UNIT 2: STRENGTH OF MATERIALS - EXAM-FOCUSED SHORT NOTES

Based on RGPV CE-305 past papers (2022-2025). Prioritized for high-frequency exam topics.

I. STRESS TRANSFORMATION & PRINCIPAL STRESSES

A. Stress at a Point & Components

• Normal Stress (σ): Force per unit area acting perpendicular to a plane. Tensile (+), Compressive (-).

• Shear Stress (τ): Force per unit area acting tangentially to a plane. Positive if it tends to rotate the element counterclockwise on the face where the normal is positive (x-face).

• Stress Tensor (2D Plane Stress): State of stress at a point defined by σ_x, σ_y, τ_xy (τ_yx = τ_xy).

$$ \begin{bmatrix} \sigma_x & \tau_{xy} \\ \tau_{xy} & \sigma_y \end{bmatrix} $$

B. Principal Stresses & Principal Planes

• Principal Stresses (σ₁, σ₂): Normal stresses on planes where shear stress is zero. σ₁ = max normal stress (major), σ₂ = min normal stress (minor).

• Principal Planes: Planes oriented at angle θ_p from the x-axis where principal stresses act.

• Equations:

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

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

> [!TIP] **Exam Focus:** Given σ_x, σ_y, τ_xy (e.g., from bending + torsion), compute σ₁, σ₂, θ_p. Remember two solutions for θ_p differ by 90°.

C. Maximum Shear Stress

• Maximum In-Plane Shear Stress (τ_max):

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

• Acts on planes oriented at 45° to principal planes.

• Average Normal Stress: (σ_x + σ_y)/2 acts on planes of max shear.

D. Mohr's Circle for Plane Stress

• Construction:

  1. Plot center C at ( (σ_x+σ_y)/2 , 0 ).

  2. Plot point A (σ_x, τ_xy) for x-face. Point B (σ_y, -τ_xy) for y-face.

  3. Circle radius R = distance CA = τ_max.

• Interpretation:

  • Principal Stresses: Coordinates of points on σ-axis where circle intersects (σ₁ at rightmost, σ₂ at leftmost).

  • Max Shear Stress: Top/bottom points of circle (coordinates: C, ±R).

  • Stresses on any plane at angle θ: Measure 2θ counterclockwise from OA on the circle.

• [!TIP] Common Pitfall: On Mohr's circle, angle is 2θ (physical plane rotation). Shear sign: positive τ_xy plots upward.

E. Applications: Combined Loading

• Combined Bending & Torsion (Shafts/Beams):

  • Bending gives σ_b = My/I (tensile on one face, compressive on other).

  • Torsion gives τ_t = Tr/J (pure shear on cross-section).

  • At a point on the surface (max y): σ_x = σ_b, σ_y = 0, τ_xy = τ_t.

  • Use principal stress equations to find σ₁, σ₂.

• Combined Direct & Shear: Same procedure. State of stress defined on a specific face.

II. BEAM THEORY: SF, BM & STRESSES

A. Definitions & Sign Conventions

• Shear Force (V): Algebraic sum of vertical forces left/right of a section. Positive if left side tends to move upward.

• Bending Moment (M): Algebraic sum of moments about a section. Positive if it causes sagging (concave up, compression on top).

• Load Types: Point load (P), UDL (w), UVL (w varies), Couple (M).

B. Relationship: Loading, SF, BM

• Rate of Loading: w = -dV/dx (negative sign per sign convention).

• Shear Force: V = dM/dx.

• Integration: M = ∫V dx, V = -∫w dx.

• [!TIP] Derivation (DEC 2023, JUN 2023): Consider equilibrium of a small element dx. Sum F_y = 0 → V + dV - (w dx) - V = 0 → dV/dx = -w. Sum M about left face = 0 → M + (V dx) - (w dx)(dx/2) - (M + dM) = 0 → dM/dx = V.

C. SF & BM Diagrams

• Procedure:

  1. Find reactions.

  2. Draw V diagram: Start from left, add/subtract loads. Jump down by P at point load, slope changes by -w for UDL.

  3. Draw M diagram: Integrate V diagram. M is constant where V=0 (max/min). Jump by couple. Slope of M = V.

• Point of Contraflexure (POC): Point where M = 0 and changes sign (sagging to hogging or vice versa). Found by setting M(x)=0.

