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Stress-strain, bending, torsion, shear and failure theories for mechanical and civil core prep.
| Stress Type | Symbol | Formula | Description |
|---|---|---|---|
| Tensile | σ_t | F/A (stretching) | Pulling apart — rod elongates |
| Compressive | σ_c | F/A (crushing) | Pushing together — rod shortens |
| Shear | τ | F/A (parallel to face) | Sliding force across a plane |
| Bearing | σ_b | F/(d×t) | Local compression at pin holes |
| Thermal | σ_th | E·α·ΔT | Constrained expansion/contraction |
| Bending | σ_b | My/I | Tension on one side, compression on other |
| Torsional | τ | T·r/J | Shear stress from twisting |
| Strain Type | Symbol | Formula | Description |
|---|---|---|---|
| Linear (tensile/compressive) | ε | ΔL/L₀ | Change in length / original length |
| Shear strain | γ | tan(φ) ≈ φ (rad) | Angular deformation |
| Lateral strain | ε_lat | −ν·ε_axial | Poisson effect — lateral contraction |
| Volumetric strain | ε_v | ΔV/V₀ = ε₁+ε₂+ε₃ | Change in volume / original |
| Thermal strain | ε_th | α·ΔT | Free expansion: ε = α·ΔT |
─── Engineering Stress-Strain Curve for Mild Steel ───
σ (stress)
│ B (UTS)
│ / │ / Necking begins
│ / │ A───
│ │ Upper yield D ── Fracture
│ │ Lower yield │
│ │_______________| C
│ O P E L ε (strain)
│ │ │ │ │
│ │ │ │ └── Elastic limit
│ │ │ └─────── Yield point
│ │ └──────────── Proportional limit
│ └───────────────── Origin
Key Points:
O–P : Linear elastic region (Hooke's law applies)
P–E : Non-linear elastic (still recoverable)
E–A : Yielding (upper & lower yield points)
A–B : Strain hardening (stress increases with strain)
B–C : Necking (localized deformation)
C–D : Fracture
σ_y = Yield strength (235 MPa for mild steel)
σ_u = Ultimate tensile strength (400–550 MPa for mild steel)
% elongation = (L_f − L₀)/L₀ × 100
% reduction in area = (A₀ − A_f)/A₀ × 100─── Hooke's Law (Linear Elastic Region) ───
Uniaxial stress:
σ = E × ε
E = Young's modulus (Modulus of Elasticity)
E_steel ≈ 200 GPa
E_aluminum ≈ 70 GPa
E_copper ≈ 120 GPa
E_concrete ≈ 25–30 GPa
E_timber ≈ 10–12 GPa
Shear (modulus of rigidity):
τ = G × γ
G = Shear modulus
G_steel ≈ 80 GPa
G_aluminum ≈ 26 GPa
Bulk modulus:
K = −V × dP/dV = E / (3(1 − 2ν))
K_steel ≈ 160 GPa
K_water ≈ 2.2 GPa
Poisson's ratio:
ν = −(lateral strain) / (axial strain)
ν_steel ≈ 0.30
ν_aluminum ≈ 0.33
ν_rubber ≈ 0.50 (nearly incompressible)
Relationship between elastic constants:
E = 2G(1 + ν)
E = 3K(1 − 2ν)
E = 9KG / (3K + G)import math
# Composite bar under axial load P
# Bar 1: E1, A1 Bar 2: E2, A2
E1, A1 = 200e3, 500 # MPa, mm² (steel)
E2, A2 = 70e3, 1000 # MPa, mm² (aluminum)
P = 150e3 # N
# Same deformation: δ1 = δ2
# σ1/E1 = σ2/E2 → σ1 = σ2 × E1/E2
# σ1×A1 + σ2×A2 = P
sigma2 = P / (A1 * E1/E2 + A2) # Pa
sigma1 = sigma2 * E1 / E2
# σ₂ = 150000 / (500×200/70 + 1000) = 150000/2428.6
# σ₂ = 61.76 MPa
# σ₁ = 61.76 × 200/70 = 176.47 MPa
# Thermal stress (fully constrained bar)
E = 200e3 # MPa
alpha = 12e-6 # per °C (steel)
dT = 50 # °C
sigma_thermal = E * alpha * dT
# σ_th = 200000 × 12e-6 × 50 = 120 MPaδ = 4PL / (π·E·d₁·d₂). For a uniform bar: δ = PL/(AE). Always check units: if E is in GPa and A in mm², convert consistently.─── Euler-Bernoulli Bending Equation ───
M / I = σ / y = E / R
M = Bending moment at the section (N·mm)
