Strength of Material (ME-304) - Unit 2 Short Notes

Unit 2: Strength of Materials – Exam-Focused Short Notes

1. Axial Deformation and Stress Analysis

Composite Bars (Different Materials)

• Principle: Bars in parallel (same deformation, total load = sum of individual loads) or series (same load, total deformation = sum of individual deformations).

• Key Formula (Parallel/Same Δ):

$$P = \sigma_1 A_1 + \sigma_2 A_2 + ...$$

$$\Delta = \frac{\sigma_1}{E_1} L = \frac{\sigma_2}{E_2} L = ...$$

• Key Formula (Series/Same P):

$$\Delta_{\text{total}} = \frac{P L_1}{A_1 E_1} + \frac{P L_2}{A_2 E_2} + ...$$

• [!TIP] For composite bars with stress limits, calculate allowable P from each material's stress limit (P_max = σ_allow * A) and deformation limit (P_max = (Δ_allow * A * E)/L). The smallest P governs.

Axially Loaded Bars with Varying Cross-Section (Tapering)

• Circular Tapering Bar: Diameter varies linearly: d(x) = d₁ + (d₂-d₁)(x/L).

• Elongation:

$$\Delta = \int_0^L \frac{P}{E A(x)} dx = \frac{4P}{\pi E} \int_0^L \frac{dx}{[d(x)]^2}$$

For linear taper, integrate to get expression in terms of `d₁` and `d₂`.

• [!TIP] For a bar with varying area A(x), always set up the integral Δ = ∫(P dx)/(E A(x)).

Thermal Stresses in Composite Systems

• Free Expansion: ΔL = α L ΔT

• If Expansion is Restricted: Thermal strain is prevented, inducing stress.

$$\sigma = E \alpha \Delta T$$

• Composite System (Rigidly Connected): Total strain compatibility gives:

$$\frac{\sigma_1}{E_1} + \alpha_1 \Delta T = \frac{\sigma_2}{E_2} + \alpha_2 \Delta T$$

Solve with equilibrium `σ₁A₁ = σ₂A₂` (for parallel).

Axial Loads with Rigid Constraints (Fixed Ends/Plates)

• Reactions: Deformation at support is zero. Use compatibility: Δ_due_to_P + Δ_due_to_R = 0.

• Example: Bar fixed at both ends, load P at mid. Symmetry gives R_A = R_B = P/2. Each half shortens by Δ under P/2, so fixed end reactions induce equal opposite elongation.

Stress-Strain Relations, Poisson’s Ratio, Lateral Deformation

• Hooke’s Law (1D): σ = E ε

• Poisson’s Ratio (ν): ν = - (lateral strain) / (axial strain)

• Volumetric Strain (ε_v) for 3D Stress:

$$\epsilon_v = \frac{\sigma_x + \sigma_y + \sigma_z}{E} (1 - 2\nu)$$

• Lateral Strain: ε_lat = -ν ε_axial

Strain Energy under Axial Loading

• Gradual Loading (P applied slowly):

$$U = \int_0^P \frac{P d\delta}{2} = \frac{1}{2} P \delta = \frac{P^2 L}{2 A E} = \frac{\sigma^2}{2E} (AL)$$

• Impact/Sudden Loading (Load P drops from height h):

$$U = \frac{1}{2} P \delta_{\text{max}} = P (\delta_{\text{static}} + h)$$

$$\Rightarrow \delta_{\text{max}} = \delta_{\text{static}} \left(1 + \sqrt{1 + \frac{2h}{\delta_{\text{static}}}} \right)$$

*`δ_static = PL/(AE)`*

• Resilience: Strain energy per unit volume up to yield point.

  Proof Resilience: u_r = σ_y² / (2E)

2. Beam Bending and Shear

Shear Force (SF) & Bending Moment (BM) Diagrams

• Sign Convention (Standard):

  • SF: Positive if left side tends to move up.

  • BM: Positive if beam sags (concave up, tension at bottom).

• Fundamental Relations:

$$\frac{dM}{dx} = V, \quad \frac{dV}{dx} = -w$$

• Key Diagrams:

  • Simply Supported (UDL): BM parabolic, max at center. SF linear, zero at center.

  • Cantilever (UDL): BM parabolic, max at fixed end. SF linear, max at fixed end.

  • Point Load on SSB: BM triangular, max under load. SF step change at load.

