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 125 mm pulley at 960 rpm drives a 500 mm pulley by a belt with 2 % slip. Find the speed n2 of the driven pulley.
n1 = 960 rpmd1 = 125 mmd2 = 500 mms = 2 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects an 80 mm pulley and a 450 mm pulley whose centres are 600 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 80 mmd2 = 450 mmC = 600 mm
θ = °
3.Torsion of shaftsMedium
A solid steel shaft 3 m long carries a torque of 1000 N·m. The angle of twist between its ends must not exceed 3°. Find the smallest diameter d that meets this limit.
T = 1000 N·mL = 3 mG = 80 GPaφallow = 3°
d = mm
4.Torsion of shaftsHard
A solid steel shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 1000 N·m is applied at C, 2.5 m from A and 0.5 m from B. Find the reaction torque TA at wall A.
d = 60 mmT0 = 1000 N·ma = 2.5 mb = 0.5 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 90 MPa and σᵧ = 50 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 100 MPa?
σx = 90 MPaσy = 50 MPaσ1 = 100 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −30 MPa, σᵧ = 80 MPa and τₓᵧ = 40 MPa (σz = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = −30 MPaσy = 80 MPaτxy = 40 MPaσY = 350 MPa
n =
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 75 kNd = 30 mmL = 0.5 mE = 100 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 10 kNsegment 1: Ø35 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø16 mm, 1.5 m, copper (E = 117 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)d = 30 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is pinned at both ends. It must carry 55 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 55 kNn = 3L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 2 cycles later by a peak of 14.6 mm. Find the damping ratio ζ.
x0 = 40 mmx2 = 14.6 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 12 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 5 kgk1 = 40 kN/mk2 = 12 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: n2 = 235.2 rpm
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.
Speed without slip
960·125/500 = 240 rpm
With slip
n2 = 240·0.98 = 235.2 rpm
Answer
n2 = 235.2 rpm
2.Gears & gear trains
Answer: θ = 144.1 °
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(0.3083) = 17.96°
Angle of wrap
θ = 180 − 2·17.96 = 144.1°
Answer
θ = 144.1°
3.Torsion of shafts
Answer: d = 51.97 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.
Required polar moment
T = 1000 N·m = 1 × 10⁶ N·mm
L = 3 m = 3000 mm
G = 80 GPa = 80000 MPa
φ = 3·π/180 = 0.05236 rad
J = TL/(Gφ) = 1 × 10⁶·3000/(80000·0.05236) = 716197 mm⁴
Diameter
d = (32J/π)1/4 = 51.97 mm
Answer
d = 51.97 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: TA = 166.7 N·m
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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).
Equilibrium and compatibility
TA + TB = 1000
TA·a = TB·b
TA = 1000·0.5/3 = 166.7 N·m
TB = 1000·2.5/3 = 833.3 N·m
Answer
TA = 166.7 N·m
5.Mohr’s circle & stress transformation
Answer: τxy = 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.
Required radius
C = (90 + 50)/2 = 70 MPa
R = σ1 − C = 100 − 70 = 30 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(30² − 20²) = 22.36 MPa
Answer
τxy = 22.36 MPa
6.Mohr’s circle & stress transformation
Answer: n = 2.573
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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = 25 MPa, R = 68.01 MPa
σ1 = 93.01 MPa, σ2 = −43.01 MPa, σ3 = 0
Tresca
σmax − σmin = 136 MPa
n = 350/136 = 2.573
Answer
n = 2.573
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: ε = 1061 µε
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·30²/4 = 706.9 mm²
σ = P/A = 75·1000/706.9 = 106.1 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 106.1/100000 = 0.001061 = 1061 µε
Answer
ε = 1061 µε
8.Axial stress & strain
Answer: δ = 0.7709 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·35²/4 = 962.1 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 10.39 MPa
σ2 = 49.74 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 10.39·1500/117000 = 0.1333 mm
δ2 = σ2L2/E2 = 49.74·1500/117000 = 0.6376 mm
δ = δ1 + δ2 = 0.7709 mm
Answer
δ = 0.7709 mm
9.Column buckling
Answer: Pcr = 50.23 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·30⁴/64 = 39761 mm⁴
Effective length
KL = 0.5·2500 = 1250 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·39761/1250² = 50230 N = 50.23 kN
Answer
Pcr = 50.23 kN
10.Column buckling
Answer: d = 57.12 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 = 3·55000 = 165000 N
KL = 1·2500 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·165000·2500²/(π³·200000))^(1/4) = 57.12 mm
Answer
d = 57.12 mm
Check: the slenderness KL/r is 175.1, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.07995
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 = (1/2)·ln(40/14.6) = 0.5039
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5039/√(39.48 + 0.2539) = 0.07995
Answer
ζ = 0.07995
12.Mechanical vibrations
Answer: keq = 52 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 = 40 + 12 = 52 kN/m
Answer
keq = 52 kN/m
ExamProblems · examproblems.com/courses/machine-design