The angled beam ABC in the figure is composed of segments AB and BC joined at B at a 90˚ angle. Both segments have length L and uniform section stiffness EI. The beam is fixed at C and loaded by a downward concentrated force, P, applied at its free end A.
You will use Castiglano's theorem to determine the vertical downward deflection vA of the free end A and the horizontal deflection uB of section B (towards A) as indicated in the figure below.
1.
Use Castigliano's theorem to obtain a symbolic expression for the downward vertical deflection vA of the free end A, in terms of P, L, and EI ( enter this as EI without the multiplication sign).
2.
Use Castigliano's theorem to obtain a symbolic expressions for the horizontal deflection uB of section B (towards A) in terms of P, L, and EI ( enter this as EI without the multiplication sign). Hint: You will have to introduce a dummy load that is work-conjugate to the desired deflection uB
To use Castigliano's theorem to determine the vertical downward deflection vA of the free end A and the horizontal deflection uB of section B, we need to follow these steps:
1. Calculate the partial derivative of the total potential energy (U) with respect to the respective deflection:
- For vA, we calculate ∂U/∂vA.
- For uB, we introduce a dummy load (W) that is work-conjugate to the desired deflection uB and calculate ∂U/∂W.
2. Set the partial derivatives to zero.
3. Solve the resulting equations to obtain the symbolic expressions for vA and uB in terms of P, L, and EI.
Let's go through each step:
1. Calculating ∂U/∂vA:
The total potential energy (U) of the beam is the sum of the strain energy (Ue) and the work done by external forces (Ue).
- Strain Energy (Ue):
For each segment AB and BC, the strain energy is given by (1/2) * (EI) * (slope of the beam)^2.
The vertical deflection vA of the free end A will affect the slopes of both segments AB and BC.
The slopes of AB and BC are given by (θA/2) and (θB/2), respectively, where θA and θB are the angles that AB and BC make with the horizontal axis.
- Work Done by External Forces (Ue):
The work done by the concentrated force P at point A is given by P * vA.
Now, by summing the strain energy and the work done by external forces, we get the expression for U in terms of vA:
U = (1/2) * (EI) * [(θA/2)^2 + (θB/2)^2] - P * vA
Taking the partial derivative of U with respect to vA, we obtain:
∂U/∂vA = -P
2. Calculating ∂U/∂W (introducing dummy load):
To obtain the expression for uB, we introduce a dummy load W that is work-conjugate to the desired deflection uB. The potential energy associated with this dummy load is given by (1/2) * (EI) * W^2.
Adding the potential energy of the dummy load to the expression for U, we get:
U = (1/2) * (EI) * [(θA/2)^2 + (θB/2)^2] - P * vA + (1/2) * (EI) * W^2
Taking the partial derivative of U with respect to W, we obtain:
∂U/∂W = (EI) * W
Setting this derivative to zero, we have:
(EI) * W = 0
W = 0
3. Solving for vA and uB:
Setting ∂U/∂vA = 0, we have -P = 0, which implies P = 0. However, this contradicts the given downward concentrated force P at point A. Hence, this equation does not provide any useful information.
Since the dummy load W is zero (∂U/∂W = 0), we can ignore its contribution.
The deformed shape of the beam under P can be approximated by the deflection uB of section B.
Therefore, for the symbolic expression for the horizontal deflection uB, we have:
uB = (1/2) * (EI) * [(θA/2)^2 + (θB/2)^2]
Please note that the above calculation assumes small deflections and linear elastic behavior of the beam material. It may not be applicable for large deformations or non-linear materials.