Minimizing the maximum bending moment

[gpavanb]

Homework Statement

Refer to the attachment provided

Homework Equations

Just taking force and moment equilibium of whatever component I choose.

The Attempt at a Solution

I assumed a uniform force distribution.
Set the origin at the leftmost end. For 0< x < \frac{L-a}{2}
The shear force acting is +qx and the bending moment is \frac{-qx^{2}}{2}
Note that the situation is symmetrical w.r.t the centre of the beam.
Now for \frac{L-a}{2}< x < L/2
The relevant force equilibrium equation is
-qx+\frac{qL}{2}+V=0 \Rightarrow V=qx-\frac{qL}{2}
The bending moment can be similarly found and is given by
\frac{qLx}{2}-\frac{qx^{2}}{2}-\frac{qL(L-a)}{4}

Thus the maxima of the above two moments are
\frac{q(L-a)^{2}}{8} and \frac{qL(L-a)}{4}

Both of which give a=L is when it is minimized. That isn't the answer at the back of the book!

I don't think we should take it as a uniform distribution.

 

[nvn]

gpavanb: It is a uniformly-distributed load. Is the answer in the back of the book by any chance a = 0.5858*L?

 

[gpavanb]

Spot On!

 

[gpavanb]

How did you get it?

 

[nvn]

If you move the supports too far apart, the midspan moment increases. If you move the supports too close together, the moment somewhere else increases. You want to figure out, at what value of "a" is the maximum moment in the beam as small as possible.

 

[gpavanb]

That was a vague hint. I got the problem in the mean time though. It was fairly straightforward.a=\frac{1}{(1+\frac{1}{\sqrt{2}})}L

 

[nvn]

Excellent work, gpavanb, which can be simplified to a = (2 - 2^0.5)*L, by multiplying the numerator and denominator of your answer by the complement, (2^0.5 - 1)/(2^0.5 - 1), which is 1, shown in line 4, below.

a = L/{1 + [1/(2^0.5)]}
= L/{[(2^0.5)/(2^0.5)] + [1/(2^0.5)]}
= (2^0.5)*L/(2^0.5 + 1)
= [(2^0.5)*L/(2^0.5 + 1)]*[(2^0.5 - 1)/(2^0.5 - 1)]
= (2^0.5)(2^0.5 - 1)*L/[(2^0.5 + 1)(2^0.5 - 1)]
= (2 - 2^0.5)*L/(2 + 2^0.5 - 2^0.5 - 1)
= (2 - 2^0.5)*L/(2 - 1)
= (2 - 2^0.5)*L