Degree of Static Indeterminacy
DSI Formula for Beams, Frames & Trusses — Internal Hinges, Releases & 10+ Solved GATE Examples
Last Updated: March 2026
📌 Key Takeaways
- The degree of static indeterminacy (DSI) is the number of unknown forces and moments that cannot be found using the equations of static equilibrium alone. It equals the number of additional (compatibility) equations needed to solve the structure.
- DSI = 0 → Determinate (solvable by equilibrium alone). DSI > 0 → Indeterminate (needs compatibility equations). DSI < 0 → Mechanism/Unstable (cannot carry load).
- For beams and frames: DSI = (3m + r) − 3j − c, where m = members, r = reactions, j = joints, c = condition equations (internal hinges).
- For pin-jointed trusses: DSI = m + r − 2j.
- Each internal hinge in a frame introduces one condition equation (M = 0 at that hinge), reducing DSI by 1.
- DSI has two components — external indeterminacy (extra support reactions beyond equilibrium) and internal indeterminacy (extra internal members beyond what’s needed for a simple frame).
- This is the most frequently tested single concept in GATE CE Structural Analysis — a DSI question appears in almost every GATE CE paper.
1. What is Static Indeterminacy?
When a structure is loaded, it develops internal forces (bending moments, shear forces, axial forces) and support reactions. To find these unknowns, we use the equations of static equilibrium. If the number of unknowns exactly equals the number of independent equilibrium equations, the structure is statically determinate — every unknown can be found directly from equilibrium, without any knowledge of the material’s stiffness or the cross-sectional dimensions.
However, many real structures have more unknowns than equilibrium equations. A continuous beam over three supports, a fixed-ended beam, a multi-storey building frame — all of these have more unknown reactions and internal forces than can be solved from equilibrium alone. These are statically indeterminate structures. The difference between the number of unknowns and the number of equilibrium equations is the degree of static indeterminacy (DSI), also called the degree of redundancy.
To solve an indeterminate structure, the DSI number of additional equations must come from compatibility conditions — geometric constraints on how the structure deforms. These conditions (for example: the deflection at a propped support is zero; the slope at a fixed end is zero) involve the material’s stiffness and the geometry of the cross-section. This is why solving indeterminate structures requires knowledge of EI — unlike determinate structures, where reactions and internal forces are independent of the material.
Why does indeterminacy matter in practice? Indeterminate structures are generally more efficient — they distribute loads through multiple paths, reducing peak stresses and deflections. However, they are sensitive to support settlement (which does not affect determinate structures at all) and temperature changes, both of which induce additional stresses. Understanding the degree of indeterminacy is therefore not just an academic exercise — it directly informs how a structure will behave and what must be checked in design.
2. Equations of Static Equilibrium
The number of available equilibrium equations determines how many unknowns can be found. In two-dimensional (planar) structural analysis — which covers all standard beam, frame, and truss problems — there are exactly three independent equilibrium equations per free body:
Three Equilibrium Equations — 2D Structures
ΣFx = 0 — Sum of all horizontal forces = 0
ΣFy = 0 — Sum of all vertical forces = 0
ΣM = 0 — Sum of all moments about any point = 0
These three equations are the maximum available for any single 2D free body. Any structure with more than 3 unknown reactions (or more than 3 total unknowns per free body) is indeterminate.
For a 3D structure: 6 equilibrium equations per free body (ΣFx, ΣFy, ΣFz, ΣMx, ΣMy, ΣMz). DSI = total unknowns − 6 × (number of free bodies). 3D analysis is beyond standard GATE CE scope.
Condition equations: An internal hinge within a structure provides an additional condition — the bending moment at an internal hinge is always zero (because a hinge cannot transmit moment). This extra equation reduces the effective number of unknowns that need compatibility to solve. Each internal hinge contributes one condition equation, and therefore reduces the DSI by 1.
3. External vs Internal Indeterminacy
The total DSI of a structure has two distinct components, each with a different physical meaning:
External Indeterminacy
DSIexternal = r − 3 (for a 2D structure with no condition equations)
Where r = total number of external support reactions.
