Chapter 13, Problem 2P

To determine

Find the moment at the supports B and C of the beam using approximate analysis.

Sketch the shear and moment diagrams of the beam.

Answer to Problem 2P

Case 1: Length $L_1=3\text{m}$.

The moment at the supports B and C of the beam using approximate analysis are $\underset{\_}{-12.86\text{kN}\cdot \text{m}}$ and $\underset{\_}{-16.88\text{kN}\cdot \text{m}}$.

The moment at the supports B and C of the beam using moment distribution method are $\underset{\_}{-13.5\text{kN}\cdot \text{m}}$ and $\underset{\_}{-15.75\text{kN}\cdot \text{m}}$.

Case 2: Length $L_1=12\text{m}$.

The moment at the supports B and C of the beam using approximate analysis are $\underset{\_}{-24.36\text{kN}\cdot \text{m}}$ and $\underset{\_}{-6.3\text{kN}\cdot \text{m}}$.

The moment at the supports B and C of the beam using moment distribution method are $\underset{\_}{-36.84\text{kN}\cdot \text{m}}$ and $\underset{\_}{-4.1\text{kN}\cdot \text{m}}$.

Explanation of Solution

Given information:

The value of EI is constant.

Case 1: Length $L_1=3\text{m}$.

Case 2: Length $L_1=12\text{m}$.

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:

Case 1: Length $L_1=3\text{m}$.

For span AB,

Let consider the location of a point of inflection at $0.3L$ to the left of support B which means 0.9 m to the left of support B or 2.1 m from the support A to the right.

For span BC,

Let consider the location of a point of inflection at $0.23L$ to the left of support C which means 1.38 m to the left of support B or 4.62 m from the support B to the right.

Show the location of a point of inflection of span AB and BC as in Figure 1.

Refer to Figure 1.

Consider P.I. of span AB,

Find the reaction $R_A$ at support A:

Summation of moments about P.I. is equal to 0.

$\begin{array}{l}\sum M_{P.I}=0\\ R_A\left(2.1\right)-20\left(0.6\right)=0\\ R_A=5.71\text{kN}\uparrow \end{array}$

Consider P.I. of span BC,

Find the reaction $R_B$ at support B:

Summation of moments about P.I. is equal to 0.

$\begin{array}{l}\sum M_{P.I}=0\\ -14.29\left(5.52\right)+R_B\left(4.62\right)-23.1\left(2.31\right)=0\\ R_B=28.62\text{kN}\uparrow \end{array}$

Consider entire structure,

Find the reaction $R_C$ at support C:

Summation of moments about P.I. is equal to 0.

$\begin{array}{l}\sum M_{P.I}=0\\ 5.71-20+28.62-30+R_C=0\\ R_C=15.67\text{kN}\uparrow \end{array}$

Find the shear values of the beam:

$F_A=5.71\text{kN}$

$F_{\text{at}20\text{kN,}\text{JL}}=5.71\text{kN}$

$\begin{array}{c}{F}_{\text{at}20\text{kN,}\text{JR}}=5.71-20\\ =-14.29\text{kN}\end{array}$

$\begin{array}{c}{F}_{B,JL}=5.71-20\\ =-14.29\text{kN}\end{array}$

$\begin{array}{c}{F}_{B,JR}=5.71-20+28.62\\ =14.33\text{kN}\end{array}$

$F_C=-15.67\text{kN}$

Find the point of zero shear force for span BC,

$\begin{array}{l}\sum F_{xx}=0\\ 15.67-5x=0\\ x=3.134\text{m}\text{from}\text{point}\text{C}\end{array}$

Find the moment values of the beam:

$M_A=0$

$\begin{array}{c}{M}_{\text{at}\text{20}\text{kN}\text{load}}=5.71\left(1.5\right)\\ =8.57\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{M}_B=5.71\left(3\right)-20\left(1.5\right)\\ =-12.87\text{kN}\cdot \text{m}\end{array}$

$M_C=-16.88\text{kN}\cdot \text{m}$

$\begin{array}{c}{M}_{x=3.134\text{m}\text{from}\text{point}\text{C}}=15.67\left(3.134\right)-5\left(3.134\right)\left(\frac{3.134}{2}\right)-16.88\\ =7.67\text{kN}\cdot \text{m}\end{array}$

Sketch the shear and moment diagram of the beam using approximate analysis as shown in Figure 2.

