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 an 80 mm driving pulley at 720 rpm. The tension is 1000 N on the tight side and 300 N on the slack side. Find the power P transmitted, in kW.
d1 = 80 mmn1 = 720 rpmT1 = 1000 NT2 = 300 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 200 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 = 200 mmd2 = 450 mmC = 600 mm
θ = °
3.Torsion of shaftsMedium
A solid brass 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 = 2.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 2000 N·m is applied at C, 0.8 m from A and 3 m from B. Find the angle of rotation φC of section C, in degrees.
d = 60 mmT0 = 2000 N·ma = 0.8 mb = 3 mG = 39 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 140 MPa, σᵧ = 10 MPa and τₓᵧ = −60 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 = 140 MPaσy = 10 MPaτxy = −60 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 60 MPa, σᵧ = −70 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° clockwise from the x-axis.
σx = 60 MPaσy = −70 MPaτxy = 30 MPaθ = -30°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 80 mm, inner diameter 60 mm, E = 120 GPa). Both are 600 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.
P = 80 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 120 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 300 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 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: brass, Ø30 mm, 200 mm, E = 100 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)d = 40 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2 mE = 70 GPa (aluminium)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 2 kg mass on a spring with a stiffness of 50 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.
m = 2 kgk = 50 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 15 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 10 kgk1 = 40 kN/mk2 = 15 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: P = 2.111 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 = 3.016 m/s
Power
P = (1000 − 300)·3.016 = 2111 W = 2.111 kW
Answer
P = 2.111 kW
2.Gears & gear trains
Answer: θ = 156 °
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(0.2083) = 12.02°
Angle of wrap
θ = 180 − 2·12.02 = 156°
Answer
θ = 156°
3.Torsion of shafts
Answer: φ = 0.5986 °
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) = 100·1000·2500/(39000·613592) = 0.01045 rad
φ = 0.01045·180/π = 0.5986°
Answer
φ = 0.5986°
4.Torsion of shafts
Answer: φC = 1.459 °
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 = 2000
TA·a = TB·b
TA = 2000·3/3.8 = 1579 N·m
TB = 2000·0.8/3.8 = 421.1 N·m
Rotation of C
Use segment AC: it is fixed at A and carries TA.
J = π·60⁴/32 = 1.272 × 10⁶ mm⁴
G = 39 GPa = 39000 MPa
φC = TA·a/(GJ) = 1.579 × 10⁶·800/(39000·1.272 × 10⁶) = 0.02546 rad = 1.459°
Answer
φC = 1.459°
5.Mohr’s circle & stress transformation
Answer: θs = −66.35 °
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 = −60/88.46 = −0.6783, sin 2θs = −65/88.46 = −0.7348
2θs = atan2(−65, −60) = −132.7°
Angle on the element
θs = −66.35°
check: θs − θp = −66.35 − (−21.35) = −45°
Answer
θs = −66.35°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 1.519 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 = −5 + 65·0.5 + 30·−0.866 = 1.519 MPa
Answer
σθ = 1.519 MPa
7.Axial stress & strain
Answer: δ = 0.09316 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·40²/4 = 1257 mm²
At = π(80² − 60²)/4 = 2199 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 120·10³·2199 = 5.152 × 10⁸ N
Shortening
δ = 80000·600/5.152 × 10⁸ = 0.09316 mm
Answer
δ = 0.09316 mm
8.Axial stress & strain
Answer: σ2 = 58.76 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 = 200000·490.9/300 = 327249 N/mm
k₂ = 100000·706.9/200 = 353429 N/mm
Share of segment 2
F2 = 80000·353429/680678 = 41538 N
Stress
σ2 = 41538/706.9 = 58.76 MPa (compression)
Answer
σ2 = 58.76 MPa
9.Column buckling
Answer: Pcr = 5.426 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 = 2·2000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·125664/4000² = 5426 N = 5.426 kN
Answer
Pcr = 5.426 kN
10.Column buckling
Answer: σcr = 38.86 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.5·2000 = 1000 mm
KL/r = 1000/7.5 = 133.3
Critical stress
σcr = π²E/(KL/r)² = π²·70000/133.3² = 38.86 MPa
Answer
σcr = 38.86 MPa
11.Mechanical vibrations
Answer: fd = 21.79 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 = √(50000/2) = 158.1 rad/s
fn = 158.1/(2π) = 25.16 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 25.16·√(1 − 0.5²) = 21.79 Hz
Answer
fd = 21.79 Hz
12.Mechanical vibrations
Answer: keq = 55 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 + 15 = 55 kN/m
Answer
keq = 55 kN/m
ExamProblems · examproblems.com/courses/machine-design