Strength of Materials (CE-305) - Unit 1 Short Notes

I. FUNDAMENTAL CONCEPTS

Stress and Strain

• Stress (σ): Internal force per unit area.

  Types:

  • Tensile/Compressive: Normal to area.

  • Shear: Tangential to area.

  Unit: Pa (N/m²) or MPa (N/mm²).

• Strain (ε): Dimensionless deformation.

  Types:

  • Normal: ΔL/L.

  • Shear: γ = tanθ ≈ θ (rad).

  • Volumetric: ε_v = ΔV/V.

• Stress-Strain Curve: Shows elastic limit, yield point, ultimate strength, fracture.

Elastic Constants

Relationships:

1. E, G, ν:

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

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

2. E, K, ν:

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

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

3. E, G, K:

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

Volumetric Strain under Multi-axial Loading

For 3D stresses (σ_x, σ_y, σ_z):

$$ \varepsilon_v = \varepsilon_x + \varepsilon_y + \varepsilon_z = \frac{1}{E} \left[ (\sigma_x + \sigma_y + \sigma_z) - 2\nu(\sigma_y + \sigma_z + \sigma_x) \right] $$

Simplified:

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

For hydrostatic stress (σ_x = σ_y = σ_z = σ):

$$ \varepsilon_v = \frac{3\sigma(1-2\nu)}{E}, \quad K = \frac{\sigma}{\varepsilon_v/3} $$

Thermal Stresses

• Unrestrained: Free expansion, no stress.

  ΔL = α L ΔT, where α = coefficient of thermal expansion.

• Restrained (ends fixed):

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

(Compressive if heated, tensile if cooled).

• Partial restraint (ends yield by δ):

$$ \sigma = E \left( \alpha \Delta T - \frac{\delta}{L} \right) $$

Impact Loading

• Stress due to falling weight (mass m, height h, static deflection δ_static):

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

where $$\displaystyle \sigma_{static} = \frac{P}{A} $$, $$\displaystyle P = mg $$, $$\displaystyle \delta_{static} = \frac{PL}{AE} $$.

Self-weight of Bars

• Uniform bar (weight W, length L, area A):

$$ \delta = \frac{W L}{2 A E} = \frac{\rho g L^2}{2E} $$

(Linear variation from zero at free end to maximum at fixed end).

• Tapering bar: Integrate δ = ∫ (ρ g A(x) dx) / (A(x) E) from 0 to L.

[!TIP]

• Common Pitfall: In thermal stress, sign convention: heating → compression if restrained.

• Impact Loading: Always check if h >> δ_static; if h=0, reduces to static.

• Self-weight: Treat as distributed load; elongation is half of that due to same force applied at end.

II. AXIAL LOADING

Uniformly Tapering Rod

• Diameter varies linearly from d to D over length L.

• Elongation:

$$ \delta = \int_0^L \frac{P}{E A(x)} dx, \quad A(x) = \frac{\pi}{4} \left( d + \frac{D-d}{L}x \right)^2 $$

Solving:

\boxed{\delta = \frac{4PL}{\pi E d D}}

(d and D are end diameters).

Composite Bars (Different Materials)

• Assumptions: Same strain (perfect bonding), load parallel to axis.

• Compatibility: ε₁ = ε₂ = ε.

• Stress distribution:

$$ \sigma_1 = \frac{P E_1}{E_1 A_1 + E_2 A_2}, \quad \sigma_2 = \frac{P E_2}{E_1 A_1 + E_2 A_2} $$

• Total deformation:

$$ \delta = \frac{\sigma_1 L}{E_1} = \frac{\sigma_2 L}{E_2} $$

Stepped Bars

• Different cross-sections in series.

• Total extension: Sum of extensions in each segment:

$$ \delta = \sum_{i=1}^n \frac{P L_i}{E A_i} $$

[!TIP]

• Composite Bars: Always check strain compatibility first.

• Stepped Bars: Identify segments with constant area; apply δ = PL/(AE) per segment.

