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 15 kW motor running at 2900 rpm drives a pinion with 15 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
P = 15 kWn1 = 2900 rpmN1 = 15N2 = 60η = 95%
T1 = N·m
2.Gears & gear trainsHard
An open belt connects a 120 mm pulley and a 500 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 120 mmd2 = 500 mmC = 800 mm
θ = °
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 10 mm is twisted by a torque of 30 N·m. The two ends of a gauge length of 300 mm rotate 6.57° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 30 N·md = 10 mmL = 300 mmφ = 6.57°
G = GPa
4.Torsion of shaftsHard
A solid steel shaft of diameter 60 mm is fixed at A. A torque of 200 N·m acts at B, 2 m from A, and a torque of 2000 N·m in the opposite direction acts at the free end C, a further 1.2 m along. Find the magnitude of the angle of twist φC of end C, in degrees.
d = 60 mmTB = 200 N·mTC = 2000 N·mAB = 2 mBC = 1.2 mG = 80 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −40 MPa, σᵧ = 70 MPa and τₓᵧ = −40 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = −40 MPaσy = 70 MPaτxy = −40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 400 mm and wall thickness 5 mm holds gas at a gauge pressure of 1 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 400 mmt = 5 mmp = 1 MPaα = 55°
σn = MPa
7.Axial stress & strainMedium
Find the elongation δ of the brass bar under the axial load.
P = 80 kNd = 12 mmL = 2.5 mE = 100 GPa
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
P = 45 kNsegment 1: Ø40 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 1.5 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 70 GPa (aluminium)b × h = 25 × 60 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1.5 mE = 200 GPa (steel)d = 20 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 5 cycles later by a peak of 19.5 mm. Find the damping ratio ζ.
x0 = 50 mmx5 = 19.5 mmn = 5
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 80 kN/m and k₂ = 12 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 20 kgk1 = 80 kN/mk2 = 12 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 49.39 N·m
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Motor torque
ω₁ = 2π·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 15000/303.7 = 49.39 N·m
Answer
T1 = 49.39 N·m
2.Gears & gear trains
Answer: θ = 152.5 °
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(0.2375) = 13.74°
Angle of wrap
θ = 180 − 2·13.74 = 152.5°
Answer
θ = 152.5°
3.Torsion of shafts
Answer: G = 79.95 GPa
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 moment and angle
J = π·10⁴/32 = 981.7 mm⁴
φ = 6.57·π/180 = 0.1147 rad
Shear modulus
T = 30 N·m = 30000 N·mm
G = TL/(Jφ) = 30000·300/(981.7·0.1147) = 79947 MPa
G = 79.95 GPa
Answer
G = 79.95 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: φC = 3.377 °
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 = 2000 N·m
TAB = TC − TB = 2000 − 200 = 1800 N·m
Twist of each segment
J = π·60⁴/32 = 1.272 × 10⁶ mm⁴
G = 80 GPa = 80000 MPa
φAB = 1.8 × 10⁶·2000/(80000·1.272 × 10⁶) = 0.03537 rad
φBC = 2 × 10⁶·1200/(80000·1.272 × 10⁶) = 0.02358 rad
Twist of end C
φC = 0.03537 + 0.02358 = 0.05895 rad → |φC| = 3.377°
Answer
φC = 3.377°
5.Mohr’s circle & stress transformation
Answer: θp = −71.99 °
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.
Angle on the circle
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·−40, −40 − 70) = −144°
Angle on the element
θp = −71.99°
Answer
θp = −71.99°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σn = 26.58 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 = 400/2 = 200 mm
σh = pr/t = 1·200/5 = 40 MPa
σa = pr/(2t) = 20 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 = 20·sin²55° + 40·cos²55° = 26.58 MPa
Answer
σn = 26.58 MPa
7.Axial stress & strain
Answer: δ = 17.68 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 = 80·1000/113.1 = 707.4 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 707.4/100000 = 0.007074 = 7074 µε
Elongation
δ = ε·L = 0.007074·2500 mm = 17.68 mm
Answer
δ = 17.68 mm
8.Axial stress & strain
Answer: σmax = 397.9 MPa
Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.
Same force in both segments
The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.
A1 = π·40²/4 = 1257 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 35.81 MPa
σ2 = 397.9 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 397.9 MPa
Answer
σmax = 397.9 MPa
9.Column buckling
Answer: Pcr = 23.99 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 = 60·25³/12 = 78125 mm⁴
Effective length
KL = 1·1500 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·78125/1500² = 23989 N = 23.99 kN
Answer
Pcr = 23.99 kN
10.Column buckling
Answer: σcr = 44.76 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·20⁴/64 = 7854 mm⁴
Radius of gyration and slenderness
A = 314.2 mm²
r = √(I/A) = √(7854/314.2) = 5 mm
KL = 0.7·1500 = 1050 mm
KL/r = 1050/5 = 210
Critical stress
σcr = π²E/(KL/r)² = π²·200000/210² = 44.76 MPa
Answer
σcr = 44.76 MPa
11.Mechanical vibrations
Answer: ζ = 0.02996
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.
Logarithmic decrement
δ = (1/5)·ln(50/19.5) = 0.1883
Damping ratio
ζ = δ/√(4π² + δ²) = 0.1883/√(39.48 + 0.03547) = 0.02996
Answer
ζ = 0.02996
12.Mechanical vibrations
Answer: keq = 92 kN/m
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
Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 80 + 12 = 92 kN/m
Answer
keq = 92 kN/m
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