• [!TIP] JUN 2025 Short Note: POC is where curvature changes. On diagram, it's where BM curve crosses the axis.

D. Theory of Simple Bending

• Assumptions:

  1. Material is homogeneous, isotropic, obeys Hooke's law.

  2. Beam is initially straight.

  3. Cross-sections remain plane and perpendicular to neutral axis (Bernoulli's hypothesis).

  4. No shear deformation effect on bending.

• Bending Equation:

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

Where:

*   M = Bending moment at section

*   I = Moment of inertia about NA

*   σ = Bending stress at distance y from NA

*   E = Modulus of elasticity

*   R = Radius of curvature

• Section Modulus (Z): $$\displaystyle Z = I/y_{\text{max}} $$. Bending stress: $$\displaystyle \sigma_{\text{max}} = M/Z $$.

  • Formulas:

    • Rectangle (b×h): $$\displaystyle Z = bh^2/6 $$

    • Circle (d): $$\displaystyle Z = \pi d^3/32 $$

    • Hollow Circle (D,d): $$\displaystyle Z = \pi (D^4 - d^4)/(32D) $$

    • I-section: $$\displaystyle Z = I/y_{\text{max from NA}} $$

• [!TIP] DEC 2023: Deduce Z for hollow circle: $$\displaystyle I = \pi(D^4-d^4)/64 $$, $$\displaystyle y_{\text{max}} = D/2 $$ → $$\displaystyle Z = I/y_{\text{max}} = \pi(D^4-d^4)/(32D) $$.

E. Shear Stress in Beams

• Shear Stress Formula (Derivation DEC 2023):

$$ \tau = \frac{VQ}{Ib} $$

Where:

*   V = SF at section

*   I = Total I about NA

*   b = Width at the level where τ is calculated

*   Q = First moment of area **above/below** the point about NA: $$\displaystyle Q = \bar{y} A' $$

• Distribution:

  • Rectangular: Parabolic, τ_max = 1.5 τ_avg at NA (y=0), τ=0 at top/bottom.

  • Circular: Parabolic, τ_max = (4/3) τ_avg at NA.

  • I-section: τ mainly in web. τ_flange << τ_web. Max τ at NA in web.

  • T-section: Similar, but NA not at centroid. Calculate Q carefully.

• [!TIP] JUN 2023 (I-section), DEC 2023 (T-section): For I-section, Q at NA = A_web * y_bar + 2 * A_flange * y_bar_flange. For T, find NA first.

III. DEFLECTION OF BEAMS

A. Methods Overview

• Double Integration: Most fundamental. Uses EI d²y/dx² = M(x).

• Macaulay's: For discontinuous loads (UDL, point loads). Uses step functions in M(x).

• Area Moment (Mohr's Theorems): Graphical/analytical using M diagrams. No integration needed.

B. Double Integration Method

• Governing Equation: $$\displaystyle EI \frac{d^2 y}{dx^2} = M(x) $$

• Procedure:

  1. Find reactions, write M(x) for each segment.

  2. Integrate twice: $$\displaystyle EI \frac{dy}{dx} = \int M(x) dx + C_1 $$, $$\displaystyle EI y = \int \int M(x) dx^2 + C_1 x + C_2 $$.

  3. Apply boundary conditions (deflection/slope zero at supports, symmetry) to find C₁, C₂.

• [!TIP] DEC 2024: Derive from curvature: 1/R = M/EI, and for small slopes, 1/R ≈ d²y/dx².

C. Macaulay's Method

• Step Functions: Write M(x) using <x-a>ⁿ notation. For load at x=a, use <x-a>.

  • Example: Point load P at x=a → M(x) includes term -P< x-a >.

  • UDL from x=a to x=b → -w/2 <x-a>² + w/2 <x-b>².

• Integration: Integrate term-by-term. <x-a>ⁿ integrates to <x-a>ⁿ⁺¹/(n+1) for x>a, else 0.

• Apply boundary conditions after integration.

• [!TIP] JUN 2025: Essential for beams with overhangs or multiple point loads. Write one continuous M(x) equation.

D. Area Moment Method (Mohr's Theorems)

• First Theorem (Slope):

$$ \theta_{AB} = \frac{1}{EI} \times (\text{Area of M diagram between A and B}) $$

*Sign:* Area above axis → +ve slope (counterclockwise), below → -ve.