• Contraflexure Point: Point where BM = 0 (change in sign). Occurs in overhanging/partially loaded beams.

Bending Stress in Beams (Simple Bending Theory)

• Assumptions:

  1. Material homogeneous, isotropic, σ ∝ ε (Hooke’s).

  2. Plane sections remain plane & perpendicular to neutral axis.

  3. Stress uniaxial along beam length.

  4. Radius of curvature R >> depth.

  5. σ_max << E.

• Bending Formula:

$$\frac{M}{I} = \frac{\sigma}{y} = \frac{E}{R}$$

$$\sigma = \frac{M y}{I}$$

`I` = Moment of Inertia about NA, `y` = distance from NA.

• Section Modulus (Z): Z = I / y_max

$$\sigma_{max} = \frac{M}{Z}$$

• Common Sections (Z values):

  • Rectangle: Z = bd²/6

  • Square: Z = a³/6

  • Circle: Z = πd³/32

  • Hollow Circle: Z = π(D⁴ - d⁴)/(32D)

Shear Stress Distribution in Beams

• General Formula (Joist’s Formula):

$$\tau = \frac{V Q}{I b}$$

`V` = SF, `Q` = statical moment of area *above/below* the point about NA, `I` = total I, `b` = width at point.

• Rectangular Section:

$$\tau = \frac{3V}{2bd} \left(1 - \frac{4y^2}{d^2}\right)$$

Parabolic, max at NA: `τ_max = 1.5 τ_avg`.

• Circular Section:

$$\tau = \frac{4V}{3\pi r^2} \sqrt{1 - \frac{y^2}{r^2}}$$

Parabolic, max at NA: `τ_max = (4/3) τ_avg`.

• I-Section: Max shear in web, τ_max ≈ V / (A_web) (approx). Flange shear negligible.

Beam Deflection Methods

1. Double Integration Method:

   EI d²y/dx² = M(x). Integrate twice, apply boundary conditions.

   Advantage: Direct. Limitation: Repeated integration for complex loads.

2. Moment Area Method (Theorems):

   • First Theorem: Slope between A & B = (1/EI) * (Area of M diagram between A & B).

   • Second Theorem: Deflection of B relative to tangent at A = (1/EI) * (Moment of M diagram area between A & B about B).

   Advantage: No integration, uses geometry. Limitation: Requires correct M diagram & centroid.

3. Macaulay’s Method (Step Functions):

   Write M(x) using ⟨x-a⟩ⁿ (zero for x<a). Integrate EI dy/dx and EI y directly. Apply boundary conditions at x=0 (usually fixed end).

   Advantage: Single integration for multiple point loads. Limitation: Requires careful handling of step functions.

Deflection Comparisons

• Same Span & Load: Cantilever > Simply Supported > Fixed-Fixed.

• Same Material & Max Stress: For rectangular section, δ ∝ 1/depth³. Depth is more effective than width.

• Same Cross-Section & Load: δ ∝ 1/EI. Material & section matter.

3. Torsion of Circular Shafts

Solid vs. Hollow Shafts (Weight Comparison for Same Torque & Stress)

• Polar Moment of Inertia:

  • Solid: J_s = πd⁴/32

  • Hollow: J_h = π(D⁴ - d⁴)/32

• Torsional Shear Stress: τ = Tρ/J. Max at outer surface: τ_max = T / (J / R) = T / Z_p, where Z_p = J/R.

• For Same τ_max and T: Z_p must be equal.

  Z_{p,s} = Z_{p,h} ⇒ πd³/16 = π(D⁴ - d⁴)/(16D)

• Weight Ratio (Same Length & Material): W_h / W_s = (D² + d²) / d²

  If d = (3/4)D, then W_h / W_s = (D² + 0.5625D²)/0.5625D² = 1.5625/0.5625 ≈ 2.78.

  Conclusion: Hollow shaft of same weight can carry √(weight ratio) times more torque.

  [!TIP] Proof: T ∝ J ∝ (D⁴ - d⁴). For same weight, D² + d² = constant. Maximize J w.r.t. d/D. Optimal d/D ≈ 0.6.

Angle of Twist & Torsional Rigidity

• Angle of Twist (φ):

$$\phi = \frac{T L}{G J} \quad (\text{radians})$$

`G` = Modulus of rigidity, `J` = Polar moment of inertia.

