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 1000 N on the tight side and 250 N on the slack side. Find the power P transmitted, in kW.
d1 = 150 mmn1 = 1450 rpmT1 = 1000 NT2 = 250 N
P = kW
2.Gears & gear trainsHard
A 7.5 kW motor at 1450 rpm drives gear A (32 teeth), which meshes with gear B (48 teeth). Gear C (45 teeth) on the same shaft as B drives gear D (60 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque Tout.
P = 7.5 kWnin = 1450 rpmA = 32B = 48C = 45D = 60ηmesh = 95%
Tout = N·m
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 = 1000 N·md = 50 mmL = 1.2 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 1500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 20 kWN = 1500 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −50 MPa, σᵧ = 0 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 = −50 MPaσy = 0 MPaτxy = −40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 25 mm carries a bending moment of 300 N·m together with a torque of 1000 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ1.
d = 25 mmM = 300 N·mT = 1000 N·m
σ1 = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 85 kNd = 16 mmL = 3 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 95 kNsegment 1: Ø25 mm, 0.5 m, copper (E = 117 GPa)segment 2: Ø16 mm, 0.5 m, aluminium (E = 70 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 200 GPa (steel)b × h = 30 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium 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 = 70 GPa (aluminium)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed one cycle later by a peak of 17.1 mm. Find the damping ratio ζ.
x0 = 25 mmx1 = 17.1 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 15 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 100 kgk1 = 25 kN/mk2 = 15 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: P = 8.541 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 = (1000 − 250)·11.39 = 8541 W = 8.541 kW
Answer
P = 8.541 kW
2.Gears & gear trains
Answer: Tout = 89.15 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.
Output speed
nout = 1450·(32/48)·(45/60) = 725 rpm
ωout = 2π·725/60 = 75.92 rad/s
Output power and torque
Pout = 7500·0.95² = 6769 W
Tout = Pout/ωout = 6769/75.92 = 89.15 N·m
Answer
Tout = 89.15 N·m
3.Torsion of shafts
Answer: φ = 4.31 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 1000·1000·1200/(26000·613592) = 0.07522 rad
φ = 0.07522·180/π = 4.31°
Answer
φ = 4.31°
4.Torsion of shafts
Answer: d = 20.09 mm
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.
Torque
ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s
T = P/ω = 20000/157.1 = 127.3 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·127324/(π·80)) = 20.09 mm
Answer
d = 20.09 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = −61 °
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, −50 − 0) = −122°
Angle on the element
θp = −61°
Answer
θp = −61°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σ1 = 438.1 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.
Stresses at the surface
σ = 32M/(πd³) = 32·300000/(π·25³) = 195.6 MPa
τ = 16T/(πd³) = 16·1000000/(π·25³) = 325.9 MPa
Mohr’s circle
C = σ/2 = 97.78 MPa
R = √(97.78² + 325.9²) = 340.3 MPa
σ1 = C + R = 438.1 MPa
σ2 = C − R = −242.5 MPa
Answer
σ1 = 438.1 MPa
7.Axial stress & strain
Answer: δ = 6.341 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·16²/4 = 201.1 mm²
σ = P/A = 85·1000/201.1 = 422.8 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 422.8/200000 = 0.002114 = 2114 µε
Elongation
δ = ε·L = 0.002114·3000 mm = 6.341 mm
Answer
δ = 6.341 mm
8.Axial stress & strain
Answer: δ = 4.202 mm
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·25²/4 = 490.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 193.5 MPa
σ2 = 472.5 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 193.5·500/117000 = 0.8271 mm
δ2 = σ2L2/E2 = 472.5·500/70000 = 3.375 mm
δ = δ1 + δ2 = 4.202 mm
Answer
δ = 4.202 mm
9.Column buckling
Answer: Pcr = 284.2 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 = 100·30³/12 = 225000 mm⁴
Effective length
KL = 0.5·2500 = 1250 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·225000/1250² = 284245 N = 284.2 kN
Answer
Pcr = 284.2 kN
10.Column buckling
Answer: σcr = 35.25 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.7·1500 = 1050 mm
KL/r = 1050/7.5 = 140
Critical stress
σcr = π²E/(KL/r)² = π²·70000/140² = 35.25 MPa
Answer
σcr = 35.25 MPa
11.Mechanical vibrations
Answer: ζ = 0.06034
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 = (1/1)·ln(25/17.1) = 0.3798
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3798/√(39.48 + 0.1442) = 0.06034
Answer
ζ = 0.06034
12.Mechanical vibrations
Answer: keq = 40 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 = 25 + 15 = 40 kN/m
Answer
keq = 40 kN/m
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