• Second Theorem (Deflection):

$$ \delta_B = \frac{1}{EI} \times (\text{Moment of area of M diagram about point B}) $$

*Sign:* Moment of area about B → +ve deflection upward.

• Procedure:

  1. Draw M diagram (to scale).

  2. For slope at A w.r.t. B: Take area of M diagram between A & B.

  3. For deflection at B: Take moment of that area about B.

• Standard Results (DEC 2023):

  • Simply supported with central load P: δ_max = PL³/(48EI), θ_support = PL²/(16EI).

  • Simply supported with UDL w: δ_max = 5wL⁴/(384EI), θ_support = wL³/(24EI).

  • Cantilever with end load P: δ_free = PL³/(3EI), θ_free = PL²/(2EI).

  • Cantilever with UDL w: δ_free = wL⁴/(8EI), θ_free = wL³/(6EI).

• [!TIP] JUN 2025: "State & Explain" → Memorize theorems with diagrams. For overhangs, use conjugate beam method or divide into segments.

IV. TORSION OF CIRCULAR SHAFTS

A. Theory of Pure Torsion

• Assumptions: Uniform twist, cross-sections remain plane, material linear elastic, shear stress proportional to radius.

• Torsional Equation:

$$ \frac{T}{J} = \frac{\tau}{r} = \frac{G\theta}{L} $$

Where:

*   T = Torque

*   J = Polar moment of inertia

*   τ = Shear stress at radius r

*   G = Modulus of rigidity

*   θ = Angle of twist (radians)

*   L = Length

• Polar Modulus (Z_p): $$\displaystyle Z_p = J/r $$. Max τ = T/Z_p.

  • Solid circular (d): $$\displaystyle J = \pi d^4/32 $$, $$\displaystyle Z_p = \pi d^3/16 $$

  • Hollow circular (D,d): $$\displaystyle J = \pi (D^4 - d^4)/32 $$, $$\displaystyle Z_p = J/(D/2) $$

B. Shear Stress & Angle of Twist

• Max Shear Stress: τ_max = T * (D/2) / J = T / Z_p.

• Angle of Twist: $$\displaystyle \theta = \frac{TL}{GJ} $$ (in radians). Convert to degrees: θ° = θ × (180/π).

• [!TIP] JUN 2024: Derive max torque for solid shaft: T_max = τ_allow * Z_p = τ_allow * (π d³/16).

C. Combined Bending & Torsion in Shafts

• Stress State at Critical Point (surface, max y):

  • σ_x = M*y/I (bending, tensile on one side)

  • σ_y = 0

  • τ_xy = Tr/J = T/(2Z_p) [since J = 2I for circular section? NO! J and I are different. For solid circle, J = πd⁴/32, I = πd⁴/64 → J = 2I. So τ_xy = Tr/J = T*(d/2)/(πd⁴/32) = 16T/(πd³) = T/Z_p.

• Principal Stresses:

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

• Max Shear Stress (3D): For pure torsion + bending, max shear = τ_max (torsion) if bending stress is zero at center? Actually:

$$ \tau_{\text{max}} = \sqrt{ \left( \frac{\sigma_1 - \sigma_2}{2} \right)^2 } = \frac{\sigma_1 - \sigma_2}{2} \quad \text{(since σ₃=0)} $$

But also from Mohr's circle: τ_max = √[(σ_x/2)² + τ_xy²].

• [!TIP] JUN 2025 Short Note: Key is to identify the stress state. For a shaft, at the point of max bending stress, σ_y=0, τ_xy from torsion.

D. Design of Shafts

• Strength Criterion (Based on Theory of Failure):

  • Ductile material (Tresca/von Mises): τ_max ≤ τ_allow.

  • Brittle (Rankine): σ₁ ≤ σ_allow.

  • Often simplified: τ_max = √[(σ_b/2)² + τ_t²] ≤ τ_allow.

• Stiffness Criterion: θ = TL/(GJ) ≤ θ_allow.

• Power Transmission: $$\displaystyle P = \frac{2\pi NT}{60} $$ → T = 60P/(2πN).

• Variable Torque: If T_max = T_mean (1 + k), use T_max for strength, T_mean for stiffness? Usually design for T_max.

