The A-frame shown stands on a pin at A and a roller at B, 2 m apart, with its apex pin C 4 m high and the horizontal tie DE 1.8 m above the ground. A vertical load of 20 kN acts on leg AC at K, 0.75 m (horizontally) from A. Find the tension T in the tie.
P = 20 kNAB = 2 mheight of C = 4 mheight of DE = 1.8 mhorizontal distance A to K = 0.75 m
T = kN
8.Frames & machinesHard
A uniform horizontal boom AB of weight 1.5 kN is pinned to a wall at A and held by a cable from its end B to the point C on the wall, 1 m above A. A load of 8 kN hangs at B. Find the tension T in the cable.
W = 8 kNWb = 1.5 kNAB = 4 mAC = 1 m
T = kN
9.Shear force & bending momentMedium
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.
25 kN @ 6 m8 kN/m, 0–12 m
1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)
10.Shear force & bending momentHard
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.
30 kN @ 0 m35 kN @ 12 m35 kN @ 4 m
1A (x = 0 m)
2B (x = 2 m)
3C (x = 4 m)
4D (x = 6 m)
5E (x = 8 m)
6F (x = 10 m)
7G (x = 12 m)
11.Centroid & second moment of areaMedium
A circular tube has outside diameter D = 60 mm and inside diameter d = 32 mm. Find its radius of gyration r about the horizontal centroidal axis.
D = 60 mmd = 32 mm
r = mm
12.Centroid & second moment of areaHard
A 200 mm × 150 mm plate has a rectangular cut-out 80 mm wide and 50 mm high, centred on the vertical centre line with its centre 97.5 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 200 mmh = 150 mmcut-out: 80 × 50 mmhole centre: y = 97.5 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: θ = 105.8 °
Dot product:a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.
Dot product
a·b = (−4)·(−2) + (−6)·5 + (−1)·(−6) = −16
Magnitudes
|a| = √((−4)² + (−6)² + (−1)²) = 7.28
|b| = √((−2)² + 5² + (−6)²) = 8.062
Angle
cos θ = −16/(7.28·8.062) = −0.2726
θ = arccos(−0.2726) = 105.8°
Answer
θ = 105.8°
2.Vectors: dot & cross product
Answer: n̂y = 0.8452
Cross product:a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.
Cross product
(a × b)x = aybz − azby = 2·0 − (−2)·1 = 2
(a × b)y = azbx − axbz = (−2)·5 − 4·0 = −10
(a × b)z = axby − aybx = 4·1 − 2·5 = −6
Normalise
The z-component of a × b is negative, so use −(a × b).
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.
Free-body diagram
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
ΣMA = 0: RB·8 = 15·3 + 35·4
RB = 185/8 = 23.13 kN
ΣFy = 0: RA = 50 − 23.13 = 26.88 kN
Answer
RA = 26.88 kN
4.Beam support reactions
Answer: RA = 24.33 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.
Free-body diagram
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Equilibrium
Moments about A, counter-clockwise positive. Then vertical force balance.
UDL resultant: W = w·ℓ = 12·4 = 48 kN, acting at x̄ = 4 m
ΣMA = 0: RB·6 = 50·5 + 12·4·4
RB = 442/6 = 73.67 kN
ΣFy = 0: RA = 98 − 73.67 = 24.33 kN
Answer
RA = 24.33 kN
5.Truss: method of joints
Answer: member DH
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.
Joint D
CD and DE lie on one straight line and the joint carries no load. Balancing forces across that line leaves only DH.
ΣF⊥ = 0 → FDH = 0
Check with the method of joints
Solving all joints gives these member forces (T = tension, C = compression).
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.
Joint G
FG and GH lie on one straight line and the joint carries no load. Balancing forces across that line leaves only CG.
ΣF⊥ = 0 → FCG = 0
Check with the method of joints
Solving all joints gives these member forces (T = tension, C = compression).
Frames and machines: unlike a truss, a frame contains at least one member that carries forces at more than two points, or a load between its pins. First look for two-force members: a member loaded only at two pins has its force along the line through them, so it adds one unknown, not two. Then use the whole structure for the support reactions when there are no more than three unknowns. Finally take the frame apart: draw each member separately, with the pin forces equal and opposite on the two members they connect, and write ΣFx = 0, ΣFy = 0 and ΣM = 0 for it. Taking moments about a pin removes both components of that pin force from the equation.
Whole frame: moments about A
By = 20·0.75/2 = 7.5 kN
Leg BC: moments about C
7.5·1 = T·2.2
T = 3.409 kN
Answer
T = 3.409 kN
8.Frames & machines
Answer: T = 36.08 kN
Frames and machines: unlike a truss, a frame contains at least one member that carries forces at more than two points, or a load between its pins. First look for two-force members: a member loaded only at two pins has its force along the line through them, so it adds one unknown, not two. Then use the whole structure for the support reactions when there are no more than three unknowns. Finally take the frame apart: draw each member separately, with the pin forces equal and opposite on the two members they connect, and write ΣFx = 0, ΣFy = 0 and ΣM = 0 for it. Taking moments about a pin removes both components of that pin force from the equation.
Cable angle
BC = √(4² + 1²) = 4.123 m, sin β = 0.2425
Moments about A
T·0.2425·4 = 8·4 + 1.5·2
T = 36.08 kN
Answer
T = 36.08 kN
9.Shear force & bending moment
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.
Support reactions
Take moments about one support, then use the vertical force balance.
R1 = 60.5 kN, R2 = 60.5 kN
Bending moment at every mark
Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.
MA = 0, MB = 105, MC = 178, MD = 219 kN·m, ME = 178, MF = 105, MG = 0
Answer
Mark D: MD = 219 kN·m
10.Shear force & bending moment
Answer: F (x = 10 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.
Support reactions
Take moments about one support, then use the vertical force balance.
R1 = 55 kN, R2 = 45 kN
Bending moment at every mark
Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.
MA = 0, MB = −60, MC = −10, MD = −30, ME = −50, MF = −70 kN·m, MG = 0
Answer
Mark F: MF = −70 kN·m
11.Centroid & second moment of area
Answer: r = 17 mm
Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = Ic + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.
Net area and second moment of area
A = π(D² − d²)/4 = 2023 mm²
I = π(D⁴ − d⁴)/64 = 584701 mm⁴
Radius of gyration
r = √(I/A) = √(584701/2023) = 17 mm
Answer
r = 17 mm
12.Centroid & second moment of area
Answer: Ix = 53.08 × 10⁶ mm⁴
Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has Ic = bh³/12 and a circle Ic = πd⁴/64.
Centroid
ȳ = (30000·75 − 4000·97.5)/(30000 − 4000) = 71.54 mm