Chapter 10, Problem 10.1P

To determine

Determine the moments at the supports and draw bending moment diagram.

Answer to Problem 10.1P

Explanation of Solution

Given information:

For AB,

In the member AB fixed end is subjected to the udl load with 25kN/m of the first half length.

Determine the fixed end moment for beam span AB, shown in the figure below, using the following formula.

$M_{FAB}=-{\displaystyle \underset{a}{\overset{b}{\int }}\frac{w_x\times dx\times x\times {(l-x)}^2}{l^2}}=-\frac{11w{l}^2}{192}$

$M_{FAB}=-\frac{11\times 25\times 6^2}{192}=-51.56kNm$

And,

$M_{FBA}={\displaystyle \underset{b}{\overset{l}{\int }}\frac{w_x\times dx\times x^2\times {(l-x)}^{}}{l^2}}=\frac{5w{l}^2}{192}$

$M_{FBA}=\frac{5\times 25\times 6^2}{192}=23.44kNm$

In the member BC, where B is roller support and C is fixed end support. It is subjected to point loads of with 15 kN, 15 kN and 15 kN.

Determine the fixed end moment for beam span BC, shown in the figure below, using the following formula.

$M_{FBC}=-\frac{5P{l}^{}}{16}$

Where, P is the concentrated load acting on the beam BC.

$M_{FBC}=-\frac{5\times 15\times 8}{16}=37.5kNm$

By using slope deflection formula, calculate the moment at the end of the beam

$\begin{array}{l}{M}_{AB}=M_{FAB}+\frac{2EI}{l}\left(2{\theta }_A+{\theta }_B+\frac{3\Delta }{l}\right)\\ M_{AB}=-51.56+\frac{2EI}{6}\left(2\times 0+{\theta }_B+\frac{3\times 0}{6}\right)\\ =-51.56+\frac{2EI}{6}{\theta }_B(1)\end{array}$

$\begin{array}{l}{M}_{BA}=M_{FBA}+\frac{2EI}{l}\left(2{\theta }_B+{\theta }_A+\frac{3\Delta }{l}\right)\\ M_{AB}=23.44+\frac{2EI}{6}\left(2{\theta }_B+0+\frac{3\times 0}{6}\right)\\ =23.44+\frac{4EI}{6}{\theta }_B(2)\end{array}$

Similarly for the beam BC,

$\begin{array}{l}{M}_{BC}=M_{FBC}+\frac{2EI}{l}\left(2{\theta }_B+{\theta }_C-\frac{3\Delta }{l}\right)\\ M_{BC}=-37.5+\frac{2EI}{8}\left(2\times {\theta }_B+0-\frac{3\times 0}{6}\right)\\ =-37.5+\frac{4EI}{8}{\theta }_B(3)\end{array}$ $\begin{array}{l}{M}_{CB}=M_{FCB}+\frac{2EI}{l}\left(2{\theta }_C+{\theta }_B-\frac{3\Delta }{l}\right)\\ M_{CB}=37.5+\frac{2EI}{8}\left(2\times 0+{\theta }_B-\frac{3\times 0}{6}\right)\\ =-37.5+\frac{2EI}{8}{\theta }_B(4)\end{array}$

Since the moment equilibrium at support B.

${\displaystyle \sum M_B}=0$

$M_{BA}+M_{BC}=0$

$\begin{array}{l}23.44+\frac{4EI}{6}{\theta }_B-37.5+\frac{4EI}{8}{\theta }_B=0\\ {\theta }_B=\frac{12.05}{EI}\end{array}$

Now substitute the value of ${\theta }_B$ in equation 2, 3, 4 and 5, we get

$\begin{array}{l}{M}_{AB}=-51.56+\frac{2EI}{6}\left(\frac{12.05}{EI}\right)=-47.54kN.m\\ M_{BA}=23.44+\frac{4EI}{6}\left(\frac{12.05}{EI}\right)=31.47kN.m\\ M_{BC}=-37.5+\frac{2EI}{4}\left(\frac{12.05}{EI}\right)=-31.47kN.m\\ M_{CB}=37.5+\frac{2EI}{8}\left(\frac{12.05}{EI}\right)=-40.51kN.m\end{array}$

Now determine the shear reaction at the end, of span AB and BC.

For span AB

Moment about A

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ 31.47-R_B\times 6+25\times 3\times \frac{3}{2}-47.54=0\\ R_{B1}=16.07\end{array}$

And,

$\begin{array}{l}{R}_A+R_B=25\times 3=75kN\\ Substitute{R}_B\\ R_A=75-16.07=58.93kN\end{array}$

For Span BC

Moment about B

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ 40.51-R_C\times 8+15\times 6+15\times 6+15\times 6-31.47=0\\ R_C=23.63kN\end{array}$

And

$\begin{array}{l}{R}_B+R_C=15+15+15=45kN\\ Substitute{R}_B\\ R_{B2}=45-23.63=21.37kN\end{array}$

Shear force diagram for the whole beam.

At point A = $58.93kN$

At right of point D = $58.93kN-75kN=-16.07kN$

At point B = $21.37kN$

At right of point B = $21.37kN$

At point E = $21.37kN-15kN=6.37kN$

At right of point E = $6.37kN$

At point F = $6.37kN-15kN=-8.63kN$

At right of point F = $-8.63kN$

At point G $-8.63kN-15kN=-23.63kN$

At right of point F = $-23.63kN$

The maximum Positive sagging moment in span AB, where the shear force is equal to zero.

$\begin{array}{l}58.93-25x=0\\ x=2.3572m\end{array}$

The maximum positive sagging moment is determine by the below formula.

$\begin{array}{l}{M}_{\max 1}=\frac{w{l}^2}{2}-M_A\\ =\frac{25\times {2.3572}^2}{2}-47.54\\ =21.91kN.m\end{array}$

Bending moment at point E.

$M_E=-31.47+21.27\times 2=11.27kNm$

Bending moment at point F.

$M_F=-31.47+21.27\times 4-15\times 2=24.01kNm$

Bending moment at point G.

$M_G=-31.47+21.27\times 4-15\times 4-15\times 2=6.75kNm$