Chapter 11, Problem 11.1P

Determine the moments at A,B, and C, then draw the moment diagram for the beam. The moment of inertia of each span is indicated in the figure. Assume the support at B is a roller and A and C are fixed. E = 29(10³) ksi.

To determine

The moments at A, B, and C, and to draw the moment diagram for the beam.

Answer to Problem 11.1P

The moment at A for member AB is, $-38.4\text{k}\cdot \text{ft}$.

The moment at B for member BA is, $45.52\text{k}\cdot \text{ft}$.

The moment at B for member BC is, $-45.52\text{k}\cdot \text{ft}$.

The moment at C for member CB is, $67.23\text{k}\cdot \text{ft}$.

The following figure shows the moment diagram of the beam.

  

Explanation of Solution

Calculation:

The following figure shows the beam diagram.

  

        Figure-(1)

Calculate the fixed end moments at every end as shown below.

Calculate fixed end moments of member AB.

   $FE{M}_{AB}=\frac{-w{l}^2}{12}$

Here, given load is $w$ and length is $l$.

Substitute $800\text{lb}/\text{ft}$ for $w$ and $24\text{ft}$ for $l$.

   $\begin{array}{c}FE{M}_{AB}=\frac{-\left(800\text{lb}/\text{ft}\right){\left(24\text{ft}\right)}^2}{12}\left(\frac{1\text{k}}{1000\text{lb}}\right)\\ =-\frac{460.8}{12}\\ =-38.4\text{k}\cdot \text{ft}\end{array}$

Calculate fixed end moments of member BA.

   $FE{M}_{BA}=\frac{w{l}^2}{12}$

Substitute $800\text{lb}/\text{ft}$ for $w$ and $24\text{ft}$ for $l$.

   $\begin{array}{c}FE{M}_{BA}=\frac{\left(800\text{lb}/\text{ft}\right){\left(24\text{ft}\right)}^2}{12}\left(\frac{1\text{k}}{1000\text{lb}}\right)\\ =\frac{460.8}{12}\\ =38.4\text{k}\cdot \text{ft}\end{array}$

Calculate fixed end moments of member BC.

   $FE{M}_{BC}=-\frac{wl}{8}$

Substitute $30\text{k}$ for $w$ and $16\text{ft}$ for $l$.

   $\begin{array}{c}FE{M}_{BC}=-\frac{\left(30\text{k}\right)\left(16\text{ft}\right)}{8}\\ =-\frac{480}{8}\text{k}\cdot \text{ft}\\ =-60\text{k}\cdot \text{ft}\end{array}$

Calculate fixed end moments of member CB.

   $FE{M}_{CB}=\frac{wl}{8}$

Substitute $30\text{k}$ for $w$ and $16\text{ft}$ for $l$.

   $\begin{array}{c}FE{M}_{BC}=\frac{\left(30\text{k}\right)\left(16\text{ft}\right)}{8}\\ =\frac{480}{8}\text{k}\cdot \text{ft}\\ =60\text{k}\cdot \text{ft}\end{array}$

Calculate the member relative stiffness factors on either side of B.

Calculate relative stiffness of member BA.

   $K_{BA}=\frac{4E{I}_{BA}}{L_{BA}}\left(\text{For}\text{far}\text{end}\text{fixed}\right)$

Here, $K_{BA}$ is relative stiffness of member BA, E is modulus of elasticity, I is the moment of inertia and L is length.

Substitute $29\times {10}^3\text{ksi}$ for E, $24\text{ft}$ for L.

   $\begin{array}{c}{K}_{BA}=\frac{4\left(29\times {10}^3\text{ksi}\right)I_{BA}}{24\text{ft}}\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =402.78I_{BA}\end{array}$

Calculate relative stiffness of member BC.

   $K_{BC}=\frac{4E{I}_{BC}}{L_{BC}}\left(\text{For}\text{far}\text{end}\text{fixed}\right)$

Substitute $29\times {10}^3\text{ksi}$ for E, $16\text{ft}$ for L.

   $\begin{array}{c}{K}_{BC}=\frac{4\left(29\times {10}^3\text{ksi}\right)I_{BA}}{16\text{ft}}\left(\frac{1\text{ft}}{12\text{in}}\right)\\ =604.16I_{BC}\end{array}$

Calculate the distribution factors for different members.

Calculate distribution factors for AB.

   ${\left(DF\right)}_{AB}=0$ (Since, far end is fixed)

Calculate distribution factors for BA.

   ${\left(DF\right)}_{BA}=\frac{K_{BA}}{K_{BA}+K_{BC}}$ .....(I)

Substitute $402.78I_{BA}$ for $K_{BA}$ and $604.16I_{BC}$ for $K_{BC}$ in Equation (I).

   ${\left(DF\right)}_{BA}=\frac{402.78I_{BA}}{402.78I_{BA}+604.16I_{BC}}$ .....(II)

Substitute $900{\text{in}}^4$ for $I_{BA}$ and $1200{\text{in}}^4$ for $I_{BC}$ in Equation (II).

   $\begin{array}{c}{\left(DF\right)}_{BA}=\frac{402.78\times 900{\text{in}}^4}{402.78\times 900{\text{in}}^4+604.16\times 1200{\text{in}}^4}\\ =\frac{362502}{362502+724992}\\ =0.33\end{array}$

Calculate distribution factors for BC.

   ${\left(DF\right)}_{BC}=\frac{K_{BC}}{K_{BA}+K_{BC}}$ .....(III).

