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. It carries a point load P at 1 m from A and a clockwise couple M at 2.5 m from A. Find the vertical reaction RB, taking upward as positive.
A compound beam is fixed to the wall at A, has an internal hinge at C (3 m from A) and rests on a roller at D (8 m from A). It carries a point load P at 6 m from A, a uniformly distributed load w on the whole part CD and a point load Q at 1 m from A. Find the magnitude of the reaction moment MA at the wall.
The simply supported beam carries a clockwise couple M0 = 20 kN·m at x = 2 m from A, and no other load. Find the bending moment Mx just to the right of the couple. 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 2 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. Find the deflection δ at the free end.
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 fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
A solid round column of length L made of steel is pinned at both ends. It must carry 65 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
Answer: RB = 28.75 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.
A negative reaction acts downward.
Answer: MA = 162 kN·m
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. An internal hinge cannot carry a bending moment, which gives one extra equation: the moment about the hinge of the forces on either side is zero.
The hinge has no moment, so moments about C on CD alone must balance.
The hinge force VC acts downward on AC (equal and opposite to its action on CD).
Answer: Mx = 15 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. An applied couple makes the bending moment diagram jump by the size of the couple.
Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.
The left part now contains RA and the couple.
Answer: xc = 3 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.
FG and GH 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 = 17.78 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 = 41.03 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 = 15 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 = 9.922 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: d = 41.68 mm
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.
Check: the slenderness KL/r is 143.9, so the column is slender and Euler applies.
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