Deriving the simply supported beam deflection formula

The starting point

The beam has span L, uniform load w and bending stiffness E I. Taking moments about a support, the bending moment at a distance x is M(x) = w L x / 2 - w x^2 / 2. With the deflection y measured in the direction of the load, the bending equation is E I y'' = -M(x). Integrating twice gives the deflection curve, and the two supports fix the two constants of integration.

The four steps

1. Integrate -(w L x / 2 - w x^2 / 2) / (E I) with respect to x. The dialog returns the slope, -L w x^2 / (4 E I) + w x^3 / (6 E I), with no constant: add C1 yourself.
2. Integrate again, giving -L w x^3 / (12 E I) + w x^4 / (24 E I) + C1 x, and add C2.
3. The supports give y(0) = 0 and y(L) = 0. Solve those two equations for C1 and C2: C1 = L^3 w / (24 E I) and C2 = 0.
4. Substitute x = L / 2 into the curve and simplify: 5 L^4 w / (384 E I).

Each result is exact: fractions stay fractions and symbols stay symbols, so there is no rounding to hide a slip. The dialog also differentiates and evaluates, so the slope at a support or the deflection for given numbers can be checked on the spot.

Numbers, and a check

The sheet then puts in w = 12 kN/m, L = 6 m, E = 200 GPa and I = 250,000,000 mm^4, and computes the deflection twice: once from the formula, once from the curve of step 3 evaluated at midspan. Results from the engine:

The ratio is 1, so the derivation and the textbook formula agree. The sheet also plots the curve along the span, with its sign flipped so the beam sags downward on the page.

What the dialog can and cannot do

The symbolic side is deliberately small. It simplifies, expands, factors, differentiates, integrates a small class of integrands, solves one equation of degree 1 or 2 or a linear system of up to four unknowns, and substitutes. It knows sin, cos, ln, exp and sqrt, has no complex numbers and treats symbols as positive real numbers. A result that does not fit its exact arithmetic is refused with a message rather than rounded.

The same four steps apply to any load whose bending moment is a polynomial in x: change M(x) and the support conditions. A cantilever fixed at x = 0, for example, has y(0) = 0 and y'(0) = 0. For anything beyond the dialog, use the numerical tools of the engine: root finding, numerical integration and linear solvers.

The sheet source

The sample sheet holds the steps as typeset formulas and the check as quantity lines. Open it with the button on this page to see it calculated.

# Beam deflection derived symbolically

The midspan deflection of a simply supported beam, worked out with the **Symbolic** dialog (Functions, then Symbolic). Each step below is an exact result from the dialog. The numbers come in at the end, so the derived formula can be checked against the engine.

## Step 1: integrate the bending equation

Take $y$ in the direction of the load, so $E I \, \dfrac{d^2 y}{dx^2} = -M(x)$ with $M(x) = \dfrac{w L x}{2} - \dfrac{w x^2}{2}$. Operation **integrate** with respect to $x$, adding a constant after each integration, gives

$$y(x) = -\frac{L w x^3}{12 E I} + \frac{w x^4}{24 E I} + C_1 x + C_2$$

## Step 2: boundary conditions

The beam sits on supports at both ends, so $y(0) = 0$ and $y(L) = 0$. Operation **solve** on those two equations for $C_1$ and $C_2$ gives

$$C_1 = \frac{L^3 w}{24 E I} \qquad C_2 = 0$$

## Step 3: midspan

Operation **substitute** $x = L/2$, then **simplify**, gives

$$y\left(\frac{L}{2}\right) = \frac{5 L^4 w}{384 E I}$$

## Numbers

w: 12 kN/m | design load
L: 6 m | span
E: 200 GPa | modulus of elasticity
I: 250000000 mm^4 | second moment of area
delta: 5 * w * L^4 / (384 * E * I) -> mm | midspan deflection from the formula

## Check against the derived deflection curve

y(x): -L * w * x^3 / (12 * E * I) + w * x^4 / (24 * E * I) + L^3 * w * x / (24 * E * I) | the curve from step 2 with its constants
y_mid: y(L / 2) -> mm | the curve at midspan
ratio: y_mid / delta | 1 when the formula and the curve agree

plot -y(x * 1 m) / (1 mm) for x in 0..L / (1 m)