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Consistent deformation method and its formula

Short answer

Choose redundants equal to the degree of indeterminacy, release them to a determinate primary structure, compute load and unit-redundant displacements, then impose compatibility: Δ_i0 + Σ f_ij X_j = δ_i. Solve for X; final actions are M = M_0 + Σ X_j m_j.

What is the consistent deformation method? Derive the formula.

Meaning

The consistent deformation method (also called the method of consistent displacements or the force/flexibility method) analyses a statically indeterminate structure by:

  1. choosing a suitable number of reactions or internal forces equal to the degree of indeterminacy as redundants,
  2. removing them to get a stable, statically determinate primary (released) structure,
  3. finding the displacement of the released structure at each redundant due to the loads, and
  4. writing that the redundant forces must restore the displacements to the value that actually exists in the real structure (zero at a rigid support, equal to the settlement if the support settles). These are the compatibility, or consistent-deformation, conditions.

Derivation (one redundant)

Let a beam be indeterminate to the first degree, with the support reaction RR at B as redundant.

  1. Released structure under the loads. Remove the support at B. The loads cause a deflection ΔB0\Delta_{B0} at B (call it Δ10\Delta_{10}, positive in the direction of RR).
  2. Released structure under a unit load at B in the direction of RR. The deflection at B is the flexibility coefficient f11f_{11} (deflection per unit force). If the force is RR, the deflection is f11Rf_{11}R (linear elasticity, superposition).
  3. Superpose. The actual deflection at B is
ΔB=Δ10+f11R\Delta_B=\Delta_{10}+f_{11}R
  1. Compatibility. In the actual structure the support prevents deflection (or allows a known settlement δB\delta_B):
Δ10+f11R=δB(δB=0 for an unyielding support)\Delta_{10}+f_{11}R=\delta_B\quad(\delta_B=0\text{ for an unyielding support}) R=δB−Δ10f11R=\frac{\delta_B-\Delta_{10}}{f_{11}}

where f11=∫m12EIdxf_{11}=\int\dfrac{m_1^2}{EI}dx and Δ10=∫m1M0EIdx\Delta_{10}=\int\dfrac{m_1M_0}{EI}dx, with m1m_1 the moment due to unit load at B, and M0M_0 the moment due to the loads in the released structure.

Several redundants

For nn redundants X1,…,XnX_1,\ldots,X_n:

Δi0+∑j=1nfijXj=δi(i=1,…,n)⟹[F]{X}={δ}−{Δ0}\Delta_{i0}+\sum_{j=1}^{n}f_{ij}X_j=\delta_i\quad(i=1,\ldots,n)\qquad\Longrightarrow\qquad [F]\{X\}=\{\delta\}-\{\Delta_0\}

fij=fjif_{ij}=f_{ji} (Maxwell). After finding XX, the final moment is M=M0+∑XjmjM=M_0+\sum X_jm_j, and the remaining reactions follow from equilibrium.

Example

Propped cantilever of span LL with UDL ww, redundant RBR_B at the roller: Δ10=−wL48EI\Delta_{10}=-\dfrac{wL^4}{8EI}, f11=L33EIf_{11}=\dfrac{L^3}{3EI}, so RB=3wL8R_B=\dfrac{3wL}{8}.

✓ Answer checked by Claude Opus 5.5 (high effort)

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