Strength of Materials (CE-305) - Unit 3 Short Notes

1.0 COMBINED STRESSES AND STRESS TRANSFORMATION

1.1 Principal Stresses and Principal Planes

• Definition: At a point in a stressed body, principal stresses are the maximum and minimum normal stresses acting on planes where shear stress is zero. These orthogonal planes are principal planes.

• Significance: They represent the extreme normal stress values, critical for failure analysis.

• Calculation (2D State: σₓ, σᵧ, τₓᵧ):

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

Where:

*   $$\displaystyle \sigma_1 $$ = Major principal stress (maximum)

*   $$\displaystyle \sigma_2 $$ = Minor principal stress (minimum)

• Orientation (θₚ): The angle θₚ that principal planes make with the x-axis is given by:

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

Solve for θₚ. The two principal planes are orthogonal (2θₚ and 2θₚ + 90°).

[!TIP]

Common Pitfall: Remember that θₚ from the formula gives the angle for σ₁. To find the plane for σ₂, add 45° to θₚ. Always verify shear stress on the calculated plane is zero.

1.2 Mohr's Circle for 2D Stress State

• Construction:

  1. Plot point A(σₓ, τₓᵧ) and point B(σᵧ, -τₓᵧ) on σ-τ axes.

  2. Line AB is the diameter. Center C is at $$\displaystyle (\frac{\sigma_x+\sigma_y}{2}, 0) $$.

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

  4. Draw circle with center C and radius R.

• Determination:

  • Principal Stresses (σ₁, σ₂): Points on σ-axis where circle intersects (τ=0).

  • Maximum Shear Stress (τ_max): Points on circle with maximum τ-value. $$\displaystyle \tau_{max} = R $$.

  • Stress on Inclined Plane: Measure angle 2θ (counter-clockwise from AB) on circle. Coordinates of point on circle give (σₙ, τₙ).

• Sign Convention: τₓᵧ positive if it causes positive rotation on x-face (counter-clockwise).

[!TIP]

Exam Key: On Mohr's circle, 2θ on the circle corresponds to θ on the physical element. Principal planes are where the circle crosses the σ-axis.

1.3 Stress Transformation Equations

For a plane inclined at angle θ from x-axis:

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

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

These equations are used to find normal ($$\displaystyle \sigma_n $$) and shear ($$\displaystyle \tau_n $$) stress on any inclined plane.

1.4 Applications: Combined Bending and Torsion

• Stress State in Shaft: For a shaft under bending moment M and torque T:

  • Bending stress (tensile/compressive): $$\displaystyle \sigma_b = \frac{M y}{I} = \frac{M}{Z} $$, where y is distance from NA, Z is section modulus.

  • Torsional shear stress: $$\displaystyle \tau_t = \frac{T r}{J} = \frac{T}{2A t} $$ (for thin tubes) or $$\displaystyle \frac{16T}{\pi d^3} $$ (solid circular).

• At a point on the surface (y = c = d/2):

  • $$\displaystyle \sigma_x = \sigma_b $$ (tensile on one side, compressive on other)

  • $$\displaystyle \tau_{xy} = \tau_t $$

  • $$\displaystyle \sigma_y = 0 $$ (for circular shaft under pure torsion/bending)

• Principal Stresses & Max Shear: Use formulas from 1.1 with $$\displaystyle \sigma_x = \sigma_b $$, $$\displaystyle \sigma_y = 0 $$, $$\displaystyle \tau_{xy} = \tau_t $$.

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

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

2.0 BEAM DEFLECTION METHODS

2.1 Moment Area Method (Area-Moment Theorems)

• Theorem 1: The change in slope between two points A and B on a beam is equal to the area of the M/EI diagram between those points.

$$\theta_B - \theta_A = \int_A^B \frac{M_x}{EI} dx$$

• Theorem 2: The deflection of point B relative to the tangent at point A is equal to the first moment of area of the M/EI diagram between A and B about B.

$$\delta_B = \int_A^B \frac{M_x}{EI} \bar{x} dx$$

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

• Sign Convention:

  • Slope: Counter-clockwise (CCW) positive.

  • Deflection: Upward positive.

  • M/EI Area: Area above axis positive, below negative.

• Application: Highly effective for simply supported and cantilever beams with standard loadings.

