Structural Equilibrium Applications

A roof truss is bolted steel with real weight, but the truss model sets that aside: members are straight and weightless, joints are frictionless pins, and every load and support force acts at a joint. The payoff shows up in one member. It is touched only at its two pins, so ΣF = 0 makes the two end forces equal and opposite, and Στ = 0 about one pin puts the other force's line of action through that pin. Both forces therefore lie along the member, and what could have been four unknown components collapses to one signed number: tension if the member pulls on its pins, compression if it pushes. Then count. Each pin gives two equations, so j joints supply 2j, against m member forces and r reaction components. A triangle on a pin and a roller has m + r = 3 + 3 = 6 = 2j. A square of four bars on the same supports has 7 unknowns against 8 equations and folds into a parallelogram. One diagonal makes it 8 = 8; a second makes 9 against 8, and equilibrium can no longer decide how the load divides between the two diagonals.

Start at a joint where no more than two member forces are unknown, and draw every unknown member force pulling away from the pin, as if in tension. Take a triangle rising at 30° with P = 12 kN on its apex. By symmetry each support pushes up with 6 kN. At the left support, vertical balance gives F_rafter sin 30° + 6 kN = 0, so F_rafter = −12 kN, and horizontal balance gives Fₜᵢₑ + F_rafter cos 30° = 0, so Fₜᵢₑ = +10.4 kN. The minus sign is the answer, not a slip: the rafter is in compression. Keeping the all-tension convention at every joint lets a force found at one pin carry its sign into the next. The same two equations expose zero-force members on sight: at an unloaded joint where only two non-collinear members meet, both carry nothing, and where three meet with two of them collinear, the third carries nothing. Geometry matters as much as load. Lower the rise to 15° and the same 12 kN drives 23.2 kN into each rafter.

To reach a member in the middle of a long truss, solving joint after joint from one end is slow and lets errors pile up. Cut the truss instead, through no more than three members whose forces are unknown, and treat either piece as a rigid body. A rigid body gives three equations, and one of them is a moment equation. Take moments about the point where two of the cut members meet: those two forces pass through it and drop out, leaving the third alone. In a truss with parallel top and bottom chords this has a clean reading. A chord force equals the moment, about the joint opposite it, of every load and reaction on one side of the cut, divided by the depth of the truss. A 2.5 m deep truss with 150 kN m about that joint puts 60 kN into the chord; make it 5.0 m deep and the chord carries 30 kN, which is why long spans use deep trusses. The diagonal, the only cut member with a vertical component, then follows from vertical force balance on the same piece.

A member pinned at three points, or loaded between its ends, is not a two-force member, and nothing makes its pin forces run along it. A structure containing such members is a frame, and it is solved by taking it apart. Draw each member on its own, give every pin an unknown x and y component, and pair them with Newton's third law: a pin pushes on one member exactly as hard as it is pushed by the other. Find the two-force members first, since their force directions are already known. Take a shelf arm 0.80 m long, pinned to a wall, carrying 200 N at its tip and propped at its midpoint by a strut rising at 45° from the wall. The strut is a two-force member, so moments about the wall pin give its push directly: F × 0.40 m × sin 45° = 200 N × 0.80 m, and F = 566 N. That push is 400 N out from the wall and 400 N up, so the wall pin must pull the arm 400 N towards the wall and 200 N down: a 447 N force aimed 27° off the arm's own line. The arm is being bent, which the truss model never allows.

Equilibrium says what force a member must carry; its material and shape say what it can carry, and the safety factor is the ratio, n = capacity ÷ working force. The margin pays for everything the calculation could not know: loads are estimates, material strengths scatter, joints are not frictionless pins, and steel corrodes. The size of the margin follows consequence and confidence, around 1.5 on ultimate load for aircraft structures, where weight is costly and loads are well measured, and of order 10 for elevator hoist ropes. In a determinate truss every member force is proportional to the load, Fᵢ = cᵢP, so each member has its own ratio and the structure's safety factor is the smallest of them. Compression members often govern, because a slender bar buckles sideways at a force far below what it holds in tension. For the 30° triangle with 12 kN on top, rafters rated at 30 kN in buckling give n = 2.5 and a tie rated at 45 kN gives 4.3, so the truss has n = 2.5, and requiring n = 2.0 caps the apex load at 15 kN.

Each idealisation fails somewhere, and a good analysis says which failure matters. Bolted and welded joints resist rotation, so members also bend; for slender members this secondary bending is usually small, for short stocky ones it is not. Members have weight: sending half of each member's weight to each end joint keeps the axial forces right but ignores the sag between joints, and a load placed mid-member turns that member into a beam. Equilibrium was also written on the undeformed shape, which is safe only while the angles barely change. For the triangle, F_rafter = P/(2 sin θ) gives dF/F = −cot θ dθ. If sagging lowers θ by 0.5°, the rafter force rises by about 1.5% at θ = 30° but by roughly 10% at θ = 5° (11% exactly), so shallow trusses and taut cables must be solved on their deflected shape. Finally, a redundant structure with m + r > 2j has no equilibrium answer at all until the stiffness of each member is known, and that is the question stress and strain answer next.