Beam deflection by integration

A beam can be strong enough and still be unsatisfactory. A floor that sags visibly, bounces underfoot, cracks the plaster or tile below it, or makes doors stick is a failure of serviceability, not strength. Long-span roofs can pond rainwater if they deflect too much, which adds more load and more deflection. Bridges must stay within deflection limits so that joints, bearings and drainage work. Deflection also matters for analysis: the indeterminate beams of lesson 11 are solved by using deflections as compatibility conditions. This lesson derives the differential equation of the deflected beam and solves it by direct integration.

The elastic curve

When a beam bends, its neutral axis takes up a curved shape called the elastic curve. Let \(v(x)\) be the deflection of the neutral axis at position \(x\), positive upward, and let \(\theta(x) = dv/dx\) be its slope. Lesson 5 showed that the curvature is related to the bending moment by

From geometry, the exact curvature of a curve \(v(x)\) is \(v''/(1 + v'^2)^{3/2}\). For real beams the slopes are tiny, typically less than 0.01 radians, so \(v'^2\) is negligible next to 1 and the curvature is simply \(v''\). That gives the differential equation of the elastic curve:

With the sign convention of lesson 4, a positive (sagging) moment gives positive curvature, a curve that is concave upward. Differentiating twice more, using \(dM/dx = V\) and \(dV/dx = -w\), gives two equivalent forms: \(EI\,v''' = V(x)\) and \(EI\,v'''' = -w(x)\). The second-order form is used most often in hand calculations, because the moment equation is usually easy to write.

Boundary and continuity conditions

Integrating \(EIv'' = M\) once gives the slope and twice gives the deflection, with two constants of integration. The constants come from what the supports enforce.

• At a pin or roller, the deflection is zero: \(v = 0\).
• At a fixed support, both the deflection and the slope are zero: \(v = 0\) and \(v' = 0\).
• At a free end, the deflection and slope are unknown, but the moment and shear are zero, which the moment equation already reflects.
• At a plane of symmetry, the slope is zero.

When the moment equation changes from one segment to the next, at a point load for example, each segment has its own two constants. The extra equations come from continuity: at the junction between segments, the beam cannot have a kink or a break, so the slope and deflection computed from both sides must be equal.

Cantilever with an end load

Simply supported beam with a uniform load

The fourth power of the span in that formula is the reason serviceability so often governs long beams. A beam that is adequate for strength at a 6 m span may be far too flexible at 9 m: the moment increases by \((9/6)^2 = 2.25\), but the deflection by \((9/6)^4 = 5.1\).

Beams with several segments and singularity functions

When a point load sits partway along a beam, the moment equation changes at the load, and direct integration needs two sets of constants plus continuity conditions. For several loads this becomes tedious. Singularity functions, also called Macaulay brackets, let you write a single moment equation valid along the whole beam. The bracket \(\langle x - a \rangle^n\) is defined as \((x - a)^n\) when \(x \ge a\) and zero when \(x < a\). It is integrated like an ordinary power:

Using brackets, a point load \(P\) at \(x = a\) contributes \(-P\langle x - a\rangle\) to the moment, a uniform load starting at \(a\) contributes \(-w\langle x - a\rangle^2/2\), and a clockwise couple \(M_0\) at \(a\) contributes \(M_0\langle x - a\rangle^0\). Because each term switches on only beyond its load, one equation and two constants cover the whole beam.

Deflection limits in practice

Building codes set deflection limits as fractions of the span. The International Building Code tabulates limits such as span/360 under live load for floors and span/240 for roofs supporting plaster ceilings, with looser values for other roofs, and the material standards (AISC 360, ACI 318, the NDS for wood) add their own provisions. Stricter values apply where vibration or sensitive equipment matters. These are serviceability limits, checked with service (unfactored) loads.

Two further effects make real deflections larger than the elastic formulas suggest. Concrete creeps under sustained load, so long-term deflection of a concrete beam can be two or three times its immediate deflection. Cracking reduces the effective moment of inertia of reinforced concrete well below the gross value. Timber also creeps. Design standards account for both, but the starting point is always \(EIv'' = M\).

Integrating numerically

For awkward loadings and beams whose \(EI\) varies, numerical integration is the practical approach. The program below integrates \(M/(EI)\) twice using the trapezoidal rule, then corrects for the boundary conditions by adding a linear function (the role of the two constants). It checks the off-center point load example and finds the maximum deflection.

# Deflection by numerical double integration of M/EI (kN, m)
L, a, P = 6.0, 2.0, 50.0
EI = 16000.0                                 # kN*m^2
n = 6000
dx = L / n
xs = [i * dx for i in range(n + 1)]
RA = P * (L - a) / L
M = [RA * x - (P * (x - a) if x > a else 0.0) for x in xs]

# Integrate curvature twice, starting with slope 0 and deflection 0 at x = 0
slope, defl = [0.0], [0.0]
for i in range(n):
    slope.append(slope[-1] + 0.5 * (M[i] + M[i + 1]) / EI * dx)
    defl.append(defl[-1] + 0.5 * (slope[i] + slope[i + 1]) * dx)

# Boundary conditions v(0) = v(L) = 0: subtract the straight line through the end values
corr = [defl[i] - defl[-1] * xs[i] / L for i in range(n + 1)]
i_load = round(a / dx)
vmin = min(corr)
x_at = xs[corr.index(vmin)]
print(f"Deflection under the load : {corr[i_load] * 1000:.2f} mm")
print(f"Maximum deflection        : {vmin * 1000:.2f} mm at x = {x_at:.3f} m")

# Check the uniform-load formula the same way
w = 20.0
M = [w * L * x / 2 - w * x * x / 2 for x in xs]
slope, defl = [0.0], [0.0]
for i in range(n):
    slope.append(slope[-1] + 0.5 * (M[i] + M[i + 1]) / EI * dx)
    defl.append(defl[-1] + 0.5 * (slope[i] + slope[i + 1]) * dx)
mid = defl[n // 2] - defl[-1] * 0.5
print(f"UDL midspan: numerical {mid * 1000:.3f} mm, formula {-5 * w * L**4 / (384 * EI) * 1000:.3f} mm")

Deflection under the load : -11.11 mm
Maximum deflection        : -12.10 mm at x = 2.734 m
UDL midspan: numerical -21.094 mm, formula -21.094 mm

The numerical results match the hand calculation and the standard formula to the precision shown. The same approach handles tapered girders, beams with cover plates, and cracked concrete members where \(EI\) changes along the span: simply make \(EI\) a list instead of a constant.