}}\right)}^4}\\ =0.0000267\text{rad}/\text{kN}\cdot \text{m}\end{array}$

The change of slope between the two tangents due to external load $\left({\theta }_{CO\text{rel}\text{.}}\right)$ is expressed as,

$\begin{array}{c}{\theta }_{CO\text{rel}\text{.}}={\theta }_{CL}+{\theta }_{CR}\\ =0.00768+0.00336\\ =0.01104\text{rad}\end{array}$

Similarly Calculate the value of ${\theta }_{EO\text{rel}\text{.}}$ as follows:

$\begin{array}{c}{\theta }_{EO\text{rel}\text{.}}={\theta }_{EL}+{\theta }_{ER}\\ =0.00384+0.006\\ =0.00984\text{rad}\end{array}$

The flexibility coefficient is expressed as $f_{CC\text{rel}\text{.}}=f_{CCL}+f_{CCR}$.

$\begin{array}{c}{f}_{CC\text{rel}\text{.}}=0.0000333+0.0000167\\ =0.00005\text{rad}/\text{kN}\cdot \text{m}\end{array}$

Calculate the value of $f_{EE\text{rel}\text{.}}$ as follows:

$\begin{array}{c}{f}_{EE\text{rel}\text{.}}=f_{EEL}+f_{EER}\\ =0.0000166+0.0000267\\ =0.0000433\text{rad}/\text{kN}\cdot \text{m}\end{array}$

Sketch the settlement at supports as shown in Figure 2.

Refer to Figure 2.

Calculate the value ${\theta }_{CL}$ as shown below.

$\begin{array}{c}{\theta }_{CL}=\frac{0.065-0.01}{10}\\ =0.0055\text{rad}\end{array}$

Calculate the value ${\theta }_{CR}$ as shown below.

$\begin{array}{c}{\theta }_{CR}=\frac{0.065-0.04}{10}\\ =0.0025\text{rad}\end{array}$

Calculate the value ${\theta }_C$ as shown below.

$\begin{array}{c}{\theta }_C={\theta }_{CR}+{\theta }_{CL}\\ =0.0025+0.0055\\ =0.008\text{rad}\end{array}$

Calculate the value ${\theta }_{EL}$ as shown below.

$\begin{array}{c}{\theta }_{EL}=\frac{0.04-0.065}{10}\\ =-0.0025\text{rad}\end{array}$

Calculate the value ${\theta }_{ER}$ as shown below.

$\begin{array}{c}{\theta }_{ER}=\frac{0.04-0.025}{8}\\ =0.001875\text{rad}\end{array}$

Calculate the value ${\theta }_C$ as shown below.

$\begin{array}{c}{\theta }_E={\theta }_{ER}+{\theta }_{EL}\\ =0.001875-0.0025\\ =-0.000625\text{rad}\end{array}$

Calculate the reactions for the given beam:

Show the compatibility Equations of the given beam as follows:

$\begin{array}{l}{\theta }_{CO\text{rel}.}+f_{CC\text{rel}.}{M}_C+f_{CE}{M}_E={\theta }_C\\ {\theta }_{EO\text{rel}.}+f_{EC}{M}_C+f_{EE\text{rel}.}{M}_E={\theta }_E\end{array}$

Substitute $0.001104\text{rad}$ for ${\theta }_{CO\text{rel}.}$, $0.00005\text{rad}/\text{kN}\cdot \text{m}$ for $f_{CC\text{rel}.}$, $0.00000833\text{rad}/\text{kN}\cdot \text{m}$ for $f_{CE}$, $0.008\text{rad}$ for ${\theta }_C$, $0.00984\text{rad}$ for ${\theta }_{EO\text{rel}.}$, $0.00000833\text{rad}/\text{kN}\cdot \text{m}$ for $f_{EC}$, $0.0000433\text{rad}/\text{kN}\cdot \text{m}$ for $f_{EE\text{rel}.}$, and $-0.000625\text{rad}$ for ${\theta }_E$.

$\begin{array}{l}0.001,104+0.00005M_C+0.00000833M_E=0.008\\ 1,104+5M_C+0.833M_E=800\\ 5M_C+0.833M_E=-304\end{array}$        (1)

$\begin{array}{l}0.00984+0.00000833M_C+0.0000433M_E=-0.000625\\ 984+0.833M_C+4.33M_E=-62.5\\ 0.833M_C+4.33M_E=-1,046.5\end{array}$        (2)

Solve Equation (1) and Equation (2).