• [!TIP] JUN 2025, JUN 2024: Typical problem: Given P, N, τ_allow, θ_allow, G, find d. Check both criteria, take larger d.

E. Hollow vs. Solid Shafts (JUN 2025 Short Note)

• Strength: For same outer diameter D and same material (τ_allow), hollow shaft (with d = kD) can transmit more torque because Z_p,hollow > Z_p,solid? Actually, for same weight (mass), hollow is stronger. For same D, solid has larger Z_p? Let's compare:

  • Solid: Z_p,s = πD³/16

  • Hollow: Z_p,h = π(D⁴-d⁴)/(16D) = (πD³/16)(1 - k⁴) where k=d/D.

  • So for same D, Z_p,h < Z_p,s → solid stronger for same outer diameter.

  • For same weight (cross-sectional area): A_s = πD_s²/4, A_h = π(D_h²-d_h²)/4. If A_s = A_h, then D_h > D_s, and Z_p,h can be larger.

• Stiffness: J_hollow < J_solid for same D, but for same weight, J_hollow can be larger.

• Weight: Hollow lighter for same strength/stiffness.

• Conclusion: Hollow shafts are more efficient (higher strength-to-weight ratio).

V. COLUMNS & STRUTS

A. Definitions & Buckling

• Column: Vertical member carrying axial compression, prone to buckling (sudden lateral deflection).

• Strut: General compression member, may be inclined.

• Slenderness Ratio (λ): $$\displaystyle \lambda = L_e / k $$, where:

  • L_e = Equivalent length (effective length for buckling)

  • k = Radius of gyration = √(I_min/A)

• End Conditions & Equivalent Length (L_e):

  | End Condition | L_e / L (Ratio) | Buckling Mode Shape | |------------------------|-----------------|---------------------------| | Both ends pinned | 1.0 | Half sine wave | | Both ends fixed | 0.5 | Quarter sine wave | | One fixed, one free | 2.0 | Half sine with inflection at fixed end? Actually, L_e = 2L, mode is 1/4 wave? | | One fixed, one pinned | 0.7 | ~ |

• [!TIP] JUN 2024: Memorize ratios: Pinned-Pinned: L_e=L; Fixed-Fixed: L_e=0.5L; Fixed-Free: L_e=2L; Fixed-Pinned: L_e=0.7L.

B. Euler's Theory (Long Columns)

• Assumptions: Perfectly straight, elastic, homogeneous, axial load, no initial imperfections.

• Euler's Buckling Load:

$$ P_{cr} = \frac{\pi^2 EI}{L_e^2} $$

• Derivation for Pinned Ends (DEC 2024):

  1. Consider deflected shape y(x). Bending moment M = -P y.

  2. EI d²y/dx² = -P y → d²y/dx² + (P/EI) y = 0.

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

  4. BCs: y(0)=0, y(L)=0 → B=0, sin(√(P/EI) L)=0 → √(P/EI) L = nπ.

  5. Smallest load (n=1): P_cr = π²EI/L².

• Limitations: Valid only for λ > λ_critical (slender columns). For short columns, material fails by crushing before buckling.

• Critical Slenderness Ratio: $$\displaystyle \lambda_c = \pi \sqrt{\frac{E}{\sigma_c}} $$, where σ_c is crushing stress.

C. Rankine's Formula (Intermediate Columns)

• Equation:

$$ P = \frac{\sigma_c A}{1 + a \left( \frac{L_e}{k} \right)^2} $$

Where:

*   σ_c = Ultimate compressive strength (crushing stress)

*   a = Rankine's constant (depends on material, ~1/6400 to 1/8000 for steel, 1/1600 for cast iron)

• Behavior:

  • For long columns (L_e/k large): P ≈ π²EI/L_e² (Euler's).

  • For short columns (L_e/k small): P ≈ σ_c A (crushing).

• [!TIP] JUN 2025: Discuss how it bridges Euler and crushing. Given a, σ_c, find load.

D. Other Empirical Formulas

• Johnson's Straight-Line: P = σ_c A [1 - β (L_e/k)] for λ < λ_c. Less common.

E. Columns with Eccentric Loading

• Stress Distribution: Combined axial stress (P/A) and bending stress (My/I).

$$ \sigma = \frac{P}{A} \pm \frac{M e y}{I} = \frac{P}{A} \pm \frac{P e y}{I} $$

Where e = eccentricity.

