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 1450 rpm drives a 500 mm pulley by a belt with 1 % slip. Find the speed n₂ of the driven pulley.

n₁ = 1450 rpmd₁ = 125 mmd₂ = 500 mms = 1 %

n₂ = rpm

2.Gears & gear trainsHard

An open belt connects an 80 mm driving pulley at 2880 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 600 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 80 mmn₁ = 2880 rpmd₂ = 300 mmC = 800 mmμ = 0.3T₁ = 600 N

P = kW

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 750 N·m. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

T = 750 N·md_i/d_o = 0.8τ_allow = 75 MPa

d_o = mm

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 1450 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 1450 rpmd_o = 50 mmd_i = 40 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 10 MPa, σᵧ = 110 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θ_s, 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).

σ_x = 10 MPaσ_y = 110 MPaτ_xy = −50 MPa

θ_s = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 30 MPa, σᵧ = 110 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 45° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 30 MPaσ_y = 110 MPaτ_xy = 40 MPaθ = 45°

τ_θ = MPa

7.Axial stress & strainMedium

Find the elongation δ of the steel bar under the axial load.

P = 45 kNd = 25 mmL = 1.5 mE = 200 GPa

δ = mm

8.Axial stress & strainHard

A steel segment (diameter 20 mm, length 500 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, Ø20 mm, 500 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 P_cr.

L = 5 mE = 200 GPa (steel)b × h = 40 × 100 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of aluminium is fixed at the base and free 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 = 40 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 3 cycles later by a peak of 6.6 mm. Find the damping ratio ζ.

x₀ = 30 mmx₃ = 6.6 mmn = 3

ζ =

12.Mechanical vibrationsHard

A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 30 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 40 kgk₁ = 100 kN/mk₂ = 30 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: n₂ = 358.9 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

1450·125/500 = 362.5 rpm

With slip

n₂ = 362.5·0.99 = 358.9 rpm

Answer

n₂ = 358.9 rpm

2.Gears & gear trains

Answer: P = 4.174 kW

Angle of wrap

θ = 2.866 rad

Slack-side tension

e^μθ = e^0.8597 = 2.362, T₂ = 600/2.362 = 254 N

Power

P = (600 − 254)·12.06 = 4.174 kW

Answer

P = 4.174 kW

3.Torsion of shafts

Answer: d_o = 44.18 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 = 750 N·m = 750000 N·mm

1 − k⁴ = 1 − 0.8⁴ = 0.5904

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·750000/(π·75·0.5904)) = 44.18 mm

d_i = 0.8·44.18 = 35.35 mm

Answer

d_o = 44.18 mm

A solid shaft would need d = 37.07 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: τ_max = 45.45 MPa

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 100000/151.8 = 658.6 N·m

Polar second moment of area

J = π(50⁴ − 40⁴)/32 = 362265 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 658572·25/362265 = 45.45 MPa

Answer

τ_max = 45.45 MPa

5.Mohr’s circle & stress transformation

Answer: θ_s = 67.5 °

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.

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 2θ_s = −50/70.71 = −0.7071, sin 2θ_s = −(−50)/70.71 = 0.7071

2θ_s = atan2(50, −50) = 135°

Angle on the element

θ_s = 67.5°

check: θ_s − θ_p = 67.5 − (−67.5) = 135°

Answer

θ_s = 67.5°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: τ_θ = 40 MPa

Transformation equation

τ_θ = −−40·1 + 40·0 = 40 MPa

Answer

τ_θ = 40 MPa

7.Axial stress & strain

Answer: δ = 0.6875 mm

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·25²/4 = 490.9 mm²

σ = P/A = 45·1000/490.9 = 91.67 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 91.67/200000 = 0.0004584 = 458.4 µε

Elongation

δ = ε·L = 0.0004584·1500 mm = 0.6875 mm

Answer

δ = 0.6875 mm

8.Axial stress & strain

Answer: σ₂ = 12.47 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, 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.

Stiffnesses

k₁ = 200000·314.2/500 = 125664 N/mm

k₂ = 70000·706.9/500 = 98960 N/mm

Share of segment 2

F₂ = 20000·98960/224624 = 8811 N

Stress

σ₂ = 8811/706.9 = 12.47 MPa (compression)

Answer

σ₂ = 12.47 MPa

9.Column buckling

Answer: P_cr = 168.4 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 = 100·40³/12 = 533333 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·533333/2500² = 168441 N = 168.4 kN

Answer

P_cr = 168.4 kN

10.Column buckling

Answer: σ_cr = 7.676 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 2·1500 = 3000 mm

KL/r = 3000/10 = 300

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/300² = 7.676 MPa

Answer

σ_cr = 7.676 MPa

11.Mechanical vibrations

Answer: ζ = 0.08007

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(30/6.6) = 0.5047

Damping ratio

ζ = δ/√(4π² + δ²) = 0.5047/√(39.48 + 0.2547) = 0.08007

Answer

ζ = 0.08007

12.Mechanical vibrations

Answer: k_eq = 130 kN/m

Equivalent stiffness

Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k₁ + k₂.

k_eq = 100 + 30 = 130 kN/m

Answer

k_eq = 130 kN/m

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