Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

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

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 = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·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: ε_T = α·ΔT

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

Mechanical vibrations

Natural frequency: ω_n = √(k/m), f_n = ω_n/(2π)
From static deflection: ω_n = √(g/δ_st)
Critical damping: c_c = 2√(km) = 2mω_n
Damping ratio: ζ = c/c_c
Damped frequency: ω_d = ω_n√(1 − ζ²)
Logarithmic decrement: δ = (1/n)·ln(x₀/x_n), 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 2880 rpm drives a 300 mm pulley by a belt with 2.5 % slip. Find the speed n₂ of the driven pulley.

n₁ = 2880 rpmd₁ = 125 mmd₂ = 300 mms = 2.5 %

n₂ = rpm

2.Gears & gear trainsHard

In a compound gear train, gear A (18 teeth) on the input shaft turns at 2880 rpm and drives gear B (60 teeth). Gear C (36 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 388.8 rpm. How many teeth N_D must gear D have?

n_in = 2880 rpmA = 18B = 60C = 36n_out = 388.8 rpm

N_D =

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 = 200 N·md = 20 mmL = 0.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 75 kW at 500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 75 kWN = 500 rpmd_o = 100 mmd_i = 70 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = −30 MPa, σᵧ = 100 MPa and τₓᵧ = 10 MPa (tension positive). The element is rotated 35° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = −30 MPaσ_y = 100 MPaτ_xy = 10 MPaθ = 35°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = 110 MPa and τₓᵧ = −10 MPa (σ_z = 0). The material has a yield strength of 355 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = 30 MPaσ_y = 110 MPaτ_xy = −10 MPaσ_Y = 355 MPa

n =

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 50 mm, length 300 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, Ø50 mm, 300 mm, E = 100 GPaδ = 0.4 mm

P = kN

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 40 kNsegment 1: Ø40 mm, 2 m, copper (E = 117 GPa)segment 2: Ø20 mm, 0.5 m, aluminium (E = 70 GPa)

σ_max = MPa

9.Column bucklingMedium

A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 4 mE = 70 GPa (aluminium)d = 40 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 1.5 mE = 70 GPa (aluminium)d = 30 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 200 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 2190 N·s/m. Find the damping ratio ζ.

m = 200 kgk = 150 kN/mc = 2190 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 = 60 kN/m and the lower one k₂ = 100 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 25 kgk₁ = 60 kN/mk₂ = 100 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: n₂ = 1170 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 = e^μθ, 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

2880·125/300 = 1200 rpm

With slip

n₂ = 1200·0.975 = 1170 rpm

Answer

n₂ = 1170 rpm

2.Gears & gear trains

Answer: N_D = 80

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.

Intermediate shaft speed

n_B = n_C = 2880·18/60 = 864 rpm

Teeth on gear D

N_D = C·n_C/n_out = 36·864/388.8 = 80

Answer

N_D = 80

3.Torsion of shafts

Answer: φ = 14.03 °

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/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·20⁴/32 = 15708 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·500/(26000·15708) = 0.2449 rad

φ = 0.2449·180/π = 14.03°

Answer

φ = 14.03°

4.Torsion of shafts

Answer: τ_max = 9.6 MPa

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 75000/52.36 = 1432 N·m

Polar second moment of area

J = π(100⁴ − 70⁴)/32 = 7.46 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 1.432 × 10⁶·50/7.46 × 10⁶ = 9.6 MPa

Answer

τ_max = 9.6 MPa

5.Mohr’s circle & stress transformation

Answer: σ_y′ = 47.83 MPa

Transformation equation for the y′ face

σ_y′ = 35 − (−65)·0.342 − 10·0.9397 = 47.83 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = 22.17 MPa, σ_x′ + σ_y′ = 70 = σ_x + σ_y

Answer

σ_y′ = 47.83 MPa

6.Mohr’s circle & stress transformation

Answer: n = 3.55

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 σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 70 MPa, R = 41.23 MPa

σ₁ = 111.2 MPa, σ₂ = 28.77 MPa, σ₃ = 0

Von Mises

σ_vM = √(111.2² − 111.2·28.77 + 28.77²) = 100 MPa

n = 355/100 = 3.55

Answer

n = 3.55

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: P = 87.27 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, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·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 = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(100000·1963) = 1.528 × 10⁻⁶ mm/N

Load

P = 0.4/4.584 × 10⁻⁶ = 87266 N = 87.27 kN

Answer

P = 87.27 kN

8.Axial stress & strain

Answer: σ_max = 127.3 MPa

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·40²/4 = 1257 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 31.83 MPa

σ₂ = 127.3 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 127.3 MPa

Answer

σ_max = 127.3 MPa

9.Column buckling

Answer: P_cr = 21.7 kN

Euler buckling: a slender column becomes unstable at P_cr = 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 = 0.5·4000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·125664/2000² = 21704 N = 21.7 kN

Answer

P_cr = 21.7 kN

10.Column buckling

Answer: σ_cr = 35.25 MPa

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 = 0.7·1500 = 1050 mm

KL/r = 1050/7.5 = 140

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/140² = 35.25 MPa

Answer

σ_cr = 35.25 MPa

11.Mechanical vibrations

Answer: ζ = 0.1999

Free vibration of a mass–spring system: the undamped natural frequency is ω_n = √(k/m) in rad/s, and f_n = ω_n/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/c_c, with critical damping c_c = 2·√(k·m) = 2·m·ω_n. An underdamped system (ζ < 1) oscillates at ω_d = ω_n·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.

Critical damping

c_c = 2·√(k·m) = 2·√(150000·200) = 10954 N·s/m

Damping ratio

ζ = c/c_c = 2190/10954 = 0.1999

Answer

ζ = 0.1999

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: k_eq = 37.5 kN/m

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/k_eq = 1/60 + 1/100 → k_eq = 60·100/(60 + 100) = 37.5 kN/m

Answer

k_eq = 37.5 kN/m

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