Structural Analysis: Indeterminacy, Force and Displacement Methods, Cables, Arches and Influence Lines

Analysis begins with counting, not computing. D_s = 3m + r − 3j for a plane frame tells you how many unknowns the equations of statics cannot reach, and until you know that number you do not know which method you are allowed to use. Determinate and statics finishes the job; indeterminate and you need one extra equation per redundant, which is where the two great families come from. The force method — also called the flexibility or compatibility method — releases redundants until the structure is determinate, computes the displacement the release opens up, and then applies whatever redundant force closes it again; you solve for forces, and the number of simultaneous equations is D_s. The displacement method — slope-deflection, moment distribution, and the stiffness method that every piece of software runs — does the reverse: it takes the joint rotations and translations as unknowns, writes each member’s end moments in terms of them, and enforces equilibrium at the joints. The choice between them is arithmetic, not principle: pick whichever has fewer unknowns, which for a continuous beam is usually the force method and for a multi-storey frame is always the displacement method. Both rest on superposition, which is legitimate only while the material is linear and the deflections small enough that equilibrium can be written on the undeformed shape. Then two structures that carry load by shape rather than by bending — the cable, which finds its own funicular, and the arch, which is a cable turned upside down and gets the same horizontal thrust wL²/8h from the same algebra. And influence lines, which answer a different question from every diagram before them: not what does this loading do everywhere, but what does a moving unit load do here.

Counting Before Computing

For a rigid-jointed plane frame the degree of static indeterminacy is D_s = 3m + r − 3j, where m counts members, r counts reaction components and j counts joints including the supports. Each member carries three internal force components and each joint supplies three equilibrium equations, so the expression is a straight book-keeping of unknowns against equations. An internal release — a hinge inserted in a member — subtracts one from the count for each moment it releases. Run it on the portal frame that every question is built from: two columns and a beam is m = 3, four joints is j = 4, and two fixed bases give r = 6, so D_s = 9 + 6 − 12 = 3. Change the bases to hinges and r falls to 4, so D_s = 1. Put a hinge at the middle of the beam as well and it drops to 2. Nothing has been computed yet, and already the method is decided.

ℹ️ Two different counts, and the one that decides which method is cheaper
Static indeterminacy D_s counts the forces statics cannot find. Kinematic indeterminacy D_k counts the independent displacements the joints can undergo — three per free joint in a plane frame, less whatever the supports prevent, and less again if members are taken as axially rigid. The two are different numbers and they point at different methods, because D_s is the number of simultaneous equations the force method must solve and D_k is the number the displacement method must solve. For a two-span continuous beam on simple supports, D_s = 1 while D_k is 3 rotations, so the force method wins easily and that is how such problems are set. For a four-storey, three-bay frame the picture inverts completely: D_s runs into the dozens while the joints, with axial deformation neglected, contribute a rotation each plus one sway per storey. That is why every structural analysis package ever written implements the stiffness method and not the flexibility method — not because it is more accurate, since the two give identical answers, but because assembling a stiffness matrix from member contributions is systematic, while choosing a good set of redundants is a judgement a computer cannot easily make.

The Force Method, and What a Redundant Buys

The force method has three steps and the middle one is the whole idea. Release enough redundants to leave a determinate structure, which you are free to choose — the released structure is a device, not the real one. Compute the displacement the release opens at that point under the applied loads. Then apply the redundant and require it to close the gap exactly, because in the real structure that displacement is zero. One compatibility equation per redundant, and the unknowns are forces.