III. TORSION OF CIRCULAR SHAFTS

Solid Circular Shaft

• Shear stress (at radius ρ):

$$ \tau = \frac{T \rho}{J}, \quad \tau_{max} = \frac{T R}{J} = \frac{16T}{\pi D^3} $$

where $$\displaystyle J = \frac{\pi D^4}{32} $$ (polar moment of inertia).

• Angle of twist (θ in radians):

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

• Power transmission:

$$ P = \frac{2\pi N T}{60} $$

(P in Watts, N in rpm, T in N·m).

Hollow Circular Shaft

• Polar moment:

$$ J = \frac{\pi (D_o^4 - D_i^4)}{32} $$

• Max shear stress:

$$ \tau_{max} = \frac{T D_o/2}{J} = \frac{16T D_o}{\pi (D_o^4 - D_i^4)} $$

• Angle of twist: Same formula with J.

• Diameter ratio: $$\displaystyle k = D_i/D_o $$. For given τ and θ, optimize k.

Design for Strength and Rigidity

• Strength: τ_max ≤ τ_allow → $$\displaystyle T \leq \frac{\tau_{allow} J}{R} $$.

• Rigidity: θ ≤ θ_allow → $$\displaystyle T \leq \frac{\theta_{allow} G J}{L} $$.

Comparison: Hollow vs Solid Shafts

Combined Bending and Torsion

• Stresses at extreme fiber:

  Bending: $$\displaystyle \sigma_b = \frac{M c}{I} = \frac{M}{Z} $$

  Torsion: $$\displaystyle \tau_t = \frac{T c}{J} = \frac{T}{2Z} $$ (for solid circular, $$\displaystyle Z = J/c $$).

• Principal stresses (σ₁, σ₂):

$$ \sigma_{1,2} = \frac{\sigma_b}{2} \pm \sqrt{ \left( \frac{\sigma_b}{2} \right)^2 + \tau_t^2 } $$

\boxed{\sigma_{1,2} = \frac{\sigma_b}{2} \pm \sqrt{ \left( \frac{\sigma_b}{2} \right)^2 + \tau_t^2 }}

• Maximum shear stress:

$$ \tau_{max} = \sqrt{ \left( \frac{\sigma_b}{2} \right)^2 + \tau_t^2 } $$

• Location of principal planes:

$$ \tan 2\theta_p = \frac{2\tau_t}{\sigma_b} $$

(θ_p measured from axis of σ_b).

[!TIP]

• Hollow shaft: Always check both τ and θ constraints; often θ governs.

• Combined loading: For solid circular shaft, note $$\displaystyle \tau_t = T/(2Z) $$ simplifies calculations.

• Principal stresses: If σ_b = 0 (pure torsion), σ₁ = τ_t, σ₂ = -τ_t.

IV. SHEAR FORCE AND BENDING MOMENT

Types of Loads on Beams

1. Point load (concentrated): Acts at a point.

2. Uniformly Distributed Load (UDL): Constant intensity w (kN/m).

3. Varying load: Intensity varies (e.g., triangular).

4. Couple: Moment applied at a point.

Sign Conventions

Differential Relationships

• Load vs Shear:

$$ \frac{dV}{dx} = -w(x) $$

\boxed{\frac{dV}{dx} = -w}

• Shear vs Moment:

$$ \frac{dM}{dx} = V $$

\boxed{\frac{dM}{dx} = V}

• Integral forms:

$$ V = -\int w \, dx + C_1, \quad M = \int V \, dx + C_2 $$

Shear Force & Bending Moment Diagrams

General Steps:

1. Find reactions (supports).

2. Section beam; compute V and M at key points (loads, supports, points of zero shear).

3. Plot diagrams (scale).

Common Cases:

• Simply Supported Beam:

  • Central point load P:

    V: ±P/2 at supports, zero at center.

    M: Parabolic, max at center: M_max = PL/4.

  • UDL over entire span:

    V: Linear from wL/2 at supports to zero at center.

    M: Parabolic, max at center: M_max = wL²/8.

  • Combination: Superpose diagrams.

• Cantilever Beam:

  • End point load P:

    V: Constant = -P (negative by convention).

    M: Linear from 0 at free end to -PL at fixed end.

