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 100 mm pulley at 1500 rpm drives a 200 mm pulley by a belt with 1.5 % slip. Find the speed n2 of the driven pulley.
n1 = 1500 rpmd1 = 100 mmd2 = 200 mms = 1.5 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 120 mm driving pulley at 960 rpm to a 400 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 120 mmn1 = 960 rpmd2 = 400 mmC = 500 mmμ = 0.25T1 = 1000 N
P = kW
3.Torsion of shaftsMedium
A stepped solid brass shaft is fixed at A. Segment AB has diameter 40 mm and length 1 m; segment BC has diameter 20 mm and length 1.2 m. A torque of 200 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.
T = 200 N·md1 = 40 mm, L1 = 1 md2 = 20 mm, L2 = 1.2 mG = 39 GPa
φC = °
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 1500 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.
P = 10 kWN = 1500 rpmτallow = 40 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 140 MPa, σᵧ = 0 MPa and τₓᵧ = 60 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 140 MPaσy = 0 MPaτxy = 60 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 150 MPa, σᵧ = −10 MPa and τₓᵧ = −40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.
σx = 150 MPaσy = −10 MPaτxy = −40 MPaθ = 45°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 30 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.
segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: brass, Ø30 mm, 500 mm, E = 100 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 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, Ø40 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 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 = 2 mE = 200 GPa (steel)d = 25 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 65 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 65 kNn = 3L = 3 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 12 kg mass is supported by a spring with a stiffness of 80 kN/m and a viscous damper with a damping coefficient of 200 N·s/m. Find the damping ratio ζ.
m = 12 kgk = 80 kN/mc = 200 N·s/m
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 40 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 25 kgk1 = 5 kN/mk2 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 738.8 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
1500·100/200 = 750 rpm
With slip
n2 = 750·0.985 = 738.8 rpm
Answer
n2 = 738.8 rpm
2.Gears & gear trains
Answer: P = 2.862 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.
Angle of wrap
θ = 2.574 rad
Slack-side tension
eμθ = e0.6435 = 1.903, T₂ = 1000/1.903 = 525.4 N
Power
P = (1000 − 525.4)·6.032 = 2.862 kW
Answer
P = 2.862 kW
3.Torsion of shafts
Answer: φC = 23.62 °
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 moments
J1 = π·40⁴/32 = 251327 mm⁴
J2 = π·20⁴/32 = 15708 mm⁴
Twist of each segment
T = 200 N·m = 200000 N·mm
G = 39 GPa = 39000 MPa
φAB = 200000·1000/(39000·251327) = 0.0204 rad
φBC = 200000·1200/(39000·15708) = 0.3918 rad
Total twist
φC = 0.0204 + 0.3918 = 0.4122 rad = 23.62°
Answer
φC = 23.62°
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 = 10000/157.1 = 63.66 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·63662/(π·40)) = 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 = 20.3 °
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(2·60, 140 − 0) = 40.6°
Angle on the element
θp = 20.3°
Answer
θp = 20.3°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 30 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 = 70 + 80·0 + −40·1 = 30 MPa
Answer
σθ = 30 MPa
7.Axial stress & strain
Answer: P = 18.05 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 = 314.2 mm², A₂ = 706.9 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·314.2) = 9.549 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(100000·706.9) = 7.074 × 10⁻⁶ mm/N
Load
P = 0.3/1.662 × 10⁻⁵ = 18047 N = 18.05 kN
Answer
P = 18.05 kN
8.Axial stress & strain
Answer: σ2 = 5.407 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 = 200000·1257/600 = 418879 N/mm
k₂ = 70000·706.9/500 = 98960 N/mm
Share of segment 2
F2 = 20000·98960/517839 = 3822 N
Stress
σ2 = 3822/706.9 = 5.407 MPa (compression)
Answer
σ2 = 5.407 MPa
9.Column buckling
Answer: Pcr = 37.85 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·25⁴/64 = 19175 mm⁴
Effective length
KL = 0.5·2000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·19175/1000² = 37849 N = 37.85 kN
Answer
Pcr = 37.85 kN
10.Column buckling
Answer: d = 46.13 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 = 3·65000 = 195000 N
KL = 0.5·3000 = 1500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·195000·1500²/(π³·200000))^(1/4) = 46.13 mm
Answer
d = 46.13 mm
Check: the slenderness KL/r is 130.1, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1021
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.
Critical damping
cc = 2·√(k·m) = 2·√(80000·12) = 1960 N·s/m
Damping ratio
ζ = c/cc = 200/1960 = 0.1021
Answer
ζ = 0.1021
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 2.122 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/5 + 1/40 → keq = 5·40/(5 + 40) = 4.444 kN/m
Natural frequency
ωn = √(4444/25) = 13.33 rad/s
fn = 13.33/(2π) = 2.122 Hz
Answer
fn = 2.122 Hz
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