I = Second moment of area (mm⁴)
σ = Bending stress at distance y from NA (MPa)
y = Distance from neutral axis (mm)
E = Young's modulus (MPa)
R = Radius of curvature (mm)
Maximum bending stress:
σ_max = M × y_max / I = M / Z
Z = I / y_max = Section modulus (mm³)
Bending stress distribution:
• Linear from zero at NA to max at extreme fibers
• Tension on one side, compression on the other
• NA passes through centroid for symmetric sections─── Sign Conventions ───
Shear Force (SF / V):
• Left side up → Positive shear
• Right side up → Negative shear
Bending Moment (BM / M):
• Sagging (smile) → Positive BM
• Hogging (frown) → Negative BM
Key Relationships:
• dV/dx = −w(x) [load = -ve slope of SF diagram]
• dM/dx = V(x) [SF = slope of BM diagram]
• d²M/dx² = −w(x)
• Area under load diagram = change in SF
• Area under SF diagram = change in BM
SF is zero at: Maximum/minimum BM
Point load: SF changes by load magnitude
UDL: SF varies linearly
Couple: BM jumps by couple magnitude| Support | Reactions | Displacement | Moment |
|---|---|---|---|
| Simply supported (pin) | Vertical + Horizontal | Free to rotate | Zero moment at support |
| Roller support | Vertical only (perpendicular) | Free to rotate + move horizontally | Zero moment |
| Fixed (cantilever) | Vertical + Horizontal + Moment | No displacement or rotation | Non-zero fixing moment |
| Hinge (internal) | Vertical (no moment transfer) | Free to rotate | Zero moment at hinge |
| Beam / Loading | Max Shear Force | Max Bending Moment | Max Deflection |
|---|---|---|---|
| Cantilever — point load P at free end | V = P | M = −PL (at fixed end) | δ = PL³/(3EI) |
| Cantilever — UDL w over length L | V = wL | M = −wL²/2 (at fixed end) | δ = wL⁴/(8EI) |
| Simply supported — point load P at center | V = P/2 | M = PL/4 (at center) | δ = PL³/(48EI) |
| Simply supported — UDL w | V = wL/2 | M = wL²/8 (at center) | δ = 5wL⁴/(384EI) |
| SS — point load P at distance a from left | V = Pb/L | M = Pab/L (under load) | Complex formula |
| Fixed both ends — UDL w | V = wL/2 | M = wL²/12 (at ends, -ve) | δ = wL⁴/(384EI) |
| Fixed both ends — central point load P | V = P/2 | M = PL/8 (at center, +ve) | δ = PL³/(192EI) |
import math
# Simply supported beam, central point load
P = 20e3 # N (20 kN)
L = 4000 # mm (4 m)
E = 200e3 # MPa (steel)
b, h = 100, 200 # mm (rectangular section)
# Section properties
I = b * h**3 / 12 # mm⁴
Z = I / (h/2) # mm³ (section modulus)
# Maximum bending moment
M_max = P * L / 4 # N·mm
# Maximum bending stress
sigma = M_max / Z # MPa
# Maximum deflection
delta = P * L**3 / (48 * E * I) # mm
print(f"I = {I:.0f} mm⁴")
print(f"M_max = {M_max/1e6:.1f} kN·m")
print(f"σ_max = {sigma:.1f} MPa")
print(f"δ_max = {delta:.2f} mm") <x−a>ⁿ where the term is zero when x < a. This allows a single equation for BM over the entire beam. Apply boundary conditions to find constants of integration. The method works for all standard loadings.─── Shear Stress Distribution in Beams ───
τ = V × Q / (I × b)
V = Shear force at the section (N)
Q = First moment of area above the point about NA (mm³)
= ∫ y dA (from the point to the extreme fiber)
I = Second moment of area of full section (mm⁴)
b = Width at the point where τ is calculated (mm)
For rectangular section (b × h):
τ_max = (3/2) × V / A = 1.5 × V/(bh)
τ_max occurs at the neutral axis!