• Torsional Rigidity: GJ (stiffness against twist).

Shear Stress Distribution

• Solid Circular: τ ∝ ρ (linear from 0 at center to max at surface).

• Hollow Circular: τ ∝ ρ (linear, zero at inner radius, max at outer).

• Stepped Shaft: Stress changes abruptly at step. τ = T / Z_p in each segment. Discontinuity in τ at step, but φ continuous.

Combined Torsion and Bending (Shaft Design)

• Principal Stresses (at outer fiber):

$$\sigma_x = \frac{M}{Z} \quad (\text{tension or compression}), \quad \tau_{xy} = \frac{T}{Z_p}$$

*`Z` = Section modulus (bending), `Z_p` = Polar section modulus.*

• Design by Theories of Failure:

  • Max Shear Stress (Tresca): τ_max = √[(σ/2)² + τ²] ≤ τ_allow

  • Max Distortion Energy (von Mises): σ_eq = √(σ² + 3τ²) ≤ σ_allow

  [!TIP] For pure torsion (M=0), Tresca gives τ_max = τ, von Mises gives σ_eq = √3 τ. Von Mises predicts ~15% higher allowable shear.

Power Transmission & Torsional Strain Energy

• Power (P): P = T ω, where ω = angular velocity (rad/s).

  T (N·m) = (P (W) * 60) / (2π N) if N in rpm.

• Strain Energy (U) in Torsion:

$$U = \int_0^L \frac{T^2}{2 G J} dx = \frac{T^2 L}{2 G J} \quad (\text{for constant T, J})$$

4. Stress Transformation and Mohr’s Circle

Principal Stresses & Principal Planes (2D)

• Principal Stresses (σ₁, σ₂): Max & min normal stresses on planes where τ_xy = 0.

• Principal Planes: Planes on which principal stresses act, inclined at θ_p to x-axis.

• Equations:

$$\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2}$$

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}$$

• Max In-Plane Shear Stress:

$$\tau_{max} = \frac{\sigma_1 - \sigma_2}{2} = \sqrt{\left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2}$$

Acts on planes at `45°` to principal planes.

Mohr’s Circle Construction

1. Plot point X(σ_x, τ_xy) and Y(σ_y, -τ_xy).

2. Center C at ((σ_x+σ_y)/2, 0).

3. Radius R = √[((σ_x-σ_y)/2)² + τ_xy²].

4. Circle intersects σ-axis at σ₁, σ₂.

5. Any point on circle represents (σ_n, τ_n) on plane at angle 2θ from x-axis (counterclockwise on circle = clockwise on element).

• [!TIP] 2θ Rule: On Mohr's circle, angle 2θ is measured from the line CX. The physical plane angle θ is half of that, and direction is opposite.

Stresses on Inclined Plane (θ to x-axis)

• Transformation Equations:

$$\sigma_n = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos 2\theta + \tau_{xy} \sin 2\theta$$

$$\tau_n = -\frac{\sigma_x - \sigma_y}{2} \sin 2\theta + \tau_{xy} \cos 2\theta$$

• Maximum Obliquity: Max angle φ between resultant stress σ_r and normal to plane.

$$\tan \phi_{max} = \frac{\tau_{max}}{\sigma_{avg}} = \frac{\tau_{max}}{(\sigma_1+\sigma_2)/2}$$

Occurs on plane where `σ_n = σ_avg`.

5. Theories of Failure

For Ductile Materials (Yield as Failure)

1. Maximum Shear Stress Theory (Tresca):

   Failure when max shear stress reaches shear yield strength (τ_y = σ_y/2).

$$ \max |\tau| = \frac{|\sigma_1 - \sigma_3|}{2} \leq \frac{\sigma_y}{2} \quad \text{or} \quad |\sigma_1 - \sigma_3| \leq \sigma_y $$

*Conservative, safe for brittle? No.*

2. Maximum Distortion Energy Theory (von Mises):

   Failure when distortion strain energy per volume reaches yield value.

$$\sigma_{eq} = \sqrt{\frac{(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2}{2}} \leq \sigma_y$$

For 2D (`σ₃=0`): `σ_eq = √(σ₁² - σ₁σ₂ + σ₂²) ≤ σ_y`.

