Chapter 13, Problem 13.1P

To determine

The moment about point A.

The moment about point B.

The moment about point C.

Answer to Problem 13.1P

The moment about point A is $-348.48\text{ k-ft}$.

The moment about point B in member BA is $301.44\text{ k-ft}$.

The moment about point B in member BC is $-301.44\text{ k-ft}$.

The moment about point C is $348.48\text{ k-ft}$.

Explanation of Solution

Given:

The supports $A$ and $C$ are fixed and $B$ is roller support in a rigid base.

The girder has thickness of $4\text{ ft}$.

Concept Used:

Write the expression for fixed end moments.

   $FEM=\text{coefficients}\times w{L}^2$   .......(I)

Here, uniform distributed load is $w$ and length of the span is $L$.

Write the expression for absolute stiffness factor for member $BA$.

   $K_{BA}=\frac{k_{BA}E{I}_C}{L}$   .......(II)

Here, stiffness factor for member $BA$ is $k_{BA}$. moment of inertia for $C$ is $I_C$, modulus of elasticity is $E$ and length of the span is $L$.

Write the expression for absolute stiffness factor for member $BC$.

   $K_{BC}=\frac{k_{BC}E{I}_C}{L}$   .......(III)

Here, stiffness factor for member $BC$ is $k_{BC}$.

Write the expression for distribution factor for member $BC$.

   $D{F}_{BC}=\frac{K_{BC}}{K_{BA}+K_{BC}}$   .......(IV)

Write the expression for distribution factor for member $BA$.

   $D{F}_{BA}=\frac{K_{BA}}{K_{BA}+K_{BC}}$   .......(V)

Write the expression for distribution factor for member $AB$.

   $D{F}_{AB}=0$   .......(VI)

Since the supports $A$ is fixed.

Write the expression for distribution factor for member $CB$.

   $D{F}_{CB}=0$   .......(VII)

Since the supports $C$ is fixed.

Calculation:

The diagram of the beam is shown below.

  

  Figure (1)

Consider Figure (1).

Length of span is $L=20\text{ ft}$.

Length of haunch at end $A$ is $6\text{ ft}$.

Length of haunch at end $B$ is $4\text{ ft}$.

Calculate ratio of length of haunch at end $A$ to the length of the span $AB$.

   $a_A=\frac{6\text{ ft}}{\begin{array}{c}20\text{ ft}\\ =0.3\end{array}}$

Calculate ratio of length of haunch at end $B$ to the length of the span $AB$.

   $a_B=\frac{4\text{ ft}}{\begin{array}{c}20\text{ ft}\\ =0.2\end{array}}$

Calculate ratio for rectangular cross-sectional area at end $A$.

   $\begin{array}{c}{r}_A=\frac{4\text{ft}-2\text{ft}}{2\text{ft}}\\ =1\end{array}$

Calculate ratio for rectangular cross-sectional area at end $B$.

   $\begin{array}{c}{r}_B=\frac{4\text{ft}-2\text{ft}}{2\text{ft}}\\ =1\end{array}$

Calculate ratio of length of haunch at end $B$ to the length of the span $AB$.

   $a_B=\frac{4\text{ ft}}{\begin{array}{c}20\text{ ft}\\ =0.2\end{array}}$

Calculate ratio of length of haunch at end $C$ to the length of the span $BC$.

   $a_C=\frac{6\text{ ft}}{\begin{array}{c}20\text{ ft}\\ =0.3\end{array}}$

Calculate ratio for rectangular cross-sectional area at end $B$.

   $\begin{array}{c}{r}_B=\frac{4\text{ft}-2\text{ft}}{2\text{ft}}\\ =1\end{array}$

Calculate ratio for rectangular cross-sectional area at end $C$.

   $\begin{array}{c}{r}_C=\frac{4\text{ft}-2\text{ft}}{2\text{ft}}\\ =1\end{array}$

Determine the carry over factor of member $AB$ at end $A$ from straight haunches-constant width table and ratios calculated.

   $C_{AB}=0.622$

Determine carry over factor of member $AB$ at end $B$ from straight haunches-constant width table and ratios calculated.

   $C_{BA}=0.748$

Determine carry over factor of member $BC$ at end $B$ from straight haunches-constant width table and ratios calculated.

   $C_{BC}=0.748$

Determine carry over factor of member $BC$ at end $C$ from straight haunches-constant width table and ratios calculated.

   $C_{CB}=0.622$

Determine stiffness factor of member $AB$ at end $A$ from straight haunches-constant width table and ratios calculated.

   $k_{AB}=10.06$

Determine stiffness factor of member $AB$ at end $B$ from straight haunches-constant width table and ratios calculated.