🎯 A propped cantilever, and the prop that removes three quarters of the wall moment
Take a cantilever of 6 m carrying a UDL of 20 kN/m, propped on a roller at the free end. D_s = 1. Release the prop and you have a plain cantilever, whose free end drops wL⁴/8EI. Now apply the prop reaction P upward on that same cantilever, which lifts the end by PL³/3EI. In the real structure the roller holds that point still, so the two must cancel: wL⁴/8EI = PL³/3EI, and P = 3wL/8 = 45 kN — note that EI divides out, so the answer does not depend on the stiffness at all, which is true for any structure whose redundant is resisted by members of the same material. The rest follows: the fixed end takes 120 − 45 = 75 kN, and the support moment is wL²/8 = 90 kN·m. Now compare that with the same beam without the prop, where the wall carries wL²/2 = 360 kN·m. The prop cut the wall moment to exactly a quarter, and cost one roller. That is what redundancy buys, and it is worth stating the general lesson: adding a restraint to a structure does not merely share the load, it redistributes the internal forces towards the stiffer path, and usually reduces the peak demand sharply. The same fact has a sting in it — because the forces now depend on relative stiffness, an indeterminate structure develops internal forces from things a determinate one ignores entirely: a support that settles, a temperature change, a member fabricated slightly short. A determinate structure simply moves; an indeterminate one argues with itself.

Slope-Deflection, Moment Distribution, and Stiffness

Slope deflection writes each member’s end moments in terms of the rotations of its two ends and any relative sway: M_AB = (2EI/L)(2θ_A + θ_B − 3Δ/L) + FEM_AB. Every term has a job. The fixed-end moment is what the member would carry if its ends could not move at all; the rotation terms relieve that as the joints turn; the sway term adds what a storey drift contributes. Write one such expression per member end, then impose equilibrium at each joint — the moments meeting at a rigid joint must sum to zero — and you have as many equations as unknown rotations. Moment distribution is the same mathematics performed iteratively so that it can be done by hand: lock every joint, compute the fixed-end moments, then release the joints one at a time, distributing each out-of-balance moment among the members meeting there in proportion to their stiffness 4EI/L, and carrying half of what you distribute to the far end. Repeat until the corrections are negligible. The two give the same answer because they are the same equations.

Two equal 6 m spans under 20 kN/m, ends simply supported: what continuity changes
QuantityTwo continuous spansTwo separate simple spans
Moment over the middle support−90 kN·m (hogging)0
Maximum sagging moment50.625 kN·m at x = 2.25 m90 kN·m at midspan
End reactions45 kN each60 kN each
Middle support reaction150 kN120 kN
Total load carried240 kN240 kN

Read that table as a single trade. Continuity removes 44 per cent of the sagging moment — from 90 down to 50.6 — and pays for it by creating a hogging moment of 90 kN·m over the middle support that did not exist before, and by loading that support with 150 kN instead of 120. Nothing has been gained for free: the total load is 240 kN either way. What has changed is where the structure is asked to be strong, and that is the point of the method. It also explains a detail visible on any reinforced concrete slab: the steel runs near the bottom at midspan, where the moment sags, and crosses to the top over the supports, where it hogs. A continuous beam detailed with bottom steel only is being reinforced for the wrong diagram.

Cables, Arches, and Influence Lines

A cable carries load in pure tension and has no choice about its shape: it hangs in whatever curve makes the bending moment zero everywhere, which is the funicular of the load. Under a load uniform along the horizontal — a suspension bridge deck — that curve is a parabola, and the horizontal component of the cable tension is constant along the whole length at H = wL²/8h, where h is the dip. For a 40 m span with a 4 m dip under 10 kN/m, H = 500 kN, the vertical component at each support is wL/2 = 200 kN, and the greatest tension, which occurs at the supports where the cable is steepest, is √(500² + 200²) = 538.5 kN. Note what H does with the dip: it is inversely proportional, so halving the dip to 2 m doubles the thrust to 1000 kN. A flat cable is an expensive cable, and that is why suspension bridges sag as much as they do.

