The bell crank shown turns about the pin O. A vertical force of 250 N at A (150 mm from O) is balanced by a horizontal force at B (120 mm above O). Find the magnitude of the pin reaction FO.
F = 250 NOA = 150 mmOB = 120 mm
FO = N
8.Frames & machinesHard
The A-frame shown stands on a pin at A and a roller at B, 2 m apart, with its apex pin C 4.5 m high and the horizontal tie DE 1.8 m above the ground. A vertical load of 6 kN acts on leg AC at K, 0.25 m (horizontally) from A. Find the magnitude of the force C that the pin C exerts on leg BC.
P = 6 kNAB = 2 mheight of C = 4.5 mheight of DE = 1.8 mhorizontal distance A to K = 0.25 m
C = 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.
50 kN @ 8 m30 kN @ 10 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 simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment Mmax in the beam.
L = 10 mw = 20 kN/m from 0 to 7 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
Find the second moment of area Ix of the T-section about the horizontal axis x through its centroid.
bf = 120 mmtf = 20 mmtw = 20 mmhw = 200 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A welded I-section has a top flange 150 × 18 mm, a web 300 mm high and 10 mm thick, and a bottom flange 280 × 14 mm (width × thickness). Find its second moment of area Ix about the horizontal axis x through its centroid.
top flange: 150 × 18 mmweb: 10 × 300 mmbottom flange: 280 × 14 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: compb a = −0.4629
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 = 0·5 + (−5)·(−1) + 2·(−4) = −3
Length of b
|b| = √(5² + (−1)² + (−4)²) = 6.481
Scalar projection
compb a = a·b/|b| = −3/6.481 = −0.4629
Answer
compb a = −0.4629
Negative: a has a component pointing against b.
2.Vectors: dot & cross product
Answer: n̂y = −0.09667
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 = 4·2 − 1·3 = 5
(a × b)y = azbx − axbz = 1·3 − 1·2 = 1
(a × b)z = axby − aybx = 1·3 − 4·3 = −9
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·4 = 45·1 + 10·2
RB = 65/4 = 16.25 kN
ΣFy = 0: RA = 55 − 16.25 = 38.75 kN
Answer
RA = 38.75 kN
4.Beam support reactions
Answer: RB = 13.8 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 = 12·2 = 24 kN, acting at x̄ = 1 m
ΣMA = 0: RB·5 = 45·1 + 12·2·1
RB = 69/5 = 13.8 kN
ΣFy = 0: RA = 69 − 13.8 = 55.2 kN
Answer
RB = 13.8 kN
5.Truss: method of joints
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.
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).
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).
Machines: pliers, levers, linkages and clamps turn an input force into a larger or a differently directed output force. Draw one part as a free body and take moments about its pivot: the pin force drops out, and Fin·a = Fout·b gives the mechanical advantage a/b. The pin force then follows from ΣF = 0 on the same part. In a toggle, two links meet at a shallow angle θ; each link is a two-force member, so a small force F at the joint creates large link forces F/(2 sin θ) and a clamping force F/(2 tan θ).
Force at B
Q = 250·150/120 = 312.5 N
Pin reaction
Ox = 312.5 N, Oy = 250 N
FO = √(312.5² + 250²) = 400.2 N
Answer
FO = 400.2 N
8.Frames & machines
Answer: C = 0.7998 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.
Support reaction at B
By = 6·0.25/2 = 0.75 kN
Tie force
T = 0.75·1/2.7 = 0.2778 kN
Pin force on BC
Cx = 0.2778 kN, Cy = −0.75 kN
C = √(0.2778² + 0.75²) = 0.7998 kN
Answer
C = 0.7998 kN
9.Shear force & bending moment
Answer: E (x = 8 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 = 21.67 kN, R2 = 58.33 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 = 43.33, MC = 86.67, MD = 130, ME = 173.3 kN·m, MF = 116.7, MG = 0
Answer
Mark E: ME = 173.3 kN·m
10.Shear force & bending moment
Answer: Mmax = 207 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. 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.
Support reactions
Equilibrium of the whole beam.
UDL resultant: W = w·ℓ = 20·7 = 140 kN, acting at x̄ = 3.5 m
ΣMA = 0: RB·10 = 20·7·3.5
RB = 490/10 = 49 kN
ΣFy = 0: RA = 140 − 49 = 91 kN
Zero shear
The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.
Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.
Split into two rectangles
Measure every y from the bottom edge.
flange: Af = 120·20 = 2400 mm², yf = 200 + 20/2 = 210 mm
Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.
Split into three rectangles
Measure every y from the bottom edge.
top flange: A = 2700 mm², y = 323 mm
web: A = 3000 mm², y = 164 mm
bottom flange: A = 3920 mm², y = 7 mm
Centroid
ȳ = (2700·323 + 3000·164 + 3920·7)/9620 = 144.7 mm
Parallel-axis theorem
top flange: 150·18³/12 + 2700·178.3² = 8.596 × 10⁷ mm⁴