External indeterminacy means there are more support reactions than the three equilibrium equations can solve. The extra reactions are called redundant reactions. Example: a fixed beam has 6 reactions (3 at each fixed end) — external DSI = 6 − 3 = 3.
If the structure has c condition equations (internal hinges): DSIexternal = r − 3 − c
Internal Indeterminacy
DSIinternal = 3 × (number of closed loops in the frame)
Internal indeterminacy arises when a frame has closed rings or loops — making it impossible to cut any member and release the structure into a simple tree without cutting a load-carrying member. Each closed loop adds 3 to the DSI because closing a loop introduces 3 additional internal unknowns (moment, shear, and axial force at the cut).
For a portal frame (single closed loop): DSIinternal = 3.
For a two-bay portal frame (two closed loops): DSIinternal = 6.
Total DSI = DSIexternal + DSIinternal
For most GATE CE problems, the combined formula (below) is used directly rather than splitting into external and internal components. However, understanding the physical meaning of each component helps in avoiding errors for unusual structures.
4. DSI Formula — Beams and Frames
For rigid-jointed frames and beams — where joints can transmit moment, shear, and axial force — the DSI formula is:
DSI Formula — Rigid Frames and Beams
DSI = (3m + r) − (3j + c)
Where:
m = number of members (count every beam or column segment between joints)
r = number of external reactions (pin = 2, roller = 1, fixed = 3)
j = number of joints (every point where members meet, including support points and free ends)
c = number of condition equations = number of internal hinges (each internal hinge releases moment transmission, giving one condition equation M = 0)
Alternative form: DSI = (3m + r) − 3j − c
Physical meaning of each term:
3m = total unknown internal forces at the ends of all members (3 per member end, but each end shared between members, so effectively 3 per member in the combined count)
r = unknown external reactions
3j = equilibrium equations available (3 per joint)
c = additional condition equations from internal hinges
Simplified Alternative Formula
Some textbooks use an equivalent form that is often faster for hand calculation:
DSI = r − (3 + c) for structures where internal indeterminacy = 0 (beams, simple frames)
This works when the structure has no closed loops (no internal indeterminacy) — only beams and open frames. For closed frames (portal frames, multi-bay frames), always use the full formula.
Even simpler for beams only:
DSI = (total reactions) − 3 − (number of internal hinges)
Example — Propped cantilever: r = 4 (fixed end = 3, roller = 1), no internal hinges → DSI = 4 − 3 − 0 = 1 ✓
5. DSI Formula — Pin-Jointed Trusses
For pin-jointed trusses — where all joints are pins and all members carry only axial force — the formula is different because each joint provides only 2 equilibrium equations (ΣFx = 0 and ΣFy = 0 — moment equilibrium is automatically satisfied at a pin joint where all forces are concurrent).
DSI Formula — Pin-Jointed Trusses
DSI = m + r − 2j
Where:
m = number of truss members
r = number of external reactions (pin support = 2, roller = 1)
j = number of joints (including support joints)
DSI = 0 → Determinate (just-rigid) truss
DSI > 0 → Indeterminate truss (redundant members)
DSI < 0 → Mechanism (insufficient members)
For a simple (just-rigid) truss starting from one triangle: m = 2j − 3 (adding 2 members per new joint). With a pin + roller support (r = 3): m + r = 2j − 3 + 3 = 2j → DSI = 0 ✓
6. Effect of Internal Hinges and Releases
An internal hinge at a point within a structure releases the transmission of bending moment at that point. This means the bending moment is known to be zero at an internal hinge — providing one additional equation (a condition equation). Each condition equation reduces the DSI by 1.
| Type of Release | What it Releases | Condition Equations Added (c) | Effect on DSI |
|---|---|---|---|
| Internal hinge | Bending moment (M = 0 at hinge) | 1 per hinge | DSI reduced by 1 per hinge |
| Internal roller | Bending moment + one transverse force | 2 per internal roller | DSI reduced by 2 |
| Internal link (shear release) | Shear force (V = 0 at release) | 1 per shear release | DSI reduced by 1 |
Important rule — maximum useful hinges: If a structure has too many internal hinges, it becomes a mechanism (DSI goes negative). The maximum number of internal hinges that can be introduced without making the structure unstable is equal to the DSI of the structure without any hinges. Adding more hinges than this makes the structure a mechanism.
