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 200 mm pulley at 2880 rpm drives a 250 mm pulley by a belt with 1.5 % slip. Find the speed n₂ of the driven pulley.

n₁ = 2880 rpmd₁ = 200 mmd₂ = 250 mms = 1.5 %

n₂ = rpm

2.Gears & gear trainsHard

In a compound gear train, gear A (12 teeth) on the input shaft turning at 2900 rpm drives gear B (20 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (48 teeth) on the output shaft. Find the output speed n_out.

n_in = 2900 rpmA = 12B = 20C = 16D = 48

n_out = rpm

3.Torsion of shaftsMedium

A solid brass 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 = 2 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 100 kW at 500 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.

P = 100 kWN = 500 rpmτ_allow = 60 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −30 MPa, σᵧ = 60 MPa and τₓᵧ = −20 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = −30 MPaσ_y = 60 MPaτ_xy = −20 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 80 MPa, σᵧ = −30 MPa and τₓᵧ = 30 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 = 80 MPaσ_y = −30 MPaτ_xy = 30 MPaσ_Y = 350 MPa

n =

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 30 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: brass, Ø30 mm, 500 mm, E = 100 GPaδ = 0.3 mm

P = kN

8.Axial stress & strainHard

A steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) and a segment made of copper (diameter 60 mm, length 500 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: copper, Ø60 mm, 500 mm, E = 120 GPaP = 40 kN

σ₂ = MPa

9.Column bucklingMedium

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

L = 1 mE = 200 GPa (steel)d = 40 mm

P_cr = kN

10.Column bucklingHard

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

L = 5 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed one cycle later by a peak of 6.4 mm. Find the damping ratio ζ.

x₀ = 12 mmx₁ = 6.4 mmn = 1

ζ =

12.Mechanical vibrationsHard

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 80 kN/m and k₂ = 60 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 10 kgk₁ = 80 kN/mk₂ = 60 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n₂ = 2269 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

2880·200/250 = 2304 rpm

With slip

n₂ = 2304·0.985 = 2269 rpm

Answer

n₂ = 2269 rpm

2.Gears & gear trains

Answer: n_out = 580 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.

First stage

n_B = 2900·12/20 = 1740 rpm = n_C

Second stage

n_out = 1740·16/48 = 580 rpm

Answer

n_out = 580 rpm

total ratio = 5 : 1

3.Torsion of shafts

Answer: φ = 11.69 °

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) = 1000·1000·2000/(39000·251327) = 0.204 rad

φ = 0.204·180/π = 11.69°

Answer

φ = 11.69°

4.Torsion of shafts

Answer: d = 54.53 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 100000/52.36 = 1910 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·1.91 × 10⁶/(π·60)) = 54.53 mm

Answer

d = 54.53 mm

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

5.Mohr’s circle & stress transformation

Answer: θ_p = −78.02 °

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·−20, −30 − 60) = −156°

Angle on the element

θ_p = −78.02°

Answer

θ_p = −78.02°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 3.143

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 = 25 MPa, R = 62.65 MPa

σ₁ = 87.65 MPa, σ₂ = −37.65 MPa, σ₃ = 0

Von Mises

σ_vM = √(87.65² − 87.65·(−37.65) + (−37.65)²) = 111.4 MPa

n = 350/111.4 = 3.143

Answer

n = 3.143

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

7.Axial stress & strain

Answer: P = 22.75 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 = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·490.9) = 6.112 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(100000·706.9) = 7.074 × 10⁻⁶ mm/N

Load

P = 0.3/1.319 × 10⁻⁵ = 22753 N = 22.75 kN

Answer

P = 22.75 kN

8.Axial stress & strain

Answer: σ₂ = 9.544 MPa

Stiffnesses

k₁ = 200000·490.9/300 = 327249 N/mm

k₂ = 120000·2827/500 = 678584 N/mm

Share of segment 2

F₂ = 40000·678584/1.006 × 10⁶ = 26986 N

Stress

σ₂ = 26986/2827 = 9.544 MPa (compression)

Answer

σ₂ = 9.544 MPa

9.Column buckling

Answer: P_cr = 248.1 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 = 1·1000 = 1000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·125664/1000² = 248050 N = 248.1 kN

Answer

P_cr = 248.1 kN

10.Column buckling

Answer: σ_cr = 31.58 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 = 0.5·5000 = 2500 mm

KL/r = 2500/10 = 250

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/250² = 31.58 MPa

Answer

σ_cr = 31.58 MPa

11.Mechanical vibrations

Answer: ζ = 0.09955

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 = (1/1)·ln(12/6.4) = 0.6286

Damping ratio

ζ = δ/√(4π² + δ²) = 0.6286/√(39.48 + 0.3951) = 0.09955

Answer

ζ = 0.09955

12.Mechanical vibrations

Answer: f_n = 18.83 Hz

Equivalent stiffness

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

k_eq = 80 + 60 = 140 kN/m

Natural frequency

ω_n = √(140000/10) = 118.3 rad/s

f_n = 118.3/(2π) = 18.83 Hz

Answer

f_n = 18.83 Hz

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