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 200 mm driving pulley at 720 rpm. The tension is 500 N on the tight side and 200 N on the slack side. Find the power P transmitted, in kW.
d1 = 200 mmn1 = 720 rpmT1 = 500 NT2 = 200 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 120 mm pulley and a 400 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 120 mmd2 = 400 mmC = 500 mm
θ = °
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 = 250 N·md = 50 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 50 kW at 1800 rpm. Find the maximum shear stress τmax in the shaft.
P = 50 kWN = 1800 rpmdo = 80 mmdi = 64 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 120 MPa, σᵧ = −10 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 120 MPaσy = −10 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 1500 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 30° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 1500 mmt = 5 mmp = 1 MPaα = 30°
σn = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 30 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 50 mm, length 300 mm, E = 120 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, Ø30 mm, 600 mm, E = 200 GPasegment 2: copper, Ø50 mm, 300 mm, E = 120 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 60 mm, length 400 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: brass, Ø60 mm, 400 mm, E = 100 GPaP = 20 kN
σ2 = MPa
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 = 5 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
An 80 kg mass on a spring with a stiffness of 15 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency fd of its free vibration in Hz.
m = 80 kgk = 15 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 80 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 5 kN/m and k₂ = 30 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
m = 10 kgk1 = 5 kN/mk2 = 30 kN/mk3 = 80 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 2.262 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 = 7.54 m/s
Power
P = (500 − 200)·7.54 = 2262 W = 2.262 kW
Answer
P = 2.262 kW
2.Gears & gear trains
Answer: θ = 147.5 °
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(0.28) = 16.26°
Angle of wrap
θ = 180 − 2·16.26 = 147.5°
Answer
θ = 147.5°
3.Torsion of shafts
Answer: φ = 0.7295 °
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) = 250·1000·2500/(80000·613592) = 0.01273 rad
φ = 0.01273·180/π = 0.7295°
Answer
φ = 0.7295°
4.Torsion of shafts
Answer: τmax = 4.469 MPa
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·1800/60 = 188.5 rad/s
T = P/ω = 50000/188.5 = 265.3 N·m
Polar second moment of area
J = π(80⁴ − 64⁴)/32 = 2.374 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 265258·40/2.374 × 10⁶ = 4.469 MPa
Answer
τmax = 4.469 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 71.59 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.
Radius of Mohr’s circle
τmax = R = √(((120 − −10)/2)² + 30²) = 71.59 MPa
Answer
τmax = 71.59 MPa
6.Mohr’s circle & stress transformation
Answer: σn = 131.3 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 = 1500/2 = 750 mm
σh = pr/t = 1·750/5 = 150 MPa
σa = pr/(2t) = 75 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 = 75·sin²30° + 150·cos²30° = 131.3 MPa
Answer
σn = 131.3 MPa
7.Axial stress & strain
Answer: P = 72.5 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 = 706.9 mm², A₂ = 1963 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·706.9) = 4.244 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(120000·1963) = 1.273 × 10⁻⁶ mm/N
Load
P = 0.4/5.517 × 10⁻⁶ = 72498 N = 72.5 kN
Answer
P = 72.5 kN
8.Axial stress & strain
Answer: σ2 = 4.716 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 = 200000·706.9/400 = 353429 N/mm
k₂ = 100000·2827/400 = 706858 N/mm
Share of segment 2
F2 = 20000·706858/1.06 × 10⁶ = 13333 N
Stress
σ2 = 13333/2827 = 4.716 MPa (compression)
Answer
σ2 = 4.716 MPa
9.Column buckling
Answer: Pcr = 635 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Effective length
KL = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/2500² = 635009 N = 635 kN
Answer
Pcr = 635 kN
10.Column buckling
Answer: σcr = 1.687 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·50⁴/64 = 306796 mm⁴
Radius of gyration and slenderness
A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 2·4000 = 8000 mm
KL/r = 8000/12.5 = 640
Critical stress
σcr = π²E/(KL/r)² = π²·70000/640² = 1.687 MPa
Answer
σcr = 1.687 MPa
11.Mechanical vibrations
Answer: fd = 2.079 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.
Undamped natural frequency
ωn = √(15000/80) = 13.69 rad/s
fn = 13.69/(2π) = 2.179 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 2.179·√(1 − 0.3²) = 2.079 Hz
Answer
fd = 2.079 Hz
12.Mechanical vibrations
Answer: fn = 7.853 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
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 5 + 30 = 35 kN/m
keq = 35·80/(35 + 80) = 24.35 kN/m
Natural frequency
ωn = √(24348/10) = 49.34 rad/s
fn = 49.34/(2π) = 7.853 Hz
Answer
fn = 7.853 Hz
ExamProblems · examproblems.com/courses/machine-design