   $k_{BA}=8.37$

Determine stiffness factor of member $BC$ at end $B$ from straight haunches-constant width table and ratios calculated.

   $k_{BC}=8.37$

Determine stiffness factor of member $BC$ at end $C$ from straight haunches-constant width table and ratios calculated.

   $k_{CB}=10.06$

Calculate absolute stiffness factor for member $BA$.

Substitute $8.37$ for $k_{BA}$ and $20\text{ ft}$ for $L$ in Equation (III).

   $\begin{array}{c}{K}_{BA}=\frac{\left(8.37\right)E{I}_C}{20\text{ ft}}\\ =0.41875E{I}_C\end{array}$

Calculate absolute stiffness factor for member $BC$.

Substitute $8.37$ for $k_{BC}$ and $20\text{ ft}$ for $L$ in Equation (III).

   $\begin{array}{c}{K}_{BA}=\frac{\left(8.37\right)E{I}_C}{20\text{ m}}\\ =0.418E{I}_C\end{array}$

Calculate fixed end moment for member $AB$.

Determine coefficients from straight haunches-constant width table.

Substitute $8\text{ k/ft}$ for $w$, $-0.1089$ for coefficient and $20\text{ ft}$ for $L$ in Equation (I).

   $\begin{array}{c}{\left(FEM\right)}_{AB}=\left(-0.1089\right)\times \left(8\text{ k/ft}\right){\left(20\text{ ft}\right)}^2\\ =-348.48\text{ k-ft}\end{array}$

Calculate fixed end moment for member $BA$.

Determine coefficients from straight haunches-constant width table.

Substitute $8\text{ k/ft}$ for $w$, $0.0942$ for coefficient and $20\text{ ft}$ for $L$ in Equation (I).

   $\begin{array}{c}{\left(FEM\right)}_{BA}=\left(0.0942\right)\times \left(8\text{ k/ft}\right){\left(20\text{ ft}\right)}^2\\ =301.44\text{ k-ft}\end{array}$

Calculate fixed end moment for member $CB$.

Determine coefficients from straight haunches-constant width table.

Substitute $8\text{ k/ft}$ for $w$, $0.1089$ for coefficient and $20\text{ ft}$ for $L$ in Equation (I).

   $\begin{array}{c}{\left(FEM\right)}_{CB}=\left(0.1089\right)\times \left(8\text{ k/ft}\right){\left(20\text{ ft}\right)}^2\\ =348.48\text{ k-ft}\end{array}$

Calculate fixed end moment for member $BC$.

Determine coefficients from straight haunches-constant width table.

Substitute $8\text{ k/ft}$ for $w$, $-0.0942$ for coefficient and $20\text{ ft}$ for $L$ in Equation (I).

   $\begin{array}{c}{\left(FEM\right)}_{BA}=\left(-0.0942\right)\times \left(8\text{ k/ft}\right){\left(20\text{ ft}\right)}^2\\ =-301.44\text{ k-ft}\end{array}$

Calculate distribution factor for member $BC$.

Substitute $0.418E{I}_C$ for $K_{BC}$ and $0.418E{I}_C$ for $K_{BA}$ in Equation (IV).

   $\begin{array}{c}D{F}_{BC}=\frac{0.418E{I}_C}{0.418E{I}_C+0.418E{I}_C}\\ =0.5\end{array}$

Calculate distribution factor for member $BA$.

Substitute $0.418E{I}_C$ for $K_{BC}$ and $0.418E{I}_C$ for $K_{BA}$ in Equation (V).

   $\begin{array}{c}D{F}_{BA}=\frac{0.418E{I}_C}{0.418E{I}_C+0.418E{I}_C}\\ =0.5\end{array}$

The moment distribution table is shown below.

Joint	A	B		C
Member	AB	BA	BC	CB
K		$0.418E{I}_C$	$0.418E{I}_C$
D.F.	$0$	$0.5$	$0.5$	$0$
C.O.F.	$0.622$	$0.748$	$0.748$	$0.622$
F.E.M.	$-348.48\text{ k-ft}$	$301.44\text{ k-ft}$	$-301.44\text{ k-ft}$	$348.48\text{ k-ft}$
DIST		$0$	$0$
SUM	$-348.48\text{ k-ft}$	$301.44\text{ k-ft}$	$-301.44\text{ k-ft}$	$348.48\text{ k-ft}$

Table (1)

Conclusion:

The moment about point A is $-348.48\text{ k-ft}$.

The moment about point B in member BA is $301.44\text{ k-ft}$.

The moment about point B in member BC is $-301.44\text{ k-ft}$.

The moment about point C is $348.48\text{ k-ft}$.