Machine Design I: Static and Dynamic Loading, Failure Theories, Fatigue and the S–N Diagram, Bolted, Riveted and Welded Joints, and Shafts
1. Design for static and dynamic loading
Design compares a stress with a strength through a factor of safety n = strength/allowable stress, the strength being the yield stress S_y for a ductile part and the ultimate stress S_ut for a brittle one. Geometry raises the local stress above the nominal one at holes, fillets and grooves: the stress concentration factor K_t = σ_max/σ_nominal is a function of the shape only, larger for a sharper notch. Under a static load a ductile part yields locally and relieves the peak, so K_t is often ignored; for a brittle material, and for fatigue, it must be applied.
A load that is not static raises the stress by a dynamic factor. A load suddenly applied, without impact, produces twice the static stress and deflection. A weight dropped through a height h onto a member whose static deflection under the weight is δ_st gives the impact factor 1 + √(1 + 2h/δ_st): for δ_st = 0.5 mm and h = 2 mm it is 1 + √9 = 4, and for h = 0 it falls to 2. Stiffness matters as much as strength here, because a large δ_st lowers the factor: a spring in the load path cushions the blow.
2. Theories of failure
A component in a general state of stress is compared with a simple tension test through a failure theory, working from the principal stresses σ₁ ≥ σ₂ ≥ σ₃. The maximum principal stress (Rankine) theory says failure when σ₁ reaches the strength; it suits brittle materials. The maximum shear stress (Tresca) theory says a ductile material yields when the maximum shear, (σ₁ − σ₃)/2, reaches S_y/2, that is σ₁ − σ₃ = S_y. The distortion-energy (von Mises) theory says yielding when the energy of shape change equals its value at yield in tension: σ_v = √[½((σ₁ − σ₂)² + (σ₂ − σ₃)² + (σ₃ − σ₁)²)] = S_y, which for plane stress is √(σ₁² − σ₁σ₂ + σ₂²).
| Theory | Used for | Yield in pure shear |
|---|---|---|
| Maximum principal stress (Rankine) | Brittle materials | — |
| Maximum shear stress (Tresca) | Ductile materials, conservative | τ_y = 0.5 S_y |
| Distortion energy (von Mises) | Ductile materials, most accurate | τ_y = S_y/√3 = 0.577 S_y |
For σ₁ = 100 and σ₂ = 50 MPa (σ₃ = 0), σ_v = √(10 000 − 5 000 + 2 500) = 86.6 MPa, while Tresca’s σ₁ − σ₃ = 100. A shaft in pure torsion, τ = 60 MPa, has σ₁ = 60, σ₃ = −60: with S_y = 240 MPa, Tresca gives n = S_y/(2τ) = 2, von Mises n = S_y/(√3τ) = 240/103.9 = 2.31. The Tresca hexagon lies inside the von Mises ellipse, so Tresca is always the more conservative, by at most about 15%. Both theories are blind to a hydrostatic stress: a material squeezed equally in all directions does not yield.
3. Fatigue strength and the S–N diagram
Under a fluctuating stress a part can fail at a stress far below S_ut. The S–N diagram plots the stress amplitude against the number of cycles to failure on log axes. Steels show a knee near 10⁶ cycles below which the life is effectively infinite, the endurance limit: for the standard polished specimen S_e′ ≈ 0.5 S_ut (for S_ut up to about 1400 MPa), so 400 MPa for S_ut = 800 MPa. Non-ferrous metals show no knee and are given a strength at a stated life. The real part’s endurance limit is S_e = k_a k_b k_c k_d k_e S_e′, with modifying factors for surface finish, size, load type, temperature and reliability, each at most 1, and a notch enters through the fatigue notch factor K_f = 1 + q(K_t − 1), q the notch sensitivity between 0 and 1.
