Gears & gear trains
- Meshing gears
- n1·N1 = n2·N2
- Torque
- T2 = T1·(N2/N1)·η
- Power
- P = T·ω, ω = 2π·n/60
- Compound train
- total ratio = product of the stage ratios
Machine Design: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
Torsion of shafts
- Shear stress
- τ = T·r/J
- Solid shaft
- J = πd⁴/32
- Hollow shaft
- J = π(D⁴ − d⁴)/32
- Angle of twist
- φ = TL/(GJ)
- Power
- P = T·ω = 2πnT/60
Method
- Find the polar moment of area: J = πd⁴/32 for a solid shaft.
- Shear stress at the surface: τ = T·r/J, with r the outer radius.
- Angle of twist: φ = TL/(GJ) in radians.
- If power is given, find the torque first with P = T·ω, ω = 2πn/60.
Mohr’s circle & stress transformation
- Centre
- C = (σx + σy)/2
- Radius
- R = √(((σx − σy)/2)² + τxy²)
- Principal stresses
- σ1,2 = C ± R
- Principal direction
- tan 2θp = 2τxy/(σx − σy)
- Inclined plane
- σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ
Method
- Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
- Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
- The principal stresses are C ± R and the largest in-plane shear stress is R.
- Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).
Axial stress & strain
- Normal stress
- σ = F/A
- Strain
- ε = ΔL/L
- Hooke’s law
- σ = E·ε
- Elongation
- ΔL = FL/(EA)
- Thermal strain
- εT = α·ΔT
Column buckling
- Euler load
- Pcr = π²EI/(KL)²
- Effective length factor
- K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
- Critical stress
- σcr = Pcr/A
- Slenderness
- λ = KL/r, r = √(I/A)
Mechanical vibrations
- Natural frequency
- ωn = √(k/m), fn = ωn/(2π)
- From static deflection
- ωn = √(g/δst)
- Critical damping
- cc = 2√(km) = 2mωn
- Damping ratio
- ζ = c/cc
- Damped frequency
- ωd = ωn√(1 − ζ²)
- Logarithmic decrement
- δ = (1/n)·ln(x0/xn), ζ = δ/√(4π² + δ²)
- Springs in parallel
- k = k1 + k2
- Springs in series
- 1/k = 1/k1 + 1/k2
2Practice problems
1.Gears & gear trainsMedium
A belt runs on a 150 mm driving pulley at 1450 rpm. The tension is 500 N on the tight side and 200 N on the slack side. Find the power P transmitted, in kW.
d1 = 150 mmn1 = 1450 rpmT1 = 500 NT2 = 200 N
P = kW
2.Gears & gear trainsHard
An open belt connects an 80 mm pulley and a 200 mm pulley whose centres are 1200 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 80 mmd2 = 200 mmC = 1200 mm
θ = °
3.Torsion of shaftsMedium
A solid aluminium shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.
T = 100 N·md = 40 mmL = 3 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 40 mm is fixed at A. A torque of 250 N·m acts at B, 1 m from A, and a torque of 200 N·m in the opposite direction acts at the free end C, a further 0.8 m along. Find the magnitude of the angle of twist φC of end C, in degrees.
d = 40 mmTB = 250 N·mTC = 200 N·mAB = 1 mBC = 0.8 mG = 39 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 60 MPa, σᵧ = 80 MPa and τₓᵧ = 20 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 60 MPaσy = 80 MPaτxy = 20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 6 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 30° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 1000 mmt = 6 mmp = 2.5 MPaα = 30°
σn = MPa
7.Axial stress & strainMedium
Find the elongation δ of the brass bar under the axial load.
