Machine Design — Stress Analysis
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
Bending Stress: σ_b = M × y / I = M / Z (where Z = I/y = section modulus). For a circular cross-section, Z = πd³/32. Maximum bending stress at extreme fibers.
Torsional Stress: τ = T × r / J = T / Z_p (Z_p = polar section modulus = πd³/16 for solid round). Shear stress varies linearly from center (zero) to surface (maximum).
Combined Stress (Shaft): σ_max/min = σ/2 ± √((σ/2)² + τ²) where σ = bending stress, τ = torsional shear stress. Use Maximum Normal Stress Theory or Maximum Shear Stress Theory for yield criteria.
Mohr’s Circle: Construct with σ_avg = (σ₁+σ₃)/2 on x-axis, radius R = √(((σ₁-σ₃)/2)² + τ²_xy). Directly gives principal stresses and maximum shear stress.
Fatigue: S-N curve shows stress amplitude (S) vs cycles to failure (N). For steels, endurance limit ≈ 0.5 × UTS (ultimate tensile strength). For N > 10⁶–10⁷ cycles, the curve flattens (fatigue limit).
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Bending Stress in Beams
Fundamental Bending Equation
For a beam under bending moment M:
σ_b = M × y / I
- σ_b = bending stress (Pa or MPa)
- M = bending moment (Nm)
- y = distance from neutral axis (mm)
- I = second moment of area (mm⁴)
For a circular cross-section:
- I = πd⁴/64 (about neutral axis)
- y = d/2
- Section modulus Z = I/y = πd³/32
⚠️ Common GATE mistake: Using d instead of d/2 for y, or mixing up I for a hollow shaft (π(d_o⁴ – d_i⁴)/64).
Bending Stress Distribution
- Maximum tensile stress at the bottom fiber (for positive M causing convex upward bending)
- Maximum compressive stress at the top fiber
- Linear variation: stress ∝ distance from neutral axis
- Neutral axis passes through centroid of cross-section
Bending equation for round bar: σ_b = 32M / (πd³)
For a hollow circular shaft (outer diameter d_o, inner diameter d_i): σ_b = 32M / (π(d_o⁴ – d_i⁴)/d_o) × … actually I = π(d_o⁴ – d_i⁴)/64, and y = d_o/2 σ_b = 32M × (d_o/2) / (π(d_o⁴ – d_i⁴)/64) = 1024M × d_o / (π(d_o⁴ – d_i⁴))
Torsional Stress
Circular Shaft Torsion
For a shaft under torque T:
τ = T × r / J
- τ = shear stress at radius r
- T = applied torque
- r = distance from center (max at outer surface)
- J = polar moment of inertia
For solid round shaft:
- J = πd⁴/32
- τ_max = 16T / (πd³) = T / Z_p where Z_p = πd³/16
For hollow shaft (outer d_o, inner d_i): J = π(d_o⁴ – d_i⁴)/32 τ_max = T × r_max / J = 16T × d_o / (π(d_o⁴ – d_i⁴))
Power transmission: P = 2πNT/60 where N = RPM
💡 GATE often combines torsion with bending for shafts — this is the “combined stress” scenario.
Torsion of Non-Circular Sections
Only circular shafts carry pure torsion without warping. Non-circular sections (rectangular, square) experience warping and non-linear shear stress distribution.
Rectangular shaft: τ_max = T / (α × b × h²) where α depends on h/b ratio (from torsion tables).
This is rarely asked in GATE directly but understanding the concept helps.
Combined Stress Analysis
Shaft Under Bending + Torsion
This is the most common machine design problem in GATE — a rotating shaft carrying both bending loads and torque.
Given: Simultaneous bending moment M and torque T act on a shaft cross-section.