A mean stress lowers the allowable amplitude. With σ_a the stress amplitude and σ_m the mean, the Goodman line is σ_a/S_e + σ_m/S_ut = 1/n, the Soderberg line replaces S_ut by S_y (and is the more conservative), and the Gerber parabola n σ_a/S_e + (n σ_m/S_ut)² = 1 fits ductile data best and lies above the Goodman line. For σ_a = σ_m = 100 MPa, S_e = 200 and S_ut = 600 MPa, 1/n = 0.5 + 0.167 = 0.667, so n = 1.5 by Goodman. For a stress that changes level, Miner’s rule adds the damage of each block, Σ n_i/N_i = 1 at failure: 10⁴ cycles at a stress whose life is 10⁵ and 4 × 10⁴ cycles at a stress whose life is 2 × 10⁵ use up 0.1 + 0.2 = 0.3 of the life.
4. Bolted, riveted and welded joints
A bolted joint is tightened to a preload F_i, which clamps the members. An external tensile load P is shared: the bolt takes C·P and the members give up (1 − C)P of their clamping, where the joint constant C = k_b/(k_b + k_m) is the bolt’s share of the joint stiffness. The bolt’s load is F_b = F_i + CP: for F_i = 12 kN, C = 0.25 and P = 8 kN, F_b = 14 kN, only 2 kN above the preload, which is why a preloaded bolt is good in fatigue. The tensile capacity is the tensile stress area times the strength: an M20 bolt has A_t = 245 mm², and class 8.8 (S_ut 800, S_y 640 MPa) yields at 245 × 640 = 156.8 kN. Bolts loaded in shear resist by τ = F/A per shear plane and, at the plate, by bearing.
A riveted joint fails in one of three ways per pitch p, with rivet diameter d and plate thickness t: tearing of the plate between holes, P_t = (p − d) t σ_t; shearing of the rivet, P_s = n τ πd²/4 (n rivets in single shear); crushing of the plate or rivet, P_c = n d t σ_c. The efficiency is η = min(P_t, P_s, P_c)/(p t σ_t), the strength of the joint against that of the solid plate. For p = 50 mm, d = 16 mm, t = 10 mm and allowable σ_t = 100, τ = 60 and σ_c = 150 MPa: P_t = 34 000 N, P_s = 12 064 N, P_c = 24 000 N, so the rivet shear governs and η = 12 064/50 000 = 24.1%. A lap joint rivets in single shear, a double-cover butt joint in double shear.
A fillet weld fails in shear across its throat, the shortest section, which for a 45° weld of leg s is 0.707 s. The load a weld of length L carries is P = 0.707 s L τ: for s = 10 mm, L = 120 mm and τ = 75 MPa, P = 0.707 × 10 × 120 × 75 = 63.6 kN. A butt weld is stressed like the plate, P = t L σ, and with full penetration is as strong as the plate, and a weld loaded eccentrically takes a direct shear plus a moment term, whose vector sum is checked at the critical point. The weld is the only common joint that adds no holes, and it also leaves residual stresses and a heat-affected zone.
5. Design of shafts
A shaft carries a bending moment M from its gears and pulleys and a torque T from the power it transmits, and the surface stresses are σ = 32M/(πd³) and τ = 16T/(πd³). For a ductile shaft under the maximum-shear theory these combine into the equivalent torque T_e = √(M² + T²), with the diameter fixed by τ_allow = 16T_e/(πd³), and the equivalent bending moment M_e = ½[M + √(M² + T²)] for the bending-stress form. The ASME code adds shock and fatigue factors, d³ = [16/(πτ_allow)]√[(K_b M)² + (K_t T)²], with K_b and K_t at least 1 and higher for suddenly applied or shock loads. For M = 300 N·m, T = 400 N·m and τ_allow = 50 MPa: T_e = 500 N·m, d³ = 16 × 500/(π × 50 × 10⁶) = 5.09 × 10⁻⁵ m³, so d = 37.1 mm, rounded up to the next standard size.
Key takeaways
- K_t = σ_max/σ_nominal depends on shape only; impact factor 1 + √(1 + 2h/δ_st), and 2 for a sudden load.