  • UDL over entire length:

    V: Linear from 0 at free to -wL at fixed.

    M: Parabolic from 0 to -wL²/2.

  • Partial UDL: Use Macaulay or step integration.

• Overhanging Beam: Reactions may be upward/downward; BM diagram may have positive and negative regions.

Point of Contra-flexure

• Definition: Point where bending moment changes sign (M=0) and curvature reverses.

• Location: Solve M(x)=0 from BM equation or diagram.

• Significance: Inflection point in deflected shape.

[!TIP]

• Sign conventions: Consistency is key; many textbooks use opposite sign for V. Follow your convention.

• Diagram construction: Always check equilibrium (ΣV=0, ΣM=0) after reactions.

• Point of contra-flexure: Only for continuous or overhanging beams; not in simply supported with symmetric loads.

V. BENDING STRESSES AND SHEAR STRESSES IN BEAMS

Assumptions in Simple Bending

1. Material homogeneous, isotropic, obeys Hooke’s law.

2. Beam initially straight, cross-sectional area constant.

3. Plane sections remain plane (Bernoulli’s hypothesis).

4. No shear deformation effect on bending.

5. Radius of curvature >> beam depth.

6. Pure bending (no shear force) or shear effects negligible for normal stress.

Bending Stress Formula

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

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

where:

• M = bending moment at section,

• y = distance from neutral axis,

• I = second moment of area about NA.

• Max stress: $$\displaystyle \sigma_{max} = \frac{M}{Z} $$, with $$\displaystyle Z = \frac{I}{y_{max}} $$ (section modulus).

Section Modulus (Z) for Common Sections

Note: For T-section, NA not at centroid; find NA first by $$\displaystyle \bar{y} = \frac{\sum A_i y_i}{\sum A_i} $$.

Shear Stress Distribution

• General formula:

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

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

where:

• V = shear force,

• Q = first moment of area about NA of area above (or below) the point,

• I = I about NA,

• b = width at the point.

Key Distributions:

1. Rectangular section (b×h):

   • τ varies parabolically: $$\displaystyle \tau = \frac{6V}{b h^2} (h^2/4 - y^2) $$.

   • Max at neutral axis: $$\displaystyle \tau_{max} = \frac{3V}{2A} = 1.5 \times \text{average} $$.

   • Zero at top/bottom.

2. I-section:

   • Web: Carries most shear. τ ≈ V / (A_web) (approx), but exact: $$\displaystyle \tau_{web} = \frac{V}{I b_{web}} Q_{flange} $$.

   • Flange: Very low shear; changes sign at NA.

   • Max shear at NA in web.

3. T-section:

   • NA in flange or web depending on proportions.

   • Compute Q for each part; τ max usually in web at NA.

Middle Third Rule

• Applicability: Rectangular sections under combined axial load + bending.

• Statement: To avoid tensile stress anywhere in the section, the resultant force must lie within the middle third of the cross-section.

• Derivation: For rectangular section (b×h) with axial load P and eccentricity e (about depth axis):

$$ \sigma_{min} = \frac{P}{A} - \frac{M c}{I} = \frac{P}{bh} - \frac{P e (h/2)}{bh^3/12} = \frac{P}{bh} \left(1 - \frac{6e}{h}\right) \geq 0 $$

⇒ $e \leq h/6$. Similarly for breadth: $$\displaystyle e_b \leq b/6 $$.

Thus, kern is rectangle of size $b/3 \times h/3$.

• Application: Check if load eccentricity is within middle third to ensure no tension (for materials weak in tension, e.g., concrete).

[!TIP]

• Shear stress: For I-section, always compute Q correctly: for web at NA, Q = area of one flange × distance from NA to flange centroid.

• Section modulus: For built-up sections, compute I about NA first, then Z.

• Middle third: Only for rectangular sections under axial + uniaxial bending. For circular, it’s middle quarter (radius R/4).

VI. DEFLECTION OF BEAMS

Area Moment Method (Mohr’s Theorems)

• First Theorem (Slope):

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

\boxed{\theta_B = \frac{1}{EI} \int_A^B M(x) , dx}

(Area signed; clockwise positive if M diagram positive).

