Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

New problems Back to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

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 worm with 3 start(s) turns at 960 rpm and drives a worm wheel with 48 teeth. Find the speed n_w of the wheel.

z = 3N_w = 48n = 960 rpm

n_w = rpm

2.Gears & gear trainsHard

An open belt connects a 100 mm driving pulley at 960 rpm to a 200 mm pulley; the centres are 800 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1200 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 100 mmn₁ = 960 rpmd₂ = 200 mmC = 800 mmμ = 0.3T₁ = 1200 N

P = kW

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 30 N·m. The two ends of a gauge length of 200 mm rotate 0.56° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 30 N·md = 20 mmL = 200 mmφ = 0.56°

G = GPa

4.Torsion of shaftsHard

A solid shaft must transmit 30 kW at 3000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 30 kWN = 3000 rpmτ_allow = 75 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 140 MPa, σᵧ = −30 MPa and τₓᵧ = 30 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 = 140 MPaσ_y = −30 MPaτ_xy = 30 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 5 mm holds gas at a gauge pressure of 2 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 1000 mmt = 5 mmp = 2 MPaα = 55°

σ_n = MPa

7.Axial stress & strainMedium

Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).

P = 65 kNd = 20 mmL = 2 mE = 100 GPa

ε = µε

8.Axial stress & strainHard

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 300 mm, E = 70 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, Ø40 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 300 mm, E = 70 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 = 4 mE = 200 GPa (steel)d = 30 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 = 4 mE = 200 GPa (steel)d = 20 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 2 cycles later by a peak of 15.1 mm. Find the damping ratio ζ.

x₀ = 25 mmx₂ = 15.1 mmn = 2

ζ =

12.Mechanical vibrationsHard

A 100 kg mass hangs from spring k₃ = 100 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 kN/m and k₂ = 80 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 = 100 kgk₁ = 50 kN/mk₂ = 80 kN/mk₃ = 100 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n_w = 60 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.

Wheel speed

n_w = n·z/N_w = 960·3/48 = 60 rpm

Answer

n_w = 60 rpm

2.Gears & gear trains

Answer: P = 3.592 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 = 3.017 rad

Slack-side tension

e^μθ = e^0.905 = 2.472, T₂ = 1200/2.472 = 485.5 N

Power

P = (1200 − 485.5)·5.027 = 3.592 kW

Answer

P = 3.592 kW

3.Torsion of shafts

Answer: G = 39.08 GPa

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 moment and angle

J = π·20⁴/32 = 15708 mm⁴

φ = 0.56·π/180 = 0.009774 rad

Shear modulus

T = 30 N·m = 30000 N·mm

G = TL/(Jφ) = 30000·200/(15708·0.009774) = 39081 MPa

G = 39.08 GPa

Answer

G = 39.08 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: d = 18.65 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 30000/314.2 = 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: θ_p = 9.72 °

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(2·30, 140 − −30) = 19.44°

Angle on the element

θ_p = 9.72°

Answer

θ_p = 9.72°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_n = 132.9 MPa

Wall stresses

r = 1000/2 = 500 mm

σ_h = pr/t = 2·500/5 = 200 MPa

σ_a = pr/(2t) = 100 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.

σ_n = 100·sin²55° + 200·cos²55° = 132.9 MPa

Answer

σ_n = 132.9 MPa

7.Axial stress & strain

Answer: ε = 2069 µε

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 = 65·1000/314.2 = 206.9 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 206.9/100000 = 0.002069 = 2069 µε

Answer

ε = 2069 µε

8.Axial stress & strain

Answer: σ₂ = 15.99 MPa

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.

Stiffnesses

k₁ = 200000·1257/600 = 418879 N/mm

k₂ = 70000·706.9/300 = 164934 N/mm

Share of segment 2

F₂ = 40000·164934/583813 = 11300 N

Stress

σ₂ = 11300/706.9 = 15.99 MPa (compression)

Answer

σ₂ = 15.99 MPa

9.Column buckling

Answer: P_cr = 4.905 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 1·4000 = 4000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·39761/4000² = 4905 N = 4.905 kN

Answer

P_cr = 4.905 kN

10.Column buckling

Answer: σ_cr = 12.34 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 = 0.5·4000 = 2000 mm

KL/r = 2000/5 = 400

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/400² = 12.34 MPa

Answer

σ_cr = 12.34 MPa

11.Mechanical vibrations

Answer: ζ = 0.04009

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 = (1/2)·ln(25/15.1) = 0.2521

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2521/√(39.48 + 0.06355) = 0.04009

Answer

ζ = 0.04009

12.Mechanical vibrations

Answer: f_n = 3.784 Hz

Equivalent stiffness

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

k₁₂ = 50 + 80 = 130 kN/m

k_eq = 130·100/(130 + 100) = 56.52 kN/m

Natural frequency

ω_n = √(56522/100) = 23.77 rad/s

f_n = 23.77/(2π) = 3.784 Hz

Answer

f_n = 3.784 Hz

ExamProblems · examproblems.com/courses/machine-design