Substitute $402.78I_{BA}$ for $K_{BA}$ and $604.16I_{BC}$ for $K_{BC}$ in equation (III).

   ${\left(DF\right)}_{BC}=\frac{604.16I_{BC}}{402.78I_{BA}+604.16I_{BC}}$ .....(IV)

Substitute $900{\text{in}}^4$ for $I_{BA}$ and $1200{\text{in}}^4$ for $I_{BC}$ in Equation (IV).

   $\begin{array}{c}{\left(DF\right)}_{BC}=\frac{604.16\times 1200{\text{in}}^4}{402.78\times 900{\text{in}}^4+604.16\times 1200{\text{in}}^4}\\ =\frac{724992}{362502+724992}\\ =0.67\end{array}$

Calculate distribution factors for CB.

   ${\left(DF\right)}_{CB}=0$ (Since, far end is fixed)

Calculate the balance moment for each member as shown below.

Calculate balance moment foe BA.

   $\begin{array}{c}{\left(\text{Balance}\text{moment}\right)}_{BA}=-\left(38.4-60\right)\times {\left(DF\right)}_{BA}\\ =21.6\times 0.33\\ =7.128\text{k}\cdot \text{ft}\end{array}$

Calculate balance moment for BC.

   $\begin{array}{c}{\left(\text{Balance}\text{moment}\right)}_{BC}=-\left(38.4-60\right)\times {\left(DF\right)}_{BC}\\ =21.6\times 0.67\\ =14.47\text{k}\cdot \text{ft}\end{array}$

Draw the moment distribution table as shown below.

Joints	A	B		C
Member	AB	BA	BC	CB
DF	$0$	$0.33$	$0.67$	$0$
FEM	$-38.4$	$38.4$	$-60$	$60$
Balance	-	$7.128$	$14.47$	-
Carry over moment	$3.564$	$0$	$0$	$7.236$
Balance	$0$	$0$	$0$	$0$
$\Sigma M$	$-34.8\text{k}\cdot \text{ft}$	$45.52\text{k}\cdot \text{ft}$	$-45.52\text{k}\cdot \text{ft}$	$67.23\text{k}\cdot \text{ft}$

Consider the section AB as shown below.

  

        Figure-(2)

Write the Equation for sum of vertical forces.

   $\begin{array}{l}\Sigma F_y=0\\ A_y+B_y-\left(800\text{lb}/\text{ft}\times 24\text{ft}\right)\left(\frac{1\text{k}}{1000\text{lb}}\right)=0\\ A_y+B_y=19.2\text{k}\end{array}$ .....(V)

Here, vertical reaction at A and B are $A_y$ and $B_y$.

Write the Equation for sum of moment about A.

   $\begin{array}{l}\Sigma M_A=0\\ \left(34.83-0.8\times 24\times 12+B_y\times 24-45.52\right)\text{k}\cdot \text{ft}=0\\ B_y\times 24\text{ft}=241.09\text{k}\cdot \text{ft}\\ B_y=10.04\text{k}\end{array}$

Substitute $10.04\text{k}$ for $B_y$ in equation (V).

   $\begin{array}{l}{A}_y+10.04\text{k}=19.2\text{k}\\ A_y=9.16\text{k}\end{array}$

Calculate the distance at which shear force changing its sign for maximum moment.

Write the Equation for shear force at distance $x$ from A.

   $V_x=9.16\text{k}-0.8x$

Equate the value of shear force to zero for distance.

   $\begin{array}{l}9.16\text{k}-0.8x=0\\ 0.8x=9.16\text{k}\\ x=11.45\text{ft}\end{array}$

Calculate the value of maximum moment at a distance of $11.45\text{ft}$.

   $\begin{array}{c}{M}_{11.45}=-34.83+9.16\times 11.45-\frac{0.8\times {11.45}^2}{2}\\ =17.611\text{k}\cdot \text{ft}\end{array}$

Consider the section BC as shown below.

  

        Figure-(3)

Write the Equation for sum of vertical forces.

   $\begin{array}{l}\Sigma F_y=0\\ C_y+B_y-30\text{k}=0\\ C_y+B_y=30\text{k}\end{array}$ .....(VI)

Write the Equation for sum of moment about B.

   $\begin{array}{l}\Sigma M_B=0\\ \left(45.52-30\times 8+C_y\times 16-67.23\right)\text{k}\cdot \text{ft}=0\\ C_y\times 16\text{ft}=261.71\text{k}\cdot \text{ft}\\ C_y=16.35\text{k}\end{array}$

Here, vertical reaction at C is $C_y$.

Substitute $16.35\text{k}$ for $C_y$ in equation (VI).

   $\begin{array}{l}16.35\text{k}+B_y=30\text{k}\\ B_y=13.65\text{k}\end{array}$

For member BC, moment will be maximum at the center.

Calculate maximum moment at the center of BC.

   $\begin{array}{c}{M}_8=\left(-45.52+13.65\times 8\right)\text{k}\cdot \text{ft}\\ =63.68\text{k}\cdot \text{ft}\end{array}$

Conclusion:

The moment at A for member AB is, $-38.4\text{k}\cdot \text{ft}$.

The moment at B for member BA is, $45.52\text{k}\cdot \text{ft}$.

The moment at B for member BC is, $-45.52\text{k}\cdot \text{ft}$.

The moment at C for member CB is, $67.23\text{k}\cdot \text{ft}$.

The following diagram shows the moment diagram of the beam.

  

        Figure-(4)