Statically Indeterminate Truss & the Redundant Bar

Hang a load from three bars or from two. Choose the materials and areas, warm a bar or let an anchor settle, and push the load up until the bars yield. The table under the drawing does the arithmetic in the order you would do it by hand.

Worked example — Statically Indeterminate Truss & the Redundant Bar

Degree of indeterminacy	1
Centre bar stiffness	50.000 kN/mm
Outer bar stiffness	35.355 kN/mm
Joint stiffness K_xx	35.355 kN/mm
Joint stiffness K_xy	0.000 kN/mm
Joint stiffness K_yy	85.355 kN/mm
Load vector R_x	0.000 kN
Load vector R_y	+150.000 kN

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Statically Indeterminate Truss & the Redundant Bar

A structure is statically determinate when the equilibrium equations alone fix every force in it, and statically indeterminate when it has more unknown forces than equations. The surplus is the degree of indeterminacy, and each surplus member is called redundant. The missing equations come from compatibility: the members are joined, so their stretches must fit one displacement of the joint. An elastic member's stretch is its force divided by its stiffness k = EA ÷ L, plus whatever a temperature change alone would give it, which is how the materials enter a calculation that statics alone could not finish.

What each symbol means

k: a bar's axial stiffness, EA ÷ L, in kN per mm. The force it takes to stretch it by one millimetre
Δ: a bar's elongation along its own length, in mm. Compatibility is a statement about these
N_y: a bar's yield force, its area times its yield stress. It carries no more than this however far it stretches
P_y: the load at which the first bar yields. It depends on the stiffnesses and on any locked-in stress
P_c: the collapse load, at which the bars still elastic yield too. Statics alone gives it

Breaks when: The bars are assumed to go on stretching after they yield. Mild steel will, by many times its yield strain, which is what lets the reserve in the first block be used. A cast-iron bar, or any brittle one that fractures near first yield, does not redistribute anything, and for it the collapse load is the first-yield load, exactly as in the two-bar truss.

Take the centre bar away and the first yield is the collapse 🖖

Press Steel, three bars and push the load up. The centre bar yields first, at 213.39 kN, and nothing falls. It goes on holding 125 kN while it stretches, the two outer bars take every kilonewton added after that, and the truss collapses only at 301.78 kN. Now press Centre bar removed. Both bars reach yield together at 176.78 kN and there is no third bar to hand the load to. For equal bars at 45° the three-bar truss's collapse load is exactly √2 times its first-yield load, which the reserve row prints as 1.414. In the three-bar truss a yielding bar is a warning; in the two-bar truss it is the end.

A warm afternoon puts 19.88 kN into a truss with nothing on it 🖖

Press Heated, no load. There is no load, and the centre bar is 40 K warmer than the other two. Free, it would grow 0.960 mm. The outer bars will not let the joint drop that far, so the centre bar is compressed by 19.88 kN and each outer bar pulled by 14.06 kN, before anything has been hung from the truss. Then press Two bars, one heated: the same 40 K on one bar of the two-bar truss, and each bar carries 106.07 kN hot or cold, identical to the last digit. Two equations fixed those two forces, and temperature appears in neither; the joint simply moves. That is what the reserve of the first block costs. The forces in a statically indeterminate structure depend on its stiffnesses, its temperature and how its supports settle, and a determinate structure's forces depend on the load and the geometry alone. Two resistors in parallel already hit the same wall: conservation says their currents add up, and only the resistances say how the current divides. The Wheatstone bridge on the Kirchhoff page is the circuit where no series or parallel shortcut is left to hide that.

Heat moves the first yield and leaves the collapse where it was 🖖

Keep three bars and move the temperature control. Cool the centre bar by 100 K and it is already in tension before any load arrives, so it yields first at 128.54 kN. Warm it by 100 K and the load has to undo a compression before it can stretch the bar, so first yield moves out to 298.24 kN. The collapse load is 301.78 kN both times, and at every temperature and settlement the controls allow, as long as nothing else is changed. Under a vertical load all three bars are at their yield forces at collapse, however they got there, and statics alone fixes the load that balances them. A locked-in stress decides when the first bar yields and has no say in when the structure falls. This is why a ductile steel frame can be designed for collapse without anyone knowing the stresses that erection and settlement left in it. It needs bars that can stretch a long way after yielding; a brittle bar that snaps at first yield takes the reserve with it.

