Skip to main content

Boundary conditions of fixed-fixed beam

Short answer

For a beam fixed at both ends (x=0 and x=L), the boundary conditions are zero deflection and zero slope at each support: y(0)=0, y'(0)=0, y(L)=0, y'(L)=0. These four conditions determine constants, and end moments correspond to fixed-end moments for the loading.

Write down the boundary conditions for a single span beam fixed at both ends.

A beam fixed at both ends (A at x=0x=0 and B at x=Lx=L) cannot move or rotate at either support. For the elastic curve y(x)y(x):

EndDeflectionSlope
A (x=0x=0)y=0y=0dydx=0\dfrac{dy}{dx}=0
B (x=Lx=L)y=0y=0dydx=0\dfrac{dy}{dx}=0

So there are four boundary conditions: y(0)=0y(0)=0, y′(0)=0y'(0)=0, y(L)=0y(L)=0, y′(L)=0y'(L)=0. They are enough to evaluate the four constants of integration of EI d4ydx4=w(x)EI\,\dfrac{d^4y}{dx^4}=w(x), and they are the compatibility conditions used in the force method. The end moments MAM_A and MBM_B and the shears are unknown reactions (the support moments are non-zero).

In the stiffness method the same statement reads θA=θB=0\theta_A=\theta_B=0 and vA=vB=0v_A=v_B=0 (no end displacements and no end rotations), so the end moments are the fixed-end moments of the loading, for example ∓wL2/12\mp wL^2/12 for a UDL.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Theory of Structures II chapter 3 questions →