Introduction

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: $D_s = (m + r) - 2j$ and $D_k = 2j - r$
• Plane rigid frame / beam: $D_s = (3m + r) - (3j + c)$
• Frame, axial deformation allowed: $D_k = 3j - r + c'$
• Frame, members inextensible: $D_k = 3j - r - m + c'$

Here $m$ = members, $j$ = joints (including supports and free ends), $r$ = reaction components, $c$ = equations of condition from internal hinges (or hinged bar ends), $c'$ = 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.

$j = 4,\ m = 6,\ r = 3$

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

Example 2: Rigid-jointed frame

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

$D_s = 3(3) + 6 - 3(4) = 3$

$D_k = 3(4) - 6 = 6$ with axial deformation, and $6 - 3 = 3$ 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: $j=4,\ m=4,\ r=6$, and the brace has two hinged ends ($c = 2$).

$D_s = 3(4) + 6 - 3(4) - 2 = 4$

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

A hinge or a pin-ended bar lowers $D_s$ but a hinge raises $D_k$. This is why hinges favour the force method less and the displacement method more.

The figure is read as a single-storey frame on four columns (two hinged bases, two fixed bases) joined by a continuous girder having two internal hinges: $j = 8,\ m = 7,\ r = 2(2) + 2(3) = 10,\ c = 2$.

Static indeterminacy

• External: $D_{se} = r - 3 - c = 10 - 3 - 2 = 5$ (the two hinges give two extra equations of condition for the reactions).
• Internal: the frame has no closed loop, so $D_{si} = 0$.
• Total: $D_s = 3m + r - 3j - c = 21 + 10 - 24 - 2 = 5$.

Kinematic indeterminacy

Each internal hinge adds one independent rotation ($c' = 2$).

• With axial deformation: $D_k = 3j - r + c' = 24 - 10 + 2 = 16$
• Neglecting axial deformation: $D_k = 16 - m = 16 - 7 = 9$

Answer: external $D_{se} = 5$, internal $D_{si} = 0$ (total 5); kinematic $D_k = 16$ (9 if members are inextensible).

Flexibility coefficient

The flexibility coefficient $f_{ij}$ is the displacement at coordinate $i$ due to a unit force applied at coordinate $j$ (all other coordinates unloaded). Its unit is m/kN (or rad/kNm). The flexibility matrix $[F]$ relates displacements and forces: $\{\Delta\} = [F]\{P\}$.

Stiffness coefficient

The stiffness coefficient $k_{ij}$ is the force required at coordinate $i$ to produce a unit displacement at coordinate $j$ while all other coordinates are held fixed. The stiffness matrix gives $\{P\} = [K]\{\Delta\}$, and $[K] = [F]^{-1}$.

Suitable method

Read the figure as a single-storey, two-bay frame with fixed outer bases and a hinged base under the middle column: $j = 6,\ m = 5,\ r = 3 + 2 + 3 = 8$.

$D_s = 3(5) + 8 - 3(6) = 5$

$D_k$ (inextensible) $= 3(6) - 8 - 5 = 5$ (three top-joint rotations, one base-hinge rotation, one sway).

Because the base of the middle column is hinged, its far-end rotation can be removed with the modified stiffness $3EI/L$. The effective unknowns of the displacement method fall to 4, compared with 5 redundants in the force method.

Displacement method is more suitable because

1. it has fewer unknowns (4 against 5);
2. the frame is a regular framed system with several rotating joints and sway, which suits slope-deflection or moment distribution well;
3. the force method needs a stable released structure and many moment diagrams for 5 redundants, which is long hand work.

The frame is a single-storey frame with three columns: left hinged, middle fixed, right on a roller. A continuous girder joins the column tops.

$j = 6,\ m = 5\ (3\ \text{columns} + 2\ \text{beams}),\ r = 2 + 3 + 1 = 6,\ c = 0$

Static indeterminacy

$D_s = 3m + r - 3j - c = 15 + 6 - 18 = 3$

External $= r - 3 = 3$; internal $= 0$ (no closed loop).

Kinematic indeterminacy

• Axial deformation considered: $D_k = 3j - r = 18 - 6 = 12$
• Axial deformation neglected: $D_k = 3j - r - m = 18 - 6 - 5 = 7$

The 7 are: rotations of the three top joints (3), rotation of the hinged base (1), rotation of the roller base (1), horizontal movement of the roller base (1) and sway of the girder (1).

Answer: $D_s = 3$; $D_k = 12$ (7 for inextensible members).

