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 from the wall to 5 m and a point load P at 9 m from the wall. Find the vertical reaction RA at the wall.
The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is 4 kN/m at the wall and falls to zero at the free end B, plus a point load P at 3 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.
Find the bending moment Mx at the section x = 6.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
Use the method of joints to find the force FAC in member AC. Give tension as positive and compression as negative.
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 maximum deflection δmax, at midspan.
A simply supported beam carries two equal point loads P at its third points (L/3 from each support). Using superposition, find the midspan deflection δmax.
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
A column of length L made of steel 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 = 75 kN
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: MA = 234 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: D (x = 6 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: Mx = 100.8 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: FAC = 10.67 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.
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
Answer: FBC = −42.06 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: δmax = 3.172 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: δmax = 5.546 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: Pcr = 62.01 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 round section has the same I about every axis.
Answer: σcr = 39.34 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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