- #36
Mohankpvk
- 102
- 3
Yes.I agree with the point that they are not totally indeterminate.haruspex said:No, there's more to it than that.
Consider the simplest case: no load applied, weightless beam, ends at same height and fixed horizontally.
If this is statically indeterminate then there must be a nontrivial solution, i.e., one in which the end supports apply equal and opposite nonzero vertical forces. Correspondingly there will be nonzero torques so that equilibrium is maintained.
For a theoretical rigid beam that would all be possible, but this field of study considers beams to be flexible. The beam undergoes a deflection y=y(x).
Suppose the supports give a force F up at the left and down at the right, and anticlockwise torques T, T' respectively.
Taking moments at offset x from the left:
ky"=-T+Fx
Integrating, and using y(0)=0:
ky'=-Tx+Fx2/2
Integrating, and using y'(0)=0:
ky=-Tx2/2+Fx3/6
Now we use the same constraints at the right hand end (x=L):
0=-TL+FL2/2
0=-TL2/2+FL3/6
When the smoke clears:
T=F=0.
So it is not indeterminate after all.
(I was wrong to agree to that earlier without checking.)
But I think they are called statically indeterminate because we can't find the reactions just by using the static equilibrium equations.
We have to exploit the cocept of deflection(deflection at the ends=0)to solve for the reactions.