Check the shear and moment values using moment distribution method:

Find the fixed end moments in the span AB and BC:

$\begin{array}{c}{\text{FEM}}_{AB}=-\frac{PL}{8}\\ =-\frac{20\left(3\right)}{8}\\ =-7.5\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{\text{FEM}}_{BA}=-{\text{FEM}}_{AB}\\ =7.5\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{\text{FEM}}_{BC}=-\frac{w{L}^2}{12}\\ =-\frac{5{\left(6\right)}^2}{12}\\ =-15\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{\text{FEM}}_{CB}=-{\text{FEM}}_{BC}\\ =15\text{kN}\cdot \text{m}\end{array}$

Find the distribution factor at joint B:

$\begin{array}{c}{K}_{AB}=\frac{3}{4}\frac{I}{3}\\ =0.25I\end{array}$

$\begin{array}{c}{K}_{BC}=\frac{I}{6}\\ =0.167I\end{array}$

$\begin{array}{c}\sum K_s=0.25I+0.167I\\ =0.417I\end{array}$

$\begin{array}{c}D{F}_{AB}=\frac{0.25I}{0.417I}\\ =0.6\end{array}$

$\begin{array}{c}D{F}_{BC}=\frac{0.167I}{0.417I}\\ =0.4\end{array}$

Consider span AB, find the reactions at the supports:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ R_A\left(3\right)-20\left(1.5\right)+13.5=0\\ R_A=5.5\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_A+R_B-20=0\\ 5.5+R_B-20=0\\ R_B=14.5\text{kN}\uparrow \end{array}$

Consider span BC, find the reactions at the supports:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ R_C\left(6\right)+13.5-15.75-5\left(6\right)\left(\frac{6}{2}\right)=0\\ R_C=15.375\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_B+R_C-5\left(6\right)=0\\ R_B+15.375-5\left(6\right)=0\\ R_B=14.625\text{kN}\uparrow \end{array}$

Therefore, the total $R_B$ is $14.5+14.625=29.125\text{kN}$

Find the shear values of the beam:

$F_A=5.5\text{kN}$

$F_{\text{at}20\text{kN,}\text{JL}}=5.5\text{kN}$

$\begin{array}{c}{F}_{\text{at}20\text{kN,}\text{JR}}=5.5-20\\ =-14.5\text{kN}\end{array}$

$\begin{array}{c}{F}_{B,JL}=5.5-20\\ =-14.5\text{kN}\end{array}$

$\begin{array}{c}{F}_{B,JR}=5.5-20+29.125\\ =14.625\text{kN}\end{array}$

$F_C=-15.375\text{kN}$

Find the point of zero shear force for span BC,

$\begin{array}{l}\sum F_{xx}=0\\ 15.375-5x=0\\ x=3.075\text{m}\text{from}\text{point}\text{C}\end{array}$

Find the moment values of the beam:

$M_A=0$

$\begin{array}{c}{M}_{\text{at}\text{20}\text{kN}\text{load}}=5.5\left(1.5\right)\\ =8.25\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{M}_B=5.5\left(3\right)-20\left(1.5\right)\\ =-13.5\text{kN}\cdot \text{m}\end{array}$

$M_C=-15.75\text{kN}\cdot \text{m}$

$\begin{array}{c}{M}_{x=3.075\text{m}\text{from}\text{point}\text{C}}=15.375\left(3.075\right)-5\left(3.075\right)\left(\frac{3.075}{2}\right)-16.88\\ =7.89\text{kN}\cdot \text{m}\end{array}$

Sketch the moment distribution results, shear, and moment diagrams of the beam using moment distribution method as shown in Figure 3.

Compare the results of approximate analysis and moment distribution method:

The approximate analysis results compare closely with moment distribution method.

Therefore, the moment at the supports B and C of the beam using approximate analysis are $\underset{\_}{-12.86\text{kN}\cdot \text{m}}$ and $\underset{\_}{-16.88\text{kN}\cdot \text{m}}$.

Therefore, the moment at the supports B and C of the beam using moment distribution method are $\underset{\_}{-13.5\text{kN}\cdot \text{m}}$ and $\underset{\_}{-15.75\text{kN}\cdot \text{m}}$.

Case 2: Length $L_1=12\text{m}$.

For span AB,

Length of span AB is longer, so a larger counterclockwise moment is applied member AB to the left end of member BC. Thus the moment rotates B-end of member BC in the counterclockwise direction causes the P.I. to shift to the right.

For span BC,

Let consider the location of a point of inflection adjacent to B at $0.3L$ and the one on the right at $0.1L$.

Show the location of a point of inflection as in Figure 4.

Sketch the reactions, forces, and moments for the segments as in Figure 5.

Refer to Figure 5.

