Exactly one member of this truss carries no force under these loads. Which is the zero-force member? Click it in the diagram.
D: 35 kN ↓H: 30 kN ←
1member AB
2member BC
3member CD
4member DE
5member FG
6member GH
7member AF
8member HE
9member BF
10member CG
11member DH
12member BG
13member DG
7.Frames & machinesMedium
In the toggle clamp shown, two equal links AB and BC are pinned together at B, and each makes 30° with the line AC. A downward force of 300 N is applied at B. The slider C presses on a workpiece. Neglecting friction and weights, find the horizontal clamping force Q.
F = 300 Nθ = 30°
Q = N
8.Frames & machinesHard
A toggle press has two equal links AB and BC, each 80 mm long between pins. The joint B sits 15 mm above the line AC. Which downward force F at B produces a horizontal pressing force of 1500 N at the slider C? Neglect friction and weights.
Q = 1500 NAB = BC = 80 mmh = 15 mm
F = N
9.Shear force & bending momentMedium
Find the shear force Vx at the section x = 4.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
L = 6 mw = 8 kN/m from 0 to 6 mx = 4.5 m
Vx = kN
10.Shear force & bending momentHard
The simply supported beam carries a uniformly distributed load w over its whole span and a point load P. Find the maximum bending moment Mmax in the beam.
L = 4 mP = 65 kN at 2.5 mw = 4 kN/m from 0 to 4 m
Mmax = kN·m
11.Centroid & second moment of areaMedium
A 120 mm × 300 mm plate has a circular hole of diameter 50 mm on its vertical centre line, with its centre 120 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
b = 120 mmh = 300 mmd = 50 mmhole centre: y = 120 mm
Ix = × 10⁶ mm⁴
12.Centroid & second moment of areaHard
A symmetric I-section has flanges 160 mm wide and 14 mm thick, an overall depth of 220 mm and a web 12 mm thick. It carries a bending moment of 80 kN·m about the axis x. Find the size of the bending stress σj at the junction of the web and the top flange.
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)·5 + 4·(−1) + 0·(−6) = −34
Magnitudes
|a| = √((−6)² + 4² + 0²) = 7.211
|b| = √(5² + (−1)² + (−6)²) = 7.874
Angle
cos θ = −34/(7.211·7.874) = −0.5988
θ = arccos(−0.5988) = 126.8°
Answer
θ = 126.8°
2.Vectors: dot & cross product
Answer: V = 78
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 b × c
(b × c)x = bycz − bzcy = (−3)·(−5) − (−5)·5 = 40
(b × c)y = bzcx − bxcz = (−5)·3 − 5·(−5) = 10
(b × c)z = bxcy − bycx = 5·5 − (−3)·3 = 34
Dot product with a
a·(b × c) = 5·40 + (−2)·10 + (−3)·34 = 78
V = |78| = 78
Answer
V = 78
3.Beam support reactions
Answer: RB = 43.75 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 = 35·4 + 30·7
RB = 350/8 = 43.75 kN
ΣFy = 0: RA = 65 − 43.75 = 21.25 kN
Answer
RB = 43.75 kN
4.Beam support reactions
Answer: RB = 43.13 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 = 10·2 = 20 kN, acting at x̄ = 3 m
ΣMA = 0: RB·4 = 45·2.5 + 10·2·3
RB = 172.5/4 = 43.13 kN
ΣFy = 0: RA = 65 − 43.13 = 21.88 kN
Answer
RB = 43.13 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 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 30° = 300
Flink = 300 N
Clamping force
Q = Flink·cos 30° = 300/(2·tan 30°) = 259.8 N
Answer
Q = 259.8 N
The force ratio Q/F = 0.866 grows without limit as the links straighten.
8.Frames & machines
Answer: F = 572.7 N
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).
Link angle
sin θ = 15/80 = 0.1875, θ = 10.81°
Link force
Flink = Q/cos θ = 1527 N
Force at B
F = 2·Flink·sin θ = 2·1500·tan θ = 572.7 N
Answer
F = 572.7 N
9.Shear force & bending moment
Answer: Vx = −12 kN
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.
Support reactions
Equilibrium of the whole beam.
UDL resultant: W = w·ℓ = 8·6 = 48 kN, acting at x̄ = 3 m
ΣMA = 0: RB·6 = 8·6·3
RB = 144/6 = 24 kN
ΣFy = 0: RA = 48 − 24 = 24 kN
Cut at x = 4.5 m and keep the left part
Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.
V = RA − (all downward loads left of the cut)
Vx = 24 − 8·4.5 = −12 kN
Answer
Vx = −12 kN
10.Shear force & bending moment
Answer: Mmax = 68.44 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 = 4·4 = 16 kN, acting at x̄ = 2 m
ΣMA = 0: RB·4 = 65·2.5 + 4·4·2
RB = 194.5/4 = 48.63 kN
ΣFy = 0: RA = 81 − 48.63 = 32.38 kN
Where is V = 0?
V(2.5⁻) = RA − w·2.5 = 22.38 kN
V(2.5⁺) = 22.38 − 65 = −42.63 kN
The shear changes sign at the load, so the maximum is under the load.
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 = (36000·150 − 1963·120)/(36000 − 1963) = 151.7 mm
Full plate about ȳ
120·300³/12 + 36000·−1.731² = 270.1 × 10⁶ mm⁴
Hole about ȳ
π·50⁴/64 + 1963·−31.73² = 2.284 × 10⁶ mm⁴
Subtract
Ix = 267.8 × 10⁶ mm⁴
Answer
Ix = 267.8 × 10⁶ mm⁴
12.Centroid & second moment of area
Answer: σj = 140.5 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.
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.