💡 An arch is a cable upside down, and the algebra proves it
Take a three-hinged parabolic arch of the same 40 m span and a rise of 4 m, under the same 10 kN/m. Taking moments about the crown hinge for half the arch gives the horizontal thrust as H = wL²/8r = 500 kN — the same expression as the cable, with the dip replaced by the rise. That is not a coincidence, and the reason is visible in the bending moment. At any section, the arch moment is the simple-beam moment for the same span and load, less H times the height of the arch axis: M = M_beam − Hy. Evaluate it at the quarter point: y = 3.0 m, M_beam = 1500 kN·m, Hy = 500 × 3 = 1500 kN·m, so M = 0. At the crown: y = 4.0, M_beam = 2000, Hy = 2000, so M = 0 again. At the three-quarter point, zero again. The bending moment is identically zero at every section, because a parabola is the funicular shape for a uniformly distributed load and the arch has been built along it. The arch is carrying 400 kN of load with no bending whatsoever, purely in compression — which is precisely what a cable does in tension, reflected. The practical consequence is the one that governs arch design: this vanishing act holds only for the load the shape was drawn for. Put a point load on the same arch, or load only half its span, and the funicular for that loading is a different curve, the arch no longer follows it, and bending moments appear at once. That is why masonry arch bridges, which cannot take much bending, are built heavy — the dead weight is uniform and dominant, so the live load is a small perturbation about a shape that is right for the dead load.

An influence line answers a question none of the earlier diagrams answers. A bending moment diagram shows what one fixed loading does at every section; an influence line for the moment at one chosen section shows what a unit load does there as it moves across the span. The two are easy to confuse and the distinction is the whole of the topic: the horizontal axis of a bending moment diagram is where you are looking, and the horizontal axis of an influence line is where the load is. For a simply supported span the influence line for the reaction at A is a straight line from 1 at A to 0 at B; for the shear just left of a section it is a pair of parallel lines with a unit step at the section; and for the moment at midspan it is a triangle peaking at L/4. On a 20 m span that peak ordinate is 5 m, so a single 100 kN wheel produces a maximum midspan moment of 500 kN·m when it stands at midspan — and the influence line has told you both the value and the position to place the load, which is what it exists for. That second answer is the useful one for a bridge, where the load is a train of axles that can stand anywhere.

Castigliano’s second theorem

The GATE 2027 revision names Castigliano’s second theorem explicitly among the force and energy methods, and it belongs beside virtual work rather than instead of it: both are statements about strain energy, and the theorem is the shorter route whenever the quantity wanted is a displacement at a point where a load already acts. It says exactly that: the partial derivative of the total strain energy with respect to a load equals the displacement at that load, in its own direction — δ = ∂U/∂P. Nothing about the structure need be determinate for it to hold.

🎯 It reproduces `PL³/3EI` from an energy argument, and that is the proof
Work the simplest case and watch a familiar answer fall out of an unfamiliar route. A cantilever of length L = 3 m carries P = 20 kN at its free end, with EI = 20,000 kN·m². The bending moment at a distance x from the free end is M = Px, so the strain energy is U = ∫M²/(2EI) dx = ∫₀ᴸ (Px)²/(2EI) dx = P²L³/(6EI) = 0.09 kN·m. Differentiate with respect to the load: δ = ∂U/∂P = PL³/(3EI) = 20 × 27/(3 × 20,000) = 0.009 m = 9.00 mm — which is exactly the standard cantilever deflection formula, obtained without integrating a curvature or drawing a conjugate beam. Three things make the theorem worth having rather than merely elegant. It works on any structure whose strain energy can be written down, including trusses (where U = Σ N²L/2AE and the derivative is a sum, not an integral) and indeterminate frames. It gives a rotation just as readily, by differentiating with respect to a moment: θ = ∂U/∂M = PL²/(2EI) = 0.00450 rad here. And where no load acts at the point of interest, the trick is to apply a dummy load Q there, differentiate, then set Q = 0 — at which point the method has become virtual work, which is why the two belong in the same section rather than being rivals. The one restriction worth stating plainly: the theorem as used here assumes linear elastic behaviour, so it says nothing about a structure past yield.