Structural idealization

Structural idealization is the process of replacing a real structure with a simplified analytical model that keeps its essential behaviour but can be analysed by the theory of structures. Members are reduced to lines along their centroidal axes, supports to idealized types (fixed, hinged, roller), joints to rigid or pinned connections, loads to point, uniformly distributed or equivalent loads, and the material to a linear elastic one. For example, a RCC building frame is analysed as a plane frame of centre lines with fixed bases, and a roof truss as a pin-jointed truss with loads applied at the joints.

Steps in identification and formulation of the problem

1. Define the function and requirements. State what the structure must do (carry floor or traffic loads, span a gap) and the conditions it must satisfy: strength, stiffness and stability.
2. Idealize the structure. Draw the line diagram with member properties ($E$, $A$, $I$), joint types and support types.
3. Idealize the loads. Collect dead, live, wind, seismic and temperature or settlement effects and arrange them into load cases.
4. Check stability and determinacy. Check that the supports and members make a geometrically stable system, then find $D_s$ and $D_k$.
5. Select the method of analysis. Use equilibrium alone if determinate; otherwise choose the force method (unknowns are redundants, $D_s$) or the displacement method (unknowns are joint displacements, $D_k$), whichever has fewer unknowns.
6. Write the governing equations. Equilibrium, compatibility and force-displacement (constitutive) relations. Solve for the unknown forces or displacements.
7. Find the final results and check. Draw the axial force, shear force and bending moment diagrams, find deflections, and verify equilibrium and compatibility. Compare the results with the strength and stiffness limits.

Kinematic indeterminacy (degree of freedom) is the number of independent joint displacements, translations and rotations, that are unknown and must be determined to describe completely the deformed shape of a structure under load.

It is the number of unknowns in the displacement (stiffness) method. For a plane frame, $D_k = 3j - r + c'$ when axial deformation is allowed, and $D_k = 3j - r - m + c'$ when members are axially rigid. For a plane truss, $D_k = 2j - r$.

Example: A fixed-based portal frame ($j = 4,\ r = 6,\ m = 3$) has $D_k = 12 - 6 - 3 = 3$ for inextensible members: the rotations of the two top joints and one sway. A propped cantilever has $D_k = 1$, the rotation at the roller end.

(i) Two-bay, two-storey frame with an internal hinge

Reading of the figure: three column lines over two storeys with floor beams at two levels. Supports: one fixed, one hinged, one roller; one internal hinge.

$j = 9,\ m = 10\ (6\ \text{columns} + 4\ \text{beams}),\ r = 3 + 2 + 1 = 6,\ c = 1$

• Static: $D_s = 3(10) + 6 - 3(9) - 1 = 8$ (external $6-3 = 3$; internal $3 \times 2\ \text{closed loops} - 1 = 5$).
• Kinematic, axial deformation allowed: $D_k = 3(9) - 6 + 1 = 22$.
• Kinematic, inextensible members: $D_k = 22 - 10 = 12$.

(ii) Truss with double diagonals

Reading of the figure: four panels with top and bottom chords, five verticals and two diagonals in each panel (X-bracing). Hinge at one end, roller at the other.

$j = 10,\ m = 4 + 4 + 5 + 8 = 21,\ r = 3$

• Static: $D_s = m + r - 2j = 21 + 3 - 20 = 4$ (external 0, internal 4, one for each X-panel).
• Kinematic: $D_k = 2j - r = 20 - 3 = 17$.

Answer: (i) $D_s = 8$, $D_k = 22$ (12 if inextensible); (ii) $D_s = 4$, $D_k = 17$.

Reading of the figure: two-storey, two-bay frame with three column lines, a hinge at the top-left joint, and supports fixed (left), hinged (middle) and roller (right).

$j = 9,\ m = 10\ (6\ \text{columns} + 4\ \text{beams}),\ r = 3 + 2 + 1 = 6,\ c = 1$

Static indeterminacy

$D_s = 3m + r - 3j - c = 30 + 6 - 27 - 1 = 8$

Kinematic indeterminacy

• Axial deformation considered: $D_k = 3j - r + c' = 27 - 6 + 1 = 22$
• Axial deformation neglected: $D_k = 22 - m = 22 - 10 = 12$

Answer: $D_s = 8$; $D_k = 22$ (12 for inextensible members).

Reading of the figure: two-storey, two-bay frame with a hinge at the top-left joint, fixed left base, hinged middle base, and a roller at the right-hand support.