Problem solved in full

Three rods under a hanging load 7 steps
A 150 kN load hangs from three 500 mm² steel rods: a vertical one 2 m long and two at 45° on either side of it, all fixed to the same beam. Find the force in each rod, the load at which the first rod yields, and the load at which the whole assembly gives way. Take E = 200 GPa and a yield stress of 250 MPa. Press Steel, three bars to check each figure.

Open this problem in the tool
1. Statics first. Symmetry makes the two outer forces equal, so horizontal equilibrium is satisfied on its own, and the vertical equation is all that is left. It has two unknowns in it:
  
  $N_{c} + 2 N_{o} cos 4 5^{\circ} = 150 kN$
2. The third equation is compatibility. Let the joint drop by v. The centre rod stretches by v, and each outer rod by the part of v that lies along it:
  
  $Δ_{c} = v, Δ_{o} = v cos 4 5^{\circ}$
3. Each rod's stiffness is EA ÷ L. The outer rods are √2 times as long, so they are softer:
  
  $k_{c} = \frac{E A}{L} = \frac{200 \times 1 0 ^{9} \cdot 500 \times 1 0 ^{- 6}}{2} = 50.000 kN/mm, k_{o} = \frac{E A cos 4 5 ^{\circ}}{L} = 35.355 kN/mm$
4. The joint's vertical stiffness adds the centre rod to the vertical share of the two outer ones, and the drop follows:
  
  $K_{y y} = k_{c} + 2 k_{o} cos^{2} 4 5^{\circ} = 50.000 + 35.355 = 85.355 kN/mm, v = \frac{150}{85 . 355} = 1.7574 mm$
5. Multiply each stiffness by its rod's stretch. The two forces satisfy the vertical equilibrium equation, which is the check:
  
  $N_{c} = 50.000 \cdot 1.7574 = 87.87 kN, N_{o} = 35.355 \cdot 1.7574 \cdot 0.7071 = 43.93 kN$
6. Every force is proportional to the load while the rods stay elastic. The centre rod reaches its yield force of 500 mm² × 250 MPa = 125 kN first, so scale the load until it does:
  
  $P_{y} = 150 \cdot \frac{125}{87 . 87} = 213.39 kN$
7. After that the centre rod holds 125 kN and the outer rods take the rest until they reach 125 kN too. With every force known, vertical equilibrium gives the collapse load:
  
  $P_{c} = N_{y} + 2 N_{y} cos 4 5^{\circ} = 125 + 2 \cdot 125 \cdot 0.7071 = 301.78 kN$
Answer

The rods carry 87.87 kN in the centre and 43.93 kN in each outer rod. The centre rod yields first at 213.39 kN, and the assembly does not give way until 301.78 kN. Look at which steps needed the stiffnesses. The division of the 150 kN and the first-yield load did, through steps 2 to 5. The collapse load did not: step 7 is the equation of step 1 with every rod at its yield force, and E appears nowhere in it.

$N_{c} = 87.87, N_{o} = 43.93, P_{y} = 213.39, P_{c} = 301.78 kN$

Learning path

Where the load goes

References (3)

The three-bar truss is the textbook's own first example of a problem statics cannot finish, and this page's closed form for it is the one check:compute asserts against the stiffness solution: S. Timoshenko, Strength of Materials, Part I: Elementary Theory and Problems, 3rd ed. D. Van Nostrand, New York, 1955. The chapter on statically indeterminate problems in tension and compression.
Insight 3 is the plastic theorems at work: a self-stress from heat or settlement moves the first yield and not the collapse load. A modern account of compatibility, the plastic theorems and why an elastic solution is fragile where a collapse load is not: J. Heyman, Basic Structural Theory. Cambridge University Press, 2008. DOI 10.1017/cbo9780511754487
Where the four materials' stiffness, yield and expansion figures would be checked. THEY WERE NOT TAKEN FROM HERE — see the note: A standard materials-property handbook, or a design code such as EN 1993-1-1 for structural steel and EN 1999-1-1 for aluminium.

By the 42Math team · Reviewed against published sources. Corrections welcome.