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

An 80 mm pulley at 960 rpm drives a 300 mm pulley by a belt with 4 % slip. Find the speed n₂ of the driven pulley.

n₁ = 960 rpmd₁ = 80 mmd₂ = 300 mms = 4 %

n₂ = rpm

2.Gears & gear trainsHard

An open belt connects an 80 mm pulley and a 315 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d₁ = 80 mmd₂ = 315 mmC = 800 mm

θ = °

3.Torsion of shaftsMedium

A stepped solid steel shaft is fixed at A. Segment AB has diameter 25 mm and length 1.2 m; segment BC has diameter 20 mm and length 0.5 m. A torque of 100 N·m acts at the free end C. Find the angle of twist φ_C of end C, in degrees.

T = 100 N·md₁ = 25 mm, L₁ = 1.2 md₂ = 20 mm, L₂ = 0.5 mG = 80 GPa

φ_C = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 40 mm is fixed at A. A torque of 2000 N·m acts at B, 0.8 m from A, and a torque of 200 N·m in the same direction acts at the free end C, a further 1.2 m along. Find the maximum shear stress τ_max in the shaft.

d = 40 mmT_B = 2000 N·mT_C = 200 N·mAB = 0.8 mBC = 1.2 m

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 80 MPa, σᵧ = −40 MPa and τₓᵧ = 40 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 = 80 MPaσ_y = −40 MPaτ_xy = 40 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −60 MPa, σᵧ = −60 MPa and τₓᵧ = 10 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.

σ_x = −60 MPaσ_y = −60 MPaτ_xy = 10 MPaθ = 15°

σ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 60 mm, length 300 mm, E = 110 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 300 mm, E = 110 GPaδ = 0.5 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 = 20 kNsegment 1: Ø25 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 1 m, aluminium (E = 70 GPa)

δ = mm

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 = 4 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

10.Column bucklingHard

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

L = 6 mE = 70 GPa (aluminium)d = 50 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 250 kg mass is supported by a spring with a stiffness of 80 kN/m and a viscous damper with a damping coefficient of 1340 N·s/m. Find the damping ratio ζ.

m = 250 kgk = 80 kN/mc = 1340 N·s/m

ζ =

12.Mechanical vibrationsHard

A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 25 kN/m and the lower one k₂ = 40 kN/m. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 5 kgk₁ = 25 kN/mk₂ = 40 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n₂ = 245.8 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

960·80/300 = 256 rpm

With slip

n₂ = 256·0.96 = 245.8 rpm

Answer

n₂ = 245.8 rpm

2.Gears & gear trains

Answer: θ = 163.1 °

Belt angle

asin(0.1469) = 8.446°

Angle of wrap

θ = 180 − 2·8.446 = 163.1°

Answer

θ = 163.1°

3.Torsion of shafts

Answer: φ_C = 4.521 °

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 moments

J₁ = π·25⁴/32 = 38350 mm⁴

J₂ = π·20⁴/32 = 15708 mm⁴

Twist of each segment

T = 100 N·m = 100000 N·mm

G = 80 GPa = 80000 MPa

φ_AB = 100000·1200/(80000·38350) = 0.03911 rad

φ_BC = 100000·500/(80000·15708) = 0.03979 rad

Total twist

φ_C = 0.03911 + 0.03979 = 0.0789 rad = 4.521°

Answer

φ_C = 4.521°

4.Torsion of shafts

Answer: τ_max = 175.1 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 200 N·m

T_AB = T_C + T_B = 200 + 2000 = 2200 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 2200 N·m

τ_max = 16·2.2 × 10⁶/(π·40³) = 175.1 MPa

Answer

τ_max = 175.1 MPa

5.Mohr’s circle & stress transformation

Answer: θ_p = 16.85 °

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

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(2·40, 80 − −40) = 33.69°

Angle on the element

θ_p = 16.85°

Answer

θ_p = 16.85°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_θ = −55 MPa

Transformation equation

σ_θ = −60 + 0·0.866 + 10·0.5 = −55 MPa

Answer

σ_θ = −55 MPa

7.Axial stress & strain

Answer: P = 99.23 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) = 400/(200000·490.9) = 4.074 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(110000·2827) = 9.646 × 10⁻⁷ mm/N

Load

P = 0.5/5.039 × 10⁻⁶ = 99227 N = 99.23 kN

Answer

P = 99.23 kN

8.Axial stress & strain

Answer: δ = 1.727 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 = 40.74 MPa

σ₂ = 99.47 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 40.74·1500/200000 = 0.3056 mm

δ₂ = σ₂L₂/E₂ = 99.47·1000/70000 = 1.421 mm

δ = δ₁ + δ₂ = 1.727 mm

Answer

δ = 1.727 mm

9.Column buckling

Answer: P_cr = 347.3 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 = 0.5·4000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/2000² = 347270 N = 347.3 kN

Answer

P_cr = 347.3 kN

10.Column buckling

Answer: σ_cr = 6.12 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 = 0.7·6000 = 4200 mm

KL/r = 4200/12.5 = 336

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/336² = 6.12 MPa

Answer

σ_cr = 6.12 MPa

11.Mechanical vibrations

Answer: ζ = 0.1498

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·√(80000·250) = 8944 N·s/m

Damping ratio

ζ = c/c_c = 1340/8944 = 0.1498

Answer

ζ = 0.1498

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

12.Mechanical vibrations

Answer: f_n = 8.828 Hz

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 = 1/25 + 1/40 → k_eq = 25·40/(25 + 40) = 15.38 kN/m

Natural frequency

ω_n = √(15385/5) = 55.47 rad/s

f_n = 55.47/(2π) = 8.828 Hz

Answer

f_n = 8.828 Hz

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