Internal hinge in a multi-member joint: When more than two members meet at a joint, and an internal hinge is placed there, the number of condition equations depends on how many members are hinged. If all members meeting at a joint are hinged together (a common hinge), the number of condition equations = (number of members meeting at the hinge − 1). For a joint where 3 members meet with a common hinge: c = 3 − 1 = 2.
Multi-Member Hinge — Condition Equations
At a joint where n members meet with a common internal hinge:
c = n − 1 condition equations
Reason: There are n members, and the hinge releases moment between each adjacent pair — but the conditions are not all independent. The effective number of independent conditions is n − 1.
Example: 3 members meeting at a hinge → c = 2
Example: 4 members meeting at a hinge → c = 3
7. Reactions at Different Support Types
Correctly identifying the number of reactions at each support is essential for the DSI calculation. Here is a complete reference:
| Support Type | Reactions Provided | r (count) | Movements Prevented |
|---|---|---|---|
| Roller support | 1 reaction — perpendicular to rolling surface (vertical for horizontal beam) | 1 | Vertical translation only |
| Pin (hinge) support | 2 reactions — vertical + horizontal | 2 | Both vertical and horizontal translation |
| Fixed support | 3 reactions — vertical + horizontal + moment | 3 | Vertical translation + horizontal translation + rotation |
| Fixed support on a roller (vertical roller — prevents rotation and horizontal movement) | 2 reactions — horizontal + moment | 2 | Horizontal translation + rotation (free to move vertically) |
| Guided (smooth wall) support | 1 reaction + 1 moment — perpendicular force + moment (no friction force) | 2 | Perpendicular translation + rotation (free to slide along wall) |
| Link (rod) support | 1 reaction along the rod’s axis only | 1 | Translation along the rod direction only |
8. Ten Solved Examples — Beams, Frames & Trusses
Example 1 — Simply Supported Beam
Pin at A, roller at B. Single span. No internal hinges.
m = 1, r = 2 + 1 = 3, j = 2, c = 0
DSI = (3×1 + 3) − (3×2 + 0) = 6 − 6 = 0 (Determinate) ✓
Example 2 — Propped Cantilever
Fixed at A, roller at B. Single span. No internal hinges.
m = 1, r = 3 + 1 = 4, j = 2, c = 0
DSI = (3×1 + 4) − (3×2 + 0) = 7 − 6 = 1 (Indeterminate to 1st degree) ✓
Example 3 — Fixed Beam
Fixed at both ends A and B. Single span. No internal hinges.
m = 1, r = 3 + 3 = 6, j = 2, c = 0
DSI = (3×1 + 6) − (3×2 + 0) = 9 − 6 = 3 (Indeterminate to 3rd degree) ✓
Example 4 — Continuous Beam over 3 Supports
Pin at A, roller at B (intermediate), roller at C. Two spans. No internal hinges.
m = 2, r = 2 + 1 + 1 = 4, j = 3, c = 0
DSI = (3×2 + 4) − (3×3 + 0) = 10 − 9 = 1 (Indeterminate to 1st degree) ✓
Each additional intermediate roller on a continuous beam adds 1 to the DSI.
Example 5 — Continuous Beam with Internal Hinge
Fixed at A, roller at B (intermediate), roller at C. Two spans. One internal hinge between A and B.
m = 3 (hinge divides AB into two members), r = 3 + 1 + 1 = 5, j = 4, c = 1
DSI = (3×3 + 5) − (3×4 + 1) = 14 − 13 = 1 (Indeterminate to 1st degree)
Without the hinge: fixed + 2 rollers → DSI = 2. The internal hinge reduces DSI by 1 → DSI = 1.