2.2 Double Integration Method

• Fundamental Equation: From beam curvature, $$\displaystyle \frac{d^2 y}{d x^2} = \frac{M_x}{E I} $$.

• Procedure:

  1. Find reactions.

  2. Write bending moment Mₓ as function of x for each segment.

  3. Integrate twice: $$\displaystyle \frac{dy}{dx} = \int \frac{M_x}{EI} dx + C_1 $$, $$\displaystyle y = \int \frac{dy}{dx} dx + C_2 $$.

  4. Apply boundary conditions (y=0 at supports, slope known at fixed ends) and continuity conditions (y and dy/dx continuous at load points) to find constants.

• Merits: General method. Demerit: Repeated integration for complex loading.

2.3 Macaulay's Method (Step-Function Method)

• Concept: Represent distributed loads, point loads, and moments using step functions (Macaulay brackets ⟨⟩). A single Mₓ expression is written for the entire beam using ⟨x-a⟩ⁿ which equals 0 for x<a and (x-a)ⁿ for x>a.

• Procedure:

  1. Write Mₓ using Macaulay brackets for all loads from left end.

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

  3. Apply boundary conditions (e.g., y=0 at x=0 and x=L for simply supported).

  4. Important: When integrating ⟨x-a⟩ⁿ, the result is ⟨x-a⟩ⁿ⁺¹/(n+1). Constants C₁, C₂ are added outside the brackets.

• Advantage: Single integration for entire beam, avoids splitting into segments.

2.4 Applications to Standard Beam Cases

• Simply Supported Beam with Central Point Load (P):

  • Max deflection at centre: $$\displaystyle \delta_{max} = \frac{P L^3}{48 E I} $$

  • Slope at supports: $$\displaystyle \theta_A = \theta_B = \frac{P L^2}{16 E I} $$

• Simply Supported Beam with UDL (w):

  • Max deflection at centre: $$\displaystyle \delta_{max} = \frac{5 w L^4}{384 E I} $$

  • Slope at supports: $$\displaystyle \theta_A = \theta_B = \frac{w L^3}{24 E I} $$

• Cantilever Beam with UDL (w):

  • Max deflection at free end: $$\displaystyle \delta_{max} = \frac{w L^4}{8 E I} $$

  • Slope at free end: $$\displaystyle \theta_{free} = \frac{w L^3}{6 E I} $$

• Cantilever Beam with Point Load (P) at free end:

  • $$\displaystyle \delta_{max} = \frac{P L^3}{3 E I} $$, $$\displaystyle \theta_{free} = \frac{P L^2}{2 E I} $$

[!TIP]

Memory Aid: For simply supported beams, central deflection formulas have denominators 48 (point load) and 384 (UDL). For cantilevers, denominators are 3 (point load) and 8 (UDL).

3.0 TORSION OF CIRCULAR SHAFTS

3.1 Torsion Formula Derivation

• Assumptions:

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

  2. Shaft is circular and straight.

  3. Cross-sections remain plane and circular after twist (no warping).

  4. Radial strain is zero.

  5. Stress does not exceed proportional limit.

• Derivation: From geometry, $$\displaystyle \gamma = r \frac{d\phi}{dx} $$. From Hooke's law, $$\displaystyle \tau = G \gamma = G r \frac{d\phi}{dx} $$.

  Equilibrium requires $\tau \cdot 2\pi r dr$ to balance torque on annular element. Integrating:

$$T = \int_0^R \tau \cdot 2\pi r^2 dr = \int_0^R G \frac{d\phi}{dx} \cdot 2\pi r^3 dr = G \frac{d\phi}{dx} \cdot \frac{\pi R^4}{2}$$

Let **Polar Moment of Inertia** $$\displaystyle J = \frac{\pi R^4}{2} $$ (solid) or $$\displaystyle J = \frac{\pi}{32}(D^4 - d^4) $$ (hollow).

• Final Formula:

$$\tau_{max} = \frac{T r}{J} = \frac{T c}{J}$$

where c = outer radius.

$$\frac{d\phi}{dx} = \frac{T}{G J}$$

(Angle of twist per unit length)

3.2 Solid vs Hollow Shafts

• Strength (τ_max): For same material, same outer diameter D, and same torque T:

  $$\displaystyle \tau_{solid} = \frac{16T}{\pi D^3} $$, $$\displaystyle \tau_{hollow} = \frac{16T}{\pi D^3 (1 - k^4)} $$ where $$\displaystyle k = d/D $$.