Step 1: Calculate bending stress σ = 32M/(πd³) (at extreme fiber) Step 2: Calculate torsional stress τ = 16T/(πd³) (at extreme fiber) Step 3: Use one of these theories:
Maximum Normal Stress Theory (Rankine’s Theory)
σ_1, σ_2 = (σ/2) ± √((σ/2)² + τ²) σ_eq = σ_max = σ/2 + √((σ/2)² + τ²) [largest principal stress] Design criterion: σ_eq ≤ σ_allow
Maximum Shear Stress Theory (Guest’s Theory)
τ_max = √((σ/2)² + τ²) Design criterion: 2τ_max ≤ σ_allow or τ_max ≤ τ_allow
Distortion Energy Theory (von Mises Theory)
σ_eq = √(σ² + 3τ²) Most accurate for ductile materials Design criterion: σ_eq ≤ σ_allow
⚠️ GATE Pattern: Questions often ask which theory to use for ductile materials — the answer is von Mises or Maximum Shear Stress Theory (both conservative). For brittle materials, use Maximum Normal Stress Theory.
Mohr’s Circle for Stress Analysis
Mohr’s circle provides a graphical solution for transforming stresses.
Construction Steps
- Mark point X(σ_x, +τ_xy) and Y(σ_y, –τ_xy) on σ-τ plane
- Find center C at ((σ_x + σ_y)/2, 0)
- Draw circle with radius R = √(((σ_x – σ_y)/2)² + τ_xy²)
- Principal stresses: σ₁, σ₃ = center ± radius
- Maximum shear stress: τ_max = radius (on horizontal line through center)
Mohr’s Circle for Combined Bending + Torsion
For a shaft surface point (where failure initiates):
- σ_x = σ_bending = 32M/(πd³)
- τ_xy = τ_torsion = 16T/(πd³)
- σ_y = 0 (plane stress condition)
Center: (σ_b/2, 0) Radius: √((σ_b/2)² + τ²) Principal stresses: σ₁ = σ_b/2 + √((σ_b/2)² + τ²), σ₃ = σ_b/2 – √((σ_b/2)² + τ²) Maximum shear stress: τ_max = √((σ_b/2)² + τ²)
This is identical to the formula for σ_max/min above.
Fatigue Analysis — S-N Diagram
S-N Curve Features
The fatigue strength or endurance limit (S_e) is the stress level below which the material can endure infinite cycles without failure.
Key characteristics:
- S-N curve for steels flattens after 10⁶ cycles (≈N_L = 10⁶–10⁷)
- For aluminum/alloys, curve continues to drop (no true fatigue limit)
- S_e ≈ 0.5 × UTS for steels (approximate; actual ranges 0.35–0.55 UTS)
Marin Modification Factors
Actual endurance limit S_e = k × S_e’ where S_e’ is ideal endurance limit and:
| Factor | Parameter | Typical Values |
|---|---|---|
| k_a | Surface finish | Ground: 0.9, Machined: 0.8, Hot rolled: 0.6 |
| k_b | Size factor | 1.0 for d<8mm, decreases for larger |
| k_c | Loading | Axial: 0.7, Bending: 1.0, Torsion: 0.6 |
| k_d | Temperature | Decreases above 450°C |
| k_e | Reliability | 0.9 at 50% survival, 0.75 at 99.9% |
Stress Concentration Factors (K_t, K_f)
K_t (theoretical): Ratio of max stress to nominal stress from elastic analysis (from handbook tables)
K_f (fatigue stress concentration factor): K_f = 1 + q(K_t – 1) where q = sensitivity index (0 to 1)
- q ≈ 1 for brittle materials (ceramics, glass)
- q < 1 for ductile materials (metals show some yielding)
- For fillets, keyways, holes in shafts — K_t values are significant (2–4 range)
⚠️ GATE Trap: Students often forget to apply K_f when a fillet or shoulder radius exists. Always check for stress raisers in fatigue problems.
High-Cycle Fatigue Criterion
For N > 10³ (high-cycle fatigue): S_a × N^b = C (Basquin’s law) where S_a is stress amplitude.
Modified Goodman criterion (common in GATE for reversing stresses): σ_a/S_e + σ_m/UTS = 1
Where σ_a = alternating stress amplitude, σ_m = mean stress, S_e = endurance limit.
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer study timeline.
