What did I do wrong? Bending moments & deflection question

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Discussion Overview

The discussion revolves around a homework problem related to bending moments and deflection in a cantilever beam. Participants explore the mathematical approach to solving for the slope and deflection using integration and boundary conditions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents their attempt at solving the problem, including their integration steps and expressions for slope and deflection.
  • Another participant suggests that the boundary conditions must be applied to determine the constants of integration, c1 and c2.
  • There is a discussion about the boundary conditions at the fixed end of the cantilever, specifically that both slope and deflection are zero at that point.
  • Participants explore the implications of setting slope and deflection to zero at the fixed end and how to derive the values of c1 and c2 from these conditions.
  • One participant expresses confusion about the relationship between the constants and the boundary conditions, indicating uncertainty about the next steps in the solution process.
  • There are repeated attempts to clarify the correct values for c1 and c2 based on the boundary conditions, with some participants suggesting different approaches to arrive at these constants.
  • One participant mentions that they obtained a negative value for the answer, suggesting a possible typo in the original problem statement.

Areas of Agreement / Disagreement

Participants generally agree on the need to apply boundary conditions to solve for constants but express differing levels of understanding about how to do this correctly. The discussion remains unresolved regarding the final expressions for slope and deflection, with some participants still uncertain about their calculations.

Contextual Notes

There are limitations in the discussion regarding the clarity of the boundary conditions and the application of integration constants. Some participants express confusion about the relationship between the constants and the physical conditions at the beam's fixed end.

Who May Find This Useful

This discussion may be useful for students studying mechanics of materials, particularly those dealing with cantilever beams and the application of boundary conditions in solving differential equations related to deflection and slope.

Matthew Heywood
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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.
 

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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
 
  • #10
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?
 
  • #11
Would it be 0?
 
  • #12
Yes.
 
  • #13
Then surely c1 = wL3/6EI for the first one? Which will all cancel out to 0? I'm confused where to go next :(
 
  • #14
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?
 
  • #15
Then c1 = 0
 
  • #16
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.
 
  • #17
So -wx3/6EI + c1 = -wx4/24EI + c1x + c2 = 0
I'm still struggling to know where to go next.
 
  • #18
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.
 
  • #19
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?
 
  • #20
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
 
  • #21
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.
 
  • #22
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
 
  • #23
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
 
  • #24
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.
 
  • #25
Ahh okay. Well thank you for all of your help and patience :D
 

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