• Second Theorem (Deflection):

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

\boxed{\delta_B = \frac{1}{EI} \int_A^B M(x) \cdot \bar{x} , dx}

where $\bar{x}$ is distance from B to centroid of area.

• Procedure: Draw M diagram to scale; use geometry to find areas and centroids.

Double Integration Method

• Differential equation:

$$ EI \frac{d^2 y}{dx^2} = M(x) $$

\boxed{EI \frac{d^2 y}{dx^2} = M(x)}

• Integrate twice:

$$ EI \frac{dy}{dx} = \int M(x) dx + C_1 $$

$$ EI y = \int \left( \int M(x) dx \right) dx + C_1 x + C_2 $$

• Boundary conditions:

  • At supports: y=0 (simple), y=0 and dy/dx=0 (fixed).

  • At free end: M=0, V=0 (cantilever).

• Constants: Apply BCs to find C₁, C₂.

Macaulay’s Method

• For discontinuous loads (point loads at arbitrary positions).

• Step function: Use Macaulay brackets $$\displaystyle \langle x-a \rangle^n $$:

  $$\displaystyle \langle x-a \rangle^n = \begin{cases} 0 & x < a \\ (x-a)^n & x \geq a \end{cases} $$

• Procedure:

  1. Express load w(x) and moment M(x) using brackets.

  2. Integrate:

$$ EI \frac{d^2 y}{dx^2} = M(x) $$

$$ EI \frac{dy}{dx} = \int M(x) dx + C_1 $$

$$ EI y = \int \int M(x) dx^2 + C_1 x + C_2 $$

3. Integrate each term with brackets; keep brackets until final substitution.

4. Apply BCs (usually at x=0 and x=L) to find C₁, C₂.

5. For deflection at point x=a, substitute x=a (brackets become zero for terms with a>x).

Applications (Common Cases)

1. Simply Supported Beam:

   • Central point load P:

     $$\displaystyle M = \frac{P}{2}x $$ for $$\displaystyle 0<x<L/2 $$, symmetric.

     Deflection at center: $$\displaystyle \delta_{max} = \frac{P L^3}{48 EI} $$.

   • UDL over entire span (w):

     $$\displaystyle M = \frac{w}{2}x(L-x) $$.

     $$\displaystyle \delta_{max} = \frac{5 w L^4}{384 EI} $$.

   • Point load at distance a from left:

     Use Macaulay or integration; deflection under load:

$$ \delta_P = \frac{P a b^2}{6 E I L} (L + b) $$

 where $$\displaystyle b = L - a $$.

2. Cantilever Beam:

   • End point load P:

     $$\displaystyle M = -P x $$, $$\displaystyle \delta_{free} = \frac{P L^3}{3 EI} $$, $$\displaystyle \theta_{free} = \frac{P L^2}{2 EI} $$.

   • UDL over entire length (w):

     $$\displaystyle M = -\frac{w}{2} x^2 $$, $$\displaystyle \delta_{free} = \frac{w L^4}{8 EI} $$, $$\displaystyle \theta_{free} = \frac{w L^3}{6 EI} $$.

   • Partial UDL (from free end to a):

     Use Macaulay; $$\displaystyle \delta_{free} = \frac{w a^2}{24 EI} (4L^2 - a^2) $$ if a < L.

3. Overhanging Beams: Combine methods; careful with sign conventions.

[!TIP]

• Area moment method: Quick for slopes/deflections at supports and points of zero M.

• Double integration: Good for general y(x).

• Macaulay: Best for multiple point loads; keep brackets until end.

• Boundary conditions: For simply supported, y=0 at both ends; for cantilever, at fixed end y=0, dy/dx=0.

VII. STRESS TRANSFORMATION AND MOHR'S CIRCLE

2D Stress Transformation

Given stresses on x-y plane: σ_x, σ_y, τ_xy.

