Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RA.
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 4 kN/m at the free end B, plus a point load P at 6 m from the wall. Find the magnitude of the reaction moment MA at the wall.
The simply supported beam carries two point loads. Find the maximum bending moment Mmax in the beam.
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 FBC in member BC. Give tension as positive and compression as negative.
Use the method of joints to find the force FAC in member AC. Give tension as positive and compression as negative.
A cantilever of length L is fixed at A and carries a point load P at a distance a = 2.5 m from A. Find the deflection δB at the free end B.
A simply supported timber beam of rectangular section, 75 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth hreq of the section.
A column of length L made of aluminium is fixed at the base and free 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 free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
Answer: RA = 15 kN
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Moments about A, counter-clockwise positive. Then vertical force balance.
Answer: MA = 318 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: Mmax = 209.4 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. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.
Equilibrium of the whole beam.
Keep the part left of each load.
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: 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: FAC = 22.5 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: δB = 1.612 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.
Deflection at the load plus the slope at the load times the remaining length.
Answer: hreq = 227.9 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 = 4.318 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 = 4.318 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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