Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
The cantilever beam is fixed to the wall at A. It carries a uniformly distributed load w along its whole length and a point load P at 1.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.
A compound beam is fixed to the wall at A, has an internal hinge at C (2.5 m from A) and rests on a roller at D (6 m from A). It carries a point load P at 4.5 m from A, a uniformly distributed load w on the whole part CD and a point load Q at 2 m from A. Find the reaction RD at the roller.
The simply supported beam carries a clockwise couple M0 = 55 kN·m at x = 3.5 m from A, and no other load. Find the bending moment Mx just to the right of the couple. Use sagging moments positive and upward shear on the left face positive.
Find the bending moment Mx at the section x = 3.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
A simply supported beam carries a uniformly distributed load w over its whole span. Find the slope θA of the beam at support A, in radians.
A propped cantilever is fixed at A and simply supported (propped) at B. It carries a uniformly distributed load w over its whole length. EI is constant. Find the reaction RB at the prop.
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
A solid round column of length L made of steel is pinned at both ends. It must carry 25 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
Answer: MA = 217.5 kN·m
Cantilever equilibrium: a fixed support can push up and also resist rotation, so it gives a vertical reaction RA and a reaction moment MA. ΣFy = 0 gives RA, and moments about the wall give MA = Σ(load × distance from the wall). A distributed load acts as its resultant at the centroid of the load diagram.
Answer: RD = 33.82 kN
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Replace each distributed load by its resultant (its area) acting at the centroid of the load diagram, and take moments about a support to remove that support’s reaction from the equation. An internal hinge cannot carry a bending moment, which gives one extra equation: the moment about the hinge of the forces on either side is zero.
The hinge has no moment, so moments about C on CD alone must balance.
Answer: Mx = 22.92 kN·m
Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut. An applied couple makes the bending moment diagram jump by the size of the couple.
Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.
The left part now contains RA and the couple.
Answer: Mx = 50.75 kN·m
Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.
Equilibrium of the whole beam.
Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.
Answer: member BF
Zero-force members: at a joint, the forces must balance in every direction. If a joint has no load and no support, and two of its three members are in line, then the balance across that line involves only the third member, so its force is zero. If only two members meet at an unloaded joint and they are not in line, both are zero.
AB and BC lie on one straight line and the joint carries no load. Balancing forces across that line leaves only BF.
Solving all joints gives these member forces (T = tension, C = compression).
Answer: FBC = −34.36 kN
Method of joints: every pin of a truss is in equilibrium, so ΣFx = 0 and ΣFy = 0 at each joint. Start at a joint with at most two unknown member forces. A negative result means the member is in compression.
Equilibrium of the whole truss. Moments about A, counter-clockwise positive.
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
Answer: θA = 0.03896 rad
Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition). The slope θ = dv/dx is the first integral of M/(EI); in consistent units it comes out in radians.
Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.
That is 2.232°.
Answer: RB = 5.625 kN
Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition). For a statically indeterminate beam, remove a redundant support and require the deflection there to be zero (compatibility).
Without the prop the beam is a cantilever. Its free end would deflect under the load.
The prop force RB alone lifts the free end by RBL³/(3EI). The prop holds B at zero deflection, so the two must cancel. EI cancels out.
Answer: Pcr = 6.909 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
A rectangle buckles about its weak axis, using the smaller dimension cubed.
Answer: d = 46.42 mm
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Check: the slenderness KL/r is 258.5, so the column is slender and Euler applies.
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