• On plane at angle θ (from x-axis):

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

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

\boxed{\sigma_n = \sigma_{avg} + R \cos 2\theta, \quad \tau_{nt} = R \sin 2\theta}

where $$\displaystyle \sigma_{avg} = \frac{\sigma_x+\sigma_y}{2} $$, $$\displaystyle R = \sqrt{ \left( \frac{\sigma_x-\sigma_y}{2} \right)^2 + \tau_{xy}^2 } $$.

Principal Stresses and Planes

• Principal planes: Where τ_nt = 0.

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

Two solutions θ_p and θ_p + 90°.

• Principal stresses:

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

\boxed{\sigma_{1,2} = \sigma_{avg} \pm R}

• Maximum shear stress:

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

Occurs on planes at 45° to principal planes.

Mohr’s Circle

• Construction:

  1. Plot point A(σ_x, τ_xy) and B(σ_y, -τ_xy) (sign: τ_xy positive if causing clockwise rotation).

  2. Center C at (σ_avg, 0), radius R.

  3. Circle through A and B.

• Interpretation:

  • Principal stresses: intersections with σ-axis (τ=0).

  • Max shear: top/bottom of circle (τ = ±R).

  • Stress on plane at θ: from A, move 2θ counterclockwise.

• Advantage: Visual, no trigonometry needed for extremum.

[!TIP]

• Sign convention: τ_xy positive if it tends to rotate element clockwise (many textbooks use opposite; be consistent).

• Principal stresses: σ₁ is max (algebraically larger), σ₂ min.

• Mohr’s circle: Angle on circle is 2θ (not θ).

• 3D: Three Mohr’s circles; principal stresses are extremes.

VIII. COMBINED STRESSES AND FAILURE THEORIES

Theories of Failure

Notes:

• For ductile materials (steel): von Mises and Guest are conservative; Rankine non-conservative.

• For brittle (concrete, cast iron): Rankine or St. Venant.

• Combined loading (e.g., bending + torsion): Compute σ₁, σ₂ from combined state; compare with criterion.

Application Example: Shaft under Bending (σ) and Torsion (τ)

• Principal stresses:

$$ \sigma_{1,2} = \frac{\sigma}{2} \pm \sqrt{ \left( \frac{\sigma}{2} \right)^2 + \tau^2 } $$

• Check von Mises:

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

For pure shear (σ=0): σ_eq = √3 τ, so τ_allow = σ_y/√3.

• Check Guest: τ_max = √((σ/2)²+τ²) ≤ τ_ult (≈ σ_y/2).

[!TIP]

• Failure theory selection: Ductile → von Mises (most accurate) or Guest (simpler). Brittle → Rankine.

• Graphical: Draw Mohr’s circles for given σ₁, σ₂; see which envelope touches first.

• Combined loading: Always reduce to principal stresses first.

IX. COLUMNS AND BUCKLING

Euler’s Buckling Load

• Derivation (pinned-pinned):

  Differential eq: $$\displaystyle EI \frac{d^2 y}{dx^2} = -M = -P y $$.

  Solution: $$\displaystyle y = A \sin(kx) + B \cos(kx) $$, with $$\displaystyle k = \sqrt{P/EI} $$.

  BCs: y=0 at x=0,L ⇒ sin(kL)=0 ⇒ kL = nπ.

  Critical load:

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

\boxed{P_{cr} = \frac{\pi^2 E I}{L_e^2}}

where $$\displaystyle L_e $$ = effective length.

• Effective length $$\displaystyle L_e = K L $$:

  | End Conditions | K | L_e | |----------------|---|-----| | Both pinned | 1.0 | L | | Both fixed | 0.5 | L/2 | | One fixed, one free | 2.0 | 2L | | One fixed, one pinned | 0.707 | L/√2 |

Rankine’s Formula (Intermediate Columns)

• Empirical:

$$ \frac{1}{P_{cr}} = \frac{1}{P_e} + \frac{1}{P_c} $$

or

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

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

where:

• σ_c = crushing strength (material),

• a = Rankine’s constant (≈ 1/1600 for steel),

• r = radius of gyration = √(I/A).

• Limits:

  • Long column (L_e/r large): $$\displaystyle P_{cr} ≈ P_e $$ (Euler).