  Hollow shaft has lower τ_max for same D (stronger).

• Stiffness (Angle of Twist): $$\displaystyle \phi = \frac{T L}{G J} $$. For same weight (mass), hollow shaft has larger J → smaller φ (stiffer).

• Weight: For same D and length, hollow shaft is lighter. For same weight, hollow shaft can have larger D → much larger J.

• Conclusion: Hollow shafts are stronger and stiffer for the same weight.

3.3 Power Transmission

• Relationship: Power $P$ (Watts), rotational speed $N$ (rpm), torque $T$ (N.m):

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

• Design Constraints:

  1. Shear Stress: $$\displaystyle \tau_{max} = \frac{T c}{J} \leq \tau_{allow} $$

  2. Angle of Twist: $$\displaystyle \phi = \frac{T L}{G J} \leq \phi_{allow} $$ (often in degrees/meter)

• Fluctuating Loads: If maximum torque $$\displaystyle T_{max} $$ is greater than mean $$\displaystyle T_{mean} $$, design based on $$\displaystyle T_{max} $$ for stress and $$\displaystyle T_{mean} $$ for twist (or use fatigue criteria).

3.4 Combined Bending and Torsion in Shafts

• Stress State: At a point on the surface of a shaft subjected to bending moment M and torque T:

  • $$\displaystyle \sigma_x = \frac{M y}{I} = \frac{M}{Z} $$ (tensile on one side, compressive on other)

  • $$\displaystyle \tau_{xy} = \frac{T c}{J} $$ (shear)

  • $$\displaystyle \sigma_y = 0 $$

• Principal Stresses:

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

• Maximum Shear Stress:

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

• Application of Failure Theories: Use σ₁, σ₂, τ_max with appropriate theory (Maximum Shear Stress for ductile, Principal Stress for brittle).

4.0 COLUMNS AND BUCKLING

4.1 Euler's Buckling Theory

• Derivation (Pinned-Pinned): From differential equation of elastic curve: $$\displaystyle EI \frac{d^2 y}{dx^2} = -M = -P y $$.

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

  Boundary conditions (y=0 at x=0, x=L) → $$\displaystyle \sin(kL)=0 $$ → $$\displaystyle kL = n\pi $$.

  Critical Load (Euler's Load):

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

For fundamental mode (n=1, lowest load):

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

• Effective Length (Le): $$\displaystyle P_{cr} = \frac{\pi^2 E I}{L_e^2} $$. $$\displaystyle L_e = K L $$, where K depends on end conditions:

  | End Condition | K | Le | |------------------------|-----|----------| | 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 |

• Limitations: Valid only for long, slender columns where failure is by elastic buckling. Slenderness Ratio $$\displaystyle \lambda = \frac{L_e}{r} $$ (r = radius of gyration = √(I/A)). Euler's formula valid for λ > limiting slenderness.

4.2 Rankine's Formula

• Purpose: Empirical formula for intermediate columns (between long and short).

$$\frac{1}{P_{cr}} = \frac{1}{P_{Euler}} + \frac{1}{P_{crushing}} = \frac{1}{\frac{\pi^2 E I}{L_e^2}} + \frac{1}{\sigma_c A}$$

or

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

where:

*   $$\displaystyle \sigma_c $$ = crushing strength (from short column test)

*   $a$ = Rankine's constant (depends on material, $$\displaystyle a = \frac{\sigma_c}{\pi^2 E} $$ for ideal case)

• Transition: For very long columns (large Le/r), Euler term dominates. For short columns, crushing term dominates.

4.3 Eccentric Loading on Columns

• Stress Distribution: Axial load P with eccentricity e produces combined axial compression and bending.

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

Maximum compressive stress at extreme fiber:

$$\sigma_{max} = \frac{P}{A} \left(1 + \frac{e y_{max}}{r^2}\right)$$

Minimum stress (could be tensile if e large):

$$\sigma_{min} = \frac{P}{A} \left(1 - \frac{e y_{max}}{r^2}\right)$$

• Middle Third Rule (Rectangular Section): For no tensile stress in rectangular section under axial load, the line of action of P must lie within the middle third of the section. For a rectangle b×h, eccentricity limits: $$\displaystyle e_x \leq \frac{b}{6} $$, $$\displaystyle e_y \leq \frac{h}{6} $$.