Key takeaways

  • Count first: D_s = 3m + r − 3j, less one per released moment. D_s is how many equations the force method solves and D_k how many the displacement method solves — pick the smaller, which is why software always uses stiffness.
  • The force method is release, measure the gap, close it. On a 6 m propped cantilever under 20 kN/m the prop takes 3wL/8 = 45 kN — with EI cancelling — and cuts the wall moment from 360 to 90 kN·m, exactly a quarter.
  • Continuity moves demand rather than removing it: two 6 m spans drop the sagging moment from 90 to 50.6 kN·m and create −90 over the middle support. Same total load, different place to be strong — which is why slab steel crosses to the top over supports.
  • A cable and a three-hinged parabolic arch share H = wL²/8h, and under the UDL the arch carries zero bending at every section because its shape is the funicular. Change the loading and the bending returns at once.
  • An influence line plots where the load is, not where you are looking. Midspan moment on a simply supported span peaks at L/4 — 5 m on a 20 m span, so one 100 kN wheel gives 500 kN·m, and the diagram also tells you where to stand it.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A single-bay single-storey rigid portal frame has both column bases fixed. Its degree of static indeterminacy is ______

    Numerical answer — type the value.

    Show answer

    Answer: 3

    3. Use D_s = 3m + r − 3j. The frame has m = 3 members — two columns and a beam — and j = 4 joints, counting the two bases and the two corners. Two fixed bases give three reaction components each, so r = 6. Then D_s = 3(3) + 6 − 3(4) = 9 + 6 − 12 = 3. The physical reading is worth having alongside the arithmetic: cut the beam anywhere and you expose three internal forces — axial, shear and moment — none of which statics can find, which is the same 3. Change the supports and the count moves with them. Hinged bases give r = 4 and D_s = 1, which is why a portal on pinned feet is so much easier to analyse by hand and why it is the standard textbook case. Add a hinge at the middle of the beam as well and the release subtracts one more, leaving D_s = 2. Two common errors: forgetting that the supports are joints, which gives j = 2 and a nonsensical answer; and treating a fixed base as two reactions instead of three, which loses the moment and gives 1. The count is worth doing before any analysis, because it decides the method — at D_s = 3 the force method needs three simultaneous compatibility equations, while the displacement method needs only the two joint rotations and one sway.

  2. A cantilever of span 6 m carries a uniformly distributed load of 20 kN/m and is propped on a roller at its free end. The reaction at the prop, in kN, is ______

    Numerical answer — type the value.

    Show answer

    Answer: 45

    45 kN. The structure is once indeterminate, so release the prop and restore compatibility. As a plain cantilever the free end drops wL⁴/8EI; the prop reaction P acting upward at that end lifts it PL³/3EI. The roller holds the point still, so wL⁴/8EI = PL³/3EI, giving P = 3wL/8 = 3(20)(6)/8 = 45 kN. Notice that EI cancels: the answer holds whatever the section and whatever the material, provided both are the same throughout. The rest of the structure follows — the fixed end carries 120 − 45 = 75 kN, and the support moment is wL²/8 = 90 kN·m. Compare that with the same beam unpropped, where the wall carries wL²/2 = 360 kN·m: one roller has removed three quarters of the wall moment. That is the general lesson about redundancy, and it comes with a general warning. Because the forces now depend on relative stiffness, this beam develops internal forces from causes a determinate beam ignores entirely — if the prop settles by even a few millimetres, P changes; a temperature gradient or a member fabricated slightly short does the same. A determinate structure accommodates such things by moving. An indeterminate one resists them, and the resistance is a real stress.

  3. In the moment distribution method, the carry-over factor for a prismatic member whose far end is fixed is

    1. 1
    2. 1/2
    3. 0
    4. 3/4
    Show answer

    Answer: B — 1/2

    1/2. Apply a moment at one end of a prismatic member whose other end is fixed and solve: the far end develops exactly half the applied moment, with the same sign convention. That is the carry-over factor, and it is a property of the member, not of the structure, so it holds every time the far end is fixed. The rest of the method is bookkeeping around it: lock all joints, compute fixed-end moments, then release joints one at a time, distributing the out-of-balance moment among the members meeting there in proportion to their stiffness 4EI/L, carrying half of each distributed moment to the far end, and repeating until the carried-over amounts become negligible. Option C is right for a different far-end condition — a hinged far end carries nothing over, and its stiffness is correspondingly reduced to 3EI/L, which is the standard modification that shortens the analysis of a beam with a simply supported end. Option A would mean the far end feels the full applied moment, which would make the method diverge rather than converge. The reason the process terminates is exactly that each cycle carries over only half and then distributes a fraction of that, so the corrections shrink geometrically — moment distribution is an iterative solution of the same simultaneous equations slope-deflection writes directly, and it converges because the carry-over factor is less than one.

