# What did I do wrong? Bending moments & deflection question

• Matthew Heywood
Thanks so much, this has been really helpful.In summary, at the fixed end of the cantilever, there is no slope and max deflection. c1 and c2 are determined based on the boundary conditions of slope = deflection = 0.

## Homework Statement

Problem shown as image
jan2010 Q5.jpg
Q5. a)

## Homework Equations

Also shown in image

## The Attempt at a Solution

Made a cut between A and the z-line, so ∑M = 0 = M - wx(x/2) = M - (wx2)/2 ⇒ M = (wx2)/2

and d2y/dx2 = -M/EI

∴ d2y/dx2 = - (wx2)/2EI

So dy/dx = ∫ d2y/dx2 dx = -w/2EI ∫ x2 dx = -wx3/6EI + c1

I was fine up to this point. My answer here agrees with the answer in the question ( equation 5.2) as x = L.

So what I got next:

y = ∫ dy/dx dx = -w/6EI ∫ x3 dx + ∫ c1 dx = -wx4/24EI + c1x + c2

Letting x = L

dy/dx = -wL3/6EI + c1 (as in the question)

and

y = -wL4/24EI + c1L + c2 (which is different to the answer given in the question)

How/why is the answer to equation 5.2 wL4/8EI ??

I probably made a silly mistake somewhere or I'm just too tired :P

Thanks for any help.

#### Attachments

• jan2010 Q5.jpg
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Your integrations are correct from what I can see. The only thing you haven't done is to use the boundary conditions for the cantilever to calculate what c1 and c2 must be. I think once you do this, you should be able to get the answers you seek.

So boundary conditions between L and 0?

Matthew Heywood said:
So boundary conditions between L and 0?
You've got only two boundary conditions with this beam: both are located at the fixed end. Do you know what they are?

Ohh. The reaction forces?

No, the boundary conditions are not the reaction forces. These can be determined by statics.

Think about what you are trying to find here. You want to calculate expressions which give the slope and deflection of the beam. What can you say about the slope and deflection of the cantilever beam at the fixed end?

At the fixed end, there's no slope and max deflection?

Matthew Heywood said:
At the fixed end, there's no slope and max deflection?

Come again? Max. deflection at the fixed end? Is that your final answer?

Yeah I have no idea :P Sorry

Matthew Heywood said:
Yeah I have no idea :P Sorry
If the right end of the cantilever in the problem is fixed, it can't move. What must the slope and deflection both be at the fixed end?

Would it be 0?

Yes.

Then surely c1 = wL3/6EI for the first one? Which will all cancel out to 0? I'm confused where to go next :(

Matthew Heywood said:
Then surely c1 = wL3/6EI for the first one? Which will all cancel out to 0? I'm confused where to go next :(
It only cancels out when x = L. What happens when x = 0, at the free end?

Then c1 = 0

Matthew Heywood said:
Then c1 = 0
That's the trivial solution to the value of c1.

I mean, what happens to the equations for the slope and deflection when you determine c1 and c2 based on the boundary conditions of slope = deflection = 0 at x = L? Since c1 and c2 are constants of integration, they have the same values for 0 ≤ x ≤ L.

So -wx3/6EI + c1 = -wx4/24EI + c1x + c2 = 0
I'm still struggling to know where to go next.

Matthew Heywood said:
So -wx3/6EI + c1 = -wx4/24EI + c1x + c2 = 0
I'm still struggling to know where to go next.

You don't set the slope and deflection equations equal to one another. You use the fact that the slope = 0 when x = L to find c1, then you substitute this value of c1 into the deflection equation, set x = L, and knowing the deflection = 0 there, then you solve for the value of c2.

Once you have solved for the values of c1 and c2, then you substitute these into the equations for slope and deflection, such that the value of the slope or deflection of the cantilever at any point depends only on the length L, the loading w, and the position x along the cantilever from the free end.

So -wx3/6EI + c1 = 0 ⇒ c1 = wx3/6EI

∴ -wx4/24EI + c1x + c2 = -wx4/24EI + wx4/6EI + c2 = 0 ⇒ c2 = -wx4/8EI

Am I right so far?

Matthew Heywood said:
So -wx3/6EI + c1 = 0 ⇒ c1 = wx3/6EI

∴ -wx4/24EI + c1x + c2 = -wx4/24EI + wx4/6EI + c2 = 0 ⇒ c2 = -wx4/8EI

Am I right so far?
No, the slope = 0 only when x = L. The deflection = 0 only when x = L

I don't really know what else I can do. I get that the slope and deflection are 0 at the fixed point, but I don't know what else I can use.

Ohhh hang on. So dy/dx = 0 when x = L ∴ c1 = wL3/6EI
∴ y = -wx4/24EI + wL3x/6EI + c2
and y = 0 when x = L ∴ c2 = -wL4/8EI
Is this looking better? :P

and then when x = 0, y = -w(0)/24EI + wL3(0)/6EI - wL4/8EI = -wL4/8EI

Finally got closer :P although still not entirely right. I've got the negative of the answer on the sheet. Sorry for being a pain here :P

Matthew Heywood said:
and then when x = 0, y = -w(0)/24EI + wL3(0)/6EI - wL4/8EI = -wL4/8EI

Finally got closer :P although still not entirely right. I've got the negative of the answer on the sheet. Sorry for being a pain here :P
I tried working the same problem from scratch and got an extra negative in the answer as well. It could be a typo in the text.

Ahh okay. Well thank you for all of your help and patience :D