Example 6 — Portal Frame (Single Bay, Single Storey)
Two columns (pinned at base), one beam (rigid connections at top). Three members, four joints.
m = 3, r = 2 + 2 = 4 (two pin supports, 2 reactions each), j = 4, c = 0
DSI = (3×3 + 4) − (3×4 + 0) = 13 − 12 = 1 (Indeterminate to 1st degree)
With fixed bases instead of pins: r = 3 + 3 = 6
DSI = (3×3 + 6) − (3×4 + 0) = 15 − 12 = 3 (Indeterminate to 3rd degree)
Example 7 — Portal Frame with Internal Hinge at Beam Centre
Two columns (fixed bases), one beam with internal hinge at centre. The hinge splits the beam into 2 members.
m = 4 (2 columns + 2 beam halves), r = 3 + 3 = 6, j = 5, c = 1
DSI = (3×4 + 6) − (3×5 + 1) = 18 − 16 = 2 (Indeterminate to 2nd degree)
The hinge reduced DSI from 3 (fixed portal) to 2. Adding a second hinge would give DSI = 1; a third hinge in the right place would make it determinate (DSI = 0) — this is the three-hinged arch/frame.
Example 8 — Two-Bay Single-Storey Frame
Three columns (all fixed bases), two beams (rigid joints). Two closed loops.
m = 5 (3 columns + 2 beams), r = 3×3 = 9 (3 fixed supports), j = 6, c = 0
DSI = (3×5 + 9) − (3×6 + 0) = 24 − 18 = 6 (Indeterminate to 6th degree)
Check using loop method: 2 closed loops × 3 + external DSI = 6 + 0 = 6 ✓
(External DSI = 9 − 3 = 6 reactions in excess of 3; internal DSI = 6 since 2 loops × 3 = 6. Total = 6 + 6 = 12? No — these are not additive in this way. Use the direct formula for frames.)
The direct formula DSI = 24 − 18 = 6 is correct. Use the direct formula always.
Example 9 — Simple Pin-Jointed Truss
Pratt truss: 7 members, 5 joints, pin support + roller support (r = 3).
DSI = m + r − 2j = 7 + 3 − 2(5) = 10 − 10 = 0 (Determinate) ✓
Example 10 — Truss with Redundant Member
Same truss as Example 9 but with one extra diagonal member added (m = 8).
DSI = 8 + 3 − 2(5) = 11 − 10 = 1 (Indeterminate to 1st degree)
The extra diagonal is the redundant member — the truss carries load through two paths where it previously had one.
9. Quick Reference — DSI of Common Structures
| Structure | Support Conditions | Internal Hinges | DSI |
|---|---|---|---|
| Simply supported beam | Pin + Roller | 0 | 0 |
| Cantilever beam | Fixed + Free | 0 | 0 |
| Overhanging beam | Pin + Roller (with overhang) | 0 | 0 |
| Propped cantilever | Fixed + Roller | 0 | 1 |
| Fixed beam | Fixed + Fixed | 0 | 3 |
| Fixed beam with 1 internal hinge | Fixed + Fixed | 1 | 2 |
| Fixed beam with 3 internal hinges | Fixed + Fixed | 3 | 0 |
| Two-span continuous beam (pin + 2 rollers) | Pin + Roller + Roller | 0 | 1 |
| Three-span continuous beam | Pin + Roller + Roller + Roller | 0 | 2 |
| Portal frame — pinned bases | Pin + Pin | 0 | 1 |
| Portal frame — fixed bases | Fixed + Fixed | 0 | 3 |
| Three-hinged arch/frame | Pin + Pin | 1 (at crown) | 0 |
| Two-bay single-storey frame — fixed bases | 3 Fixed | 0 | 6 |
| Simple determinate truss | Pin + Roller | — | 0 |
10. Common Mistakes Students Make
- Using the truss formula (m + r − 2j) for rigid frames: The truss formula is only valid for pin-jointed trusses where all members carry only axial force and all joints are pins. For rigid frames (where joints are welded or monolithically cast, transmitting moment), the frame formula (3m + r − 3j − c) must be used. Applying the truss formula to a rigid frame gives a completely wrong answer — typically much lower than the correct DSI.
- Miscounting joints and members: Every point where two or more members meet — including support points and free ends — is a joint. Every straight segment between two joints is a member. A common error is to count a continuous beam over three supports as one member (it has three joints and two members). Another error is to forget the free end of a cantilever as a joint (even though only one member meets there, it is still a joint in the formula).
