Chapter 16, Problem 1P

To determine

Sketch the shear and moment curves of the beam using stiffness method.

Explanation of Solution

Apply the sign conventions for the 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 positive and the counter clockwise moment as negative.

Calculation:

Analysis of the restrained structure:

The rotation at joint B and C are restrained by a clamp.

Calculate the fixed end moments as shown below.

For member AB:

$\begin{array}{c}M=\pm \frac{w{L}^2}{12}\\ =\pm \frac{2\times {20}^2}{12}\\ =\pm 66.67\text{ kip}\cdot \text{ft}\\ {M^{\prime }}_{AB}=-66.67\text{ kip}\cdot \text{ft}\\ {M^{\prime }}_{BA}=66.67\text{ kip}\cdot \text{ft}\end{array}$

For member BC:

$\begin{array}{c}M=\pm \frac{w{L}^2}{12}\\ =\pm \frac{2\times {30}^2}{12}\\ =\pm 150\text{ kip}\cdot \text{ft}\\ {M^{\prime }}_{BC}=-150\text{ kip}\cdot \text{ft}\\ {M^{\prime }}_{CB}=150\text{ kip}\cdot \text{ft}\end{array}$

Sketch the details of the beam, fixed end moments in restrained structure produced by applied loads, the clamp applied, and the moment diagrams in restrained structure as shown in Figure 1.

Sketch the free body diagram of joint as shown in Figure 2.

Refer to Figure 2.

Calculate the restraining moments as shown below.

At joint B:

$\begin{array}{l}{M}_1-66.67+150=0\\ M_1=-83.33\text{ kip}\cdot \text{ft}\end{array}$

At joint B:

$\begin{array}{l}{M}_2-150=0\\ M_2=150\text{ kip}\cdot \text{ft}\end{array}$

Provide the restraining force vector as shown below.

The elements in the restraining force vectors is the value of restraining moments with its sign reversed.

$\begin{array}{c}F=\left[\begin{array}{c}{M}_1\\ M_2\end{array}\right]\\ =\left[\begin{array}{c}83.33\text{ kip}\cdot \text{ft}\\ -150\text{ kip}\cdot \text{ft}\end{array}\right]\end{array}$

Assembly of the structure stiffness matrix:

Sketch the Free Body Diagram of the unit rotation at B as shown in Figure 3.

Refer to Figure 3.

Introduce a unit rotation at joint B.

Calculate the end moments of members AB and BC as shown below.

$\left[\begin{array}{c}{M}_i\\ M_i\end{array}\right]=\frac{2EI}{L}\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{c}{\theta }_i\\ {\theta }_i\end{array}\right]$

For member AB:

$\begin{array}{c}\left[\begin{array}{c}{M}_{AB}\\ M_{BA}\end{array}\right]=\frac{2EI}{20}\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]\\ =\left[\begin{array}{c}0.1EI\\ 0.2EI\end{array}\right]\end{array}$

For member BC:

$\begin{array}{c}\left[\begin{array}{c}{M}_{BC}\\ M_{CB}\end{array}\right]=\frac{2EI}{30}\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ =\left[\begin{array}{c}0.133EI\\ 0.067EI\end{array}\right]\end{array}$

The stiffness matrix for joint B and C due to the unit rotation at B is $K_1=\left[\begin{array}{c}0.333EI\\ 0.067EI\end{array}\right]$.

Introduce a unit rotation at joint C.

Sketch the Free Body Diagram of the unit rotation at C as shown in Figure 4.

Refer to Figure 4.

Calculate the end moments of members AB and BC as shown below.

For member AB:

$\begin{array}{c}\left[\begin{array}{c}{M}_{AB}\\ M_{BA}\end{array}\right]=\frac{2EI}{20}\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{c}0\\ 0\end{array}\right]\\ =\left[\begin{array}{c}0\\ 0\end{array}\right]\end{array}$

For member BC:

$\begin{array}{c}\left[\begin{array}{c}{M}_{BC}\\ M_{CB}\end{array}\right]=\frac{2EI}{30}\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]\\ =\left[\begin{array}{c}0.067EI\\ 0.133EI\end{array}\right]\end{array}$

The stiffness matrix for joint B and C due to the unit rotation at C is $K_2=\left[\begin{array}{c}0.067EI\\ 0.133EI\end{array}\right]$.

Provide the stiffness matrix of the beam as shown below.

$K=\left[\begin{array}{cc}0.333EI& 0.067EI\\ 0.067EI& 0.133EI\end{array}\right]$

Calculate the unknown joint displacements as shown below.

