Castiglianos Theorem: Elastic Body Displacements & Swing Analysis

  • Thread starter Thread starter umarfarooq
  • Start date Start date
  • Tags Tags
    Theorem
AI Thread Summary
Castigliano's theorem relates to the calculation of displacements in elastic bodies due to applied loads, emphasizing the importance of understanding translational and rotational displacements. In the context of a swing constructed from a steel tube, the discussion focuses on deriving expressions for vertical and horizontal displacements when a downward load is applied. A user attempts to solve the problem using moment equations and integration, but expresses confusion regarding the inclusion of fictitious forces. Feedback suggests that the fictitious force should equal the applied load, highlighting the need for clarity on the load's application point. The conversation underscores the necessity of accurately applying Castigliano's theorem for effective analysis of the swing's deflection.
umarfarooq
Messages
4
Reaction score
0

Homework Statement


a) State Castiglianos theorem for translational and rotational displacements of an elastic body, stating precisely the meanings of the terms.

b) A swing in a childrens play area is constrcuted from a steel tube bent into a quarter circle of radius R. One end is rigidly fixed to the ground with the tangent to the circle vertical, and the swing attached to the other end. Assuming that the beam has a section constant EI, derive experssions for the vertical and horizontal displacements of the swing when a downwards load P is applied to it.


Homework Equations


dV=dU/dL=d(int(M^2/(2*E*I)*r*dTheta,Pi->0)/dP


The Attempt at a Solution



sorry but I am completely baffled
 
Physics news on Phys.org
Where are you stuck??

You need to show your work, may i recommend to use your moment equation in function of the angle the radius makes with the vertical.
 
okay, i think my answer is wrong but this is what I've got.
the moment is M(theta) is PRCos(theta) + F(R-RSin(theta)) where f is ficticious i know so do i disregard that.

Therefore M^2(theta) = R^2(P^2Cos^2(theta) + F^2 - F^2Sin^2(Theta).

Therefore i use that in the formula d(int(M^2/(2*E*I)*r*dTheta,Pi->0)/dP
I use the trig identities for Cos^2(theta) and Sin^2(theta) and integrate. If i ignore the ficticous force F the value of the integral is P^2(Pi/2)

This gives me a deflection of (P^2*Pi*R^3)/(4*E*I)
Is this correct
Would appreciate it alot, Thanks
 
umarfarooq said:
okay, i think my answer is wrong but this is what I've got.
the moment is M(theta) is PRCos(theta) + F(R-RSin(theta)) where f is ficticious i know so do i disregard that.

Therefore M^2(theta) = R^2(P^2Cos^2(theta) + F^2 - F^2Sin^2(Theta).

Therefore i use that in the formula d(int(M^2/(2*E*I)*r*dTheta,Pi->0)/dP
I use the trig identities for Cos^2(theta) and Sin^2(theta) and integrate. If i ignore the ficticous force F the value of the integral is P^2(Pi/2)

This gives me a deflection of (P^2*Pi*R^3)/(4*E*I)
Is this correct
Would appreciate it alot, Thanks

You need to read castigliano's theorem again, in this case the fictitious force is equal to the applied load, unless the applied load is not at the free end, if that's the case you must specify where it is, so we can actually help out!
 
Last edited:

Similar threads

Back
Top