Hi everybody! I need this quick answer, please help!
The beam AB in the figure has length 3L and uniform section stiffness EI. The beam is fixed at A(x=0) and loaded by an upward concentrated force,
P, applied at its free end B(x=3L). A concentrated moment
Q is also applied, as indicated, at distance L from the support at A.
S14Q2_3_2
Obtain a symbolic expression for the vertical deflection at the free end B, v(x=3L), in terms of the given quantities.
v(x=3L)= ?
To obtain a symbolic expression for the vertical deflection at the free end B, v(x=3L), we can use the principles of statics and mechanics of materials. Here's how you can approach it:
Step 1: Establish the equations for the bending moment along the beam.
Since the beam is subjected to a concentrated moment Q at distance L from the support, the moment equation can be represented as follows:
M(x) = Qδ(x - L)
Where M(x) is the bending moment at any point x along the beam, and δ(x - L) is the Dirac delta function which accounts for the concentrated moment.
Step 2: Determine the equation for the bending moment for the given loading condition.
The bending moment can be obtained by considering the equilibrium of forces. The beam is subjected to an upward concentrated force P at the free end B, which creates a moment:
M(x) = Px
For the section of the beam from A to L, the bending moment equation is:
M(x) = Qδ(x - L)
Step 3: Use the bending moment equation to find the equation for the deflection.
The deflection equation can be obtained by integrating the bending moment equation twice with respect to x.
For the section of the beam from A to L:
v(x) = -∫(Qδ(x - L))dx = -Q∫(δ(x - L))dx
Since δ(x - L) = 0 for x ≠ L, the integral simplifies to:
v(x) = -Q∫(δ(x - L))dx = -Q
For the section of the beam from L to B:
v(x) = ∫(Px)dx
Since P is a constant, we can pull it out of the integral:
v(x) = P∫xdx = P[(x^2)/2] + C
Where C is the constant of integration. Evaluating the integral at the limits of integration from L to 3L:
v(x=3L) = P[(3L^2)/2] + C
Step 4: Combine the deflection equations.
To find the deflection at x=3L, we need to find the value of the constant C. We can do this by applying the boundary condition that the beam is fixed at A, which implies that the deflection at A is zero, i.e. v(x=0) = 0.
Substituting this boundary condition into the deflection equation:
0 = P[(0^2)/2] + C
C = 0
Therefore, the deflection at x=3L is:
v(x=3L) = P[(3L^2)/2] + 0 = (3PL^2)/2
So, the symbolic expression for the vertical deflection at the free end B, v(x=3L), is:
v(x=3L) = (3PL^2)/2