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 7.5 kW motor running at 2900 rpm drives a pinion with 20 teeth and a module of 2 mm. Find the tangential force F_t on the pinion teeth.

P = 7.5 kWn = 2900 rpmN₁ = 20m = 2 mm

F_t = N

2.Gears & gear trainsHard

An open belt connects a 160 mm driving pulley at 720 rpm to a 315 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 160 mmn₁ = 720 rpmd₂ = 315 mmC = 1200 mmμ = 0.35T₁ = 500 N

P = kW

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 = 400 N·md = 40 mmL = 2.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft of diameter 20 mm is fixed at A. A torque of 1000 N·m acts at B, 0.5 m from A, and a torque of 400 N·m in the opposite direction acts at the free end C, a further 1 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 20 mmT_B = 1000 N·mT_C = 400 N·mAB = 0.5 mBC = 1 mG = 39 GPa

φ_C = °

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 160 MPa, σᵧ = −70 MPa and τₓᵧ = −60 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 = 160 MPaσ_y = −70 MPaτ_xy = −60 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 12 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 50° to the axis of the cylinder. Find the magnitude of the shear stress |τ_n| acting along the weld seam.

D = 1000 mmt = 12 mmp = 2.5 MPaα = 50°

|τ_n| = MPa

7.Axial stress & strainMedium

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

P = 30 kNd = 20 mmL = 1 mE = 117 GPa

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 200 MPatube: Ø120/Ø70 mm, E = 120 GPa, σ_allow = 80 MPa

P_max = kN

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 = 1 mE = 200 GPa (steel)d = 25 mm

P_cr = kN

10.Column bucklingHard

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

L = 2.5 mE = 70 GPa (aluminium)d = 20 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 20 kg mass is supported by a spring with a stiffness of 8 kN/m and a viscous damper with a damping coefficient of 40 N·s/m. Find the damping ratio ζ.

m = 20 kgk = 8 kN/mc = 40 N·s/m

ζ =

12.Mechanical vibrationsHard

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 15 kN/m and k₂ = 100 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 25 kgk₁ = 15 kN/mk₂ = 100 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: F_t = 1235 N

Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: F_t = 2T/d.

Torque on the pinion

ω = 2π·2900/60 = 303.7 rad/s

T = P/ω = 7500/303.7 = 24.7 N·m

Pitch diameter

d = m·N = 2·20 = 40 mm

Tangential force

F_t = 2T/d = 2·24.7/0.04 = 1235 N

Answer

F_t = 1235 N

2.Gears & gear trains

Answer: P = 1.965 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 = 3.012 rad

Slack-side tension

e^μθ = e^1.054 = 2.87, T₂ = 500/2.87 = 174.2 N

Power

P = (500 − 174.2)·6.032 = 1.965 kW

Answer

P = 1.965 kW

3.Torsion of shafts

Answer: φ = 5.845 °

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) = 400·1000·2500/(39000·251327) = 0.102 rad

φ = 0.102·180/π = 5.845°

Answer

φ = 5.845°

4.Torsion of shafts

Answer: φ_C = 9.353 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 400 N·m

T_AB = T_C − T_B = 400 − 1000 = −600 N·m

Twist of each segment

J = π·20⁴/32 = 15708 mm⁴

G = 39 GPa = 39000 MPa

φ_AB = −600000·500/(39000·15708) = −0.4897 rad

φ_BC = 400000·1000/(39000·15708) = 0.6529 rad

Twist of end C

φ_C = −0.4897 + 0.6529 = 0.1632 rad → |φ_C| = 9.353°

Answer

φ_C = 9.353°

5.Mohr’s circle & stress transformation

Answer: θ_p = −13.78 °

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·−60, 160 − −70) = −27.55°

Angle on the element

θ_p = −13.78°

Answer

θ_p = −13.78°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: |τ_n| = 25.65 MPa

Wall stresses

r = 1000/2 = 500 mm

σ_h = pr/t = 2.5·500/12 = 104.2 MPa

σ_a = pr/(2t) = 52.08 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| = (104.2 − 52.08)·sin 50°·cos 50° = 25.65 MPa

Answer

|τ_n| = 25.65 MPa

7.Axial stress & strain

Answer: δ = 0.8162 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·20²/4 = 314.2 mm²

σ = P/A = 30·1000/314.2 = 95.49 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 95.49/117000 = 0.0008162 = 816.2 µε

Elongation

δ = ε·L = 0.0008162·1000 mm = 0.8162 mm

Answer

δ = 0.8162 mm

8.Axial stress & strain

Answer: P_max = 764.5 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

A_r = π·40²/4 = 1257 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·7461 = 1.147 × 10⁹ N

Governing strain

ε = min(0.001, 0.0006667) = 0.0006667 (copper)

Largest load

P_max = 0.0006667·1.147 × 10⁹ = 764454 N = 764.5 kN

Answer

P_max = 764.5 kN

9.Column buckling

Answer: P_cr = 9.462 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·19175/2000² = 9462 N = 9.462 kN

Answer

P_cr = 9.462 kN

10.Column buckling

Answer: σ_cr = 11.05 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.5·2500 = 1250 mm

KL/r = 1250/5 = 250

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/250² = 11.05 MPa

Answer

σ_cr = 11.05 MPa

11.Mechanical vibrations

Answer: ζ = 0.05

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·√(8000·20) = 800 N·s/m

Damping ratio

ζ = c/c_c = 40/800 = 0.05

Answer

ζ = 0.05

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

12.Mechanical vibrations

Answer: k_eq = 115 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 = 15 + 100 = 115 kN/m

Answer

k_eq = 115 kN/m

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