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
An 11 kW motor running at 1500 rpm drives a pinion with 12 teeth, which meshes with a gear with 72 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
P = 11 kWn1 = 1500 rpmN1 = 12N2 = 72η = 96%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turning at 1500 rpm drives gear B (32 teeth). Gear C (45 teeth) is fixed to the same shaft as B and drives gear D (48 teeth) on the output shaft. Find the output speed nout.
nin = 1500 rpmA = 12B = 32C = 45D = 48
nout = rpm
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 = 50 mmL = 1 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft transmits 10 kW at 750 rpm. The shear stress must not exceed 80 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.
P = 10 kWN = 750 rpmτallow = 80 MPaθallow = 0.5°/mG = 80 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a thin plate is in plane stress: σₓ = 110 MPa, σᵧ = 60 MPa and τₓᵧ = −60 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).
σx = 110 MPaσy = 60 MPaτxy = −60 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 160 MPa, σᵧ = −70 MPa and τₓᵧ = −40 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 160 MPaσy = −70 MPaτxy = −40 MPaθ = 35°
τθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 30 mm, length 200 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: brass, Ø30 mm, 200 mm, E = 100 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 200 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: brass, Ø30 mm, 200 mm, E = 100 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 1 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 20 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 20 kNn = 3L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 4 cycles later by a peak of 4 mm. Find the damping ratio ζ.
x0 = 50 mmx4 = 4 mmn = 4
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 = 60 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 = 60 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 70.03 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·1500/60 = 157.1 rad/s
T1 = P/ω₁ = 11000/157.1 = 70.03 N·m
Answer
T1 = 70.03 N·m
2.Gears & gear trains
Answer: nout = 527.3 rpm
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.
First stage
nB = 1500·12/32 = 562.5 rpm = nC
Second stage
nout = 562.5·45/48 = 527.3 rpm
Answer
nout = 527.3 rpm
total ratio = 2.844 : 1
3.Torsion of shafts
Answer: φ = 0.3591 °
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) = 100·1000·1000/(26000·613592) = 0.006268 rad
φ = 0.006268·180/π = 0.3591°
Answer
φ = 0.3591°
4.Torsion of shafts
Answer: d = 36.92 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·750/60 = 78.54 rad/s
T = P/ω = 10000/78.54 = 127.3 N·m
Diameter for strength
T = 127.3 N·m = 127324 N·mm
dτ = ∛(16·127324/(π·80)) = 20.09 mm
Diameter for stiffness
θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm
G = 80 GPa = 80000 MPa
dφ = (32·127324/(π·80000·8.727 × 10⁻⁶))1/4 = 36.92 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(20.09, 36.92) = 36.92 mm
Answer
d = 36.92 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: τabs = 75 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.
In-plane principal stresses
C = 85 MPa, R = √(25² + (−60)²) = 65 MPa
σ1 = 150 MPa, σ2 = 20 MPa, σz = 0
Largest of the three circles
τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(130, 150, 20)/2 = 75 MPa
Answer
τabs = 75 MPa
Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.
6.Mohr’s circle & stress transformation
Answer: τθ = −121.7 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
τθ = −115·0.9397 + −40·0.342 = −121.7 MPa
Answer
τθ = −121.7 MPa
7.Axial stress & strain
Answer: P = 67.97 kN
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
A₁ = 490.9 mm², A₂ = 706.9 mm²
Flexibilities
L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(100000·706.9) = 2.829 × 10⁻⁶ mm/N
Load
P = 0.4/5.885 × 10⁻⁶ = 67967 N = 67.97 kN
Answer
P = 67.97 kN
8.Axial stress & strain
Answer: σ2 = 27.28 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 = 200000·490.9/500 = 196350 N/mm
k₂ = 100000·706.9/200 = 353429 N/mm
Share of segment 2
F2 = 30000·353429/549779 = 19286 N
Stress
σ2 = 19286/706.9 = 27.28 MPa (compression)
Answer
σ2 = 27.28 MPa
9.Column buckling
Answer: Pcr = 248.1 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·40⁴/64 = 125664 mm⁴
Effective length
KL = 1·1000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·125664/1000² = 248050 N = 248.1 kN
Answer
Pcr = 248.1 kN
10.Column buckling
Answer: d = 61.03 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 = 3·20000 = 60000 N
KL = 0.7·4000 = 2800 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·60000·2800²/(π³·70000))^(1/4) = 61.03 mm
Answer
d = 61.03 mm
Check: the slenderness KL/r is 183.5, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.09999
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 = (1/4)·ln(50/4) = 0.6314
Damping ratio
ζ = δ/√(4π² + δ²) = 0.6314/√(39.48 + 0.3987) = 0.09999
Answer
ζ = 0.09999
12.Mechanical vibrations
Answer: fn = 6.366 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 + 60 = 80 kN/m
Natural frequency
ωn = √(80000/50) = 40 rad/s
fn = 40/(2π) = 6.366 Hz
Answer
fn = 6.366 Hz
ExamProblems · examproblems.com/courses/machine-design