4.4 Equivalent Length and Effective Length Factor

• Definition: The length of an equivalent pinned-pinned column that would buckle under the same critical load.

• Relationship: $$\displaystyle L_e = K L $$, where:

  • $$\displaystyle L_e $$ = Effective length

  • $K$ = Effective length factor (depends on end restraint)

  • $L$ = Actual unsupported length

• Values: (See table in 4.1). K=1.0 for pinned-pinned, K=0.5 for fixed-fixed, K=2.0 for fixed-free.

5.0 THEORIES OF FAILURE

5.1 Maximum Principal Stress Theory (Rankine's Theory)

• Statement: Failure occurs when the maximum principal stress reaches the elastic limit stress in simple tension ($$\displaystyle \sigma_{yt} $$).

• Criterion (Brittle Materials):

$$\sigma_1 \leq \sigma_{yt} \quad (\text{tension})$$

$$\sigma_3 \geq -\sigma_{yc} \quad (\text{compression})$$

Often for brittle materials with $$\displaystyle \sigma_{yc} \approx \sigma_{yt} $$: $$\displaystyle |\sigma_1|, |\sigma_3| \leq \sigma_y $$.

• Graphical: On Mohr's circle, failure when largest circle touches the $$\displaystyle \pm \sigma_y $$ vertical lines.

5.2 Maximum Shear Stress Theory (Guest's Theory)

• Statement: Failure occurs when the maximum shear stress reaches the shear stress at yield in simple tension ($$\displaystyle \tau_y = \sigma_y/2 $$).

• Criterion (Ductile Materials):

$$\tau_{max} = \frac{\sigma_1 - \sigma_3}{2} \leq \frac{\sigma_y}{2}$$

or

$$\sigma_1 - \sigma_3 \leq \sigma_y$$

• Graphical: On Mohr's circle, failure when circle touches the $$\displaystyle \tau = \pm \sigma_y/2 $$ horizontal lines.

5.3 Maximum Principal Strain Theory (St. Venant's Theory)

• Statement: Failure occurs when the maximum principal strain reaches the strain at yield in simple tension ($$\displaystyle \epsilon_y = \sigma_y/E $$).

• Criterion:

$$\epsilon_1 \leq \epsilon_y \quad \text{or} \quad \frac{\sigma_1}{E} - \nu \frac{\sigma_2}{E} - \nu \frac{\sigma_3}{E} \leq \frac{\sigma_y}{E}$$

For 2D ($$\displaystyle \sigma_3=0 $$):

$$\sigma_1 - \nu \sigma_2 \leq \sigma_y$$

• Graphical: On Mohr's circle, failure when circle touches a line inclined at 45° to σ-axis.

5.4 Maximum Strain Energy Theory (Haigh's Theory / Distortion Energy Theory)

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

• Criterion (Von Mises):

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

For 2D ($$\displaystyle \sigma_3=0 $$):

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

• Graphical: On Mohr's circle, failure when circle touches an inscribed hexagon (von Mises hexagon).

5.5 Comparison and Application

[!TIP]

Rule of Thumb: For ductile metals (steel, Al), use Maximum Shear Stress or Distortion Energy (Von Mises). For brittle materials (cast iron, concrete), use Maximum Principal Stress.

6.0 BEAM PROPERTIES AND SHEAR STRESS

6.1 Section Modulus (Z)

• Definition: $$\displaystyle Z = \frac{I}{y_{max}} $$, where I = second moment of area about NA, y_max = distance from NA to farthest fiber.

• Significance: Directly relates bending moment to maximum bending stress: $$\displaystyle \sigma_{max} = \frac{M}{Z} $$.

• Calculation:

  • Rectangular (b×h): $$\displaystyle I = bh^3/12 $$, $$\displaystyle y_{max}=h/2 $$ → $$\displaystyle Z = \frac{b h^2}{6} $$

  • Circular (diameter D): $$\displaystyle I = \pi D^4/64 $$, $$\displaystyle y_{max}=D/2 $$ → $$\displaystyle Z = \frac{\pi D^3}{32} $$

  • I-Section: $$\displaystyle Z = \frac{I}{y_{top}} $$ (about major axis). Often given as $$\displaystyle Z_{xx} $$ for strong axis.