τ = 0 at top and bottom surfaces
Parabolic distribution:
τ(y) = (V/(2I)) × [(h²/4) − y²]
For circular section (diameter d):
τ_max = (4/3) × V / A
For I-section / wide-flange:
τ_max ≈ V / (web_area) (approximately)
Most shear is carried by the web| Section | I (about centroidal axis) | Z (section modulus) |
|---|---|---|
| Rectangle (b × h) | I = bh³/12 | Z = bh²/6 |
| Circle (radius r) | I = πr⁴/4 | Z = πr³/4 |
| Hollow circle (R, r) | I = π(R⁴−r⁴)/4 | Z = π(R⁴−r⁴)/(4R) |
| Triangle (base b, ht h) | I = bh³/36 | Z = bh²/24 |
| Semicircle (r) | I = (π/8 − 8/9π)r⁴ | — |
Parallel axis theorem: I = I_c + Ad²
Perpendicular axis theorem: J = I_x + I_y (for polar moment)
| Section | Description | Key Advantage |
|---|---|---|
| ISMB (I-beam) | Indian Standard Medium weight Beam | High I with less material |
| ISJB (Joist) | Lighter I-section | Lighter applications |
| ISLB | Light Beam | Reduced weight |
| ISWB | Wide Flange Beam | Better lateral stability |
| Channel (ISMC) | C-shaped section | Asymmetric bending, purlins |
| Angle (ISA) | L-shaped equal/unequal | Bracing, connections |
| T-section | T-shaped, cut from I-beam | Tension flange applications |
| Hollow sections | SHS, RHS, CHS | Torsion resistance, aesthetics |
import math
# I-section properties (built-up)
# Flanges: 2 × (200 × 20) mm at top and bottom
# Web: (10 × 360) mm
bf = 200; tf = 20 # flange width and thickness
bw = 10; hw = 360 # web width and height
h = hw + 2*tf # total height = 400 mm
# Moment of inertia about strong axis (centroidal)
I_web = bw * hw**3 / 12
I_flanges = 2 * (bf * tf**3 / 12 + bf * tf * ((hw + tf)/2)**2)
I_total = I_web + I_flanges
# I_web = 10 × 360³/12 = 3.888 × 10⁸ mm⁴
# I_flanges ≈ 2 × (200 × 20³/12 + 200 × 20 × 190²)
# ≈ 2 × (1.333×10⁵ + 1.444×10⁸) = 2.890 × 10⁸ mm⁴
# I_total ≈ 6.778 × 10⁸ mm⁴
Z = I_total / (h/2) # section modulus
A = 2*bf*tf + bw*hw # total area
print(f"I = {I_total:.3e} mm⁴, Z = {Z:.0f} mm³, A = {A} mm²")─── Torsion of Circular Shafts ───
T / J = τ / r = G × θ / L
T = Torque applied (N·mm)
J = Polar moment of inertia (mm⁴)
τ = Shear stress at radius r (MPa)
r = Radial distance from center (mm)
G = Modulus of rigidity (MPa)
θ = Angle of twist (radians)
L = Length of shaft (mm)
Polar moment of inertia:
Solid shaft (diameter d): J = πd⁴/32
Hollow shaft (D, d): J = π(D⁴ − d⁴)/32
Maximum shear stress:
τ_max = T × r_max / J = T / Z_p
Z_p = J / r_max = polar section modulus
Solid shaft: Z_p = πd³/16
Hollow shaft: Z_p = π(D⁴ − d⁴)/(16D)
Angle of twist:
θ = TL / (GJ) [radians]
Power transmitted:
P = T × ω = T × 2πN / 60 [Watts]
T = P × 60 / (2πN) [N·m]import math
# Design a solid shaft for given power and speed
P = 100e3 # W (100 kW)
N = 300 # RPM
tau_allow = 60 # MPa (allowable shear stress)
G = 80e3 # MPa (steel)
L = 2000 # mm
theta_allow = 1 # degree (max twist)
# Torque
T = P * 60 / (2 * math.pi * N) # N·m
T_Nmm = T * 1000 # N·mm
# T = 3183.1 N·m = 3,183,099 N·mm
# Based on strength (τ_max ≤ τ_allow)
# τ_max = T / (πd³/16)
d_strength = (16 * T_Nmm / (math.pi * tau_allow))**(1/3)
# d = (16 × 3183099 / (π × 60))^(1/3) = 63.7 mm
# Based on stiffness (θ ≤ θ_allow)
# θ = TL/(GJ) → d = [TL × 32 / (G × π × θ_rad)]^(1/4)
theta_rad = math.radians(theta_allow)