  4. A two-span continuous beam has equal spans of 6 m and carries a uniformly distributed load of 20 kN/m over both spans. The end supports are simple. The magnitude of the bending moment over the middle support, in kN·m, is ______

    Numerical answer — type the value.

    Show answer

    Answer: 90

    90 kN·m (hogging). By symmetry both spans behave identically, so each half of the beam is a propped cantilever about the middle support, and the standard result wL²/8 = 20(36)/8 = 90 kN·m applies. The three-moment equation or one cycle of moment distribution gives the same. The full picture is worth carrying: the end reactions are 3wL/8 = 45 kN each, the middle support takes 2 × 5wL/8 = 150 kN, and the three sum to 240 kN, which is the total load 2 × 120 ✓ — always run that check. Shear vanishes in the first span at x = 45/20 = 2.25 m, where the sagging moment peaks at 45(2.25) − 20(2.25²)/2 = 50.625 kN·m. Compare with two separate simply supported 6 m spans, which would carry wL²/8 = 90 kN·m at midspan and nothing over the support. Continuity has cut the sagging moment by 44 per cent and paid for it with a hogging moment of equal size over the support, plus a middle reaction of 150 kN instead of 120. Nothing was gained for nothing — the demand moved rather than shrank. And that migration is visible in any continuous slab: the reinforcement runs near the bottom in the spans and crosses to the top over the supports, because that is where the tension now is.

  5. A cable of span 40 m carries a uniformly distributed load of 10 kN/m and has a central dip of 4 m. The horizontal component of the cable tension, in kN, is ______

    Numerical answer — type the value.

    Show answer

    Answer: 500

    500 kN. For a cable under a load uniform along the horizontal the profile is parabolic and the horizontal component of tension is constant along the whole length at H = wL²/(8h) = 10(40²)/(8 × 4) = 16000/32 = 500 kN. That H is constant is the fact worth holding on to — the cable tension varies along the length because the vertical component varies, being zero at the lowest point and wL/2 = 200 kN at each support, so the greatest tension is at the supports where the cable is steepest: T_max = √(500² + 200²) = 538.5 kN, only 7.7 per cent above H. The dip is the design variable, and H is inversely proportional to it: halve the dip to 2 m and the thrust doubles to 1000 kN. A flatter cable is not a cheaper one — it is a more heavily loaded one, and its anchorages must be bigger, which is why suspension bridges are built with as much sag as clearance allows. Two related results are worth linking here. A three-hinged parabolic arch of the same span and a rise of 4 m under the same load takes H = wL²/8r = 500 kN by exactly the same algebra, because an arch is a cable inverted. And under a load uniform along the cable rather than along the horizontal — a cable carrying only its own weight — the true profile is a catenary rather than a parabola, though for shallow sags the two differ by very little.

  6. A three-hinged parabolic arch of span L and rise r carries a uniformly distributed load w over its entire span. The bending moment at any section of the arch is

    1. maximum at the crown
    2. zero everywhere
    3. maximum at the quarter points
    4. equal to wL²/8 throughout
    Show answer

