Chapter D, Problem 10P

To determine

Calculate the reactions for the given beam.

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

Answer to Problem 10P

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

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

The vertical reaction at C is $\underset{\_}{C_y=100.5\text{kN}\uparrow }$.

The vertical reaction at E is $\underset{\_}{E_y=198.3\text{kN}\uparrow }$.

The vertical reaction at G is $\underset{\_}{G_y=45.32\text{kN}\uparrow }$.

Explanation of Solution

Given information:

The structure is given in the Figure.

The settlement of support A is 10 mm, support C is 65 mm, support E is 40 mm, and support G is 25 mm.

Apply the sign conventions for calculating reactions 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:

Find 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 moment $M_C$ and $M_E$ at the interior supports C and E as redundant.

Consider the supports $A$, C, and E as l, c, and r respectively.

Express the general three-moment equation as shown below:

$\frac{M_l{L}_l}{I_{{}_l}}+2M_c\left(\frac{L_l}{I_l}+\frac{L_r}{I_r}\right)+\frac{M_r{L}_r}{I_{{}_r}}=\left[\begin{array}{l}-\sum \frac{P_l{L}_l^2{k}_l}{I_l}\left(1-k_l^2\right)-\sum \frac{P_r{L}_r^2{k}_r}{I_r}\left(1-k_r^2\right)\\ -\frac{w_l{L}_l^3}{4I_l}-\frac{w_r{L}_r^3}{4I_r}-6E\left(\frac{{\Delta }_l-{\Delta }_c}{L_l}+\frac{{\Delta }_r-{\Delta }_c}{L_r}\right)\end{array}\right]$        (1)

Here, $M_c$ is the bending moment at support c, $M_l,M_r$ are the bending moments at the adjacent supports to the left and to the right of c, E is the modulus of elasticity, $L_l,L_r$ are the lengths of the spans to the left and to the right of c, $I_l,I_r$ are the moments of inertia of the spans to the left and to the right of c, $P_l,P_r$ are the concentrated loads acting on the left and the right spans, $k_l$ or $k_r$ is the ratio of distance of $P_l$ or $P_r$ from the left or right support to the span length, $w_l,w_r$ are the uniformly distributed loads to the left and right spans, ${\Delta }_c$ is the settlement of support c, ${\Delta }_l,{\Delta }_r$ is the settlement of adjacent supports to the left and to the right of c,

The settlement of support A is 0.01 m, support C is 0.065 m, support E is 0.04 mm, and support G is 0.025 m.

Apply three-moment equation at joint C,

The moment at $A$ is $M_A=0$.

Substitute 10 m for $L_l$, 10 m for $L_r$ $I$ for $I_l$, 2I for $I_r$, 120 kN for $P_l$, 120 kN for $P_r$, 0.6 for $k_l$, 0.4 for $k_r$, 0 for $w_l$, $0$ for $w_r$, 0.01 m for ${\Delta }_l$, 0.065 m for ${\Delta }_c$, and 0.04 m for ${\Delta }_r$ in Equation (1).

$\begin{array}{l}0+2M_C\left(\frac{10}{I}+\frac{10}{2I}\right)+M_E\left(\frac{10}{2I}\right)=\left[\begin{array}{l}-\frac{120{\left(10\right)}^2\left(0.6\right)}{I}\left(1-{0.6}^2\right)\\ -\frac{120{\left(10\right)}^2\left(0.4\right)}{2I}\left(1-{0.4}^2\right)\\ -6E\left(\frac{0.01-0.065}{10}+\frac{0.04-0.065}{10}\right)\end{array}\right]\\ \frac{30M_C}{I}+\frac{5M_E}{I}=-\frac{4,608}{I}-\frac{2,016}{I}+0.048E\\ 30M_C+5M_E=-6,624+0.048EI\end{array}$

Substitute 200 GPa for E and $500\left({10}^6\right){\text{mm}}^4$ for I.

$\begin{array}{l}30M_C+5M_E=-6,624+0.048\left(200\text{GPa}\right)\left(500\left({10}^6\right){\text{mm}}^4\right)\\ 30M_C+5M_E=\left[\begin{array}{l}-6,624+0.048\left(200\text{GPa}\times \frac{{10}^6{\text{kN/m}}^{\text{2}}}{1\text{GPa}}\right)\\ \times \left(500\left({10}^6\right){\text{mm}}^4\times \frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)\end{array}\right]\\ 30M_C+5M_E=-6,624+4,800\\ 30M_C+5M_E=-1,824\end{array}$        (2)

Apply three-moment equation at joint E,

The moment at $G$ is $M_G=0$.

Substitute 10 m for $L_l$, 8 m for $L_r$ 2$I$ for $I_l$, I for $I_r$, 120 kN for $P_l$, 150 kN for $P_r$, 0.6 for $k_l$, 0.5 for $k_r$, 0 for $w_l$, $0$ for $w_r$, 0.065 m for ${\Delta }_l$, 0.04 m for ${\Delta }_c$, and 0.025 m for ${\Delta }_r$ in Equation (1).

