End rotation of a propped cantilever with a rotational spring at the roller

A beam is fixed at end A and supported at end B by a roller that is restrained against rotation by a rotational spring of stiffness \(\beta\). The beam carries a uniformly distributed load \(w\) over its full length \(L\). This article finds the rotation at B and shows how the spring changes it.

Beam with a uniform load $w$, fixed at A, and a roller at B with a rotational spring $\beta$.

Idea of the solution

Without the spring, end B is a free roller and the beam is a propped cantilever. Its end rotation under the load is known. The spring then applies a moment \(M_B\) at B that opposes the rotation. Superposition combines the two effects:

\[ \theta_B = \theta_0 - \theta_M \]

where \(\theta_0\) is the rotation caused by the load alone and \(\theta_M\) is the rotation caused by the spring moment acting on the beam.

Step 1: rotation caused by the load

For the propped cantilever (fixed at A, roller at B, no spring) under the uniform load, with \(x\) measured from A, the bending moment is

\[ M(x) = -\frac{w}{2}x^2 + \frac{5wL}{8}x - \frac{wL^2}{8} \]

Integrating \(EI\,\dfrac{d^2y}{dx^2} = -M(x)\) twice, and using \(\theta(0)=0\) at the fixed end, gives

\[ \theta(x) = \frac{1}{EI}\left(\frac{wx^3}{6} - \frac{5wL}{16}x^2 + \frac{wL^2}{8}x\right) \]

At the roller, \(x = L\):

\[ \theta_0 = \theta(L) = \frac{1}{EI}\left(\frac{wL^3}{6} - \frac{5wL^3}{16} + \frac{wL^3}{8}\right) = \frac{wL^3}{48EI} \]

The evaluation gives \(-wL^3/(48EI)\) with the sign convention of the bending moment. The negative sign only marks the sense of rotation, so the magnitude \(wL^3/(48EI)\) is used below, with the rotation taken in the direction that the spring opposes.

Step 2: rotation caused by the spring moment

The spring moment at B is \(M_B = \beta\,\theta_B\). A moment applied at the roller end of a propped cantilever rotates that end by

\[ \theta_M = \frac{M_B L}{4EI} \]

so the spring moment alone rotates B by \(\beta\theta_B L/(4EI)\), in the opposite sense to the load.

Step 3: combine the two effects

Substituting into \(\theta_B = \theta_0 - \theta_M\):

\[ \theta_B = \frac{wL^3}{48EI} - \frac{\beta L}{4EI}\,\theta_B \]

Collecting the terms in \(\theta_B\):

\[ \theta_B\left(1 + \frac{\beta L}{4EI}\right) = \frac{wL^3}{48EI} \]

Multiplying through by \(48EI\) gives the result:

\[ \boxed{\;\theta_B = \frac{wL^3}{12\,(\beta L + 4EI)}\;} \]

The spring moment follows directly:

\[ M_B = \beta\,\theta_B = \frac{\beta\, wL^3}{12\,(\beta L + 4EI)} \]

Checking the limits

Case	Spring stiffness	End rotation \(\theta_B\)	Comment
No spring	\(\beta = 0\)	\(\dfrac{wL^3}{48EI}\)	Propped cantilever; the roller end is free to rotate
Intermediate	\(0 < \beta < \infty\)	between the two limits	The spring partly restrains the end
Rigid rotational restraint	\(\beta \to \infty\)	\(\to 0\)	The end is fully restrained against rotation

The rotation falls from the propped-cantilever value to zero as the spring becomes stiffer. This matches the physical expectation.

Worked example

Take \(w = 10\ \text{kN/m}\), \(L = 6\ \text{m}\) and \(EI = 2\times10^{4}\ \text{kN·m}^2\). Then

\[ \theta_0 = \frac{wL^3}{48EI} = \frac{10 \cdot 6^3}{48 \cdot 2\times10^{4}} = 0.00225\ \text{rad} \]

For a spring with \(\beta = 4EI/L = 1.333\times10^{4}\ \text{kN·m/rad}\):

\[ \theta_B = \frac{wL^3}{12\,(\beta L + 4EI)} = \frac{10 \cdot 6^3}{12\,(8\times10^{4} + 8\times10^{4})} = 0.001125\ \text{rad} \]

The spring moment is \(M_B = \beta\theta_B = 1.333\times10^{4} \cdot 0.001125 = 15\ \text{kN·m}\). The spring halves the end rotation and carries one third of the fixed-end moment \(wL^2/8 = 45\ \text{kN·m}\) of the unsprung beam.

Rotation at B relative to the no-spring value, and spring moment relative to $wL^2/12$, against the relative stiffness $\beta L/EI$.

Assumptions

The solution assumes linear elastic material, constant \(EI\), small rotations, and a linear spring with \(M_B = \beta\theta_B\). The load is uniform and acts over the whole span. The spring stiffness \(\beta\) is in kN·m/rad and \(EI\) in kN·m², so \(\beta L/EI\) is dimensionless.

Superposition: the rotation under the load minus the rotation caused by the spring moment.