- Forgetting that a fixed support counts as r = 3: A fixed (built-in) support provides three reactions: a vertical force, a horizontal force, and a moment. Students often assign r = 2 to a fixed support (confusing it with a pin). This is perhaps the single most common numerical error in DSI problems — it reduces the calculated DSI by 1 per fixed support.
- Misapplying the internal hinge formula for multi-member joints: When n members meet at an internal hinge, the number of condition equations is c = n − 1, not n. Students often count c = n, which overcounts the conditions and gives a DSI that is 1 too low. For a joint with 2 members (the standard case), c = 1 — which is why most textbooks simply say “each internal hinge gives c = 1.” For joints with 3 or more members at a common hinge, remember c = n − 1.
- Declaring a structure stable just because DSI = 0: DSI = 0 is a necessary but not sufficient condition for a structure to be statically determinate and stable. A structure can have DSI = 0 yet still be geometrically unstable — for example, three pin supports all on the same horizontal line, or a truss where all members are parallel. Always visually check that the support arrangement and member layout can actually prevent all rigid body motions (translation and rotation).
11. Frequently Asked Questions
Why does a fixed beam have DSI = 3 and not DSI = 1?
A fixed beam has fixed supports at both ends. Each fixed support provides 3 reactions: vertical force, horizontal force, and moment. The total reactions r = 3 + 3 = 6. With 3 equilibrium equations available, the structure has 6 − 3 = 3 more unknowns than equations — DSI = 3. Students often say DSI = 1 (thinking of the moment redundancy only), but this ignores the fact that there are also 2 extra force reactions at the two fixed ends. The DSI = 3 means three compatibility equations are needed: typically, the two slope conditions (dy/dx = 0 at each fixed end) plus one deflection condition — or equivalently, the three components of relative displacement and rotation between the two ends are all zero (they are built into a rigid wall). Using the formula: m = 1, r = 6, j = 2, c = 0 → DSI = (3×1 + 6) − (3×2) = 9 − 6 = 3 ✓
Does support settlement change the degree of indeterminacy?
No — support settlement does not change the DSI. The degree of indeterminacy is a property of the structure’s geometry (member layout, support types, and internal releases) and does not depend on the loading or on whether supports settle. What settlement does change is the distribution of forces and moments in an indeterminate structure — a settling support causes a redistribution of reactions and internal forces throughout the structure. This is precisely why indeterminate structures are sensitive to support settlement, while determinate structures are completely unaffected by it (the reactions in a determinate structure are found from equilibrium alone, with no reference to the actual support positions).
What is the difference between static indeterminacy and kinematic indeterminacy?
Static indeterminacy (DSI) counts the number of unknown forces that cannot be found by equilibrium — it determines how many compatibility equations are needed. Kinematic indeterminacy (DKI), also called the degree of freedom, counts the number of independent displacement components (rotations and translations) that the structure’s joints can undergo. The two are related but different. DSI is relevant when using force methods (compatibility method) to solve indeterminate structures. DKI is relevant when using displacement methods (Slope-Deflection Method, Moment Distribution Method, Stiffness Matrix Method) — the number of unknowns in a displacement method equals the DKI. For GATE CE, DSI is the more frequently tested concept, but DKI appears in questions about choosing the appropriate analysis method.
How do you find DSI for a structure with both rigid joints and internal hinges?
Use the full frame formula: DSI = (3m + r) − (3j + c). Count members carefully — each internal hinge that divides a member into two creates an extra member and an extra joint (the hinge point). Then count c = 1 for each internal hinge (or c = n − 1 if n members meet at the hinge). For example, a fixed beam with one internal hinge mid-span: the hinge splits the beam into 2 members, creating 3 joints total (A, hinge, B). m = 2, r = 6 (two fixed supports), j = 3, c = 1. DSI = (6 + 6) − (9 + 1) = 12 − 10 = 2. Check: fixed beam has DSI = 3; one internal hinge reduces it by 1 → DSI = 2 ✓.