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 belt runs on a 100 mm driving pulley at 1500 rpm. The tension is 600 N on the tight side and 150 N on the slack side. Find the power P transmitted, in kW.

d₁ = 100 mmn₁ = 1500 rpmT₁ = 600 NT₂ = 150 N

P = kW

2.Gears & gear trainsHard

An open belt connects a 125 mm driving pulley at 1500 rpm to a 300 mm pulley; the centres are 800 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 125 mmn₁ = 1500 rpmd₂ = 300 mmC = 800 mmμ = 0.3T₁ = 1500 N

P = kW

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 = 1000 N·md = 40 mmL = 0.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 75 kW at 1450 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

P = 75 kWN = 1450 rpmd_i/d_o = 0.6τ_allow = 50 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 110 MPa, σᵧ = −60 MPa and τₓᵧ = −10 MPa (tension positive). The element is rotated 20° clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = 110 MPaσ_y = −60 MPaτ_xy = −10 MPaθ = -20°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = −40 MPa and τₓᵧ = 20 MPa (σ_z = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = −70 MPaσ_y = −40 MPaτ_xy = 20 MPaσ_Y = 350 MPa

n =

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 50 mm, length 400 mm, E = 110 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, Ø20 mm, 500 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 400 mm, E = 110 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 total elongation δ.

P = 100 kNsegment 1: Ø25 mm, 0.5 m, aluminium (E = 70 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 the base and free at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 1.5 mE = 200 GPa (steel)d = 25 mm

P_cr = kN

10.Column bucklingHard

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

L = 5 mE = 70 GPa (aluminium)d = 50 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 50 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 350 N·s/m. Find the damping ratio ζ.

m = 50 kgk = 10 kN/mc = 350 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 = 30 kN/m and the lower one k₂ = 30 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 25 kgk₁ = 30 kN/mk₂ = 30 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: P = 3.534 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 = 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.

Belt speed

v = 7.854 m/s

Power

P = (600 − 150)·7.854 = 3534 W = 3.534 kW

Answer

P = 3.534 kW

2.Gears & gear trains

Answer: P = 8.598 kW

Angle of wrap

θ = 2.922 rad

Slack-side tension

e^μθ = e^0.8767 = 2.403, T₂ = 1500/2.403 = 624.2 N

Power

P = (1500 − 624.2)·9.817 = 8.598 kW

Answer

P = 8.598 kW

3.Torsion of shafts

Answer: φ = 1.425 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 1000·1000·500/(80000·251327) = 0.02487 rad

φ = 0.02487·180/π = 1.425°

Answer

φ = 1.425°

4.Torsion of shafts

Answer: d_o = 38.66 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 75000/151.8 = 493.9 N·m

Outer diameter from the allowable stress

T = 493.9 N·m = 493929 N·mm

1 − k⁴ = 1 − 0.6⁴ = 0.8704

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·493929/(π·50·0.8704)) = 38.66 mm

d_i = 0.6·38.66 = 23.2 mm

Answer

d_o = 38.66 mm

A solid shaft would need d = 36.92 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: σ_y′ = −46.54 MPa

Transformation equation for the y′ face

σ_y′ = 25 − 85·0.766 − (−10)·(−0.6428) = −46.54 MPa

Check

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

σ_x′ = 96.54 MPa, σ_x′ + σ_y′ = 50 = σ_x + σ_y

Answer

σ_y′ = −46.54 MPa

6.Mohr’s circle & stress transformation

Answer: n = 5

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 = −55 MPa, R = 25 MPa

σ₁ = −30 MPa, σ₂ = −80 MPa, σ₃ = 0

Von Mises

σ_vM = √((−30)² − (−30)·(−80) + (−80)²) = 70 MPa

n = 350/70 = 5

Answer

n = 5

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

7.Axial stress & strain

Answer: P = 40.78 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 = 314.2 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·314.2) = 7.958 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(110000·1963) = 1.852 × 10⁻⁶ mm/N

Load

P = 0.4/9.81 × 10⁻⁶ = 40776 N = 40.78 kN

Answer

P = 40.78 kN

8.Axial stress & strain

Answer: δ = 7.832 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·25²/4 = 490.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 203.7 MPa

σ₂ = 497.4 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 203.7·500/70000 = 1.455 mm

δ₂ = σ₂L₂/E₂ = 497.4·1500/117000 = 6.376 mm

δ = δ₁ + δ₂ = 7.832 mm

Answer

δ = 7.832 mm

9.Column buckling

Answer: P_cr = 4.205 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 2·1500 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·19175/3000² = 4205 N = 4.205 kN

Answer

P_cr = 4.205 kN

10.Column buckling

Answer: σ_cr = 4.318 MPa

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 = 1·5000 = 5000 mm

KL/r = 5000/12.5 = 400

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/400² = 4.318 MPa

Answer

σ_cr = 4.318 MPa

11.Mechanical vibrations

Answer: ζ = 0.2475

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·√(10000·50) = 1414 N·s/m

Damping ratio

ζ = c/c_c = 350/1414 = 0.2475

Answer

ζ = 0.2475

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

12.Mechanical vibrations

Answer: k_eq = 15 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 = 1/30 + 1/30 → k_eq = 30·30/(30 + 30) = 15 kN/m

Answer

k_eq = 15 kN/m

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