d_stiffness = (T_Nmm * L * 32 / (G * math.pi * theta_rad))**(1/4)
# d = (3183099 × 2000 × 32 / (80000 × π × 0.01745))^(1/4)
# Use the larger diameter
d = max(d_strength, d_stiffness)
print(f"Required diameter: {d:.1f} mm")
print(f"From strength: {d_strength:.1f} mm")
print(f"From stiffness: {d_stiffness:.1f} mm")| Property | Solid Shaft | Hollow Shaft |
|---|---|---|
| Polar moment (J) | J = πd⁴/32 | J = π(D⁴−d⁴)/32 |
| Weight (same τ_max) | W | W_hollow < W_solid (lighter!) |
| Strength-weight ratio | Lower | Higher (material at outer radius) |
| Stress distribution | Zero at center, max at surface | More uniform (efficient) |
| Cost | Lower material cost | Higher manufacturing cost |
| Application | General purpose | Aircraft, automobiles, high-performance |
For the same maximum shear stress, a hollow shaft requires less material. A hollow shaft with inner diameter d = D/2 requires only ~78.4% of the weight of a solid shaft of same outer diameter D.
─── Compound Shafts ───
Shafts in series (same torque, different diameters):
θ_total = θ₁ + θ₂ + ... = Σ TᵢLᵢ/(GᵢJᵢ)
Same torque T through each section.
Shafts in parallel (same θ, torque splits):
T_total = T₁ + T₂
θ₁ = θ₂ → T₁L₁/(G₁J₁) = T₂L₂/(G₂J₂)
Torque distributes proportional to GJ/L.
Non-circular sections (approximate):
Rectangular (a × b, a > b):
τ_max ≈ T / (β × a × b²)
θ ≈ T × L / (α × G × a × b³)
β, α depend on a/b ratio (β ≈ 0.208, α ≈ 0.141 for square)
Elliptical (major a, minor b):
τ_max = 2T / (π × a × b²) [at ends of minor axis]
J = π × a × b³ / 4τ_max ≤ τ_allow) AND stiffness (θ ≤ θ_allow, typically 0.25°–1° per meter length). For precision applications, stiffness often governs the design.─── Euler's Critical Buckling Load ───
P_cr = π²EI / (KL)²
P_cr = Critical (crippling) buckling load (N)
E = Young's modulus (MPa)
I = Minimum second moment of area (mm⁴)
L = Actual length of column (mm)
K = Effective length factor (depends on end conditions)
Effective length: L_e = K × L
End Conditions:
┌───────────────────────┬─────┬──────────┐
│ Both ends pinned │ K=1 │ L_e = L │
│ Both ends fixed │K=0.5│ L_e = L/2│
│ One fixed, one free │ K=2 │ L_e = 2L │
│ One fixed, one pinned │K≈0.7│ L_e=0.7L │
└───────────────────────┴─────┴──────────┘
Slenderness ratio:
λ = KL / r
r = √(I/A) = radius of gyration (mm)
Use minimum I (weakest axis)
Euler's formula is valid when:
λ > λ_critical (long columns only)
λ_critical = π√(E/σ_y) for yielding criteria─── Rankine-Gordon Formula ───
P_Rankine = σ_c × A / (1 + a × (KL/r)²)
σ_c = Crushing strength of material (MPa)
A = Cross-sectional area (mm²)
a = Rankine constant (material dependent)
Rankine Constants:
┌─────────────┬───────────┬──────────┐
│ Material │ σ_c (MPa) │ a │
├─────────────┼───────────┼──────────┤
│ Mild steel │ 320 │ 1/7500 │
│ Cast iron │ 550 │ 1/1600 │
│ Wrought iron│ 250 │ 1/9000 │
│ Timber │ 40 │ 1/750 │
│ Aluminum │ 250 │ 1/5000 │
└─────────────┴───────────┴──────────┘
For short columns: P ≈ σ_c × A (crushing)
For long columns: P ≈ P_Euler─── Johnson's Parabolic Formula ───
P_Johnson = A × [σ_y − (σ_y/(2π))² × (KL/r)² × (σ_y/E)]
Or simplified:
σ_cr = σ_y × [1 − (σ_y × λ²)/(4π²E)]
Valid for intermediate slenderness ratios where
Euler's formula over-predicts capacity.
