In the toggle clamp shown, two equal links AB and BC are pinned together at B, and each makes 10° with the line AC. A downward force of 400 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 400 Nθ = 10°
Q = N
8.Frames & machinesHard
The frame shown has pins at A, B and C; A and C are fixed to the ground 6 m apart, and B is 2 m high and 1.8 m to the right of A. Member AB carries a vertical load of 4 kN at K, 1.35 m (horizontally) from A, and a horizontal load of 8 kN to the right acts at B on member AB. Find the size of the horizontal reaction Ax at A.
P = 4 kNQ = 8 kNL = 6 mH = 2 mhorizontal distance A to B = 1.8 mhorizontal distance A to K = 1.35 m
Ax = 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 @ 3 m15 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)
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 = 6 mw = 12 kN/m from 0 to 5 m
x0 = m
11.Centroid & second moment of areaMedium
A 120 mm × 240 mm plate has a circular hole of diameter 30 mm on its vertical centre line, with its centre 168 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 120 mmh = 240 mmd = 30 mmhole centre: y = 168 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A 120 mm × 240 mm plate has a rectangular cut-out 80 mm wide and 20 mm high, centred on the vertical centre line with its centre 156 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 120 mmh = 240 mmcut-out: 80 × 20 mmhole centre: y = 156 mm
Ix = × 10⁶ mm⁴
3Solutions
1.Vectors: dot & cross product
Answer: compb a = −1.249
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 = (−6)·0 + (−4)·(−5) + (−7)·4 = −8
Length of b
|b| = √(0² + (−5)² + 4²) = 6.403
Scalar projection
compb a = a·b/|b| = −8/6.403 = −1.249
Answer
compb a = −1.249
Negative: a has a component pointing against b.
2.Vectors: dot & cross product
Answer: A = 34.93
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 = 5·4 − 5·(−2) = 30
(a × b)y = azbx − axbz = 5·0 − (−4)·4 = 16
(a × b)z = axby − aybx = (−4)·(−2) − 5·0 = 8
Area
A = |a × b| = √(30² + 16² + 8²) = 34.93
Answer
A = 34.93
3.Beam support reactions
Answer: RB = 47.5 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·10 = 35·1 + 55·8
RB = 475/10 = 47.5 kN
ΣFy = 0: RA = 90 − 47.5 = 42.5 kN
Answer
RB = 47.5 kN
4.Beam support reactions
Answer: RB = 36.2 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 = 8·2 = 16 kN, acting at x̄ = 1 m
ΣMA = 0: RB·5 = 55·3 + 8·2·1
RB = 181/5 = 36.2 kN
ΣFy = 0: RA = 71 − 36.2 = 34.8 kN
Answer
RB = 36.2 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 C
BC and CD 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).
Equilibrium of pin B
2·Flink·sin 10° = 400
Flink = 1152 N
Clamping force
Q = Flink·cos 10° = 400/(2·tan 10°) = 1134 N
Answer
Q = 1134 N
The force ratio Q/F = 2.836 grows without limit as the links straighten.
8.Frames & machines
Answer: Ax = 0.51 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.
Direction of BC
BC = 4.652 m, cos = 0.9029, sin = 0.4299
Whole frame: moments about A
FBC·0.4299·6 = 4·1.35 + 8·2
FBC = 8.296 kN
Horizontal balance
Ax + 8 − 8.296·0.9029 = 0
Ax = −0.51 kN → |Ax| = 0.51 kN
Answer
Ax = 0.51 kN
The minus sign means Ax points to the left.
9.Shear force & bending moment
Answer: D (x = 3 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 = 68.75 kN, R2 = 56.25 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 = 61.25, MC = 107.5, MD = 138.8 kN·m, ME = 105, MF = 56.25, MG = 0
Answer
Mark D: MD = 138.8 kN·m
10.Shear force & bending moment
Answer: x0 = 2.917 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 = 12·5 = 60 kN, acting at x̄ = 2.5 m
ΣMA = 0: RB·6 = 12·5·2.5
RB = 150/6 = 25 kN
ΣFy = 0: RA = 60 − 25 = 35 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 + 35/12 = 2.917 m
Answer
x0 = 2.917 m
11.Centroid & second moment of area
Answer: Ix = 136.5 × 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 = (28800·120 − 706.9·168)/(28800 − 706.9) = 118.8 mm
Full plate about ȳ
120·240³/12 + 28800·1.208² = 138.3 × 10⁶ mm⁴
Hole about ȳ
π·30⁴/64 + 706.9·49.21² = 1.751 × 10⁶ mm⁴
Subtract
Ix = 136.5 × 10⁶ mm⁴
Answer
Ix = 136.5 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: Ix = 136 × 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 = (28800·120 − 1600·156)/(28800 − 1600) = 117.9 mm