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Bending Moments for Portal Frames

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Portal Frames.

The drawing shows a Portal Frame in which the ends A and D are fixed vertically and a distributed load is carries on BC.

If $\displaystyle M_1$ and $M_2$ are the Bending Moments at A and B, then the B.M. diagrams for AB and BC are as shown. As the joints at B and C are rigid, the angle $\phi$ is the same for AB and BC.

For an upright AB.

The intercept at B=0. i.e. using moment-areas about B ( See Bending of Beams Part 3)

$\frac{1}{2}(M_1+M_2)\times2l\times\frac{4\;l}{3}\;-\;M_2\times2\;l\timesl\;=\;0$

(1)

$\therefore\;\;\;\;\;\;M_1=\frac{1}{2}\;M_2$

(2)

As $\displaystyle \phi$ is very small

$\phi =\frac{Z_1}{2\;l} = \frac{M_2\times 2\;l\times l - \frac{1}{2}(M_1+M_2)\times 2\;l\times2\;l/3}{2\;E\;I}$

(3)

$=\frac{M_2\;l}{2\;E\;I}$

(4)

For the top of the Frame BC

$\phi =\frac{Z_2}{l} = \frac{[\frac{2}{3}(w\;l^2/8)\;l]\times l/2-M_2\;l\times l\times l/2}{E\;I\;l}$

(5)

$=\frac{W\;l^3/24 - M_2\;l/2}{E\;I}$

(6)

Equating Equations (4) and (6)

$M_2=\frac{w\;l^2}{24}$

(7)

And from Equation (2)

$M_1=\frac{w\;l^2}{48}$

(8)

The maximum Bending Moment occurs at the middle of BC and

$\hat{M}=\frac{w\;l^2}{8}-\frac{w\;l^2}{24}=\frac{w\;l^2}{12}$

(9)