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
Show 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.
Step 1: Use the standard slope formula for a point load on a simply supported beam.
Instead of the reciprocal theorem, let's use a result that is normally kept in a deflection table: for a simply supported beam of span $L$ carrying a single point load $P$ at distance $a$ from the left support A (with $b=L-a$ from the right support B), the slope at the far support B works out, from double integration with the same two conditions $y(0)=y(L)=0$, to
\[
\theta_B = \frac{Pa(L^2-a^2)}{6LEI}
\]
This is derived once, from the same idea as before (piecewise bending moment, double integration, then match slope and deflection at the load point), and then just reused as a formula.
Step 2: Substitute the moving unit load.
Here the load is a unit value ($P=1$) and its position is exactly the variable $x$ used in the influence line, so we set $a=x$:
\[
\theta_B(x) = \frac{x(L^2-x^2)}{6LEI}
\]
Step 3: Match this to the sign convention used in the figure.
The diagram defines $\theta_B(x)$ as the angle the tangent at B makes with the undeformed axis, measured in the sense that matches a beam sagging downward under load (the same convention that makes the deflection curve $y(x)$ positive in the direction the beam actually sags). With that convention, this magnitude is written with a sign flip relative to $x(L^2-x^2)$, giving
\[
\theta_B(x) = -\frac{x(L^2-x^2)}{6EIL} = \frac{x^3-L^2x}{6EIL}
\]
which is exactly the cubic shape found here, just reached from a table formula instead of from the reciprocal theorem.
Step 4: Sanity check with a famous special case.
At midspan, $x=L/2$: $\theta_B = \dfrac{(L/2)(L^2-L^2/4)}{6LEI} = \dfrac{(L/2)(3L^2/4)}{6LEI} = \dfrac{L^2}{16EI}$. This is the classical, widely used result for the slope at the support of a simply supported beam under a central point load, $PL^2/16EI$ with $P=1$ - a strong independent check that the formula is right.
Step 5: Final expression.
Both the zero values at $x=0,L$ and the $L^2/16EI$ check at midspan confirm the same cubic influence line found by the other method.
\[ \boxed{\theta_B(x) = \frac{1}{6EIL}\left(x^3-L^2x\right)} \]
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