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 worm with 3 start(s) turns at 2900 rpm and drives a worm wheel with 72 teeth. Find the speed n_w of the wheel.

z = 3N_w = 72n = 2900 rpm

n_w = rpm

2.Gears & gear trainsHard

An open belt connects a 150 mm pulley and a 300 mm pulley whose centres are 1000 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d₁ = 150 mmd₂ = 300 mmC = 1000 mm

θ = °

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 500 N·m. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

T = 500 N·md_i/d_o = 0.7τ_allow = 40 MPa

d_o = mm

4.Torsion of shaftsHard

A solid steel shaft of diameter 60 mm is fixed at A. A torque of 1000 N·m acts at B, 1 m from A, and a torque of 200 N·m in the opposite direction acts at the free end C, a further 1.2 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 60 mmT_B = 1000 N·mT_C = 200 N·mAB = 1 mBC = 1.2 mG = 80 GPa

φ_C = °

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 40 MPa, σᵧ = −10 MPa and τₓᵧ = 60 MPa (tension positive). The element is rotated 50° 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 = 40 MPaσ_y = −10 MPaτ_xy = 60 MPaθ = 50°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1500 mm and wall thickness 8 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 60° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 1500 mmt = 8 mmp = 2.5 MPaα = 60°

σ_n = MPa

7.Axial stress & strainMedium

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 70 mm, E = 110 GPa). Both are 300 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.

P = 100 kNrod: Ø20 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 110 GPaL = 300 mm

δ = mm

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 = 110 kNsegment 1: Ø25 mm, 2 m, steel (E = 200 GPa)segment 2: Ø16 mm, 1.5 m, copper (E = 117 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 = 6 mE = 70 GPa (aluminium)b × h = 30 × 80 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 = 1 mE = 70 GPa (aluminium)d = 30 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 3 cycles later by a peak of 15.6 mm. Find the damping ratio ζ.

x₀ = 40 mmx₃ = 15.6 mmn = 3

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from spring k₃ = 30 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 20 kN/m and k₂ = 25 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency f_n of vertical vibration in Hz.

m = 20 kgk₁ = 20 kN/mk₂ = 25 kN/mk₃ = 30 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n_w = 120.8 rpm

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.

Wheel speed

n_w = n·z/N_w = 2900·3/72 = 120.8 rpm

Answer

n_w = 120.8 rpm

2.Gears & gear trains

Answer: θ = 171.4 °

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 angle

asin(0.075) = 4.301°

Angle of wrap

θ = 180 − 2·4.301 = 171.4°

Answer

θ = 171.4°

3.Torsion of shafts

Answer: d_o = 43.76 mm

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.

Outer diameter from the allowable stress

T = 500 N·m = 500000 N·mm

1 − k⁴ = 1 − 0.7⁴ = 0.7599

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·500000/(π·40·0.7599)) = 43.76 mm

d_i = 0.7·43.76 = 30.63 mm

Answer

d_o = 43.76 mm

A solid shaft would need d = 39.93 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: φ_C = 0.3152 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 200 N·m

T_AB = T_C − T_B = 200 − 1000 = −800 N·m

Twist of each segment

J = π·60⁴/32 = 1.272 × 10⁶ mm⁴

G = 80 GPa = 80000 MPa

φ_AB = −800000·1000/(80000·1.272 × 10⁶) = −0.00786 rad

φ_BC = 200000·1200/(80000·1.272 × 10⁶) = 0.002358 rad

Twist of end C

φ_C = −0.00786 + 0.002358 = −0.005502 rad → |φ_C| = 0.3152°

Answer

φ_C = 0.3152°

5.Mohr’s circle & stress transformation

Answer: σ_y′ = −39.75 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 for the y′ face

σ_y′ = 15 − 25·(−0.1736) − 60·0.9848 = −39.75 MPa

Check

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

σ_x′ = 69.75 MPa, σ_x′ + σ_y′ = 30 = σ_x + σ_y

Answer

σ_y′ = −39.75 MPa

6.Mohr’s circle & stress transformation

Answer: σ_n = 146.5 MPa

Wall stresses

r = 1500/2 = 750 mm

σ_h = pr/t = 2.5·750/8 = 234.4 MPa

σ_a = pr/(2t) = 117.2 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 117.2·sin²60° + 234.4·cos²60° = 146.5 MPa

Answer

σ_n = 146.5 MPa

7.Axial stress & strain

Answer: δ = 0.03395 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, 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

A_r = π·20²/4 = 314.2 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·7461 = 8.836 × 10⁸ N

Shortening

δ = 100000·300/8.836 × 10⁸ = 0.03395 mm

Answer

δ = 0.03395 mm

8.Axial stress & strain

Answer: σ_max = 547.1 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·25²/4 = 490.9 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 224.1 MPa

σ₂ = 547.1 MPa

Largest stress

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

σ_max = 547.1 MPa

Answer

σ_max = 547.1 MPa

9.Column buckling

Answer: P_cr = 13.82 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·180000/3000² = 13817 N = 13.82 kN

Answer

P_cr = 13.82 kN

10.Column buckling

Answer: σ_cr = 38.86 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 = 1·1000 = 1000 mm

KL/r = 1000/7.5 = 133.3

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/133.3² = 38.86 MPa

Answer

σ_cr = 38.86 MPa

11.Mechanical vibrations

Answer: ζ = 0.04989

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.

Logarithmic decrement

δ = (1/3)·ln(40/15.6) = 0.3139

Damping ratio

ζ = δ/√(4π² + δ²) = 0.3139/√(39.48 + 0.09851) = 0.04989

Answer

ζ = 0.04989

12.Mechanical vibrations

Answer: f_n = 4.775 Hz

Equivalent stiffness

Only k₁ and k₂ are in parallel. That pair is in series with k₃, so combine them as reciprocals.

k₁₂ = 20 + 25 = 45 kN/m

k_eq = 45·30/(45 + 30) = 18 kN/m

Natural frequency

ω_n = √(18000/20) = 30 rad/s

f_n = 30/(2π) = 4.775 Hz

Answer

f_n = 4.775 Hz

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