Beam Deflection Problem: Determining Slope and Deflection at a Point x on a Beam

Click For Summary
SUMMARY

The discussion focuses on solving the beam deflection problem using the bending stiffness equation EIV'' = M, where E is Young's modulus, I is the second moment of area, V is deflection, and M is the moment. The user successfully determined the reactions at the supports but struggles with the integration constants, particularly the constant A, which results in a complicated fraction. The expected outcomes for the slope and deflection are V' = (ab/3)((a-b)/(a+b)) and V = -a^2b^2/3(a+b). The community suggests verifying the expression for A by testing specific values to simplify the problem.

PREREQUISITES
  • Understanding of beam mechanics and deflection analysis
  • Familiarity with the bending stiffness equation and its components
  • Knowledge of integration techniques in calculus
  • Ability to apply boundary conditions in structural analysis
NEXT STEPS
  • Review the derivation of the bending stiffness equation EIV'' = M
  • Practice solving beam deflection problems using boundary conditions
  • Learn about Macaulay's method for beam deflection analysis
  • Explore simplification techniques for complex fractions in calculus
USEFUL FOR

Students and professionals in civil engineering, structural engineering, and mechanics who are working on beam deflection problems and require a deeper understanding of bending stiffness and integration techniques.

IIan
Messages
1
Reaction score
0

Homework Statement



Determine the slope and deflection at the point x on the beam (attached image)


Homework Equations



Bending stiffness equation: EIV'' = M
where E = young's modulus, I = second moment of area, V = deflection, M = moment



The Attempt at a Solution



By taking moments about both supports I have determined the reactions at the left and right supports to be b/(a+b) and a/(a+b) respectively.

Cutting the beam between point x and the right hand support:

EIV'' = M = bx/(a+b) - <x-a> (where <> are macaulay brackets)

EIV' = bx^2/2(a+b) - (<x-a>^2)/2 + A

EIV = bx^3/6(a+b) - (<x-a>^3)/6 + Ax + B

Here's where I get confused:

Using V=0 at the boundary conditions (x=0 and x=a+b) to find the integration constants, B is found to be 0 but A always comes out as a complicated fraction that I can't seem to simplify to get anything sensible.

I know the final answers are supposed to be: V' = (ab/3)((a-b)/(a+b))
and V =-a^2b^2/3(a+b)


I have worked through the question several times and I can't figure out where I'm going wrong so any help would be much appreciated.
 

Attachments

  • beamQ.GIF
    beamQ.GIF
    1.5 KB · Views: 585
Physics news on Phys.org
Have you checked your complicated expression for "A" by putting a=b and setting the gradient to zero at the centre? It is no guarantee if it is correct, but, if it is incorrect in this special case, then it will be incorrect more generally.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 24 ·
Replies
24
Views
4K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 7 ·
Replies
7
Views
8K
Replies
6
Views
4K