Question

A simply supported, linearly elastic, homogeneous, prismatic beam of length \(L\) and flexural rigidity \(EI\) is shown in the figure.

The expression for the Influence Line Diagram (ILD) of the rotation \(\theta_B(x)\) at the support B is

• A. \(\dfrac{1}{6EI}(x^2 - L^2)\)
• B. \(\dfrac{1}{6EIL}(x^3 - L^2x)\)
• C. \(\dfrac{1}{3EI}(x^2 - Lx)\)
• D. \(\dfrac{1}{3EIL}(x^3 - L^2x)\)

Hint:

Use the Maxwell-Betti theorem: the rotation at B from a moving unit load equals the deflection curve caused by a unit moment applied at B.

Answer 1

The Correct Option is B

Step 1: Understand what an ILD for a rotation means.
The influence line \(\theta_B(x)\) gives the rotation produced at support B when a unit downward load stands at distance \(x\) from support A. As \(x\) sweeps from \(0\) to \(L\), \(\theta_B(x)\) traces out this influence line. Finding it directly (moving the load and re-solving each time) is slow, so we use a shortcut.

Step 2: Bring in the Maxwell-Betti reciprocal theorem.
The reciprocal theorem says: the rotation at B caused by a unit load at \(x\) equals the deflection at \(x\) caused by a unit moment applied at B. In symbols, \(\theta_B(x) = y_M(x)\), where \(y_M(x)\) is the deflected shape of the beam when a unit moment \(M_0=1\) is applied at end B. This turns a moving-load problem into a single, fixed-load problem, which is much easier to integrate.

Step 3: Find the bending moment when a unit moment is applied at B.
With only a moment \(M_0\) applied at B, the reactions are \(R_A = -M_0/L\) and \(R_B = M_0/L\) (found from \(\Sigma M_A=0\) and \(\Sigma F_y=0\)). The bending moment at a section a distance \(x\) from A is then a straight line:
\[ M(x) = R_A\,x = \frac{M_0}{L}x \] This is zero at A and rises linearly to \(M_0\) at B, which is exactly what happens when you apply a pure moment at one end.

Step 4: Integrate \(EIy'' = M(x)\) twice.
\[ EIy'' = \frac{M_0}{L}x \]
Integrating once: \(EIy' = \dfrac{M_0}{2L}x^2 + C_1\).
Integrating again: \(EIy = \dfrac{M_0}{6L}x^3 + C_1x + C_2\).
Apply the boundary condition \(y(0)=0\): this forces \(C_2=0\).
Apply \(y(L)=0\): \(\dfrac{M_0}{6L}L^3 + C_1L = 0\), which gives \(C_1 = -\dfrac{M_0L}{6}\).

Step 5: Write out the deflection curve and set \(M_0=1\).
\[ EIy(x) = \frac{M_0}{6L}x^3 - \frac{M_0L}{6}x = \frac{M_0}{6L}\left(x^3 - L^2x\right) \] Setting the unit moment \(M_0=1\):
\[ y(x) = \frac{1}{6EIL}\left(x^3 - L^2x\right) \] By the reciprocal theorem from Step 2, this deflection is exactly the rotation influence line we want: \(\theta_B(x) = \dfrac{1}{6EIL}(x^3-L^2x)\).

Step 6: Check the answer against the physical boundary conditions.
At \(x=0\) (load sitting right on support A) the formula gives \(\theta_B=0\): correct, because a load applied directly at a support cannot bend the span at all. At \(x=L\) (load right on support B) it also gives \(\theta_B=0\): correct, for the same reason. Option (A), \(\frac{1}{6EI}(x^2-L^2)\), gives a nonzero value at \(x=0\) (equal to \(-L^2/6EI\)), which is physically impossible, so it fails this check immediately. Options (C) and (D) use the wrong constant (they are not consistent with the two boundary conditions together), so only option (B) survives every check.

Final Answer:
\[ \boxed{\theta_B(x) = \dfrac{1}{6EIL}\left(x^3 - L^2x\right)} \]