  • Short column (L_e/r small): $$\displaystyle P_{cr} ≈ \sigma_c A $$ (crushing).

Eccentric Loading

• Direct + bending:

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

Max stress: $$\displaystyle \sigma_{max} = \frac{P}{A} \left(1 + \frac{e c}{r^2} \right) $$, min: $$\displaystyle \sigma_{min} = \frac{P}{A} \left(1 - \frac{e c}{r^2} \right) $$.

• Middle third rule (rectangular): To avoid tension, $e \leq h/6$ (for bending about depth axis).

St. Venant’s Theory (Short Columns)

• Assumes uniform stress distribution (no buckling).

  $$\displaystyle P_{cr} = \sigma_c A $$ (crushing load).

  Valid for stocky columns (L_e/r < 30 approx).

[!TIP]

• Euler valid for long columns (L_e/r > critical slenderness).

• Rankine bridges Euler and crushing; use for all lengths.

• Eccentric loading: Always compute both max and min stresses; check for tension if material weak in tension.

• Comparison: For same material, Euler load ∝ 1/L², Rankine ∝ 1/(1+aL²).

X. THIN PRESSURE VESSELS

Thin Cylindrical Shell

• Assumptions: Wall thickness t << radius r; stresses uniform across thickness.

• Hoop stress (circumferential):

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

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

• Longitudinal stress:

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

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

• Ratio: $$\displaystyle \sigma_h : \sigma_l = 2:1 $$.

Thin Spherical Shell

• Hoop stress (same in all directions):

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

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

Deformations

• Cylinder:

  • Circumferential strain:

$$ \varepsilon_c = \frac{\sigma_h}{E} - \nu \frac{\sigma_l}{E} = \frac{p d}{4 E t} (2 - \nu) $$

→ Change in diameter: $$\displaystyle \Delta d = d \varepsilon_c $$.

• Longitudinal strain:

$$ \varepsilon_l = \frac{\sigma_l}{E} - \nu \frac{\sigma_h}{E} = \frac{p d}{4 E t} (1 - 2\nu) $$

→ Change in length: $$\displaystyle \Delta L = L \varepsilon_l $$.

• Volume change:

$$ \frac{\Delta V}{V} = 2\varepsilon_c + \varepsilon_l = \frac{p d}{4 E t} (5 - 4\nu) $$

• Sphere:

  • Strain: $$\displaystyle \varepsilon = \frac{\sigma}{E} - 2\nu \frac{\sigma}{E} = \frac{p d}{4 E t} (1 - 2\nu) $$.

  • Volume change: $$\displaystyle \frac{\Delta V}{V} = 3\varepsilon = \frac{3 p d}{4 E t} (1 - 2\nu) $$.

Problems

1. Given p, find stresses and deformations: Use formulas above.

2. Given deformation, find p:

   From measured Δd or ΔV, compute p using strain formulas.

[!TIP]

• Thin assumption: t/d ≤ 1/20.

• Hoop stress is critical (higher).

• Sphere is more efficient (stress half of cylinder for same p,d,t).

XI. COMPOSITE AND COMPOUND STRUCTURES

Composite Bars (Axial Loading)

• Covered in Unit II. Same strain, different materials.

$$ \sigma_1 = \frac{P E_1}{E_1 A_1 + E_2 A_2}, \quad \sigma_2 = \frac{P E_2}{E_1 A_1 + E_2 A_2} $$

Compound Cylinders (Shrink Fit)

• Radial pressure at junction (p): Compatibility of deformations.

• Stresses in each cylinder (before pressure: zero; after pressure: radial + hoop).

• Lame’s equations for thick cylinders (if needed, but thin assumption may not hold).

• Superposition: Stresses due to shrink pressure + internal pressure.

Bolted Joints with Thermal Effects

• Pre-stress: Due to tightening (torque → tension in bolt, compression in members).

• Thermal stresses: If temperature changes, different expansions cause additional stresses.

  Example: Bolt and plate have different α; if bolt is tighter, thermal change alters pre-load.

[!TIP]

• Shrink fit: Pressure p found from interference fit δ = (p D)/(E) (1/D1 - 1/D2) for thin? Actually, for thick, use Lame.