P = 120 kNd = 12 mmL = 1 mE = 100 GPa
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) and a segment made of aluminium (diameter 40 mm, length 500 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 500 mm, E = 70 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 200 GPa (steel)b × h = 25 × 50 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 3 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 40 kg mass is supported by a spring with a stiffness of 8 kN/m and a viscous damper with a damping coefficient of 60 N·s/m. Find the damping ratio ζ.
m = 40 kgk = 8 kN/mc = 60 N·s/m
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from spring k₃ = 5 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 60 kN/m and k₂ = 60 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
m = 5 kgk1 = 60 kN/mk2 = 60 kN/mk3 = 5 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 3.416 kW
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Belt speed
v = 11.39 m/s
Power
P = (500 − 200)·11.39 = 3416 W = 3.416 kW
Answer
P = 3.416 kW
2.Gears & gear trains
Answer: θ = 174.3 °
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Belt angle
asin(0.05) = 2.866°
Angle of wrap
θ = 180 − 2·2.866 = 174.3°
Answer
θ = 174.3°
3.Torsion of shafts
Answer: φ = 2.63 °
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Polar second moment of area
J = π·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 100·1000·3000/(26000·251327) = 0.04591 rad
φ = 0.04591·180/π = 2.63°
Answer
φ = 2.63°
4.Torsion of shafts
Answer: φC = 0.643 °
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Internal torques
Sense of TC taken as positive.
TBC = TC = 200 N·m
TAB = TC − TB = 200 − 250 = −50 N·m
Twist of each segment
J = π·40⁴/32 = 251327 mm⁴
G = 39 GPa = 39000 MPa
φAB = −50000·1000/(39000·251327) = −0.005101 rad
φBC = 200000·800/(39000·251327) = 0.01632 rad
Twist of end C
φC = −0.005101 + 0.01632 = 0.01122 rad → |φC| = 0.643°
Answer
φC = 0.643°
5.Mohr’s circle & stress transformation
Answer: τmax = 22.36 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Radius of Mohr’s circle
τmax = R = √(((60 − 80)/2)² + 20²) = 22.36 MPa
Answer
τmax = 22.36 MPa
6.Mohr’s circle & stress transformation
Answer: σn = 182.3 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Wall stresses
r = 1000/2 = 500 mm
σh = pr/t = 2.5·500/6 = 208.3 MPa
σa = pr/(2t) = 104.2 MPa
Transform to the seam
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
σn = 104.2·sin²30° + 208.3·cos²30° = 182.3 MPa
Answer
σn = 182.3 MPa
7.Axial stress & strain
Answer: δ = 10.61 mm
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
Stress
A = π·12²/4 = 113.1 mm²
σ = P/A = 120·1000/113.1 = 1061 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 1061/100000 = 0.01061 = 10610 µε
Elongation
δ = ε·L = 0.01061·1000 mm = 10.61 mm
Answer
δ = 10.61 mm
8.Axial stress & strain
Answer: σ2 = 22.26 MPa
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Stiffnesses
k₁ = 200000·490.9/300 = 327249 N/mm
k₂ = 70000·1257/500 = 175929 N/mm
Share of segment 2
F2 = 80000·175929/503178 = 27971 N
Stress
σ2 = 27971/1257 = 22.26 MPa (compression)
Answer
σ2 = 22.26 MPa
9.Column buckling
Answer: Pcr = 10.49 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Smallest second moment of area
A rectangle buckles about its weak axis, using the smaller dimension cubed.
Imin = h·b³/12 = 50·25³/12 = 65104 mm⁴
Effective length
KL = 0.7·5000 = 3500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·65104/3500² = 10491 N = 10.49 kN
Answer
Pcr = 10.49 kN
10.Column buckling
Answer: σcr = 30.71 MPa
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Smallest second moment of area
A round section has the same I about every axis.
I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 0.5·3000 = 1500 mm
KL/r = 1500/10 = 150
Critical stress
σcr = π²E/(KL/r)² = π²·70000/150² = 30.71 MPa
Answer
σcr = 30.71 MPa
11.Mechanical vibrations
Answer: ζ = 0.05303
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Critical damping
cc = 2·√(k·m) = 2·√(8000·40) = 1131 N·s/m
Damping ratio
ζ = c/cc = 60/1131 = 0.05303
Answer
ζ = 0.05303
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 4.931 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Equivalent stiffness
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 60 + 60 = 120 kN/m
keq = 120·5/(120 + 5) = 4.8 kN/m
Natural frequency
ωn = √(4800/5) = 30.98 rad/s
fn = 30.98/(2π) = 4.931 Hz
Answer
fn = 4.931 Hz
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