8.6 Deflection of Beams: Double Integration, Macaulay & Moment-Area

Why Deflection, Not Just Stress

A beam may be perfectly safe in stress and still useless. A conveyor gantry that sags visibly under a loaded belt, a machine-tool bed that deflects enough to lose alignment, or a shaft whose deflection at the bearing exceeds the permissible slope will all fail in service while the material is nowhere near yield. Design codes therefore impose serviceability limits — often span over 325 or span over 500 — alongside strength limits.

The Elastic Curve Equation

For small deflections, the curvature of a beam relates to the bending moment through the flexure formula:

$$\frac{1}{R} = \frac{M}{EI} \quad\text{and}\quad \frac{1}{R} \approx \frac{d^2y}{dx^2}$$

Combining gives the governing differential equation:

$$\boxed{EI\frac{d^2y}{dx^2} = M(x)}$$

The full family of relations, obtained by successive differentiation, is worth memorising as a ladder:

Double Integration Method

Integrating once gives the slope and twice the deflection, each step introducing a constant:

$$EI\frac{dy}{dx} = \int M\,dx + C_1, \qquad EI\,y = \iint M\,dx\,dx + C_1x + C_2$$

The constants are found from boundary conditions:

Worked example: cantilever with end load

A cantilever of length $L$ carries a point load $W$ at the free end. Measuring $x$ from the fixed end, $M(x) = -W(L-x)$.

$$EI\frac{d^2y}{dx^2} = -W(L-x)$$ $$EI\frac{dy}{dx} = -W\left(Lx - \frac{x^2}{2}\right) + C_1$$

At $x = 0$ the slope is zero, so $C_1 = 0$. Integrating again:

$$EI\,y = -W\left(\frac{Lx^2}{2} - \frac{x^3}{6}\right) + C_2$$

At $x = 0$, $y = 0$, so $C_2 = 0$. At the free end, $x = L$:

$$y_{\max} = -\frac{WL^3}{3EI}, \qquad \theta_{\max} = -\frac{WL^2}{2EI}$$

The negative sign indicates downward deflection; magnitudes are what the exam asks for.

Macaulay's Method

When loading is discontinuous — several point loads, a partial UDL, an applied couple — writing a separate moment expression for each segment produces many constants. Macaulay's method avoids this by using bracket terms:

$$\langle x - a \rangle^{n} = \begin{cases} (x-a)^{n} & \text{if } x > a \\ 0 & \text{if } x \leq a\end{cases}$$

The rules that make it work:

1. Measure $x$ from one end only, conventionally the left.
2. Integrate the bracket as a whole: $\displaystyle\int\langle x-a\rangle^n dx = \frac{\langle x-a\rangle^{n+1}}{n+1}$. Never expand the bracket.
3. Discard any bracket whose contents are negative when substituting a value of $x$.
4. A UDL that stops partway must be extended to the end of the beam and cancelled by an equal upward UDL from that point on.
5. The constants $C_1$ and $C_2$ apply to the whole beam, which is the saving.

Moment-Area Method

Mohr's two theorems convert the problem into areas under the $M/EI$ diagram, which is fast for cantilevers and for beams with varying section.

Theorem I. The change in slope between two points equals the area of the $M/EI$ diagram between them:

$$\theta_{AB} = \int_A^B \frac{M}{EI}\,dx$$

Theorem II. The vertical intercept on the tangent at A, measured at B, equals the first moment of that area about B:

$$\delta = \int_A^B \frac{M}{EI}\,x\,dx$$

For a cantilever, where the tangent at the fixed end is horizontal, Theorem II gives the free-end deflection directly.

Check. For the cantilever with end load $W$, the $M/EI$ diagram is a triangle of base $L$ and height $WL/EI$, so its area is $WL^2/2EI$ — matching the slope found earlier. Its centroid lies at $2L/3$ from the free end, so the deflection is $\dfrac{WL^2}{2EI}\times\dfrac{2L}{3} = \dfrac{WL^3}{3EI}$. The two methods agree.

Energy Method: Castigliano's Second Theorem

The deflection at the point of application of a load $P$, in the direction of that load, is the partial derivative of strain energy with respect to that load:

$$\delta = \frac{\partial U}{\partial P}, \qquad U = \int \frac{M^2}{2EI}\,dx$$

so in practice

$$\delta = \int \frac{M}{EI}\frac{\partial M}{\partial P}\,dx$$

If no load acts where the deflection is wanted, apply a dummy load $Q$, differentiate, then set $Q = 0$. This is the unit load method, and it handles curved members and frames that defeat direct integration.

Standard Results Table

These must be recalled without derivation; they account for the majority of objective items on the topic.

Reading the table

Two patterns are worth internalising rather than memorising blindly.

Fixing the ends stiffens the beam dramatically. A simply supported beam with a central load deflects $WL^3/48EI$; fixing both ends reduces this to $WL^3/192EI$, a factor of four. Under UDL the factor is five. This is why machine frames use welded rather than pinned connections wherever possible.

Deflection depends on the fourth power of span for distributed load and the cube for point load. Doubling the span of a conveyor gantry under its own distributed weight multiplies the sag by sixteen. Span reduction is therefore far more effective than adding section depth, though depth helps strongly too since $I$ for a rectangle varies as $d^3$.

Worked Example

A simply supported steel beam spans 6 m and carries a UDL of 12 kN/m. Take $E = 200$ GPa and $I = 8\times10^{7}$ mm$^4$. Find the central deflection.

$$y = \frac{5wL^4}{384EI} = \frac{5\times12\times(6000)^4}{384\times200000\times8\times10^{7}}$$

Working in newtons and millimetres, $w = 12$ N/mm:

$$y = \frac{5\times12\times1.296\times10^{14}}{6.144\times10^{15}} = \frac{7.776\times10^{15}}{6.144\times10^{15}} = 1.27 \text{ mm}$$

The span-to-deflection ratio is $6000/1.27 \approx 4700$, comfortably within any normal serviceability limit.