$j = 9,\ m = 10,\ r = 3 + 2 + 1 = 6,\ c = 1$

Static indeterminacy

• External: $r - 3 = 6 - 3 = 3$
• Internal: $3 \times (\text{closed loops}) - c = 3 \times 2 - 1 = 5$ (the two upper panels are closed loops; the lower panels are open at the base)
• Total: $D_s = 3 + 5 = 8$. Check: $3m + r - 3j - c = 30 + 6 - 27 - 1 = 8$.

Kinematic indeterminacy

• With axial deformation: $D_k = 3j - r + 1 = 27 - 6 + 1 = 22$
• Without axial deformation: $D_k = 22 - 10 = 12$

Answer: external 3, internal 5 (total 8); kinematic 22 (12 if inextensible).

Static indeterminacy

A structure is statically indeterminate when the equilibrium equations ($\Sigma F_x = 0,\ \Sigma F_y = 0,\ \Sigma M = 0$) are not enough to find all reactions and member forces. The excess number of unknowns is the degree of static indeterminacy $D_s$, the number of redundants. For plane frames $D_s = (3m + r) - (3j + c)$. It has an external part ($r - 3 - c$) and an internal part (from closed loops).

Kinematic indeterminacy

It is the number of independent joint displacements (rotations and translations) that are unknown. $D_k = 3j - r + c'$, and $D_k = 3j - r - m + c'$ if axial deformation is neglected. It decides the work of the displacement method.

Frame in the figure

Reading: two-storey, two-bay frame, bases fixed, hinged and roller, no internal hinge.

$j = 9,\ m = 10,\ r = 3 + 2 + 1 = 6$

• $D_s = 3(10) + 6 - 3(9) = 9$ (external 3, internal $3 \times 2 = 6$)
• Members are inextensible: $D_k = 3(9) - 6 - 10 = 11$

Answer: $D_s = 9$, $D_k = 11$.

Structural idealization

It is the replacement of an actual structure by a simple analytical model (line members, ideal supports, pinned or rigid joints, idealized loads and linear elastic material) that can be analysed and still gives the real behaviour with acceptable accuracy. A roof truss is idealized as pin-jointed bars with loads at the joints.

Stability of a plane truss

A truss is stable when it keeps its shape and position under any load. Two conditions are checked.

1. Necessary condition (count)

• Internally, the number of members must be at least $m \ge 2j - 3$ (for a stable, internally rigid truss).
• Including supports, $m + r \ge 2j$.

If $m + r < 2j$ the truss is unstable (a mechanism). If $m + r = 2j$ it is determinate if stable, and if $m + r > 2j$ it is indeterminate.

2. Sufficient condition (arrangement) The count is not enough, so the arrangement must also be right:

• Each part must be a rigid unit, normally a chain of triangles. A panel without a diagonal is a mechanism.
• Reactions must be at least 3 and must not be all parallel or all concurrent at a point, otherwise the truss can move as a rigid body.
• No part may be a collapsible link chain, and a pair of support bars must not meet in one line with the members they support.

A truss is stable when $m + r \ge 2j$ and it is built from properly arranged triangles with supports preventing rigid-body motion. Example: a triangle with hinge and roller has $j=3,\ m=3,\ r=3$, and $m + r = 6 = 2j$, so it is stable and determinate.

Reading of the figure: trapezoidal two-storey frame with two inclined columns on each side, a floor beam and a top beam forming one closed loop, one internal hinge at each of two levels, left base hinged, right base fixed.

$j = 6,\ m = 6,\ r = 2 + 3 = 5,\ c = 2$

Static indeterminacy

• External: $r - 3 = 2$
• Internal: $3 \times 1\ \text{loop} - 2 = 1$
• Total: $D_s = 3m + r - 3j - c = 18 + 5 - 18 - 2 = 3$

Kinematic indeterminacy

• Axial deformation considered: $D_k = 3j - r + c' = 18 - 5 + 2 = 15$
• Axial deformation neglected: $D_k = 15 - 6 = 9$

Answer: external 2, internal 1 (total 3); $D_k = 15$ (9 if inextensible).

(i) Braced two-bay frame

Reading: one storey, three columns (two fixed, one pinned), two beams, one pin-ended diagonal in each bay, and two internal hinges.

$j = 6,\ m = 5 + 2 = 7,\ r = 3 + 3 + 2 = 8,\ c = 2\ \text{hinges} + 4\ \text{bar ends} = 6$

• $D_s = 3(7) + 8 - 3(6) - 6 = 5$. External: $r - 3 = 5$ (reduced by hinges in the open frame, the two braces then add internal indeterminacy). Count check: frame alone $15 + 8 - 18 - 2 = 3$, each brace adds 1, giving 5.
• Kinematic: $D_k = 3j - r + c' - m = 18 - 8 + 2 - 7 = 5$ (inextensible members, the two braces prevent sway).
• Stability: reactions are non-concurrent and non-parallel, and the bays are closed by braces. The frame is geometrically stable.

