Castigliano's method for calculating displacements

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SUMMARY

Castigliano's method for calculating displacements utilizes the second theorem, which relates strain energy to generalized forces and displacements. The discussion highlights that for non-symmetrical cross sections, such as an "N" shaped cantilever beam, applied forces may cause displacements in directions other than the applied force direction, particularly when the force does not align with the principal axes. It is established that if a force is applied along the y-axis, the resulting displacement will occur in both the y and z directions, necessitating the consideration of principal axis components for accurate displacement calculations.

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  • Understanding of Castigliano's theorem and its application in structural analysis.
  • Knowledge of strain energy concepts in linearly elastic structures.
  • Familiarity with principal axes and their significance in displacement calculations.
  • Basic principles of cantilever beam mechanics and cross-sectional analysis.
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Dell
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Castigliano's method for calculating displacements is an application of his second theorem, which states:
If the strain energy of a linearly elastic structure can be expressed as a function of generalised force Qi; then the partial derivative of the strain energy with respect to generalised force gives the generalised displacement qi in the direction of Qi.

is this only correct for forces working on the principal axes?? up till now i have not been dealing with nonsymetrical cross sections but now that i have started it is clear that if i have a nonsymetrical cross section with a force only in one direction, my displacement will not be in that direction but rather perpendicular to the neutral axis. (unless the force is on one of the principal axes)

taking for example a simple cantelever beam with an "N" shaped cross section on plane zy, if a force is applied on the y-axis the cross section will move in both y and z directions. if i used castiglianos theorem i can find a displacement for the y axis(not sure if this is the correct displacement) but to find the z displacement i would usually add a fictitious force (Q=0) compute the internal forces as a function of Q and integrate, but since Q=0, i would simply get displacement=0

do i need to divide my real force into its principal axes components, find the derivative of each one and integrate?
 
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Dell said:
taking for example a simple cantelever beam with an "N" shaped cross section on plane zy, if a force is applied on the y-axis the cross section will move in both y and z directions.

If N is the cross section and if N is in the yz plane then your load on the y-axis passes through the shear centre. Doesn't that mean that there is no displacement in the z-direction?
 

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