• Maximum Stress: $$\displaystyle \sigma_{\text{max}} = \frac{P}{A} + \frac{P e y_{\text{max}}}{I} $$.

• Middle Third Rule (Rectangular Section):

  • For no tensile stress, resultant compressive force must lie within the middle third of the section.

  • Condition: e ≤ h/6 for rectangular section (height h).

  • [!TIP] JUN 2025: Explain with stress diagram. If load is within middle third, stress is entirely compressive (linear distribution). If outside, part of section experiences tension.

VI. THIN CYLINDERS & SPHERICAL SHELLS

A. Thin-walled Assumptions

• Wall thickness t << internal radius r (usually t/r < 1/20).

• Stress through thickness is negligible (uniform).

• Only hoop (circumferential) and longitudinal stresses considered.

B. Cylindrical Vessel

• Hoop Stress (σ_h): Circumferential. Greater than longitudinal.

$$ \sigma_h = \frac{p d}{2t} $$

*Derivation (DEC 2023):* Cut cylinder longitudinally, balance force: 2(σ_h * t * L) = p * d * L → σ_h = pd/(2t).

• Longitudinal Stress (σ_l): Axial.

$$ \sigma_l = \frac{p d}{4t} $$

*Derivation:* Cut across, balance force: σ_l * (π d t) = p * (π d²/4) → σ_l = pd/(4t).

• Ratio: σ_h : σ_l = 2 : 1.

C. Spherical Vessel

• Hoop Stress (σ): Same in all directions in wall.

$$ \sigma = \frac{p d}{4t} $$

*Derivation (DEC 2023):* Cut sphere through diameter, balance force: 2(σ * π r t) = p * π r² → σ = pr/(2t) = pd/(4t).

• Advantage: For same p, d, t, spherical vessel has half the stress of cylindrical (hoop) → thinner wall possible.

D. Change in Dimensions & Volume

• Hoop Strain: ε_h = (σ_h - νσ_l)/E.

• Change in Diameter: Δd = d * ε_h.

• Change in Length: ΔL = L * ε_l = L * (σ_l - νσ_h)/E.

• Change in Volume (ΔV):

$$ \Delta V = V \left( \frac{p d}{2t E} (1 - \frac{\nu}{2}) \right) \quad \text{(for cylinder)} $$

More precisely: ΔV/V = ε_l + 2ε_h.

• [!TIP] JUN 2023, DEC 2023: Numerical problems: Given p, d, t, E, ν, find Δd, ΔL, or extra fluid volume (ΔV). Remember sign conventions: internal pressure → tensile stresses → expansion.

VII. ELASTIC CONSTANTS & COMPOSITE BARS

A. Elastic Constants

• E (Young's Modulus): σ/ε (tensile/compressive).

• G (Modulus of Rigidity): τ/γ (shear).

• K (Bulk Modulus): p / (ΔV/V) (hydrostatic pressure).

• ν (Poisson's Ratio): -lateral strain / longitudinal strain. For isotropic: -0.5 ≤ ν ≤ 0.5.

• Relationships (DEC 2024):

  1. E and G: $$\displaystyle E = 2G(1 + \nu) $$

  2. E and K: $$\displaystyle E = 3K(1 - 2\nu) $$

  3. Combined: $$\displaystyle E = \frac{9KG}{3K + G} $$

• [!TIP] Derivation E-G-ν: Consider a cube under shear τ → γ = τ/G. Also under hydrostatic pressure p → volumetric strain = p/K. Combine with uniaxial test.

B. Composite/S Compound Bars

• Parallel System (Bars in parallel, same deformation):

  • Compatibility: δ₁ = δ₂ = ... = δ (same strain if same length).

  • Equilibrium: P = Σ P_i.

  • Stress: σ_i = P_i / A_i, and since ε_i = σ_i/E_i = same → σ₁/E₁ = σ₂/E₂.

  • Total Deformation: δ = (P L)/(Σ (A_i E_i)).

• Series System (Bars in series, same force):

  • Equilibrium: P₁ = P₂ = ... = P.

  • Compatibility: δ = Σ δ_i.

  • Strain: ε_i = σ_i/E_i, but σ_i may differ if A_i differs? Actually, force same, so σ_i = P/A_i. Then δ_i = σ_i L_i / E_i.