- Ductile: Tresca σ₁ − σ₃ = S_y (conservative), von Mises √(σ₁² − σ₁σ₂ + σ₂²) = S_y; brittle: maximum principal stress; pure shear yields at 0.5 S_y or 0.577 S_y.
- S_e′ ≈ 0.5 S_ut; Goodman σ_a/S_e + σ_m/S_ut = 1/n, Soderberg the most conservative; Miner Σ n/N = 1.
- Bolt load F_i + CP; rivet joint fails by tearing, shearing or crushing, η = min/(ptσ_t); fillet weld P = 0.707 s L τ.
- Shaft: T_e = √(M² + T²) and τ = 16T_e/(πd³); ASME adds K_b and K_t for shock and fatigue.
Practice questions (15)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
The theoretical stress concentration factor K_t of a notched member depends on:
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Answer: A — the geometry of the notch only
K_t = σ_max/σ_nominal is a purely geometric elastic factor. How much of it a real part feels in fatigue depends on the material through the notch sensitivity q, giving K_f = 1 + q(K_t − 1).A weight is dropped from a height of 2 mm onto a member whose static deflection under the same weight is 0.5 mm. The impact factor, the ratio of the maximum stress to the static stress, is ____.
Numerical answer — type the value.
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Answer: 4
Impact factor = 1 + √(1 + 2h/δ_st) = 1 + √(1 + 8) = 1 + 3 = 4. A suddenly applied load (h = 0) gives 2, and the factor tends to 1 only for an infinitely compliant member.At a point in a component, the principal stresses in plane stress are 100 MPa and 50 MPa. The von Mises equivalent stress, correct to one decimal place, is ____ MPa.
Numerical answer — type the value.
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Answer: 86.6
σ_v = √(σ₁² − σ₁σ₂ + σ₂²) = √(10 000 − 5 000 + 2 500) = √7500 = 86.6 MPa. The maximum-shear theory’s equivalent, σ₁ − σ₃ = 100 MPa, is larger, which is what makes it conservative.A ductile shaft with yield strength 240 MPa is in pure torsion with a shear stress of 60 MPa. The factor of safety by the distortion-energy (von Mises) theory, correct to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.31
For pure shear σ_v = √3 τ = 103.9 MPa, so n = 240/103.9 = 2.31. The maximum-shear theory gives σ₁ − σ₃ = 2τ = 120 and n = 2, more conservative.For a brittle material such as cast iron, the most appropriate of the classical failure theories is the:
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Answer: A — maximum principal stress theory
Brittle materials fracture, without yielding, when the largest tensile stress reaches the ultimate strength, which is exactly Rankine’s criterion. Tresca and von Mises are yield criteria for ductile metals.A steel has an ultimate tensile strength of 800 MPa. The endurance limit of the standard polished rotating-beam specimen, using S_e′ = 0.5 S_ut, is ____ MPa.
Numerical answer — type the value.
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Answer: 400
S_e′ = 0.5 × 800 = 400 MPa. The rule applies to steels up to about 1400 MPa; above that the endurance limit levels off near 700 MPa. The real part’s limit is lower after the surface, size and reliability factors.A component carries a stress with amplitude 100 MPa and mean 100 MPa. Its endurance limit is 200 MPa and its ultimate strength 600 MPa. The factor of safety by the Goodman criterion is ____.
Numerical answer — type the value.
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Answer: 1.5
1/n = σ_a/S_e + σ_m/S_ut = 100/200 + 100/600 = 0.5 + 0.1667 = 0.6667, so n = 1.5. Using the amplitude alone, 200/100 = 2, ignores the mean stress.A part is subjected to 10⁴ cycles at a stress for which its life is 10⁵ cycles, followed by 4 × 10⁴ cycles at a stress for which its life is 2 × 10⁵ cycles. By Miner’s rule, the fraction of life used up is ____.
Numerical answer — type the value.
