Beam ABC in the figure is composed of segment AB of length L and uniform section stiffness EI joined to segment BC of length L and uniform section stiffness 2EI. The beam is fixed at C (x=2L) and loaded by a downward concentrated force, P, applied at its free end A (x=0).
You will use Castiglano's theorem to determine the vertical downward deflection vA and the rotation (slope) ϑA of the beam at its free end A .
1.
Obtain symbolic expressions for the integrands in the two expressions below, which give the complementary energy in the two segments of the beam, 𝔼∗AB and 𝔼∗BC, in terms of P, x, and EI ( enter this as EI without the multiplication sign). NOTE that you do not need to carry out the integration, just write the integrands: e.g., if the expression for 𝔼∗AB were to be 𝔼∗AB=∫L0(ax+b)dx, you would just write a∗x+b in the answer box.
2.
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).
3.
Use Castigliano's theorem to obtain a symbolic expression for the rotation ϑA (the slope) of the beam at 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 rotation ϑA
1. To obtain the integrands for the complementary energies, we need to consider the strain energy in each segment of the beam.
For segment AB:
The strain energy in this segment is given by the equation:
𝔼∗AB = ∫(M/M₀)²(EI)²/(2∫M₀²/EI dx)
Where M is the bending moment, M₀ is the reference bending moment, EI is the flexural rigidity, and the integral is taken over the segment AB.
Since the beam is simply supported at C, we know that the bending moment at C is zero. Therefore, the reference bending moment M₀ can be taken as zero.
The strain energy integrand for segment AB is:
𝔼∗AB = (1/(2EI)) * (M)²
For segment BC:
Similarly, the strain energy in this segment is given by:
𝔼∗BC = ∫(M/M₀)²(2EI)²/(2∫M₀²/(2EI) dx)
Where M is the bending moment, M₀ is the reference bending moment, and the integral is taken over the segment BC.
Again, since the beam is simply supported at C, we know that the bending moment at C is zero. Therefore, the reference bending moment M₀ can be taken as zero.
The strain energy integrand for segment BC is:
𝔼∗BC = (1/(2(2EI))) * (M)²
2. Using Castigliano's theorem, we can express the vertical downward deflection vA at the free end A as:
vA = ∂(𝔼∗AB)/∂P
Substituting the integrand for segment AB, we have:
vA = ∂((1/(2EI)) * (M)²)/∂P
Differentiating with respect to P gives:
vA = (1/(2EI)) * 2M * (∂M/∂P)
Since the bending moment M is related to the applied force P by M = P*x, where x is the distance from the fixed end C, we can substitute this expression into the equation:
vA = (1/(2EI)) * 2P*x * (∂(P*x)/∂P)
Simplifying further:
vA = (1/EI) * P*x²
Therefore, the symbolic expression for the vertical downward deflection vA is:
vA = (P*x²)/(EI)
3. Similarly, we can use Castigliano's theorem to obtain the expression for the rotation ϑA (slope) of the beam at A. Using the same approach as in step 2, the symbolic expression for the rotation ϑA is:
ϑA = (1/EI) * ∂(𝔼∗AB)/∂P
Substituting the integrand for segment AB, we have:
ϑA = (1/EI) * ∂((1/(2EI)) * (M)²)/∂P
Differentiating with respect to P gives:
ϑA = (1/EI) * 2M * (∂M/∂P)
Using the relationship M = P*x, we can substitute this expression into the equation:
ϑA = (1/EI) * 2P*x * (∂(P*x)/∂P)
Simplifying further:
ϑA = (2P*x²)/(EI)
Therefore, the symbolic expression for the rotation ϑA (slope) of the beam at A is:
ϑA = (2P*x²)/(EI)