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 4 start(s) turns at 2900 rpm and drives a worm wheel with 40 teeth. Find the speed n_w of the wheel.

z = 4N_w = 40n = 2900 rpm

n_w = rpm

2.Gears & gear trainsHard

In a compound gear train, gear A (18 teeth) on the input shaft turning at 1450 rpm drives gear B (24 teeth). Gear C (24 teeth) is fixed to the same shaft as B and drives gear D (36 teeth) on the output shaft. Find the output speed n_out.

n_in = 1450 rpmA = 18B = 24C = 24D = 36

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 = 30 mmL = 0.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 60 mm is fixed at A. A torque of 250 N·m acts at B, 1 m from A, and a torque of 500 N·m in the same direction acts at the free end C, a further 1 m along. Find the maximum shear stress τ_max in the shaft.

d = 60 mmT_B = 250 N·mT_C = 500 N·mAB = 1 mBC = 1 m

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 70 MPa, σᵧ = 40 MPa and τₓᵧ = 40 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 70 MPaσ_y = 40 MPaτ_xy = 40 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = −60 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 Tresca (maximum shear stress) criterion.

σ_x = −10 MPaσ_y = −60 MPaτ_xy = 30 MPaσ_Y = 350 MPa

n =

7.Axial stress & strainMedium

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 60 mm, E = 110 GPa). Both are 500 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 110 GPaL = 500 mm

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø100/Ø50 mm, E = 110 GPa, σ_allow = 80 MPa

P_max = kN

9.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 3 mE = 70 GPa (aluminium)b × h = 20 × 60 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 = 4 mE = 70 GPa (aluminium)d = 60 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 2 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 170 N·s/m. Find the damping ratio ζ.

m = 2 kgk = 10 kN/mc = 170 N·s/m

ζ =

12.Mechanical vibrationsHard

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 50 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 = 50 kN/mk₂ = 60 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n_w = 290 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 = 2900·4/40 = 290 rpm

Answer

n_w = 290 rpm

2.Gears & gear trains

Answer: n_out = 725 rpm

First stage

n_B = 1450·18/24 = 1088 rpm = n_C

Second stage

n_out = 1088·24/36 = 725 rpm

Answer

n_out = 725 rpm

total ratio = 2 : 1

3.Torsion of shafts

Answer: φ = 9.237 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 1000·1000·500/(39000·79522) = 0.1612 rad

φ = 0.1612·180/π = 9.237°

Answer

φ = 9.237°

4.Torsion of shafts

Answer: τ_max = 17.68 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 500 N·m

T_AB = T_C + T_B = 500 + 250 = 750 N·m

Maximum shear stress

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

|T|_max = 750 N·m

τ_max = 16·750000/(π·60³) = 17.68 MPa

Answer

τ_max = 17.68 MPa

5.Mohr’s circle & stress transformation

Answer: τ_max = 42.72 MPa

Radius of Mohr’s circle

τ_max = R = √(((70 − 40)/2)² + 40²) = 42.72 MPa

Answer

τ_max = 42.72 MPa

6.Mohr’s circle & stress transformation

Answer: n = 4.481

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 = −35 MPa, R = 39.05 MPa

σ₁ = 4.051 MPa, σ₂ = −74.05 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 78.1 MPa

n = 350/78.1 = 4.481

Answer

n = 4.481

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

7.Axial stress & strain

Answer: δ = 0.2436 mm

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·20²/4 = 314.2 mm²

A_t = π(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·5027 = 6.158 × 10⁸ N

Shortening

δ = 300000·500/6.158 × 10⁸ = 0.2436 mm

Answer

δ = 0.2436 mm

8.Axial stress & strain

Answer: P_max = 574.1 kN

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 110·10³·5890 = 7.893 × 10⁸ N

Governing strain

ε = min(0.00075, 0.0007273) = 0.0007273 (bronze)

Largest load

P_max = 0.0007273·7.893 × 10⁸ = 574055 N = 574.1 kN

Answer

P_max = 574.1 kN

9.Column buckling

Answer: P_cr = 6.266 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 = 60·20³/12 = 40000 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·40000/2100² = 6266 N = 6.266 kN

Answer

P_cr = 6.266 kN

10.Column buckling

Answer: σ_cr = 19.83 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²

r = √(I/A) = √(636173/2827) = 15 mm

KL = 0.7·4000 = 2800 mm

KL/r = 2800/15 = 186.7

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/186.7² = 19.83 MPa

Answer

σ_cr = 19.83 MPa

11.Mechanical vibrations

Answer: ζ = 0.601

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·√(10000·2) = 282.8 N·s/m

Damping ratio

ζ = c/c_c = 170/282.8 = 0.601

Answer

ζ = 0.601

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

12.Mechanical vibrations

Answer: f_n = 16.69 Hz

Equivalent stiffness

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

k_eq = 50 + 60 = 110 kN/m

Natural frequency

ω_n = √(110000/10) = 104.9 rad/s

f_n = 104.9/(2π) = 16.69 Hz

Answer

f_n = 16.69 Hz

ExamProblems · examproblems.com/courses/machine-design