  • Total Deformation: δ = P Σ (L_i/(A_i E_i)).

• [!TIP] DEC 2024 (steel-copper), JUN 2024 (steel bolt-copper tube): Often a bolt/nut system where tightening introduces pre-stress. Use compatibility: δ_steel + δ_copper = initial tightening displacement (pitch × turns).

C. Thermal Stresses

• Free Expansion: ΔL = α L ΔT.

• If Ends Fixed (No expansion): Thermal strain prevented → thermal stress σ = E α ΔT (compressive if heated).

• If Ends Yield/Partially Restrained: σ = E (α ΔT - ΔL/L_actual).

• [!TIP] JUN 2025 Short Note: Key: Compare actual deformation with free thermal expansion. If restrained, stress develops.

VIII. THEORIES OF FAILURE

A. Need

• Ductile materials (steel) fail in shear (yielding). Brittle materials (cast iron) fail in tension (fracture).

• Uniaxial tensile test gives σ_ut or σ_yt. For multiaxial stress, need criteria to predict failure.

B. Major Theories (JUN 2025: Explain any three with graphs)

1. Maximum Principal Stress Theory (Rankine's):

   • Statement: Failure occurs when max principal stress reaches σ_ult (tensile) or σ_uc (compressive).

   • For ductile: Not accurate (ignores shear).

   • For brittle: Good.

   • Graph: Square in σ₁-σ₂ plane (sides parallel to axes).

2. Maximum Shear Stress Theory (Tresca/Guest's):

   • Statement: Failure occurs when max shear stress reaches shear stress at yield in simple tension (τ_y = σ_y/2).

   • For ductile: Conservative, good.

   • Graph: Hexagon in σ₁-σ₂ plane (inscribed in von Mises circle).

3. Distortion Energy Theory (von Mises/Hencky):

   • Statement: Failure occurs when distortion strain energy per unit volume reaches that at yield in simple tension.

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

   • For plane stress (σ₃=0): $$\displaystyle \sigma_{\text{eq}} = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2} \leq \sigma_y $$.

   • For ductile: Most accurate.

   • Graph: Circle in σ₁-σ₂ plane (radius = σ_y).

4. Maximum Principal Strain Theory (St. Venant):

   • Statement: Failure when max principal strain reaches strain at failure in simple tension.

   • $$\displaystyle \epsilon_1 = \frac{\sigma_1}{E} - \nu\frac{\sigma_2}{E} - \nu\frac{\sigma_3}{E} \leq \frac{\sigma_y}{E} $$.

   • Graph: Parabola.

   • Rarely used.

C. Comparison & Application

• Ductile (steel): von Mises (best), Tresca (safe but conservative).

• Brittle (cast iron, concrete): Rankine (max principal stress).

• Application (JUN 2024): Given principal stresses and σ_yt, check against theory. For Tresca: τ_max = (σ₁-σ₂)/2 ≤ σ_yt/2. For von Mises: use equivalent stress formula.

IX. SPECIAL TOPICS & MISCELLANEOUS

A. Section Modulus (JUN 2025 Short Note)

• Definition: $$\displaystyle Z = I / y_{\text{max}} $$. Measures bending resistance.

• Units: mm³, cm³, m³.

• Significance: Bending stress σ_max = M/Z. For given M, larger Z → smaller σ.

• Strongest Rectangular Beam from Circular Log:

  • For circular section diameter D: Z = πD³/32.

  • For rectangular section (b×h) inscribed: b = D cosθ, h = D sinθ? Actually, diagonal = D → b² + h² = D².

  • Z = b h²/6 = (h²/6) √(D² - h²). Maximize Z w.r.t h → h/D = √(2/3) ≈ 0.816, b/D = √(1/3) ≈ 0.577 → h/b = √2 ≈ 1.414.

  • [!TIP] DEC 2023: Prove h/b = √2. Differentiate Z w.r.t h, set dZ/dh=0.

B. Point of Contraflexure (POC) (JUN 2025 Short Note)

• Definition: Point in a beam where bending moment is zero and changes sign (from sagging +ve to hogging -ve or vice versa).

• Significance: Indicates a point of inflection in the deflected shape. Shear force may or may not be zero.

• Location: Solve M(x) = 0 for x within beam span.

