Bending moment

• Mar 14th 2009, 08:36 AM
Jaffa
Bending moment
Hello .the question is ,The bending moment,M.at a position x m from the end of a simply supported beam of lenght l m carrying a uniformly distrbuted load
of wkNm-1 is given by
M= w/2 l (l -x)-w/2 (l-x)^2
Show using the above expression, that the maximum bending moment occurs atthe mid point of beam and determines its value interms of w and l
I have not got a clue !
• Mar 14th 2009, 08:48 AM
skeeter
Quote:

Originally Posted by Jaffa
Hello .the question is ,The bending moment,M.at a position x m from the end of a simply supported beam of lenght l m carrying a uniformly distrbuted load
of wkNm-1 is given by
M= w/2 l (l -x)-w/2 (l-x)^2
Show using the above expression, that the maximum bending moment occurs atthe mid point of beam and determines its value interms of w and l
I have not got a clue !

your notation leaves much to be desired. Is the equation ...

$\displaystyle M = \frac{w}{2} L(L-x) - \frac{w}{2} (L-x)^2$

or

$\displaystyle M = \frac{w}{2L(L-x)} - \frac{w}{(L-x)^2}$

or something else?
• Mar 14th 2009, 09:22 AM
Jaffa
Sorry , the first one is how it is meant to be, hope you can help
• Mar 14th 2009, 09:44 AM
skeeter
$\displaystyle M = \frac{w}{2} L(L-x) - \frac{w}{2} (L-x)^2$

$\displaystyle \frac{dM}{dx} = -\frac{w}{2}L + w(L-x)$

set $\displaystyle \frac{dM}{dx} = 0$

$\displaystyle w(L-x) = \frac{w}{2}L$

$\displaystyle L-x = \frac{L}{2}$

$\displaystyle x = \frac{L}{2}$

$\displaystyle \frac{d^2M}{dx^2} = -w < 0$ ... $\displaystyle x = \frac{L}{2}$ is where $\displaystyle M$ will be a maximum

substitute $\displaystyle \frac{L}{2}$ for $\displaystyle x$ in the original equation and determine the maximum value of $\displaystyle M$.
• Mar 16th 2009, 12:30 PM
Jaffa
Thanks for your help,makes things clearer.