(ii) Double-diagonal truss

Reading: four panels, chords, five verticals, both diagonals in each panel, hinge and roller.

$j = 10,\ m = 4 + 4 + 5 + 8 = 21,\ r = 3$

• $D_s = m + r - 2j = 21 + 3 - 20 = 4$ (internal)
• $D_k = 2j - r = 17$
• Stability: $m + r = 24 > 2j = 20$, all panels are braced triangles, and the three reactions are neither parallel nor concurrent. The truss is stable.

Answer: (i) $D_s = 5$, $D_k = 5$, stable; (ii) $D_s = 4$, $D_k = 17$, stable.

(i) Truss with X-bracing in two panels and a cantilever arm

Reading: top and bottom chords with three verticals and two X-braced panels, hinge at one end and roller at the other, plus a cantilever arm of two bars carrying a joint.

$j = 6 + 1 = 7,\ m = 4 + 3 + 4 + 2 = 13,\ r = 3$

• $D_s = m + r - 2j = 13 + 3 - 14 = 2$ (internal 2, one per X-panel; external 0)
• $D_k = 2j - r = 14 - 3 = 11$
• Stable: triangulated panels, the cantilever arm is attached to the rigid truss by two bars, and the reactions are non-concurrent.

(ii) Braced single-storey frame

Reading: three columns (fixed, hinged, roller), two beams and two pin-ended diagonals.

$j = 6,\ m = 7,\ r = 3 + 2 + 1 = 6,\ c = 4\ \text{(bar ends)}$

• $D_s = 3(7) + 6 - 18 - 4 = 5$ (external 3, internal 2 from the two closed braced bays)
• $D_k = 18 - 6 - 7 = 5$ (inextensible)
• Stable: braced panels are rigid and reactions prevent all rigid-body motion.

Answer: (i) $D_s = 2,\ D_k = 11$; (ii) $D_s = 5,\ D_k = 5$. Both stable.

Axial deformation is neglected, so every member is axially rigid.

(i) Double-diagonal truss with hinged supports

Reading: four panels, chords, five verticals, X-diagonals, both supports hinged.

$j = 10,\ m = 21,\ r = 4$

• Static: $D_s = m + r - 2j = 21 + 4 - 20 = 5$ (external 1, internal 4)
• Kinematic: with rigid bars and a stable arrangement no joint can move, so $D_k = 0$.
• Stability: every panel is triangulated and four reactions (non-concurrent) prevent rigid-body motion. The truss is stable.

(ii) Two-bay braced frame with an internal hinge

Reading: two storeys, three column lines, X-bracing in all four panels, one internal hinge, three base supports (fixed, hinged, roller).

$j = 9,\ m_{frame} = 10,\ \text{braces} = 8,\ r = 6,\ c = 1\ \text{(hinge)} + 16\ \text{(bar ends)}$

• Static: $D_s = 3(18) + 6 - 3(9) - 17 = 16$ (frame alone 8, plus 8 braces)
• Kinematic: rigid X-braces stop all joint translations, so only rotations remain: six upper joints (6), hinged base (1), roller base (1), and the extra rotation at the hinge (1). $D_k = 9$.
• Stability: braced panels and well-placed reactions give a stable frame.

Answer: (i) $D_s = 5$, $D_k = 0$, stable; (ii) $D_s = 16$, $D_k = 9$, stable.

All members are extensible, so $D_k$ counts all joint displacements.

(i) Truss with hinge and roller supports and internal hinges

Reading: a three-panel truss with top and bottom chords, four verticals and three diagonals (no redundant member). The internal hinges are pin joints of the truss and do not change the count.

$j = 8,\ m = 3 + 3 + 4 + 3 = 13,\ r = 3$

• $D_s = m + r - 2j = 13 + 3 - 16 = 0$ (determinate)
• $D_k = 2j - r = 16 - 3 = 13$

(ii) Beam supported by cables

Reading: a horizontal beam fixed at one end and held by two cables (tension-only bars) anchored above, attached at two points along the beam.

Unknowns: 3 fixed-end reactions + 2 cable tensions = 5. Equations = 3.

$D_s = 5 - 3 = 2$

Free beam joints: two cable-attachment points and the free end = 3 joints, each with 2 translations and 1 rotation, with the cable anchors fixed:

$D_k = 3 \times 3 = 9$

Answer: (i) $D_s = 0$, $D_k = 13$; (ii) $D_s = 2$, $D_k = 9$.