Energy & Matrix Methods

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 matrices

Classical methods become unwieldy for large structures. Matrix methods write the same principles — equilibrium, compatibility and member force–displacement laws — in a systematic form that a computer can solve for thousands of unknowns. Every structural analysis program (STAAD, ETABS, SAP) uses the stiffness method.

Flexibility and stiffness

For a structure with coordinates (directions of forces/displacements) $1\dots n$:

• Flexibility coefficient $f_{ij}$ = displacement at coordinate $i$ due to a unit force at coordinate $j$ (all other forces zero).
• Stiffness coefficient $k_{ij}$ = force at coordinate $i$ required to produce a unit displacement at coordinate $j$ with all other displacements held at zero.

Force (flexibility) method in matrix form

Choose redundants $\{X\}$. With the released structure:

The unknowns equal the degree of static indeterminacy. Selection of redundants is not unique and affects convenience.

Stiffness (displacement) method

Unknowns are joint displacements — their number equals the degree of kinematic indeterminacy. The procedure is automatic and does not require choosing redundants, which is why it dominates software.

Element stiffness matrices (local coordinates)

Axial (truss) element, length $L$, area $A$:

Beam element (coordinates: transverse displacement and rotation at each end, $v_1,{\theta }_1,v_2,{\theta }_2$):

Reading the rotation terms: $4EI/L$ at the near end and $2EI/L$ at the far end (carry-over ½); translation gives $6EI/L^2$ moments and $12EI/L^3$ shears — the same numbers as in slope deflection.

Plane frame element — 6×6: combines the axial matrix (terms $AE/L$) and the beam matrix.

Beam element for rotations only (translations restrained, common in hand problems):

Transformation to global coordinates

For an inclined truss member at angle $\theta$ with $c=\cos \theta$, $s=\sin \theta$:

In general $[k{]}_{global}=[T{]}^T[k{]}_{local}[T]$.

Assembly, boundary conditions and solution

1. Number joints and degrees of freedom; list each element's global DOF numbers.
2. Assemble $[K]$ by adding each element's global stiffness terms into the rows and columns of its DOFs (direct stiffness method).
3. Form the load vector $\{P\}$: joint loads plus equivalent joint loads from member loads (negative of fixed-end forces).
4. Apply boundary conditions: delete rows and columns of restrained DOFs (or use a large number on the diagonal).
5. Solve $[K]\{\Delta \}=\{P\}$.
6. Member forces: $\{p\}=[k]\{\delta \}+\{p_F\}$ (fixed-end forces added back).
7. Reactions from the restrained rows.