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 2.2 kW motor running at 1450 rpm drives a pinion with 20 teeth, which meshes with a gear with 24 teeth. Ignore losses. Find the torque T1 on the motor shaft.
P = 2.2 kWn1 = 1450 rpmN1 = 20N2 = 24η = 100%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (24 teeth) on the input shaft turns at 960 rpm and drives gear B (80 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 57.6 rpm. How many teeth ND must gear D have?
nin = 960 rpmA = 24B = 80C = 12nout = 57.6 rpm
ND =
3.Torsion of shaftsMedium
A solid steel 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 = 30 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 100 kW at 3000 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.
P = 100 kWN = 3000 rpmdi/do = 0.6τallow = 60 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −50 MPa, σᵧ = −40 MPa and τₓᵧ = −10 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = −50 MPaσy = −40 MPaτxy = −10 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −40 MPa, σᵧ = −10 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.
σx = −40 MPaσy = −10 MPaτxy = −10 MPaθ = 35°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 50 mm, E = 70 GPa). Both are 500 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.
P = 80 kNrod: Ø30 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 70 GPaL = 500 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) and a segment made of brass (diameter 60 mm, length 300 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø20 mm, 300 mm, E = 200 GPasegment 2: brass, Ø60 mm, 300 mm, E = 100 GPaP = 40 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1 mE = 200 GPa (steel)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is pinned at both ends. It must carry 50 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 50 kNn = 3L = 6 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 20 mm is followed one cycle later by a peak of 13.7 mm. Find the damping ratio ζ.
x0 = 20 mmx1 = 13.7 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 20 kN/m and k₂ = 5 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 50 kgk1 = 20 kN/mk2 = 5 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 14.49 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·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 2200/151.8 = 14.49 N·m
Answer
T1 = 14.49 N·m
2.Gears & gear trains
Answer: ND = 60
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.
Intermediate shaft speed
nB = nC = 960·24/80 = 288 rpm
Teeth on gear D
ND = C·nC/nout = 12·288/57.6 = 60
Answer
ND = 60
3.Torsion of shafts
Answer: φ = 4.503 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 1000·1000·500/(80000·79522) = 0.0786 rad
φ = 0.0786·180/π = 4.503°
Answer
φ = 4.503°
4.Torsion of shafts
Answer: do = 31.43 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·3000/60 = 314.2 rad/s
T = P/ω = 100000/314.2 = 318.3 N·m
Outer diameter from the allowable stress
T = 318.3 N·m = 318310 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·318310/(π·60·0.8704)) = 31.43 mm
di = 0.6·31.43 = 18.86 mm
Answer
do = 31.43 mm
A solid shaft would need d = 30.01 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: θs = 76.72 °
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
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = −10/11.18 = −0.8944, sin 2θs = −(−5)/11.18 = 0.4472
2θs = atan2(5, −10) = 153.4°
Angle on the element
θs = 76.72°
check: θs − θp = 76.72 − (−58.28) = 135°
Answer
θs = 76.72°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = −39.53 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.
Transformation equation
σθ = −25 + −15·0.342 + −10·0.9397 = −39.53 MPa
Answer
σθ = −39.53 MPa
7.Axial stress & strain
Answer: δ = 0.1124 mm
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.
Areas
Ar = π·30²/4 = 706.9 mm²
At = π(80² − 50²)/4 = 3063 mm²
Combined axial stiffness
ΣEA = 200·10³·706.9 + 70·10³·3063 = 3.558 × 10⁸ N
Shortening
δ = 80000·500/3.558 × 10⁸ = 0.1124 mm
Answer
δ = 0.1124 mm
8.Axial stress & strain
Answer: σ2 = 11.57 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 = 200000·314.2/300 = 209440 N/mm
k₂ = 100000·2827/300 = 942478 N/mm
Share of segment 2
F2 = 40000·942478/1.152 × 10⁶ = 32727 N
Stress
σ2 = 32727/2827 = 11.57 MPa (compression)
Answer
σ2 = 11.57 MPa
9.Column buckling
Answer: Pcr = 313.9 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 round section has the same I about every axis.
I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴
Effective length
KL = 2·1000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·636173/2000² = 313939 N = 313.9 kN
Answer
Pcr = 313.9 kN
10.Column buckling
Answer: d = 86.4 mm
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.
Required critical load
Pcr = n·P = 3·50000 = 150000 N
KL = 1·6000 = 6000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·150000·6000²/(π³·200000))^(1/4) = 86.4 mm
Answer
d = 86.4 mm
Check: the slenderness KL/r is 277.8, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.06011
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(20/13.7) = 0.3783
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3783/√(39.48 + 0.1431) = 0.06011
Answer
ζ = 0.06011
12.Mechanical vibrations
Answer: fn = 3.559 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
Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 20 + 5 = 25 kN/m
Natural frequency
ωn = √(25000/50) = 22.36 rad/s
fn = 22.36/(2π) = 3.559 Hz
Answer
fn = 3.559 Hz
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