  • T-Section: Calculate I about NA, then find y_max from NA to top/bottom flange.

6.2 Point of Contraflexure (Inflection Point)

• Definition: Point in a beam where bending moment changes sign (M=0) and curvature changes direction.

• Significance: Marks transition from sagging to hogging (or vice versa). Deflection curve has a point of inflection there.

• Location: Found by setting Bending Moment equation to zero and solving for x. Common in overhanging beams and beams with mixed loading.

6.3 Shear Stress Distribution in Beams

• General Formula (Shear Formula):

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

where:

*   V = Shear force at section

*   Q = First moment of area **above (or below)** the point where τ is calculated, about the NA. $$\displaystyle Q = \bar{y} A' $$

*   I = Total second moment of area about NA

*   b = Width (thickness) of material at the point in the horizontal direction.

6.3.1 Rectangular Section (b × h)

• Q at distance y from NA: $$\displaystyle A' = b (h/2 - y) $$, $$\displaystyle \bar{y} = h/4 - y/2 $$

  $$\displaystyle Q = b (h/2 - y)(h/4 - y/2) $$

• τ distribution is parabolic.

• Max shear at NA (y=0): $$\displaystyle \tau_{max} = \frac{3}{2} \tau_{avg} $$, where $$\displaystyle \tau_{avg} = V/(b h) $$.

$$\tau_{max} = \frac{3V}{2 b h}$$

6.3.2 I-Section

• Web: Carries majority of shear (flanges carry very little). τ in web is nearly constant (since b=web thickness t_w is constant, Q varies linearly).

• Max shear at NA: Calculate Q for area in one flange + half web. τ_max occurs at NA.

• Flanges: τ is very small because b (flange width) is large.

6.3.3 T-Section

• τ distribution is not symmetric about NA.

• Calculate Q for area above the point.

• Max τ usually occurs at the NA, but verify by checking at junction of flange and web (where b changes abruptly).

[!TIP]

Key Point: Shear stress is zero at the top and bottom fibers (y = ±h/2, Q=0). It is maximum at the neutral axis for symmetric sections.

7.0 THERMAL STRESSES AND COMPOSITE BARS

7.1 Thermal Stresses

• Free Thermal Expansion: $$\displaystyle \delta = \alpha L \Delta T $$

  • α = coefficient of linear expansion

  • ΔT = temperature change

• Thermal Stress (Expansion Prevented): If expansion is completely restricted, compressive stress develops:

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

(Tension if temperature decrease).

• Cases:

  1. Both ends fixed: Stress = $E \alpha \Delta T$ (if ends rigid).

  2. One end fixed, other free: No stress (free expansion).

  3. Expansion partially restricted: Find actual expansion δ, then stress from $$\displaystyle \sigma = E \epsilon = E (\delta_{actual} - \delta_{free})/L $$.

7.2 Composite Bars (Axial Loading)

• Assumptions: Bars of different materials, same length L, perfectly bonded, strain is same in both (ε₁ = ε₂).

• Load Distribution:

$$\frac{P_1}{A_1 E_1} = \frac{P_2}{A_2 E_2} = \epsilon$$

Total load: $$\displaystyle P = P_1 + P_2 $$.

• Total Deformation:

$$\delta = \frac{P_1 L}{A_1 E_1} = \frac{P_2 L}{A_2 E_2} = \frac{P L}{A_1 E_1 + A_2 E_2}$$

where $$\displaystyle A_{eq} = A_1 + A_2 \frac{E_2}{E_1} $$ (or similar, depending on knowns).

• Stresses: $$\displaystyle \sigma_1 = \frac{P_1}{A_1} $$, $$\displaystyle \sigma_2 = \frac{P_2}{A_2} $$.

7.3 Applications

• Bolted Connections with Pretightening: Initial tension in bolt creates clamping force. Thermal effects or external loads change stresses.

• Bimetallic Strips: Two metals with different α bonded together. Temperature change causes bending.

• Reinforced Concrete: Steel bars (high E) and concrete (low E) act as composite bar under compression/tension.

8.0 THIN SHELLS (PRESSURE VESSELS)

8.1 Thin Cylindrical Shells

• Assumption: Wall thickness t is much smaller than radius R (D/t > 20). Stress through thickness is negligible (membrane stress).