    Answer: B — zero everywhere

    Zero everywhere. The arch moment at any section is the simple-beam moment for the same span and loading, less the horizontal thrust times the height of the arch axis: M = M_beam − Hy. Under a full-span UDL, M_beam = wx(L−x)/2, the parabolic axis is y = 4rx(L−x)/L², and the thrust is H = wL²/8r. Substituting, Hy = (wL²/8r)(4rx(L−x)/L²) = wx(L−x)/2, which is M_beam exactly — so the two terms cancel identically and M = 0 at every x. Check it numerically on a 40 m span with r = 4 m and w = 10 kN/m, where H = 500 kN: at the quarter point y = 3 m and M_beam = 1500 kN·m against Hy = 1500; at the crown y = 4 and M_beam = 2000 against Hy = 2000. Zero both times. The reason is that a parabola is the funicular shape for a uniformly distributed load, so the arch is following the line of thrust and carries the whole 400 kN in pure compression. The essential qualification, and what the examiner is usually testing: this holds only for the loading the shape suits. Put a concentrated load on the arch, or load half the span, and the funicular for that case is a different curve, the arch no longer lies along it, and bending moments appear immediately. That is why masonry arches, which have little bending capacity, are built massive — a dominant uniform dead load keeps the real thrust line close to the shape that was drawn.

  7. The influence line for the bending moment at the midspan of a simply supported beam of span L is

    1. a parabola with a peak ordinate of L²/8
    2. a triangle with a peak ordinate of L/4 at midspan
    3. a rectangle of constant ordinate L/4
    4. a triangle with a peak ordinate of 1 at midspan
    Show answer

    Answer: B — a triangle with a peak ordinate of `L/4` at midspan

    A triangle peaking at L/4. Put a unit load at distance a from the left support. The reaction at the left is (L − a)/L, and while the load is to the right of midspan the moment at midspan is (L−a)/L × L/2, which rises linearly from zero at a = 0 to L/4 when the load reaches midspan; by symmetry it falls linearly back to zero. Hence a triangle with a peak ordinate of L/4 — a length, because an influence line for a moment has units of length per unit load. On a 20 m span the peak is 5 m, so a single 100 kN wheel produces 500 kN·m when it stands at midspan. Option A is the bending moment diagram for a UDL of unit intensity, and confusing the two is the mistake this question exists to catch: a bending moment diagram plots what one fixed loading does at every section, while an influence line plots what a moving unit load does at one section. The horizontal axes mean different things. Option D confuses the moment influence line with the one for a reaction, which does peak at 1 and is dimensionless. Option C would say the position of the load does not matter, which is exactly what an influence line exists to deny. And the practical value is the second answer it gives: for a train of axle loads on a bridge, the diagram tells you not only the effect but where to stand the vehicle to maximise it.

  8. Which of the following are true of an indeterminate structure but not of a determinate one? (One or more)

    1. Internal forces are developed by a settlement of one support
    2. Internal forces are developed by a uniform temperature rise with free expansion prevented
    3. The distribution of internal forces depends on the relative stiffness of the members
    4. The total load carried must equal the sum of the reactions
    Show answer

    Answer: A — Internal forces are developed by a settlement of one support; B — Internal forces are developed by a uniform temperature rise with free expansion prevented; C — The distribution of internal forces depends on the relative stiffness of the members

    A, B and C. All three describe the same underlying property from different directions. In a determinate structure the internal forces follow from equilibrium alone, so anything that merely changes the shape without changing the loads — a support settling, a temperature change, a member fabricated a little short — is accommodated by the structure simply moving, and produces no stress at all. In an indeterminate structure those displacements are resisted by the redundant restraints, and resistance means force. A is the classic case: settle one support of a continuous beam and the moment diagram changes throughout, with no change in the applied load. B is the same mechanism through temperature, and it is why long bridges are given expansion joints and bearings — not because the material cannot take the strain, but because preventing the movement creates forces that nothing else asked for. C is the general statement of both, and it is what makes indeterminate analysis harder: the answer depends on EI ratios, so you cannot solve for forces without knowing something about the sections, whereas in a determinate structure the forces come out before any section is chosen. D is false as a distinguishing statement, and it is the trap: overall equilibrium holds for every structure that is standing, determinate or not. Indeterminacy means statics is not sufficient to find all the unknowns, never that it stops being true — which is why summing the reactions against the total load remains a valid arithmetic check on an indeterminate analysis, as it was for the 240 kN two-span beam in this chapter.