$\begin{array}{l}{M}_C\left(\frac{10}{2I}\right)+2M_E\left(\frac{10}{2I}+\frac{8}{I}\right)=\left[\begin{array}{l}-\frac{120{\left(10\right)}^2\left(0.6\right)}{2I}\left(1-{0.6}^2\right)\\ -\frac{150{\left(8\right)}^2\left(0.5\right)}{I}\left(1-{0.5}^2\right)\\ -6E\left(\frac{0.065-0.04}{10}+\frac{0.025-0.04}{8}\right)\end{array}\right]\\ \frac{5M_C}{I}+\frac{26M_E}{I}=-\frac{2,304}{I}-\frac{3,600}{I}-0.00375E\\ 5M_C+26M_E=-5,904-0.00375EI\end{array}$

Substitute 200 GPa for E and $500\left({10}^6\right){\text{mm}}^4$ for I.

$\begin{array}{l}5M_C+26M_E=-5,904-0.00375\left(200\text{GPa}\right)\left(500\left({10}^6\right){\text{mm}}^4\right)\\ 5M_C+26M_E=\left[\begin{array}{l}-5,904-0.00375\left(200\text{GPa}\times \frac{{10}^6{\text{kN/m}}^{\text{2}}}{1\text{GPa}}\right)\\ \times \left(500\left({10}^6\right){\text{mm}}^4\times \frac{{10}^{-12}{\text{m}}^4}{1{\text{mm}}^4}\right)\end{array}\right]\\ 5M_C+26M_E=-5,904-375\\ 5M_C+26M_E=-6,279\end{array}$        (3)

Solve Equation (2) and Equation (3).

$\begin{array}{l}{M}_C=-21.2\text{kN}\cdot \text{m}\\ M_E=-237.4\text{kN}\cdot \text{m}\end{array}$

Sketch the span end moments and shears for span ABC, CDE, and EFG as shown in Figure 1.

Use equilibrium equations:

For span ABC,

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

$\begin{array}{l}\sum M_A=0\\ C_y\left(10\right)-21.2-120\left(6\right)=0\\ C_y=74.12\text{kN}\end{array}$

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.

$\begin{array}{l}\sum M_C=0\\ E_y\left(10\right)+21.2-237.4-120\left(6\right)=0\\ E_y=93.62\text{kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ C_y-120+E_y=0\\ C_y-120+93.62=0\\ C_y=26.38\text{kN}\end{array}$

For span EFG,

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

$\begin{array}{l}\sum M_E=0\\ G_y\left(8\right)+237.4-150\left(4\right)=0\\ G_y=45.32\text{kN}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ E_y-150+G_y=0\\ E_y-150+45.32=0\\ E_y=104.68\text{kN}\end{array}$

Sketch the reactions for the given beam as shown in Figure 2.

Find 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,

$\begin{array}{c}{S}_{C,L}=45.88-120\\ =-74.12\text{kN}\\ S_{C,R}=45.88-120+100.5\\ =26.38\text{kN}\end{array}$

At point D,

$\begin{array}{c}{S}_{D,L}=45.88-120+100.5\\ =26.38\text{kN}\\ S_{D,R}=45.88-120+100.5-120\\ =-93.62\text{kN}\end{array}$

At point E,

$\begin{array}{c}{S}_{E,L}=45.88-120+100.5-120\\ =-93.62\text{kN}\\ S_{E,R}=45.88-120+100.5-120+198.3\\ =104.68\text{kN}\end{array}$

At point F,

$\begin{array}{c}{S}_{F,L}=45.88-120+100.5-120+198.3\\ =104.68\text{kN}\\ S_{F,R}=45.88-120+100.5-120+198.3-150\\ =-45.32\text{kN}\end{array}$

At point G,

$S_G=-45.32\text{kN}$

Sketch the shear diagram for the given beam as shown in Figure 3.

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

At point A,

$M_A=0$

At point B,

$\begin{array}{c}{M}_B=45.88\left(6\right)\\ =275.3\text{kN}\cdot \text{m}\end{array}$

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,

$\begin{array}{c}{M}_D=45.88\left(16\right)-120\left(10\right)+100.5\left(6\right)\\ =137.1\text{kN}\cdot \text{m}\end{array}$

At point E,

$\begin{array}{c}{M}_E=45.32\left(8\right)-150\left(4\right)\\ =-237.4\text{kN}\cdot \text{m}\end{array}$

At point F,

$\begin{array}{c}{M}_F=45.32\left(4\right)\\ =181.3\text{kN}\cdot \text{m}\end{array}$

At point G,

$M_G=0$

Sketch the bending moment diagram for the given beam as shown in Figure 4.