• Bolted joints: Assume bolt takes all tension; members in compression. Thermal: ΔT increases tension if bolt expands more.

XII. ELASTIC CONSTANTS AND MATERIAL PROPERTIES

Derivation of Relations

1. E, G, ν:

   Consider pure shear: τ = Gγ. Also, from uniaxial tension: ε_x = σ/E, ε_y = -νσ/E.

   But shear strain γ = ε_x - ε_y = σ/E (1+ν).

   Since τ = Gγ, and τ = σ (on 45° planes), so σ = G σ (1+ν)/E ⇒ E = 2G(1+ν).

2. E, K, ν:

   Hydrostatic stress: σ_h = p, ε_v = 3ε = p/K.

   But ε = (p/E) - 2ν(p/E) = p(1-2ν)/E.

   So 3p(1-2ν)/E = p/K ⇒ K = E/(3(1-2ν)).

Poisson’s Ratio Determination

• From tension test: Measure axial strain ε and lateral strain ε_lat; ν = -ε_lat/ε.

• From torsion test:

  For circular shaft, angle of twist θ, torque T:

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

Also, from tension test get E. Then ν = (E/(2G)) - 1.

Material Behavior

• Ductile: Large plastic deformation before fracture (steel, aluminum).

  Yield point distinct; failure theories: von Mises, Guest.

• Brittle: Little plastic deformation (cast iron, concrete).

  Failure theories: Rankine, St. Venant.

[!TIP]

• ν range: For most metals, 0.25-0.35; for rubber, ~0.5.

• Physical constraint: ν < 0.5 (from K>0).

• Relation check: If ν=0.5, material incompressible (K→∞).

XIII. SPECIAL TOPICS

Middle Third Rule (Detailed)

• For rectangular section under axial load + uniaxial bending, to avoid tension: resultant must lie within central rectangle of size b/3 × h/3.

• Proof: For bending about depth h, eccentricity e along h:

  σ_min = P/A - P e c/I ≥ 0 ⇒ e ≤ I/(A c) = (bh³/12)/(bh·h/2) = h/6.

  Similarly for breadth: e_b ≤ b/6.

• Application: Check if load point is within middle third; if outside, tensile stress develops.

Section Modulus (Z)

• Definition: $$\displaystyle Z = I / y_{max} $$; measure of beam’s resistance to bending.

• Significance: σ_max = M/Z. For given M, larger Z → lower stress.

• Design: Choose section with adequate Z.

• Calculation: Compute I about NA, divide by max y.

Thermal Stresses in Composite Systems

• Two materials bonded, temperature change ΔT.

• Free expansions: δ₁ = α₁ L ΔT, δ₂ = α₂ L ΔT.

• Compatibility: Actual δ same → stresses develop.

• Equations:

  σ₁/E₁ + α₁ ΔT = σ₂/E₂ + α₂ ΔT (if no external load).

  Solve with force equilibrium: σ₁ A₁ + σ₂ A₂ = 0.

• Result:

$$ \sigma_1 = \frac{E_1 E_2 (\alpha_2 - \alpha_1) \Delta T}{E_1 A_1 + E_2 A_2} A_1 $$

(sign depends on α difference).

St. Venant’s Principle

• Statement: The difference between the actual stress distribution and the statically equivalent, simplified distribution becomes negligible at distances large compared to the dimensions of the loaded region.

• Implication: Stress concentrations localize; far field uniform.

• Application: Replace complex load with equivalent force-couple; stress analysis simplified away from load.

[!TIP]

• Middle third: Only for rectangular sections under axial + bending about one axis.

• Section modulus: For built-up sections, compute I about NA carefully.

• Thermal composite: If α₁ > α₂, heating causes σ₁ compressive, σ₂ tensile.

• St. Venant: Used to justify using average stress in members away from concentrated loads.

Final Note: This summary covers all topics from Unit 1 frequently appearing in RGPV exams. Focus on derivations (E-G-ν, Euler, shear stress), formula applications, and diagram construction. Practice numerical problems from past papers for mastery.