Transition slenderness ratio:
λ_t = π × √(2E/σ_y) = √2 × λ_criticalimport math
# Steel column: I-section
E = 200e3 # MPa
sigma_y = 250 # MPa
L = 3000 # mm
K = 1.0 # pinned-pinned
A = 4000 # mm²
I_min = 5e6 # mm⁴ (minimum I)
r = math.sqrt(I_min / A) # radius of gyration
lam = K * L / r # slenderness ratio
lam_cr = math.pi * math.sqrt(E / sigma_y) # critical λ
print(f"r = {r:.2f} mm, λ = {lam:.1f}, λ_cr = {lam_cr:.1f}")
if lam > lam_cr:
# Euler buckling
P_cr = math.pi**2 * E * I_min / (K * L)**2
sigma_cr = P_cr / A
print(f"Euler: P_cr = {P_cr/1e3:.1f} kN, σ_cr = {sigma_cr:.1f} MPa")
else:
# Rankine formula
sigma_c = 320 # MPa
a = 1/7500
P_rankine = sigma_c * A / (1 + a * lam**2)
print(f"Rankine: P_cr = {P_rankine/1e3:.1f} kN")| Loading | Strain Energy (U) | Notes |
|---|---|---|
| Direct axial load | U = σ²V/(2E) = F²L/(2AE) | V = volume = A × L |
| Bending | U = ∫M²dx/(2EI) | Integrate over beam length |
| Torsion | U = T²L/(2GJ) | Circular shafts |
| Shear | U = τ²V/(2G) | V = volume |
| Transverse shear (beam) | U = ∫kV²dx/(2GA) | k = form factor (rect: 3/2) |
| General (Castigliano) | U = ∫(σ²/(2E) + τ²/(2G)) dV | Combined loading |
─── Castigliano's Theorems ───
First Theorem (for deflections):
δᵢ = ∂U/∂Pᵢ
The deflection at point i in the direction of
force Pᵢ equals the partial derivative of total
strain energy U with respect to Pᵢ.
θᵢ = ∂U/∂Mᵢ (for rotation/angle)
Second Theorem (for forces):
Pᵢ = ∂U/∂δᵢ
Procedure:
1. Write strain energy U in terms of loads
2. Take partial derivative ∂U/∂P (treat P as variable)
3. Substitute P = actual load value
4. If finding deflection where no load exists, apply
a fictitious load P₀, find ∂U/∂P₀, then set P₀ = 0
Example: Cantilever with tip load P
M(x) = −P(L − x) → 0 ≤ x ≤ L
U = ∫₀ᴸ M²/(2EI) dx = ∫₀ᴸ P²(L−x)²/(2EI) dx
= P²L³/(6EI)
δ = ∂U/∂P = PL³/(3EI) ✓─── Maxwell's Reciprocal Theorem ───
In a linearly elastic structure, the deflection at
point A due to a unit load at point B equals the
deflection at point B due to a unit load at point A.
