Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
The beam has a pin at A and a roller at B, 6 m from A, and overhangs B to the free end C. Find the vertical reaction RA, taking upward as positive.
The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is zero at the wall and rises to 6 kN/m at the free end B, plus a point load P at 1.5 m from the wall. Find the magnitude of the reaction moment MA at the wall.
The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.
The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.
Use the method of joints to find the force FAB in member AB. Give tension as positive and compression as negative.
Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.
A cantilever carries a uniformly distributed load w over its whole length and a point load P at its free end. Using superposition, find the deflection δ at the free end.
A simply supported timber beam of rectangular section, 125 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth hreq of the section.
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
Answer: RA = 49.17 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.
Counter-clockwise positive.
Answer: MA = 92 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: C (x = 4 m)
Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.
Take moments about one support, then use the vertical force balance.
Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.
Answer: E (x = 4 m)
Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.
Take moments about one support, then use the vertical force balance.
Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.
Answer: FAB = −6.67 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.
Assume every member is in tension (pulling away from the joint). Only AB has a vertical component at A.
Answer: member CG
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.
BC and CD lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.
Solving all joints gives these member forces (T = tension, C = compression).
Answer: δ = 75 mm
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).
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.
Answer: hreq = 233.8 mm
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).
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.
In practice, choose the next standard depth above this value.
Answer: Pcr = 3.06 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: σcr = 24.48 MPa
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 round section has the same I about every axis.
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