Three-Dimensional Stress State and Transformation
General Stress Transformation
For a general 3D stress state with σ_x, σ_y, σ_z, τ_xy, τ_yz, τ_zx:
Principal stresses are eigenvalues of the stress tensor: σ₁ ≥ σ₂ ≥ σ₃
Invariant equations: I₁ = σ_x + σ_y + σ_z I₂ = σ_xσ_y + σ_yσ_z + σ_zσ_x – τ_xy² – τ_yz² – τ_zx² I₃ = det(stress tensor)
Maximum shear stress: τ_max = (σ₁ – σ₃)/2 (acts on planes at 45° to principal planes)
Theories of Failure for Ductile Materials
| Theory | Criterion | Application |
|---|---|---|
| Maximum Normal Stress | σ_max ≤ σ_allow | Brittle materials |
| Maximum Shear Stress (Tresca) | τ_max ≤ τ_allow = σ_allow/2 | Conservative, ductile |
| Distortion Energy (von Mises) | σ_eq = √(((σ₁-σ₂)²+(σ₂-σ₃)²+(σ₃-σ₁)²)/2) ≤ σ_allow | Most accurate for ductile |
| Maximum Principal Strain | ε_max ≤ ε_allow | Special cases |
Tresca vs von Mises for shaft combined stress:
- Tresca: τ_allow ≥ √((σ/2)² + τ²)
- von Mises: σ_allow ≥ √(σ² + 3τ²)
For the same shaft under same loading, von Mises is always less conservative (higher allowable stress by factor ≈1.155 at pure torsion where σ=0).
Notch Sensitivity and Fatigue Life Estimation
Notch Sensitivity Index (q)
q = (K_f – 1)/(K_t – 1)
| Material | q (approximately) |
|---|---|
| Aluminum alloys | 0.1–0.3 |
| Steel (low strength) | 0.3–0.5 |
| Steel (high strength) | 0.6–0.9 |
| Titanium | 0.4–0.6 |
High-strength steels are more notch-sensitive — their high hardness makes them more susceptible to crack initiation at stress concentrations.
Soderberg and Modified Goodman Relations
Soderberg (Most Conservative)
σ_a/S_e + σ_m/σ_yt = 1
Modified Goodman (Common in Design)
σ_a/S_e + σ_m/UTS = 1
Gerber (Parabolic, Optimistic)
(σ_a/S_e) + (σ_m/UTS)² = 1
ASME Elliptic (Moderately Conservative)
(σ_a/S_e)² + (σ_m/σ_yt)² = 1
GATE typically uses Modified Goodman for design problems involving combined mean and alternating stresses.
Finite Element Validation
While FE analysis is used in industry, hand calculations using the above formulas remain the basis for GATE questions. Understanding the stress concentration at a fillet or keyway is essential:
Fillet radius to shaft diameter ratio: r/d
- Small r/d (sharp fillet) → high K_t
- Large r/d → lower K_t
For r/d = 0.1, K_t ≈ 1.6 for a stepped shaft in bending. For r/d = 0.05, K_t ≈ 2.0.
Example Problem
GATE 2019: A solid circular shaft of diameter d = 50 mm is subjected to a bending moment M = 1.5 kNm and a torque T = 1 kNm. Using von Mises criterion, find the required diameter if yield strength σ_yt = 300 MPa and factor of safety = 3.
Solution:
Step 1: Calculate stresses at outer surface σ_b = 32M/(πd³) = 32 × 1.5 × 10³/(π × 50³) = 48000/(π × 125000) = 48000/392699 ≈ 122 MPa τ = 16T/(πd³) = 16 × 1.0 × 10³/(π × 125000) = 16000/392699 ≈ 40.7 MPa
Step 2: von Mises equivalent stress σ_eq = √(σ_b² + 3τ²) = √(122² + 3×40.7²) = √(14884 + 4973) = √(19857) ≈ 141 MPa
Step 3: Allowable stress σ_allow = σ_yt/FS = 300/3 = 100 MPa
Step 4: Required diameter σ_eq ∝ M/d³ for bending and T/d³ for torsion — both scale as 1/d³. Required d_new = d × (σ_eq/σ_allow)^(1/3) = 50 × (141/100)^(1/3) = 50 × 1.41^(0.333) ≈ 50 × 1.12 ≈ 56 mm
Check: Since σ_eq > σ_allow at d=50mm, diameter must increase. d_required ≈ 56 mm. Round to standard size.
Content adapted based on your selected roadmap duration. Switch tiers using the selector above.