Simply supported beam—second-order ODE

Here’s the first of the derivations for the center deflection of a simply supported beam with a uniform load.

Beam and moment diagram

We start with the differential relationship between the bending moment, M, and the deflection, y:

M=EId2ydx2

The second derivative of y is the curvature of the beam (for small deflections, which is one of the fundamental assumptions of beam theory), and the negative sign is there to account for the usual sign conventions for moment and displacement.

M is a parabola that passes through 0 at each end of the beam and peaks at wL2/8 at the center. Its formula is

M=wL2xw2x2

Therefore,

y=wEI(12x2L2x)

where I’ve started using primes for differentiation.

Integrating once gives

y=wEI(16x314Lx2)+C1

Symmetry tells us the slope at the center of the beam is zero, so

y(L2)=wEI(148L3116L3)+C1=0

Which means

C1=wL324EI

Plugging this result in and integrating again gives us

y=wEI(124x4112Lx3+124L3x)+C2

Because the deflection is zero at x=0,

C2=0

and the deflection at the center of the beam is

y(L2)=wL4EI(1384196+148)=5wL4384EI

which is the answer we were expecting.