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 simply supported beam has a pin at A and a roller at B. A load P acts at 3 m from A, inclined at 40° to the horizontal, and a vertical load Q acts at 1.5 m from A. Find the angle α between the total reaction force at the pin A and the horizontal.
Find the bending moment Mx at the section x = 1.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
The beam rests on a pin at A and a roller at B and overhangs B by 1.5 m to the free end C. A uniformly distributed load w acts over the whole length A–C. Find the distance xc from A to the point of contraflexure between the supports (where the bending moment changes sign).
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 FAC in member AC. Give tension as positive and compression as negative.
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 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 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 = 17.5 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: α = 35.04 °
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.
Answer: Mx = 7.5 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: xc = 7.719 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, including the load on the overhang.
Between the supports M(x) = RA·x − w·x²/2. Apart from x = 0 it is zero at x = 2RA/w.
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: FAC = 4.444 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: δ = 48.46 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 = 6.42 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 = 39.69 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 = 2.763 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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