δ_AB = δ_BA
Useful for:
• Verifying structural analysis results
• Simplifying deflection calculations
• Influence line analysis
Bettil's Law (unit load method):
δ = Σ (F_i × f_i × L_i) / (A_i × E)
F_i = force in member due to actual loading
f_i = force in member due to unit load at desired point
L_i = length of member i─── Strain Energy Under Impact ───
Suddenly applied load:
δ_sudden = 2 × δ_static
σ_sudden = 2 × σ_static
Impact load (mass m falling through height h):
Using energy conservation:
m·g·(h + δ) = (1/2) × P_eq × δ
For elastic member (P_eq = P_static = kδ):
δ_impact = δ_static × [1 + √(1 + 2h/δ_static)]
Impact factor: IF = 1 + √(1 + 2h/δ_static)
If h = 0 (sudden load): IF = 2
If h >> δ_static: IF ≈ √(2h/δ_static)
σ_impact = IF × σ_static─── Close-Coiled Helical Spring ───
Assumptions: Helix angle < 10° (wire mainly torsion)
Deflection:
δ = 8PD³n / (Gd⁴)
P = Axial load (N)
D = Mean coil diameter (mm)
n = Number of active coils
G = Modulus of rigidity (MPa)
d = Wire diameter (mm)
Spring rate (stiffness):
k = P / δ = Gd⁴ / (8D³n)
Maximum shear stress (Wahl corrected):
τ_max = K_w × 8PD / (πd³)
K_w = Wahl's correction factor
K_w = (4C − 1)/(4C − 4) + 0.615/C
C = Spring index = D/d
Recommended C = 4 to 12
C < 4: Manufacturing difficult, high curvature
C > 12: Spring tends to buckle, coil tangles─── Springs in Series ───
Same force P through each spring.
δ_total = δ₁ + δ₂ + ...
1/k_eq = 1/k₁ + 1/k₂ + ...
(Same formula as resistors in parallel!)
─── Springs in Parallel ───
Same deflection δ, force divides.
P_total = P₁ + P₂ + ...
k_eq = k₁ + k₂ + ...
(Same formula as resistors in series!)
─── Example ───
k₁ = 2 N/mm, k₂ = 3 N/mm
Series: 1/k_eq = 1/2 + 1/3 = 5/6 → k_eq = 1.2 N/mm
Parallel: k_eq = 2 + 3 = 5 N/mm─── Leaf (Semi-elliptic) Spring ───
Maximum bending stress:
σ = 3PL / (2nbt²)
Deflection:
δ = 3PL³ / (8Enbt³)
P = Load at center, L = Span length
n = Number of leaves, b = Width, t = Thickness
─── Energy Stored in Springs ───
Helical spring:
U = (1/2) × P × δ = P²/(2k) = kδ²/2
Leaf spring:
U = σ² × V / (6E) = σ² × 2nbtL / (6E)
Spring efficiency = U/Volume (higher = better)
┌────────────────────┬──────────────────┐
│ Spring type │ U/Volume │
├────────────────────┼──────────────────┤
│ Helical spring │ τ²/(4G) │
│ Leaf spring │ σ²/(6E) │
│ Spiral spring │ τ²/(4G) │
│ Belleville (disc) │ Variable │
└────────────────────┴──────────────────┘import math
# Design a helical compression spring
P = 500 # N (axial load)
G = 80e3 # MPa (steel)
tau_allow = 300 # MPa
delta = 40 # mm (required deflection)
C = 8 # Spring index (D/d)
# Wahl's factor
Kw = (4*C - 1)/(4*C - 4) + 0.615/C
# Kw = 31/28 + 0.615/8 = 1.107 + 0.077 = 1.184
# Wire diameter from stress: τ = Kw × 8PD/(πd³)
# Since C = D/d → D = C×d
# τ = Kw × 8P×C/(πd²) → d² = Kw × 8PC/(π×τ)
d = math.sqrt(Kw * 8 * P * C / (math.pi * tau_allow))
# d = √(1.184 × 8 × 500 × 8 / (π × 300)) = √40.17 = 6.34 mm
# Use standard wire: d = 6.5 mm
D = C * 6.5 # Mean diameter = 52 mm
# Number of coils from deflection
# δ = 8PD³n/(Gd⁴) → n = δGd⁴/(8PD³)
n = delta * G * d**4 / (8 * P * D**3)
print(f"d = {d:.2f} mm, D = {D:.1f} mm, n = {n:.2f} coils")─── Thin-Walled Cylindrical Pressure Vessel ───
Hoop (circumferential) stress:
σ_h = Pd / (2t) = Pr / t
Longitudinal (axial) stress:
σ_l = Pd / (4t) = Pr / (2t)
σ_h = 2 × σ_l (hoop stress is twice the longitudinal!)
