The bell crank shown turns about the pin O. A vertical force of 120 N at A (150 mm from O) is balanced by a horizontal force at B (60 mm above O). Find the magnitude of the pin reaction FO.
F = 120 NOA = 150 mmOB = 60 mm
FO = N
8.Frames & machinesHard
A uniform horizontal boom AB of weight 1 kN is pinned to a wall at A and held by a cable from B to the point C on the wall, 1.5 m above A. A load of 8 kN hangs at B. Find the magnitude of the pin reaction A.
W = 8 kNWb = 1 kNAB = 2.5 mAC = 1.5 m
A = kN
9.Shear force & bending momentMedium
The simply supported beam carries a clockwise couple M0 = 40 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.
L = 5 mM0 = 40 kN·m clockwise, at 2 m
Mx = kN·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.
40 kN @ 6 m50 kN @ 1 m12 kN/m, 0–5 m
1A (x = 0 m)
2B (x = 1 m)
3C (x = 2 m)
4D (x = 3 m)
5E (x = 4 m)
6F (x = 5 m)
7G (x = 6 m)
11.Centroid & second moment of areaMedium
A circular tube has outside diameter D = 210 mm and inside diameter d = 190 mm. Find its radius of gyration r about the horizontal centroidal axis.
D = 210 mmd = 190 mm
r = mm
12.Centroid & second moment of areaHard
The T-section is used as a beam with the flange on top and carries a sagging bending moment of 10 kN·m. Find the bending stress σbottom at the bottom edge.
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 equals zero
a·b = (−4)·(−3) + (−4)·k + 6·5 = 0
42 + (−4)·k = 0
Solve for k
k = −42/(−4) = 10.5
Answer
k = 10.5
2.Vectors: dot & cross product
Answer: A = 25.81 m²
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.
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.
Moments about A
Counter-clockwise positive.
ΣMA = 0: RB·2.5 = 40·2 + 45·4
RB = 260/2.5 = 104 kN
Vertical equilibrium
RA = 40 + 45 − 104 = −19 kN
Answer
RA = −19 kN
RA is negative: the pin at A has to hold the beam down.
4.Beam support reactions
Answer: RB = 61.13 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 = 6·7 = 42 kN, acting at x̄ = 4.5 m
ΣMA = 0: RB·8 = 50·6 + 6·7·4.5
RB = 489/8 = 61.13 kN
ΣFy = 0: RA = 92 − 61.13 = 30.88 kN
Answer
RB = 61.13 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 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).
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 = 120·150/60 = 300 N
Pin reaction
Ox = 300 N, Oy = 120 N
FO = √(300² + 120²) = 323.1 N
Answer
FO = 323.1 N
8.Frames & machines
Answer: A = 14.18 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 tension
T = (8 + 1/2)/0.5145 = 16.52 kN
Components at A
Ax = T·cos β = 14.17 kN
Ay = 8 + 1 − T·sin β = 0.5 kN
Magnitude
A = √(14.17² + 0.5²) = 14.18 kN
Answer
A = 14.18 kN
9.Shear force & bending moment
Answer: Mx = 24 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.
Support reactions
Moments about A, counter-clockwise positive: RB·L − M0 = 0. Vertical balance: RA + RB = 0.
RB = M0/L = 40/5 = 8 kN (upward)
RA = −RB = −8 kN (downward)
Cut just right of the couple
The left part now contains RA and the couple.
Mx = RA·a + M0 = −8·2 + 40 = 24 kN·m
Answer
Mx = 24 kN·m
10.Shear force & bending moment
Answer: B (x = 1 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 = 62 kN, R2 = 88 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 = 56 kN·m, MC = 50, MD = 32, ME = 2, MF = −40, MG = 0
Answer
Mark B: MB = 56 kN·m
11.Centroid & second moment of area
Answer: r = 70.8 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 = 6283 mm²
I = π(D⁴ − d⁴)/64 = 3.149 × 10⁷ mm⁴
Radius of gyration
r = √(I/A) = √(3.149 × 10⁷/6283) = 70.8 mm
Answer
r = 70.8 mm
12.Centroid & second moment of area
Answer: σbottom = 44.87 MPa
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 = 140·25 = 3500 mm², yf = 200 + 25/2 = 212.5 mm