• Hoop (Circumferential) Stress (σₕ): From force balance on half-cylinder:

$$2 \sigma_h (t L) = p (D L) \quad \Rightarrow \quad \sigma_h = \frac{p D}{2 t}$$

**Hoop stress is twice the longitudinal stress.**

• Longitudinal Stress (σₗ): From force balance on end cap:

$$\sigma_l (2\pi R t) = p (\pi R^2) \quad \Rightarrow \quad \sigma_l = \frac{p R}{2 t} = \frac{p D}{4 t}$$

• Radial Stress: Negligible (≈0) compared to σₕ and σₗ.

8.2 Thin Spherical Shells

• Stress (σ): Same in all directions (hoop = longitudinal). From force balance on hemisphere:

$$\sigma (2\pi R t) = p (\pi R^2) \quad \Rightarrow \quad \sigma = \frac{p R}{2 t} = \frac{p D}{4 t}$$

Spherical shell is **more efficient** (stress half of cylinder's hoop stress for same p, D, t).

8.3 Deformation Under Pressure

• Cylinder:

  • Change in Diameter: $$\displaystyle \delta_D = \frac{D}{E} (\sigma_h - \nu \sigma_l) = \frac{p D^2}{4 t E} (2 - \nu) $$

  • Change in Length: $$\displaystyle \delta_L = \frac{L}{E} (\sigma_l - \nu \sigma_h) = \frac{p D L}{4 t E} (1 - 2\nu) $$

• Sphere:

  • Change in Diameter: $$\displaystyle \delta_D = \frac{D}{E} (\sigma - \nu \sigma) = \frac{p D^2}{4 t E} (1 - \nu) $$ (since σ same in all directions).

9.0 ADVANCED TOPICS

9.1 St. Venant's Theory (Stress Concentration)

• Concept: At geometric discontinuities (holes, notches, fillets), local stress is much higher than nominal (average) stress calculated by simple formulas.

• Stress Concentration Factor (Kt): $$\displaystyle K_t = \frac{\sigma_{max}}{\sigma_{nom}} $$

• Importance: Kt depends on geometry, not material. Governs brittle material failure. For ductile materials, local yielding may reduce effective Kt.

9.2 Compound Cylinders (Shrink Fit)

• Setup: Two cylinders (inner and outer) assembled by shrinking (interference fit). Initial radial pressure $$\displaystyle p_i $$ at junction.

• Stress Distribution: Stresses are radial (σ_r) and hoop (σ_θ), varying across thickness.

• Analysis: Use Lame's equations for thick cylinders, applying boundary conditions:

  • At inner radius: σ_r = -p₁ (internal pressure, positive inward)

  • At outer radius: σ_r = -p₂ (external pressure)

  • At junction (r = b): Radial pressure $$\displaystyle p_i $$ continuous, radial displacement continuous.

• Under Internal Pressure: Superimpose pressure on shrink-fit stresses.

9.3 Impact Loading

• Suddenly Applied Load (P₀): Load applied in negligible time, imparting kinetic energy.

• Stress vs Static: Impact stress is greater than static stress.

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

where:

*   h = height of fall of load

*   $$\displaystyle \delta_{static} = \frac{P_0 L}{A E} $$ (static deformation under load P₀)

• If h=0 (load applied gradually): $$\displaystyle \sigma_{impact} = \sigma_{static} $$.

• If h >> δ_static: $$\displaystyle \sigma_{impact} \approx \sigma_{static} \sqrt{\frac{2h}{\delta_{static}}} $$.

9.4 Self-Weight of Bars

• Elongation under Self-Weight: For a vertical bar of length L, area A, density ρ, weight W = ρgAL.

  Consider a small element dx at distance x from top. Force on element due to weight below = ρgA(L-x).

$$d\delta = \frac{[\rho g A (L-x)] dx}{A E} = \frac{\rho g}{E} (L-x) dx$$

Total elongation:

$$\delta = \int_0^L \frac{\rho g}{E} (L-x) dx = \frac{\rho g L^2}{2E} = \frac{W L}{2 A E}$$

• Comparison: Equivalent to applying half the total weight (W/2) as a point load at the bottom for elongation calculation. $$\displaystyle \delta_{self-weight} = \frac{1}{2} \delta_{if\ W\ applied\ at\ bottom} $$.