*More accurate for ductile metals.*

For Brittle Materials (Ultimate Strength as Failure)

1. Maximum Principal Stress Theory (Rankine):

   Failure when max tensile OR max compressive principal stress reaches ultimate strength.

$$\sigma_1 \leq \sigma_{ut} \quad \text{and} \quad \sigma_3 \geq -\sigma_{uc}$$

*Suitable for brittle materials (cast iron, concrete).*

Comparison & Graphical Representation (Haigh’s Diagram)

• Tresca: Hexagon in σ₁-σ₃ plane.

• von Mises: Ellipse.

• Rankine: Rectangle.

• Order of Conservatism (for ductile, σ_t=σ_c): Tresca > von Mises > Rankine.

• [!TIP] For pure shear (σ₁ = -σ₃ = τ): Tresca: 2τ ≤ σ_y → τ_allow = σ_y/2. von Mises: √3 τ ≤ σ_y → τ_allow = σ_y/√3 ≈ 0.577 σ_y.

Factor of Safety (FOS)

• FOS = (Yield/Ultimate Strength) / (Permissible/Working Stress).

• Based on Tresca: FOS = σ_y / (σ₁ - σ₃).

• Based on von Mises: FOS = σ_y / σ_eq.

• Based on Rankine: FOS = σ_ut / σ₁ (tension) or σ_uc / |σ₃| (compression).

6. Elastic Constants and Strain Energy

Relationships between E, G, K, ν

• Basic Definitions:

  • E = Young’s Modulus (tension/compression).

  • G = Modulus of Rigidity (shear).

  • K = Bulk Modulus (volumetric).

  • ν = Poisson’s ratio.

• Key Relationships (Isotropic Material):

$$G = \frac{E}{2(1+\nu)}$$

$$K = \frac{E}{3(1-2\nu)}$$

$$E = \frac{9KG}{3K+G}$$

$$\nu = \frac{3K - 2G}{2(3K + G)}$$

• [!TIP] For most metals, ν ≈ 0.3. Then G ≈ 0.385E, K ≈ 1.67E.

Volumetric Strain and Bulk Modulus

• Volumetric Strain (ε_v): ε_v = (ΔV)/V = ε_x + ε_y + ε_z.

• Hydrostatic Pressure (p): p = -K ε_v (compressive pressure positive).

  For hydrostatic stress state (σ_x = σ_y = σ_z = -p):

$$\epsilon_v = -\frac{3p}{E}(1-2\nu) \quad \Rightarrow \quad K = \frac{E}{3(1-2\nu)}$$

Strain Energy

• Gradual Loading (Axial): U = P²L/(2AE).

• Impact Loading (Sudden Load P from height h):

$$\delta_{max} = \delta_{static} \left(1 + \sqrt{1 + \frac{2h}{\delta_{static}}} \right)$$

$$U = \frac{1}{2} P \delta_{max}$$

• Resilience (u_r): Strain energy per unit volume up to elastic limit.

$$u_r = \frac{\sigma_y^2}{2E}$$

• Proof Resilience: Same as resilience.

7. Columns and Buckling

Euler’s Buckling Theory

• Assumptions:

  1. Column is perfectly straight, homogeneous, isotropic.

  2. Load is axial, centroidal.

  3. Material obeys Hooke’s law.

  4. Deformations small.

  5. Column fails by buckling (elastic range).

• Differential Equation: EI d²y/dx² = -M = -P y

  Solution: y = A sin(√(P/EI) x) + B cos(√(P/EI) x)

• Euler’s Crippling Load (P_cr):

$$P_{cr} = \frac{\pi^2 E I}{(KL)^2}$$

`K` = effective length factor, `L_e = KL` = effective length.

• Effective Length (L_e) for End Conditions:

  | End Condition | K | L_e | |------------------------|----|-----------| | Both ends pinned | 1.0| L | | Both ends fixed | 0.5| L/2 | | One fixed, one free | 2.0| 2L | | One fixed, one pinned | 0.7| 0.7L |

• Slenderness Ratio (λ): λ = L_e / r, where r = √(I/A) is radius of gyration.

• Limitation: Valid only if σ_cr ≤ σ_proportional_limit (elastic buckling).

Rankine’s Formula (Empirical, All Lengths)

• Formula:

$$P_{cr} = \frac{f_c A}{1 + a (L_e/r)^2}$$

`f_c` = compressive strength (yield/ultimate), `a` = Rankine constant (`a = 1/(π² E / σ_prop)`).