$\begin{array}{l}K\Delta =F\\ \left[\begin{array}{cc}0.333EI& 0.067EI\\ 0.067EI& 0.133EI\end{array}\right]\left[\begin{array}{c}{\theta }_1\\ {\theta }_2\end{array}\right]=\left[\begin{array}{c}83.33\\ -150\end{array}\right]\\ \left[\begin{array}{c}{\theta }_1\\ {\theta }_2\end{array}\right]=\frac{1}{EI}{\left[\begin{array}{cc}0.333& 0.067\\ 0.067& 0.133\end{array}\right]}^{-1}\left[\begin{array}{c}83.33\\ -150\end{array}\right]\\ =\frac{1}{EI}\left[\begin{array}{c}531\\ -1395\end{array}\right]\end{array}$

Evaluation of the effects of joint displacements:

Calculate the final moment of the spans due to the unit rotation as shown below.

For span AB:

$\begin{array}{c}{M}_{AB}={M^{\prime }}_{AB}+{M^{″}}_{AB}\\ =-66.67+\left(0.1EI{\theta }_1+\left(0\right){\theta }_2\right)\\ =-66.67+0.1EI\times \frac{531}{EI}\\ =-13.6\text{ kip}\cdot \text{ft}\end{array}$

$\begin{array}{c}{M}_{BA}={M^{\prime }}_{BA}+{M^{″}}_{BA}\\ =66.67+\left(0.2EI{\theta }_1+\left(0\right){\theta }_2\right)\\ =66.67+0.2EI\times \frac{531}{EI}\\ =172.8\text{ kip}\cdot \text{ft}\end{array}$

For span BC.

$\begin{array}{c}{M}_{BC}={M^{\prime }}_{BC}+{M^{″}}_{BC}\\ =-150+\left(0.133EI{\theta }_1+0.067EI{\theta }_2\right)\\ =-150+0.133EI\times \frac{531}{EI}+0.067EI\times \left(-\frac{1395}{EI}\right)\\ =-172.8\text{ kip}\cdot \text{ft}\end{array}$

$\begin{array}{c}{M}_{CB}={M^{\prime }}_{CB}+{M^{″}}_{CB}\\ =150+\left(0.067EI{\theta }_1+0.133EI{\theta }_2\right)\\ =150+0.067EI\times \frac{531}{EI}+0.133EI\times \left(-\frac{1395}{EI}\right)\\ =0\end{array}$

Sketch the Free Body Diagram of span AB and BC as shown in Figure 5.

Refer to Figure 5.

Consider span AB.

Use equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -V_{BA}\times 20+2\times 20\times \frac{20}{2}+172.8-13.6=0\\ V_{BA}=27.96\text{ kips}(\uparrow )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ V_{AB}+27.96-2\times 20=0\\ V_{AB}=12.04\text{ kips}(\uparrow )\end{array}$

Consider span BC.

Use equilibrium equations:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ -V_{CB}\times 30+2\times 30\times \frac{30}{2}-172.8=0\\ V_{CB}=24.24\text{ kips}(\uparrow )\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ V_{BC}+24.24-2\times 30=0\\ V_{BC}=35.76\text{ kips}(\uparrow )\end{array}$

The reaction at support B.

$\begin{array}{c}{R}_B=V_{BA}+V_{BC}\\ =27.96+35.76\\ =63.72\text{ kips}\end{array}$

Calculate the shear at each point of the beam as shown below.

$\begin{array}{l}{V}_A=12.04\\ V_{@\text{ left of }B}=12.04-2\times 20\\ =-27.96\\ V_B=12.04-2\times 20+63.72\\ =35.76\end{array}$

$\begin{array}{l}{V}_{@\text{ left of }C}=12.04-2\times 50+63.72\\ =-24.24\\ V_C=12.04-2\times 50+63.72-24.24\\ =0\end{array}$

Calculate the point of zero shear as shown below.

Consider a section at a distance x from A in span AB.

$\begin{array}{l}{V}_x=0\\ 12.04-2\times x=0\\ x=6.02\text{ ft from }A\end{array}$

Consider a section at a distance x from B in span BC.

$\begin{array}{l}{V}_x=0\\ 12.04-2\times \left(20+x\right)+63.72=0\\ x=17.88\text{ ft from }B\end{array}$

Calculate the moment at each point of the beam as shown below.

$\begin{array}{l}{M}_A=-13.6\text{ kip}\cdot \text{ft}\\ M_{@\text{ 6}\text{.02 ft from }A}=12.04\times 6.02-2\times 6.02\times \frac{6.02}{2}-13.6\\ =22.6\text{ kip}\cdot \text{ft}\\ M_B=12.04\times 20-2\times 20\times \frac{20}{2}-13.6\\ =-172.8\text{ kip}\cdot \text{ft}\end{array}$

$\begin{array}{l}{M}_{@17.88\text{ ft from }B}=12.04\times 37.88-2\times 37.88\times \frac{37.88}{2}-13.6+63.72\times 17.88\\ =146.9\text{ kip}\cdot \text{ft}\\ M_C=12.04\times 50-2\times 50\times \frac{50}{2}-13.6+63.72\times 30\\ =0\end{array}$

Sketch the shear and moment diagram for the beam as shown in Figure 6.