The bell crank shown turns about the pin O. A vertical force of 50 N at A (150 mm from O) is balanced by a horizontal force at B (100 mm above O). Find the magnitude of the pin reaction FO.
F = 50 NOA = 150 mmOB = 100 mm
FO = N
8.Frames & machinesHard
A uniform horizontal boom AB of weight 2 kN is pinned to a wall at A and held by a cable from B to the point C on the wall, 1 m above A. A load of 8 kN hangs at B. Find the magnitude of the pin reaction A.
W = 8 kNWb = 2 kNAB = 6 mAC = 1 m
A = kN
9.Shear force & bending momentMedium
The cantilever is fixed at A. It carries a uniformly distributed load w over its whole length and a point load P at 1 m from A. Find the bending moment MA at the fixed support A. Use sagging moments positive and upward shear on the left face positive.
L = 4 mP = 10 kN at 1 mw = 10 kN/m from 0 to 4 m
MA = kN·m
10.Shear force & bending momentHard
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.
L = 4 mw = 6 kN/m from 0 to 3 m
x0 = m
11.Centroid & second moment of areaMedium
A symmetric I-section has flanges 110 mm wide and 20 mm thick, an overall depth of 200 mm and a web 8 mm thick. Find its elastic section modulus Sx for bending about the horizontal centroidal axis x.
B = 110 mmH = 200 mmtf = 20 mmtw = 8 mm
Sx = × 10³ mm³
12.Centroid & second moment of areaHard
A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 35 MPa in tension and 90 MPa in compression. Find the largest hogging bending moment Mallow the beam can carry (give its size).
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.
Negative: the angle between the vectors is more than 90°.
2.Vectors: dot & cross product
Answer: (a × b)z = 12
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.
Check: a × b is perpendicular to both vectors, so its dot product with a and with b is zero.
3.Beam support reactions
Answer: RB = 45 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.
ΣMA = 0: RB·4 = 60·1 + 60·2
RB = 180/4 = 45 kN
ΣFy = 0: RA = 120 − 45 = 75 kN
Answer
RB = 45 kN
4.Beam support reactions
Answer: RA = 35 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 = 5·2 = 10 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 40·1 + 5·2·2
RB = 60/4 = 15 kN
ΣFy = 0: RA = 50 − 15 = 35 kN
Answer
RA = 35 kN
5.Truss: method of joints
Answer: FAC = 9.722 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.
Support reactions
Equilibrium of the whole truss. Moments about A, counter-clockwise positive.
ΣMA = 0: Cy·6 = 35·1 → Cy = 5.833 kN
ΣFy = 0: Ay = 35 − 5.833 = 29.17 kN
Joint C
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
sin γ = 3/5.831 = 0.5145, cos γ = 5/5.831 = 0.8575
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 = 50·150/100 = 75 N
Pin reaction
Ox = 75 N, Oy = 50 N
FO = √(75² + 50²) = 90.14 N
Answer
FO = 90.14 N
8.Frames & machines
Answer: A = 54.01 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 + 2/2)/0.1644 = 54.74 kN
Components at A
Ax = T·cos β = 54 kN
Ay = 8 + 2 − T·sin β = 1 kN
Magnitude
A = √(54² + 1²) = 54.01 kN
Answer
A = 54.01 kN
9.Shear force & bending moment
Answer: MA = −90 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.
Keep the free-end part
P: −10·1 = −10 kN·m
w: −10·4²/2 = −80 kN·m
MA = −10 − 80 = −90 kN·m
Answer
MA = −90 kN·m
Negative means hogging: tension on the top face.
10.Shear force & bending moment
Answer: x0 = 1.875 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 = 6·3 = 18 kN, acting at x̄ = 1.5 m
ΣMA = 0: RB·4 = 6·3·1.5
RB = 27/4 = 6.75 kN
ΣFy = 0: RA = 18 − 6.75 = 11.25 kN
Zero shear
The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.
RA − w·(x0 − 0) = 0
x0 = 0 + 11.25/6 = 1.875 m
Answer
x0 = 1.875 m
11.Centroid & second moment of area
Answer: Sx = 385.2 × 10³ mm³
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.
Outer rectangle minus the two side cut-outs
Both cut-outs together are (B − tw) wide and hw = H − 2tf high, and share the axis x.
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.