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Static and kinematic indeterminacy with examples

Short answer

Static indeterminacy (Ds) is the number of extra unknown forces beyond equilibrium equations; kinematic indeterminacy (Dk) is the number of independent joint displacements. Plane truss: Ds=(m+r)-2j, Dk=2j-r; rigid frame: Ds=3m+r-3j-c, Dk=3j-r. Example: square truss with both diagonals (j=4,m=6,r=3) has Ds=1, Dk=5.

Define degree of static and kinematic indeterminacies with suitable expressions and give suitable examples (pin jointed, rigid jointed and hybrid structures) to explain the concept.

Static indeterminacy (DSI) is the number of unknown forces (reactions and internal forces) in excess of the available equilibrium equations. It is the number of redundants that must be removed to make the structure statically determinate. Kinematic indeterminacy (DKI) is the number of independent joint displacements (translations and rotations) that must be known to fix the deformed shape of the structure; it is the number of unknowns in the displacement method.

Expressions

  • Plane truss: Ds=(m+r)−2jD_s = (m + r) - 2j and Dk=2j−rD_k = 2j - r
  • Plane rigid frame / beam: Ds=(3m+r)−(3j+c)D_s = (3m + r) - (3j + c)
  • Frame, axial deformation allowed: Dk=3j−r+c′D_k = 3j - r + c'
  • Frame, members inextensible: Dk=3j−r−m+c′D_k = 3j - r - m + c'

Here mm = members, jj = joints (including supports and free ends), rr = reaction components, cc = equations of condition from internal hinges (or hinged bar ends), c′c' = extra independent rotations created by internal hinges.

Example 1: Pin-jointed truss

Square panel ABCD with both diagonals, hinge at A and roller at B.

j=4, m=6, r=3j = 4,\ m = 6,\ r = 3

Ds=6+3−2(4)=1D_s = 6 + 3 - 2(4) = 1 (internal, one diagonal is redundant); Dk=2(4)−3=5D_k = 2(4) - 3 = 5.

Example 2: Rigid-jointed frame

Portal frame with both bases fixed: j=4, m=3, r=6j = 4,\ m = 3,\ r = 6.

Ds=3(3)+6−3(4)=3D_s = 3(3) + 6 - 3(4) = 3

Dk=3(4)−6=6D_k = 3(4) - 6 = 6 with axial deformation, and 6−3=36 - 3 = 3 if axial deformation is neglected (rotations of the two top joints and one side sway).

   B ________ C
   |          |
   |          |
  ///        ///
  (fixed)  (fixed)

Example 3: Hybrid (rigid joints plus pin-ended bar)

The same portal frame with a pin-ended diagonal brace BD: j=4, m=4, r=6j=4,\ m=4,\ r=6, and the brace has two hinged ends (c=2c = 2).

Ds=3(4)+6−3(4)−2=4D_s = 3(4) + 6 - 3(4) - 2 = 4

Dk=12−6=6D_k = 12 - 6 = 6 (extensible); with inextensible members Dk=6−4=2D_k = 6 - 4 = 2, because the brace stops sway and only the rotations of B and C remain.

A hinge or a pin-ended bar lowers DsD_s, but a hinge raises DkD_k. So hinges make the force method easier (fewer redundants) and the displacement method harder (more unknown displacements).

✓ Answer checked by Claude Opus 5.5 (high effort)

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