• For Short Columns (L_e/r → 0): P_cr → f_c A (material failure).

• For Long Columns (L_e/r → ∞): P_cr → π² E I / L_e² (Euler).

• Safe Load: P_safe = P_cr / FOS.

Crippling Load & Safe Load

• Crippling Load (P_cr): Theoretical buckling load.

• Safe Load: P_safe = P_cr / FOS.

Columns with Composite Sections

• Use I of transformed section if materials differ (use n = E1/E2).

• Euler’s formula still applies with composite I and appropriate E (use material at outer fiber for stress check).

Comparison: Euler vs. Rankine

• Euler: Only for long, slender, elastic columns. Predicts P_cr ∝ 1/L².

• Rankine: For all lengths. Matches Euler for long columns, material strength for short. Intermediate lengths give conservative P_cr.

• [!TIP] In exams, if σ_cr (Euler) < σ_proportional_limit, use Euler. If not, use Rankine or direct material failure.

8. Special Topics (From Past Papers)

Curved Beams (Circular Section)

• Neutral Axis (NA): Does NOT pass through centroid. For circular ring of mean radius R_m and depth h:

$$R_n = \frac{R_m}{1 \pm \frac{h}{2R_m}}$$

(sign depends on inner/outer radius).

• Bending Stress (at radius r):

$$\sigma_\theta = \frac{M}{A e (R_n \pm y)} \quad \text{where} \quad e = R_m - R_n, \quad y = r - R_n$$

`+` for inner fiber, `-` for outer fiber (for hogging moment).

• [!TIP] Stress is NOT linear with y. Highest stress at inner radius for given M (smaller R_n ± y).

Shear Stress in Circular Sections (Derivation & Plot)

• Derivation: From τ = VQ/(Ib). For circle of radius R at distance y from NA:

  Q = (2/3)(R² - y²)^{3/2}, b = 2√(R² - y²), I = πR⁴/4.

$$\tau = \frac{4V}{3\pi R^2} \sqrt{1 - \frac{y^2}{R^2}}$$

• Plot: Parabolic, zero at y = ±R, max at y=0 (τ_max = 4V/(3A)).

Composite Beams (Transformed Section Method)

• Principle: Transform materials to equivalent section of one material using n = E2/E1.

• Steps:

  1. Transform width of material 2: b₂' = n b₂.

  2. Find neutral axis of transformed section.

  3. Calculate I_transformed about NA.

  4. Bending Stress in material 1: σ₁ = M y / (I_transformed).

  5. Bending Stress in material 2: σ₂ = n M y / (I_transformed) (or σ₂ = (E₂/E₁) σ₁).

• Shear Stress: Calculate VQ/(I b) using actual width b and transformed I.

Pressure Vessels (Thin-Walled, Internal Pressure)

• Assumptions: t << R, stress uniform through thickness.

• Hoop Stress (Circumferential):

$$\sigma_h = \frac{p R}{t} \quad (\text{or } \frac{p d}{2t})$$

• Longitudinal Stress (Axial):

$$\sigma_l = \frac{p R}{2t} \quad (\text{or } \frac{p d}{4t})$$

• Failure Theories for Thickness:

  • Max Principal Stress (Rankine): σ_h ≤ σ_allow → t ≥ pR/σ_allow.

  • Max Shear Stress (Tresca): τ_max = (σ_h - σ_l)/2 = σ_h/3 ≤ τ_allow → t ≥ pR/(3τ_allow).

  • von Mises: σ_eq = √(σ_h² + 3σ_l²) = √(σ_h² + 3(σ_h/2)²) = √(1.75) σ_h ≤ σ_allow → t ≥ √(1.75) pR / σ_allow ≈ 1.323 pR/σ_allow.

  [!TIP] Von Mises gives thickest wall, Tresca thinnest, Rankine in between for internal pressure vessels.

Final Exam Strategy:

1. Draw Diagrams: Always sketch SF/BM, Mohr’s circle, column buckled shape.

2. State Assumptions: Especially for bending, torsion, Euler.

3. Units: Consistent (N, mm, MPa) or (N, m, Pa).

4. Formulas: Box final expressions. Derive if asked (Euler, curved beam stress, shear in circle).

5. Theory Comparison: Know which theory for ductile/brittle, and why (yield vs. fracture).

6. Composite Systems: Compatibility (Δ same) & Equilibrium (ΣF=0) are keys.