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 160 mm pulley at 720 rpm drives a 300 mm pulley by a belt with 4 % slip. Find the speed n₂ of the driven pulley.

n₁ = 720 rpmd₁ = 160 mmd₂ = 300 mms = 4 %

n₂ = rpm

2.Gears & gear trainsHard

An open belt connects a 200 mm driving pulley at 960 rpm to a 300 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 600 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 200 mmn₁ = 960 rpmd₂ = 300 mmC = 1000 mmμ = 0.25T₁ = 600 N

P = kW

3.Torsion of shaftsMedium

A solid steel 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 = 750 N·md = 25 mmL = 1 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 750 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 50 kWN = 750 rpmτ_allow = 40 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 140 MPa, σᵧ = 70 MPa and τₓᵧ = 20 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 = 70 MPaτ_xy = 20 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −80 MPa, σᵧ = 60 MPa and τₓᵧ = −10 MPa (σ_z = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = −80 MPaσ_y = 60 MPaτ_xy = −10 MPaσ_Y = 460 MPa

n =

7.Axial stress & strainMedium

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 70 mm, E = 100 GPa). Both are 300 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.

P = 100 kNrod: Ø20 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 100 GPaL = 300 mm

δ = mm

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 45 kNsegment 1: Ø40 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 1 m, aluminium (E = 70 GPa)

σ_max = 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 = 3 mE = 200 GPa (steel)d = 50 mm

P_cr = kN

10.Column bucklingHard

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

L = 2 mE = 70 GPa (aluminium)d = 50 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 10 kg mass is supported by a spring with a stiffness of 200 kN/m and a viscous damper with a damping coefficient of 1410 N·s/m. Find the damping ratio ζ.

m = 10 kgk = 200 kN/mc = 1410 N·s/m

ζ =

12.Mechanical vibrationsHard

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 20 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 = 25 kgk₁ = 12 kN/mk₂ = 20 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n₂ = 368.6 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

720·160/300 = 384 rpm

With slip

n₂ = 384·0.96 = 368.6 rpm

Answer

n₂ = 368.6 rpm

2.Gears & gear trains

Answer: P = 3.212 kW

Angle of wrap

θ = 3.042 rad

Slack-side tension

e^μθ = e^0.7604 = 2.139, T₂ = 600/2.139 = 280.5 N

Power

P = (600 − 280.5)·10.05 = 3.212 kW

Answer

P = 3.212 kW

3.Torsion of shafts

Answer: φ = 14.01 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·1000/(80000·38350) = 0.2445 rad

φ = 0.2445·180/π = 14.01°

Answer

φ = 14.01°

4.Torsion of shafts

Answer: d = 43.28 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 50000/78.54 = 636.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·636620/(π·40)) = 43.28 mm

Answer

d = 43.28 mm

In practice you would round up to the next standard size.

5.Mohr’s circle & stress transformation

Answer: θ_p = 14.87 °

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(2·20, 140 − 70) = 29.74°

Angle on the element

θ_p = 14.87°

Answer

θ_p = 14.87°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 3.253

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 = −10 MPa, R = 70.71 MPa

σ₁ = 60.71 MPa, σ₂ = −80.71 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 141.4 MPa

n = 460/141.4 = 3.253

Answer

n = 3.253

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

7.Axial stress & strain

Answer: δ = 0.03708 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(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·7461 = 8.09 × 10⁸ N

Shortening

δ = 100000·300/8.09 × 10⁸ = 0.03708 mm

Answer

δ = 0.03708 mm

8.Axial stress & strain

Answer: σ_max = 143.2 MPa

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·40²/4 = 1257 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 35.81 MPa

σ₂ = 143.2 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 143.2 MPa

Answer

σ_max = 143.2 MPa

9.Column buckling

Answer: P_cr = 67.29 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·306796/3000² = 67288 N = 67.29 kN

Answer

P_cr = 67.29 kN

10.Column buckling

Answer: σ_cr = 6.747 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 = 2·2000 = 4000 mm

KL/r = 4000/12.5 = 320

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/320² = 6.747 MPa

Answer

σ_cr = 6.747 MPa

11.Mechanical vibrations

Answer: ζ = 0.4985

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·√(200000·10) = 2828 N·s/m

Damping ratio

ζ = c/c_c = 1410/2828 = 0.4985

Answer

ζ = 0.4985

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

12.Mechanical vibrations

Answer: f_n = 5.694 Hz

Equivalent stiffness

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

k_eq = 12 + 20 = 32 kN/m

Natural frequency

ω_n = √(32000/25) = 35.78 rad/s

f_n = 35.78/(2π) = 5.694 Hz

Answer

f_n = 5.694 Hz

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