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 100 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 100 Nθ = 10°
Q = 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 1.5 m above the ground. A vertical load of 20 kN acts on leg AC at K, 0.6 m (horizontally) from A. Find the magnitude of the force C that the pin C exerts on leg BC.
P = 20 kNAB = 2 mheight of C = 4.5 mheight of DE = 1.5 mhorizontal distance A to K = 0.6 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.
35 kN @ 2 m60 kN @ 6 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 = 8 mw = 10 kN/m from 4 to 8 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
A hollow rectangular (box) section has outer width B = 240 mm, outer height H = 350 mm and a uniform wall thickness t = 12 mm. Find its radius of gyration r about the horizontal centroidal axis.
B = 240 mmH = 350 mmt = 12 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 5 kN·m. Find the size of the bending stress σtop at the top 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 = 2·k + 0·(−2) + 1·(−1) = 0
−1 + 2·k = 0
Solve for k
k = −(−1)/2 = 0.5
Answer
k = 0.5
2.Vectors: dot & cross product
Answer: (a × b)z = −16
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: RA = 31.88 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·8 = 55·4 + 35·7
RB = 465/8 = 58.13 kN
ΣFy = 0: RA = 90 − 58.13 = 31.88 kN
Answer
RA = 31.88 kN
4.Beam support reactions
Answer: RA = 11 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 = 5·1 = 5 kN, acting at x̄ = 1.5 m
ΣMA = 0: RB·5 = 15·2.5 + 5·1·1.5
RB = 45/5 = 9 kN
ΣFy = 0: RA = 20 − 9 = 11 kN
Answer
RA = 11 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 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).
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° = 100
Flink = 287.9 N
Clamping force
Q = Flink·cos 10° = 100/(2·tan 10°) = 283.6 N
Answer
Q = 283.6 N
The force ratio Q/F = 2.836 grows without limit as the links straighten.
8.Frames & machines
Answer: C = 6.325 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 = 20·0.6/2 = 6 kN
Tie force
T = 6·1/3 = 2 kN
Pin force on BC
Cx = 2 kN, Cy = −6 kN
C = √(2² + 6²) = 6.325 kN
Answer
C = 6.325 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 = 59.17 kN, R2 = 35.83 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 = 118.3, MC = 166.7, MD = 215 kN·m, ME = 143.3, MF = 71.67, MG = 0
Answer
Mark D: MD = 215 kN·m
10.Shear force & bending moment
Answer: Mmax = 45 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 = 10·4 = 40 kN, acting at x̄ = 6 m
ΣMA = 0: RB·8 = 10·4·6
RB = 240/8 = 30 kN
ΣFy = 0: RA = 40 − 30 = 10 kN
Zero shear
The shear force starts at RA, stays constant over any unloaded part and falls with slope w inside the load.
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
b = 216 mm, h = 326 mm
A = BH − bh = 240·350 − 216·326 = 13584 mm²
I = (BH³ − bh³)/12 = 2.339 × 10⁸ mm⁴
Radius of gyration
r = √(I/A) = √(2.339 × 10⁸/13584) = 131.2 mm
Answer
r = 131.2 mm
12.Centroid & second moment of area
Answer: σtop = 10.47 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 = 260·15 = 3900 mm², yf = 200 + 15/2 = 207.5 mm