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 2 m and a point load P at 3.5 m from the wall. Find the vertical reaction RA at the wall.
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RA.
Find the bending moment Mx at the section x = 4.5 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 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).
Use the method of joints to find the force FBC in member BC. 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 simply supported beam carries a point load P at a = 1 m from support A. Find the deflection δP under the load.
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 steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
Answer: RA = 52 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: RA = 41.88 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: Mx = 10.13 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 = 5.833 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: FBC = −15.93 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: 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: δP = 1.667 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: RB = 13.5 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 = 8.721 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 = 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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