$\begin{array}{l}{M}_C=-21.22\text{kN}\cdot \text{m}\\ M_E=-237.6\text{kN}\cdot \text{m}\end{array}$

Sketch the span end moments and shears for span ABC, CDE, and EFG as shown in Figure 3.

Refer to Figure 3.

Use equilibrium equations:

For span ABC,

Summation of moments of all forces about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ C_y\left(10\right)-21.22-120\left(6\right)=0\\ C_y=74.12\text{kN}\end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y-120+C_y=0\\ A_y-120+74.12=0\\ A_y=45.88\text{kN}\end{array}$

For span CDE,

Summation of moments of all forces about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ E_y\left(10\right)+21.22-237.6-120\left(6\right)=0\\ E_y=93.64\text{kN}\end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ C_y-120+E_y=0\\ C_y-120+93.64=0\\ C_y=26.36\text{kN}\end{array}$

For span EFG,

Summation of moments of all forces about E is equal to 0.

$\begin{array}{l}\sum M_E=0\\ G_y\left(8\right)+236.7-150\left(4\right)=0\\ G_y=45.41\text{kN}\end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ E_y-150+G_y=0\\ E_y-150+45.41=0\\ E_y=104.59\text{kN}\end{array}$

Provide the reaction at supports C and E as shown below.

$\begin{array}{c}{C}_y=74.12+26.36\\ =100.48\text{kN}\\ E_y=93.64+104.59\\ =198.23\text{kN}\end{array}$

Sketch the reactions for the given beam as shown in Figure 4.

Refer to Figure 4.

Calculate the shear force $\left(S\right)$ for the given beam:

At point A,

$S_A=45.88\text{kN}$

At point B,

$\begin{array}{c}{S}_{B,L}=45.88\text{kN}\\ S_{B,R}=45.88-120\\ =-74.12\text{kN}\end{array}$

At point C,

$\begin{array}{c}{S}_{C,L}=45.88-120\\ =-74.12\text{kN}\\ S_{C,R}=45.88-120+100.48\\ =26.36\text{kN}\end{array}$

At point D,

$\begin{array}{c}{S}_{D,L}=45.88-120+100.48\\ =26.36\text{kN}\\ S_{D,R}=45.88-120+100.48-120\\ =-93.64\text{kN}\end{array}$

At point E,

$\begin{array}{c}{S}_{E,L}=45.88-120+100.48-120\\ =-93.64\text{kN}\\ S_{E,R}=45.88-120+100.48-120+198.23\\ =104.59\text{kN}\end{array}$

At point F,

$\begin{array}{c}{S}_{F,L}=45.88-120+100.48-120+198.23\\ =104.59\text{kN}\\ S_{F,R}=45.88-120+100.48-120+198.23-150\\ =-45.41\text{kN}\end{array}$

At point G,

$\begin{array}{c}{S}_{G,L}=45.88-120+100.48-120+198.23-150\\ =-45.41\text{kN}\\ S_G=45.88-120+100.48-120+198.23-150+45.41\\ =0\end{array}$

Sketch the shear diagram for the given beam as shown in Figure 5.

Refer to Figure 4.

Calculate the bending moment $\left(M\right)$ for the given beam:

At point A,

$M_A=0$

At point B,

$\begin{array}{c}{M}_B=45.88\left(6\right)\\ =275.28\text{kN}\cdot \text{m}\end{array}$

At point C,

$\begin{array}{c}{M}_C=45.88\left(10\right)-120\left(4\right)\\ =-21.2\text{kN}\cdot \text{m}\end{array}$

At point D,

$\begin{array}{c}{M}_D=45.88\left(16\right)-120\left(10\right)+100.48\left(6\right)\\ =136.96\text{kN}\cdot \text{m}\end{array}$

At point E,

$\begin{array}{c}{M}_E=45.88\left(20\right)-120\left(14\right)+100.48\left(10\right)-120\left(4\right)\\ =-237.6\text{kN}\cdot \text{m}\end{array}$

At point F,

$\begin{array}{c}{M}_F=45.88\left(24\right)-120\left(18\right)+100.48\left(14\right)-120\left(8\right)+198.23\left(4\right)\\ =180.76\text{kN}\cdot \text{m}\end{array}$

At point G,

$M_G=0$

Sketch the bending moment diagram for the given beam as shown in Figure 6.