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Answer: 0.3
Damage = Σ n_i/N_i = 10⁴/10⁵ + 4 × 10⁴/(2 × 10⁵) = 0.1 + 0.2 = 0.3. Failure is predicted when the sum reaches 1, so 70% of the life remains, with no account of the order of the blocks.An M20 bolt of property class 8.8 has a tensile stress area of 245 mm² and a yield strength of 640 MPa. The axial load at which it yields is ____ kN.
Numerical answer — type the value.
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Answer: 156.8
P_y = A_t S_y = 245 × 640 = 156 800 N = 156.8 kN. The tensile stress area, not the shank’s cross-section, is used because the threads reduce the area that carries the load.A bolt is tightened to a preload of 12 kN. The joint constant is 0.25 and an external tensile load of 8 kN is then applied to the joint. The total load in the bolt is ____ kN.
Numerical answer — type the value.
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Answer: 14
F_b = F_i + C P = 12 + 0.25 × 8 = 14 kN; the members keep 12 − 0.75 × 8 = 6 kN of clamping. Adding the whole 8 kN gives 20 kN, which forgets that the stiff members take most of it.A single-riveted lap joint has pitch 50 mm, rivet diameter 16 mm and plate thickness 10 mm, one rivet per pitch. The allowable stresses are 100 MPa in tension of the plate, 60 MPa in shear of the rivet and 150 MPa in crushing. The efficiency of the joint, correct to one decimal place, is ____ %.
Numerical answer — type the value.
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Answer: 24.1
P_t = (50 − 16) × 10 × 100 = 34 000 N, P_s = 60 × π × 16²/4 = 12 064 N, P_c = 16 × 10 × 150 = 24 000 N. The least, 12 064 N, governs, and η = 12 064/(50 × 10 × 100) = 0.2413, that is 24.1%.A single transverse fillet weld of leg size 10 mm and length 120 mm is loaded in shear along its throat with an allowable shear stress of 75 MPa. Taking the throat as 0.707 times the leg, the load it can carry is ____ kN, correct to one decimal place.
Numerical answer — type the value.
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Answer: 63.6
P = 0.707 s L τ = 0.707 × 10 × 120 × 75 = 63 630 N = 63.6 kN. Using the leg instead of the throat gives 90 kN, overstating the strength by 41%.A solid shaft carries a bending moment of 300 N·m and a torque of 400 N·m. The allowable shear stress is 50 MPa, and the maximum-shear-stress theory applies. The minimum diameter, correct to one decimal place, is ____ mm.
Numerical answer — type the value.
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Answer: 37.1
T_e = √(300² + 400²) = 500 N·m, and d³ = 16T_e/(πτ) = 16 × 500/(π × 50 × 10⁶) = 5.093 × 10⁻⁵ m³, so d = 37.07 mm ≈ 37.1 mm. Using T alone gives 34.4 mm, and M + T = 700 gives 41.5 mm.Which of the following statements about fatigue design are true?
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Answer: A — For steels the endurance limit of the standard specimen is about half the ultimate tensile strength; B — A ground or polished surface gives a higher endurance limit than a machined or hot-rolled one; C — The Soderberg line is more conservative than the Goodman line
(A) The rule S_e′ ≈ 0.5 S_ut. (B) The surface factor is below 1 for rougher finishes. (C) Soderberg uses S_y in place of S_ut, and S_y < S_ut. (D) False: a compressive mean stress usually helps, which is why shot peening, which leaves compressive residual stress, improves fatigue life.Which of the following statements about the failure theories of ductile materials are true?
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Answer: A — The maximum shear stress theory is more conservative than the distortion energy theory; B — The distortion energy theory predicts a yield stress in pure shear of S_y/√3; C — The maximum shear stress theory predicts a yield stress in pure shear of 0.5 S_y
(A) The Tresca hexagon lies inside the von Mises ellipse. (B) τ_y = 0.577 S_y. (C) τ_y = 0.5 S_y. (D) False: equal principal stresses give σ₁ − σ₃ = 0 and σ_v = 0, so neither yields, however large the pressure.