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 125 mm driving pulley at 1440 rpm. The tension is 400 N on the tight side and 200 N on the slack side. Find the power P transmitted, in kW.

d₁ = 125 mmn₁ = 1440 rpmT₁ = 400 NT₂ = 200 N

P = kW

2.Gears & gear trainsHard

An open belt connects a 150 mm driving pulley at 960 rpm to a 300 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 2000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 150 mmn₁ = 960 rpmd₂ = 300 mmC = 500 mmμ = 0.35T₁ = 2000 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 = 500 N·md = 30 mmL = 0.8 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 5 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 5 kWN = 500 rpmτ_allow = 75 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 20 MPa, σᵧ = 80 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 60° 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 = 20 MPaσ_y = 80 MPaτ_xy = 30 MPaθ = 60°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 60 MPa, σᵧ = 30 MPa and τₓᵧ = −50 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 = 60 MPaσ_y = 30 MPaτ_xy = −50 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 60 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, Ø60 mm, 300 mm, E = 100 GPaδ = 0.4 mm

P = kN

8.Axial stress & strainHard

A steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 60 mm, length 400 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: brass, Ø60 mm, 400 mm, E = 100 GPaP = 60 kN

σ₂ = MPa

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 = 6 mE = 200 GPa (steel)d = 80 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel 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 mE = 200 GPa (steel)d = 80 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 40 kg mass on a spring with a stiffness of 80 kN/m has a viscous damper giving a damping ratio of 0.35. Find the damped natural frequency f_d of its free vibration in Hz.

m = 40 kgk = 80 kN/mζ = 0.35

f_d = Hz

12.Mechanical vibrationsHard

A 10 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 10 kN/m and k₂ = 15 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 = 10 kgk₁ = 10 kN/mk₂ = 15 kN/mk₃ = 10 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: P = 1.885 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 = 9.425 m/s

Power

P = (400 − 200)·9.425 = 1885 W = 1.885 kW

Answer

P = 1.885 kW

2.Gears & gear trains

Answer: P = 9.5 kW

Angle of wrap

θ = 2.84 rad

Slack-side tension

e^μθ = e^0.9942 = 2.702, T₂ = 2000/2.702 = 740.1 N

Power

P = (2000 − 740.1)·7.54 = 9.5 kW

Answer

P = 9.5 kW

3.Torsion of shafts

Answer: φ = 3.603 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·800/(80000·79522) = 0.06288 rad

φ = 0.06288·180/π = 3.603°

Answer

φ = 3.603°

4.Torsion of shafts

Answer: d = 18.65 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 5000/52.36 = 95.49 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·95493/(π·75)) = 18.65 mm

Answer

d = 18.65 mm

In practice you would round up to the next standard size.

5.Mohr’s circle & stress transformation

Answer: σ_y′ = 9.019 MPa

Transformation equation for the y′ face

σ_y′ = 50 − (−30)·(−0.5) − 30·0.866 = 9.019 MPa

Check

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

σ_x′ = 90.98 MPa, σ_x′ + σ_y′ = 100 = σ_x + σ_y

Answer

σ_y′ = 9.019 MPa

6.Mohr’s circle & stress transformation

Answer: n = 3.515

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 = 45 MPa, R = 52.2 MPa

σ₁ = 97.2 MPa, σ₂ = −7.202 MPa, σ₃ = 0

Von Mises

σ_vM = √(97.2² − 97.2·(−7.202) + (−7.202)²) = 101 MPa

n = 355/101 = 3.515

Answer

n = 3.515

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

7.Axial stress & strain

Answer: P = 97.16 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 = 2827 mm²

Flexibilities

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

L₂/(E₂A₂) = 300/(100000·2827) = 1.061 × 10⁻⁶ mm/N

Load

P = 0.4/4.117 × 10⁻⁶ = 97163 N = 97.16 kN

Answer

P = 97.16 kN

8.Axial stress & strain

Answer: σ₂ = 17.36 MPa

Stiffnesses

k₁ = 200000·314.2/400 = 157080 N/mm

k₂ = 100000·2827/400 = 706858 N/mm

Share of segment 2

F₂ = 60000·706858/863938 = 49091 N

Stress

σ₂ = 49091/2827 = 17.36 MPa (compression)

Answer

σ₂ = 17.36 MPa

9.Column buckling

Answer: P_cr = 27.56 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·6000 = 12000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/12000² = 27561 N = 27.56 kN

Answer

P_cr = 27.56 kN

10.Column buckling

Answer: σ_cr = 197.4 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 2·1000 = 2000 mm

KL/r = 2000/20 = 100

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/100² = 197.4 MPa

Answer

σ_cr = 197.4 MPa

11.Mechanical vibrations

Answer: f_d = 6.667 Hz

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.

Undamped natural frequency

ω_n = √(80000/40) = 44.72 rad/s

f_n = 44.72/(2π) = 7.118 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 7.118·√(1 − 0.35²) = 6.667 Hz

Answer

f_d = 6.667 Hz

12.Mechanical vibrations

Answer: f_n = 4.254 Hz

Equivalent stiffness

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

k₁₂ = 10 + 15 = 25 kN/m

k_eq = 25·10/(25 + 10) = 7.143 kN/m

Natural frequency

ω_n = √(7143/10) = 26.73 rad/s

f_n = 26.73/(2π) = 4.254 Hz

Answer

f_n = 4.254 Hz

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