C. Middle Third Rule (JUN 2025 Explanation)

• For Rectangular Section under Axial Load:

  • Stress distribution: σ = P/A ± M*y/I = P/(bh) ± P e y / (bh³/12).

  • Max stress at extreme fiber: σ_max = P/(bh) [1 ± (6e/h)].

  • To avoid tension (σ_min ≥ 0), need 1 - (6e/h) ≥ 0 → e ≤ h/6.

  • Similarly, to avoid compression on other side? Actually, if e ≤ h/6, both stresses are compressive (σ_min = P/(bh)(1 - 6e/h) ≥ 0).

• Middle Third: Load must lie within central third of width (for rectangular, within h/6 from centroid in both directions).

• Significance: Prevents tensile stresses in brittle materials (concrete, masonry) which are weak in tension.

D. St. Venant's Principle (NOV 2022)

• Statement: The difference between the effects of two different but statically equivalent loads becomes negligible at sufficiently large distances from the load application region.

• Significance: Allows replacing a complex load distribution with a simpler equivalent one (e.g., point load vs. distributed) for stress analysis away from the load. Stress "evens out" as we move away.

E. Compound Cylinders (Shrunk Fits) (JUN 2023)

• Setup: Two cylinders (inner & outer) assembled by shrinking (heating outer, cooling inner) or pressing.

• Pressure at Junction (p): Radial pressure developed at interface due to interference fit.

• Stress Analysis:

  • Inner cylinder: Subjected to internal pressure p (from outer) and possibly external pressure p (from fit)? Actually, after assembly, inner cylinder has compressive radial pressure p on its outer surface. Outer cylinder has tensile radial pressure p on its inner surface.

  • Use Lame's equations for thick cylinders:

    • Radial: σ_r = A - B/r²

    • Hoop: σ_θ = A + B/r²

  • Apply boundary conditions for each cylinder separately.

• When subjected to internal pressure P: Superimpose stresses from fit and from internal pressure.

• [!TIP] JUN 2023: Given diameters, interference, E, ν, find p first from compatibility (radial displacement continuity). Then find final stresses.

F. Impact Loading (Sudden Load) (DEC 2023)

• Stress due to Load Falling from Height h:

  • Gradual Application: δ = PL/(AE).

  • Sudden Load: Energy conservation: $$\displaystyle P_h (h + \delta) = \frac{1}{2} P_{max} \delta $$, where P_h = weight.

  • Solve: $$\displaystyle P_{max} = P_h \left(1 + \sqrt{1 + \frac{2AE}{P_h h}}\right) $$. For h=0 (load dropped from zero height): $$\displaystyle P_{max} = 2P_h $$.

  • Stress: σ_max = P_max / A.

• [!TIP] DEC 2023: Compare with static load. Impact factor = P_max / P_h.

G. Poisson's Ratio Determination (DEC 2023)

• From Combined Tension & Torsion:

  • Tension test: σ = E ε_long, ν = -ε_lat / ε_long.

  • Torsion test: τ = G γ, where γ = angle of twist * r / L.

  • Given: Axial load P → σ = P/A, measure elongation ΔL → ε_long = ΔL/L.

  • Given: Torque T → τ = T r / J, measure angle of twist θ → γ = θ r / L.

  • From E = 2G(1+ν) → ν = E/(2G) - 1.

  • Compute E from tension, G from torsion, then ν.

Final Exam Strategy:

1. Stress Transformation & Mohr's Circle: Master principal stress calculation and Mohr's construction. (JUN 2025, DEC 2024)

2. Beam Deflection: Be fluent in Area Moment method for standard cases and Macaulay's for discontinuous loads. (JUN 2025, DEC 2023)

3. Columns: Distinguish Euler (long) vs. Rankine (intermediate). Know equivalent lengths. (JUN 2024, DEC 2024)

4. Combined Loading: Shafts under M+T → find principal stresses. (JUN 2025)

5. Theories of Failure: Know graphs and applications for Tresca & von Mises (ductile) and Rankine (brittle). (JUN 2025)

6. Thin Cylinders: Remember hoop = pd/2t, longitudinal = pd/4t. (DEC 2023, JUN 2023)

7. Composite Bars & Thermal Stresses: Compatibility is key. (DEC 2024, JUN 2023)

\boxed{\text{Revise derivations: Euler's, Torsion, Bending Equation, Shear Stress, Elastic Constants Relations.}}