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 2.2 kW motor at 960 rpm drives a worm with 2 start(s). The worm wheel has 50 teeth and the drive has an efficiency of 65%. Find the output torque T_w on the wheel shaft.

z = 2N_w = 50n = 960 rpmP = 2.2 kWη = 65%

T_w = N·m

2.Gears & gear trainsHard

An open belt connects a 150 mm driving pulley at 960 rpm to a 250 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 800 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 150 mmn₁ = 960 rpmd₂ = 250 mmC = 1200 mmμ = 0.25T₁ = 800 N

P = kW

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 250 N·m. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

T = 250 N·md_i/d_o = 0.6τ_allow = 50 MPa

d_o = mm

4.Torsion of shaftsHard

A solid aluminium shaft transmits 30 kW at 1500 rpm. The shear stress must not exceed 50 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.

P = 30 kWN = 1500 rpmτ_allow = 50 MPaθ_allow = 0.5°/mG = 26 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 10 MPa, σᵧ = 70 MPa and τₓᵧ = −50 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 = 10 MPaσ_y = 70 MPaτ_xy = −50 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 400 mm and wall thickness 15 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 20° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 400 mmt = 15 mmp = 1.5 MPaα = 20°

|τ_n| = MPa

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 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 0.25 mm in total.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 300 mm, E = 110 GPaδ = 0.25 mm

P = kN

8.Axial stress & strainHard

A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of copper (diameter 50 mm, length 200 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, 400 mm, E = 200 GPasegment 2: copper, Ø50 mm, 200 mm, E = 120 GPaP = 40 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 = 3 mE = 200 GPa (steel)b × h = 30 × 40 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 115 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 115 kNn = 2L = 2 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

An 8 kg mass on a spring with a stiffness of 150 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency f_d of its free vibration in Hz.

m = 8 kgk = 150 kN/mζ = 0.45

f_d = Hz

12.Mechanical vibrationsHard

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 30 kN/m and k₂ = 50 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 25 kgk₁ = 30 kN/mk₂ = 50 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: T_w = 355.6 N·m

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 = 960·2/50 = 38.4 rpm

Output torque

ω_w = 2π·38.4/60 = 4.021 rad/s

T_w = η·P/ω_w = 0.65·2200/4.021 = 355.6 N·m

Answer

T_w = 355.6 N·m

2.Gears & gear trains

Answer: P = 3.224 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.

Angle of wrap

θ = 3.058 rad

Slack-side tension

e^μθ = e^0.7646 = 2.148, T₂ = 800/2.148 = 372.4 N

Power

P = (800 − 372.4)·7.54 = 3.224 kW

Answer

P = 3.224 kW

3.Torsion of shafts

Answer: d_o = 30.81 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 = 250 N·m = 250000 N·mm

1 − k⁴ = 1 − 0.6⁴ = 0.8704

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·250000/(π·50·0.8704)) = 30.81 mm

d_i = 0.6·30.81 = 18.49 mm

Answer

d_o = 30.81 mm

A solid shaft would need d = 29.42 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 54.11 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 30000/157.1 = 191 N·m

Diameter for strength

T = 191 N·m = 190986 N·mm

d_τ = ∛(16·190986/(π·50)) = 26.89 mm

Diameter for stiffness

θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm

G = 26 GPa = 26000 MPa

d_φ = (32·190986/(π·26000·8.727 × 10⁻⁶))^1/4 = 54.11 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(26.89, 54.11) = 54.11 mm

Answer

d = 54.11 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: θ_p = −60.48 °

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·−50, 10 − 70) = −121°

Angle on the element

θ_p = −60.48°

Answer

θ_p = −60.48°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: |τ_n| = 3.214 MPa

Wall stresses

r = 400/2 = 200 mm

σ_h = pr/t = 1.5·200/15 = 20 MPa

σ_a = pr/(2t) = 10 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.

|τ_n| = (20 − 10)·sin 20°·cos 20° = 3.214 MPa

Answer

|τ_n| = 3.214 MPa

7.Axial stress & strain

Answer: P = 35.33 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) = 600/(200000·490.9) = 6.112 × 10⁻⁶ mm/N

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

Load

P = 0.25/7.076 × 10⁻⁶ = 35330 N = 35.33 kN

Answer

P = 35.33 kN

8.Axial stress & strain

Answer: σ₂ = 16.86 MPa

Stiffnesses

k₁ = 200000·490.9/400 = 245437 N/mm

k₂ = 120000·1963/200 = 1.178 × 10⁶ N/mm

Share of segment 2

F₂ = 40000·1.178 × 10⁶/1.424 × 10⁶ = 33103 N

Stress

σ₂ = 33103/1963 = 16.86 MPa (compression)

Answer

σ₂ = 16.86 MPa

9.Column buckling

Answer: P_cr = 78.96 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 = 40·30³/12 = 90000 mm⁴

Effective length

KL = 0.5·3000 = 1500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·90000/1500² = 78957 N = 78.96 kN

Answer

P_cr = 78.96 kN

10.Column buckling

Answer: d = 60.38 mm

Required critical load

P_cr = n·P = 2·115000 = 230000 N

KL = 0.7·2000 = 1400 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·230000·1400²/(π³·70000))^(1/4) = 60.38 mm

Answer

d = 60.38 mm

Check: the slenderness KL/r is 92.74, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: f_d = 19.46 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 = √(150000/8) = 136.9 rad/s

f_n = 136.9/(2π) = 21.79 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 21.79·√(1 − 0.45²) = 19.46 Hz

Answer

f_d = 19.46 Hz

12.Mechanical vibrations

Answer: k_eq = 80 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 = 30 + 50 = 80 kN/m

Answer

k_eq = 80 kN/m

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