Find the reaction $R_B$ at support B:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ R_B\left(12\right)-20\left(6\right)-9\left(0.9+12\right)-9\left(13.8\right)=0\\ R_B=30.03\text{kN}\uparrow \end{array}$

Find the shear values of the beam:

$F_A=7.97\text{kN}$

$F_{\text{at}20\text{kN,}\text{JL}}=7.97\text{kN}$

$\begin{array}{c}{F}_{\text{at}20\text{kN,}\text{JR}}=7.97-20\\ =-12.03\text{kN}\end{array}$

$\begin{array}{c}{F}_{B,JL}=7.97-20\\ =-12.03\text{kN}\end{array}$

$\begin{array}{c}{F}_{B,JR}=7.97-20+30.03\\ =18\text{kN}\end{array}$

$F_C=-12\text{kN}$

Find the point of zero shear force for span BC,

$\begin{array}{l}\sum F_{xx}=0\\ 12-5x=0\\ x=2.4\text{m}\text{from}\text{point}\text{C}\end{array}$

Find the moment values of the beam:

$M_A=0$

$\begin{array}{c}{M}_{\text{at}\text{20}\text{kN}\text{load}}=7.97\left(6\right)\\ =47.82\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{M}_B=7.97\left(12\right)-20\left(6\right)\\ =-24.36\text{kN}\cdot \text{m}\end{array}$

$M_C=-6.3\text{kN}\cdot \text{m}$

$\begin{array}{c}{M}_{x=2.4\text{m}\text{from}\text{point}\text{C}}=12\left(2.4\right)-5\left(2.4\right)\left(\frac{2.4}{2}\right)-6.3\\ =8.1\text{kN}\cdot \text{m}\end{array}$

Sketch the shear and moment diagram of the beam using approximate analysis as shown in Figure 6.

Check the shear and moment values using moment distribution method:

Find the fixed end moments in the span AB and BC:

$\begin{array}{c}{\text{FEM}}_{AB}=-\frac{PL}{8}\\ =-\frac{20\left(12\right)}{8}\\ =-30\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{\text{FEM}}_{BA}=-{\text{FEM}}_{AB}\\ =30\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{\text{FEM}}_{BC}=-\frac{w{L}^2}{12}\\ =-\frac{5{\left(6\right)}^2}{12}\\ =-15\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{\text{FEM}}_{CB}=-{\text{FEM}}_{BC}\\ =15\text{kN}\cdot \text{m}\end{array}$

Consider span AB, find the reactions at the supports:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ R_A\left(12\right)-20\left(6\right)+36.84=0\\ R_A=6.93\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_A+R_B-20=0\\ 6.93+R_B-20=0\\ R_B=13.07\text{kN}\uparrow \end{array}$

Consider span BC, find the reactions at the supports:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ R_C\left(6\right)+36.84-4.1-5\left(6\right)\left(\frac{6}{2}\right)=0\\ R_C=9.54\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_B+R_C-5\left(6\right)=0\\ R_B+9.54-5\left(6\right)=0\\ R_B=20.46\text{kN}\uparrow \end{array}$

Therefore, the total $R_B$ is $13.07+20.46=33.53\text{kN}$

Find the shear values of the beam:

$F_A=6.93\text{kN}$

$F_{\text{at}20\text{kN,}\text{JL}}=6.93\text{kN}$

$\begin{array}{c}{F}_{\text{at}20\text{kN,}\text{JR}}=6.93-20\\ =-13.07\text{kN}\end{array}$

$\begin{array}{c}{F}_{B,JL}=6.93-20\\ =-13.07\text{kN}\end{array}$

$\begin{array}{c}{F}_{B,JR}=6.93-20+33.53\\ =20.46\text{kN}\end{array}$

$F_C=-9.54\text{kN}$

Find the point of zero shear force for span BC,

$\begin{array}{l}\sum F_{xx}=0\\ 9.54-5x=0\\ x=1.908\text{m}\text{from}\text{point}\text{C}\end{array}$

Find the moment values of the beam:

$M_A=0$

$\begin{array}{c}{M}_{\text{at}\text{20}\text{kN}\text{load}}=6.93\left(6\right)\\ =41.58\text{kN}\cdot \text{m}\end{array}$

$\begin{array}{c}{M}_B=6.93\left(12\right)-20\left(6\right)\\ =-36.84\text{kN}\cdot \text{m}\end{array}$

$M_C=-4.1\text{kN}\cdot \text{m}$

$\begin{array}{c}{M}_{x=1.908\text{m}\text{from}\text{point}\text{C}}=9.4\left(1.908\right)-5\left(1.908\right)\left(\frac{1.908}{2}\right)-4.1\\ =4.73\text{kN}\cdot \text{m}\end{array}$

Sketch the shear, and moment diagrams of the beam using moment distribution method as shown in Figure 7.

Compare the results of approximate analysis and moment distribution method:

Differences exist between the approximate analysis and the moment distribution method results.

The moment distribution method provides exact results.

Therefore, the moment at the supports B and C of the beam using approximate analysis are $\underset{\_}{-24.36\text{kN}\cdot \text{m}}$ and $\underset{\_}{-6.3\text{kN}\cdot \text{m}}$.

Therefore, the moment at the supports B and C of the beam using moment distribution method are $\underset{\_}{-36.84\text{kN}\cdot \text{m}}$ and $\underset{\_}{-4.1\text{kN}\cdot \text{m}}$.