Deflections of Determinate Structures

Why energy methods

Geometric methods (double integration, moment-area) become tedious for frames, curved members and trusses. Energy methods find displacements from the work done by forces and the strain energy stored, using simple integrals or sums. They also give the extra equations needed for statically indeterminate structures.

All results here assume linear elastic behaviour and small deformations.

Strain energy expressions

For beams and frames, bending energy dominates; axial and shear terms are usually ignored. For trusses, only axial energy exists.

Principle of virtual work

For a deformable body in equilibrium, if a system of forces in equilibrium undergoes a compatible set of small virtual displacements, the external virtual work equals the internal virtual work.

Two applications:

• Virtual displacements applied to real forces → equilibrium equations (e.g. finding reactions, plastic collapse loads).
• Virtual forces applied through real displacements → deflections (the unit load method).

Unit load (dummy load) method

To find the displacement $\Delta$ at a point in a given direction:

1. Analyse the structure for the real loads: find $M$ (or $N$ for trusses).
2. Remove the real loads and apply a unit load at the point in the direction of the required displacement: find $m$ (or $k$).
3. Equate the external work of the unit load with the internal work:

Product integrals (Vereshchagin's rule)

When one of the two diagrams is linear, $\int Mmdx=A_M\times m_c$, where $A_M$ is the area of the $M$ diagram and $m_c$ is the ordinate of the linear $m$ diagram under the centroid of the $M$ diagram. This speeds up frame problems greatly.

Temperature and lack of fit in trusses

• Member heated by $T$: $\Delta =\sum k\alpha TL$.
• Member fabricated too long by $\lambda$: $\Delta =\sum k\lambda$.
• Beam with temperature gradient (top $T_1$, bottom $T_2$, depth $d$): curvature $=\alpha (T_2-T_1)/d$, and $\Delta =\int m\frac{\alpha (T_2-T_1)}{d}dx$.

Castigliano's theorems

First theorem: $\frac{\partial U}{\partial {\Delta }_i}=P_i$ (valid for non-linear elastic too).

Second theorem (linear structures):

Note that $\partial M/\partial P_i$ is exactly the moment $m$ due to a unit load at $i$ — so Castigliano's second theorem and the unit load method give identical working. Where there is no load at the point, apply a fictitious load $Q$, differentiate, and set $Q=0$.

Theorem of least work

For an indeterminate structure with redundant $R$ (and supports that do not yield), the redundant takes the value that makes the strain energy a minimum:

With several redundants, one equation per redundant. If a support yields by $\delta$ in the direction of $R$, use $\partial U/\partial R=-\delta$ (sign as per direction).

Why deflection matters

A beam can be strong enough and still be unserviceable: excessive sag cracks plaster and partitions, causes ponding on roofs, misaligns machinery and alarms occupants. Design therefore checks deflection as well as stress. Deflection analysis is also the basis of solving statically indeterminate beams.

• Deflection $y$ — vertical displacement of a point on the neutral axis.
• Slope $\theta =dy/dx$ — rotation of the tangent to the elastic curve (radians).
• Elastic curve — the deflected shape of the neutral axis.

Differential equation of the elastic curve

From the bending equation, curvature $=1/R=M/EI$. For small slopes $1/R\approx d^{2}y/d{x}^2$. Taking sagging moment positive and deflection $y$ positive upward:

Double integration method

Write $M(x)$ for the whole span, integrate twice, apply boundary conditions. Straightforward when one expression for $M$ covers the beam (UDL over full span, end loads).

Cantilever with end load $W$ (fixed at $x=L$, free at $x=0$): $M=-Wx$.

$EI{y}^{\prime }=-\frac{W{x}^2}{2}+C_1$; at $x=L$, $y^{\prime }=0$ → $C_1=\frac{W{L}^2}{2}$

$EIy=-\frac{W{x}^3}{6}+\frac{W{L}^{2}x}{2}+C_2$; at $x=L$, $y=0$ → $C_2=-\frac{W{L}^3}{3}$

At the free end ($x=0$): $y=-\frac{W{L}^3}{3EI}$ and $\theta =\frac{W{L}^2}{2EI}$.

Macaulay's method

When loads are discontinuous (point loads part-way along, partial UDLs), Macaulay writes one moment equation using brackets $[x-a]$ that are taken as zero when negative. Rules:

1. Measure $x$ from one end, taking a section in the last segment so all loads appear.
2. Write each term as $[x-a{]}^n$ and integrate the bracket as a whole: $\int [x-a]dx=[x-a{]}^2/2$.
3. A UDL that stops before the end must be extended to the end and cancelled by an equal upward load.
4. A couple $M_0$ at $x=a$ is written $M_0[x-a{]}^0$.

Moment-area method (Mohr's theorems)

Draw the $M/EI$ diagram.

Very efficient for cantilevers (tangent at the fixed end is horizontal) and symmetric simply supported beams (tangent at mid-span is horizontal).

Conjugate beam method

Load an imaginary conjugate beam (same span) with the $M/EI$ diagram as its load. Then:

• Slope at a point of the real beam = shear force at that point of the conjugate beam.
• Deflection at a point of the real beam = bending moment at that point of the conjugate beam.

Superposition

Within the elastic range, the slope or deflection due to several loads is the sum of those due to each load alone. Combined with the standard table below, this solves most exam problems in a few lines.

Standard results

Cantilever (span $L$; values at the free end)

Simply supported beam (span $L$)

Fixed and propped beams