Maximum shear stress:
τ_max = σ_h / 2 = Pd / (4t)
Volumetric strain:
ε_v = ΔV/V = (Pd/(4tE)) × (5 − 4ν)
= (Pd/(tE)) × (2.5 − 2ν) for ν = 0.3:
ε_v = 1.9 × Pd/(tE)
Change in dimensions:
ΔL/L = σ_l/E − ν·σ_h/E = Pd(1−2ν)/(4tE)
Δd/d = (σ_h − ν·σ_l)/E = Pd(2−ν)/(4tE)─── Thin-Walled Spherical Pressure Vessel ───
Membrane stress (same in all directions):
σ = Pd / (4t) = Pr / (2t)
Only HALF the hoop stress of a cylinder!
Volumetric strain:
ε_v = 3σ(1−ν)/E = 3Pd(1−ν)/(4tE)
Comparison (same P, d, t, E):
┌─────────────────────┬──────────────┬────────────┐
│ Property │ Cylinder │ Sphere │
├─────────────────────┼──────────────┼────────────┤
│ Max stress │ Pd/(2t) │ Pd/(4t) │
│ Vol. strain │ ≈1.9Pd/(tE) │ ≈1.5Pd/(tE)│
│ Volume/S.A ratio │ Higher │ Lower │
│ Preferred for │ Pipelines │ Storage │
└─────────────────────┴──────────────┴────────────┘─── Lamé's Equations for Thick Cylinders ───
Radial stress:
σ_r = A − B/r²
Circumferential (hoop) stress:
σ_θ = A + B/r²
Where A and B are constants from boundary conditions.
For solid shaft (internal pressure only):
BC: σ_r(r_i) = −P_i, σ_r(r_o) = 0
A = P_i × r_i² / (r_o² − r_i²)
B = P_i × r_i² × r_o² / (r_o² − r_i²)
σ_θ(r_i) = P_i(r_o² + r_i²)/(r_o² − r_i²) [max, at inner surface]
σ_θ(r_o) = 2P_i × r_i²/(r_o² − r_i²) [min, at outer surface]
Note: Hoop stress is MAXIMUM at inner radius!
Radial displacement:
u = (1/E)[(1−ν)Ar + (1+ν)B/r]
Maximum shear stress (at r = r_i):
τ_max = σ_θ(r_i)/2 = P_i × r_o²/(r_o² − r_i²)─── Compound Cylinder (Shrink / Press Fit) ───
Two cylinders assembled by heating outer / cooling inner.
Junction pressure (interface pressure): P_j
For the inner cylinder:
Internal: P_i (working pressure)
External: P_j (junction pressure)
For the outer cylinder:
Internal: P_j (junction pressure)
External: P_o (working pressure or 0)
Radial interference (shrinkage allowance):
δ = 2 × P_j × r_j / E × [r_j²/(r_j²−r_i²) + (r_o²−r_j²)/(r_o²−r_j²)]
Benefit:
Combined cylinder can withstand higher internal pressure
than a single thick cylinder of same dimensions, because
residual compressive hoop stress reduces peak stress.─── Wire-Wound Thin Cylinder ───
Winding thin wire under tension around a thin cylinder
creates initial compressive hoop stress in the cylinder.
When internal pressure is applied:
Net hoop stress = σ_h (from pressure) − σ_pre-compression (from winding)
Result: Can handle much higher internal pressure for
the same cylinder thickness.
Wire tension (T):
σ_wire = (P_i × d)/(2t_wire × n)
where n = number of wire layerst/d ≤ 1/20(or t/r ≤ 1/10), use thin-walled theory (uniform stress through thickness). For thicker walls, use Lamé's equations where stress varies with radius. Hoop stress in a thick cylinder is always maximum at the inner surface and decreases towards the outer surface.