Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
The beam has a pin at A and a roller at B, 5 m from A, and overhangs B to the free end C. Find the vertical reaction RA, taking upward as positive.
The cantilever beam is fixed to the wall at A. It carries a linearly varying load that is 6 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.
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.
The simply supported beam carries a uniformly distributed load over part of its span. Find the distance x0 from A to the section where the bending moment is largest.
Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.
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 uniformly distributed load w over its whole span. Find the maximum deflection δmax, at midspan.
A simply supported beam carries a uniformly distributed load w. Its deflection may not exceed span/360. Find the smallest second moment of area Ireq that meets this limit.
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
A solid round column of length L made of steel is fixed at both ends. It must carry 65 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
Answer: RA = −3 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.
RA is negative: the pin at A has to hold the beam down.
Answer: MA = 259 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: Mx = 91 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: x0 = 2.1 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.
The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.
Answer: member BF
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.
AB and BC lie on one straight line and the joint carries no load. Balancing forces across that line leaves only BF.
Solving all joints gives these member forces (T = tension, C = compression).
Answer: member BF
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.
AB and BC lie on one straight line and the joint carries no load. Balancing forces across that line leaves only BF.
Solving all joints gives these member forces (T = tension, C = compression).
Answer: δmax = 7.706 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: Ireq = 1.808 × 10⁶ mm⁴
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).
Answer: Pcr = 197.7 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 = 53.81 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 185.8, so the column is slender and Euler applies.
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