Strength of Material (ME-304) - Unit 1 Short Notes

UNIT 1: STRENGTH OF MATERIALS - EXAM-FOCUSED SHORT NOTES

1.0 FUNDAMENTAL CONCEPTS

1.1 Stress and Strain

• Stress (σ): Internal resistance force per unit area. Normal stress acts perpendicular to the area. Shear stress (τ) acts tangentially.

  • Tensile stress: $$\displaystyle \sigma = \frac{P}{A} $$ (pulling)

  • Compressive stress: $$\displaystyle \sigma = -\frac{P}{A} $$ (pushing)

• Strain (ε): Dimensionless measure of deformation.

  • Longitudinal strain: $$\displaystyle \epsilon = \frac{\delta L}{L} $$

  • Shear strain (γ): Angular deformation, $$\displaystyle \gamma = \tan \theta \approx \theta $$ (in radians).

• Hooke's Law: For linearly elastic materials, stress is proportional to strain.

$$ \sigma = E \epsilon \quad \text{(Normal stress)} $$

$$ \tau = G \gamma \quad \text{(Shear stress)} $$

> [!TIP] **Exam Alert:** Hooke's Law is the foundation. Remember $E$ (Young's modulus) applies to normal stress/strain, $G$ (Shear modulus) to shear.

1.2 Poisson's Ratio (ν)

• Definition: Ratio of lateral strain to longitudinal strain within elastic limit.

$$ \nu = -\frac{\text{lateral strain}}{\text{longitudinal strain}} $$

Negative sign indicates opposite deformation.

• Volumetric Strain (εᵥ): Change in volume per unit volume for a cubical element.

$$ \epsilon_v = \epsilon_x + \epsilon_y + \epsilon_z $$

For isotropic material under tri-axial stress: $$\displaystyle \epsilon_v = \frac{1}{E}[\sigma_x + \sigma_y + \sigma_z - 2\nu(\sigma_y + \sigma_z + \sigma_x)] $$.

> [!TIP] **Common Pitfall:** Volumetric strain is the **sum** of the three mutually perpendicular linear strains. This is a frequent 7-mark derivation.

1.3 Elastic Moduli Relationships

For a homogeneous, isotropic material:

1. Shear Modulus (G): $$\displaystyle G = \frac{\text{Shear stress}}{\text{Shear strain}} $$

2. Bulk Modulus (K): $$\displaystyle K = \frac{\text{Hydrostatic pressure}}{\text{Volumetric strain}} = -\frac{p}{\epsilon_v} $$

3. Key Relationships:

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

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

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

> [!TIP] **Memory Aid:** If you know any **two** of (E, G, K, ν), you can find the others. These are **direct formula** questions.

1.4 Strain Energy

• Strain Energy (U): Work done by external loads in deforming a body, stored as potential energy.

• Gradual Loading (Load applied slowly from zero):

$$ U = \frac{1}{2} P \delta = \frac{P^2 L}{2AE} \quad \text{(For axial bar)} $$

$$ U = \frac{1}{2} \int \frac{M^2}{EI} dx \quad \text{(For bending)} $$

$$ U = \frac{1}{2} \int \frac{T^2}{GJ} dx \quad \text{(For torsion)} $$

• Impact/Shock Loading (Load applied suddenly/immediately):

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

where $h$ is height of fall. **Instantaneous stress** $$\displaystyle \sigma_{\text{max}} = E \cdot \epsilon_{\text{max}} $$.

> [!TIP] **Exam Pattern:** Derivation of strain energy for gradual loading (4 marks) and numerical on impact loading (7 marks) are very common. Remember the "sudden load" stress is **double** the static stress if $$\displaystyle h=0 $$.

2.0 AXIAL LOADING

2.1 Uniform Bars

• Axial Stress: $$\displaystyle \sigma = \frac{P}{A} $$ (constant if A constant).

• Elongation/Shortening:

$$ \boxed{\delta = \frac{PL}{AE}} $$

> [!TIP] **Sign Convention:** $$\displaystyle \delta > 0 $$ for elongation (tension), $$\displaystyle \delta < 0 $$ for shortening (compression).

2.2 Composite Bars (Sections of Different Materials)

• Principle: Compatibility of deformations (δ₁ = δ₂ = ... = δ) and Equilibrium of forces (ΣP = Applied load).

• Solution Steps:

  1. Assume a common deformation $\delta$.

  2. For each material: $$\displaystyle \delta = \frac{P_i L_i}{A_i E_i} \Rightarrow P_i = \frac{\delta A_i E_i}{L_i} $$.

  3. Apply force equilibrium: $$\displaystyle \sum P_i = P_{\text{applied}} $$.

  4. Solve for $\delta$, then find individual stresses $$\displaystyle \sigma_i = P_i/A_i $$.

• Maximum Load: Calculate stress in each material for a given P. The governing condition is the one that reaches its allowable stress first.

  [!TIP] Critical Point: In composite bars, strain is equal in series (same length change), stress is equal in parallel (same force). This is a classic 7-mark problem.

2.3 Thermal Stresses

• Free Expansion: $$\displaystyle \delta_{\text{thermal}} = \alpha L \Delta T $$

  where $\alpha$ = coefficient of linear expansion, $\Delta T$ = temperature change.

• Restrained Expansion (Bar rigidly fixed):

  • Thermal stress develops because expansion is prevented.

  • Compressive stress: $$\displaystyle \sigma_{\text{thermal}} = E \alpha \Delta T $$ (if heated).

  • Tensile stress if cooled.

• Composite Bars with Temperature Change:

  • Case 1 (Rigidly fixed ends): Total strain = 0.

$$ \epsilon_{\text{total}} = \frac{\sigma_1}{E_1} + \alpha_1 \Delta T + \frac{\sigma_2}{E_2} + \alpha_2 \Delta T = 0 $$

    Use $$\displaystyle \sum P = 0 $$ to solve for $$\displaystyle \sigma_1, \sigma_2 $$.

*   **Case 2 (Free to expand but connected)**: Deformations are compatible ($$\displaystyle \delta_1 = \delta_2 $$), but net force may not be zero.

$$ \frac{P_1 L}{A_1 E_1} + \alpha_1 L \Delta T = \frac{P_2 L}{A_2 E_2} + \alpha_2 L \Delta T $$

    Use $$\displaystyle \sum P = P_{\text{external}} $$ to solve.

> [!TIP] **Golden Rule:** Always write the **total strain = mechanical strain + thermal strain**. For a fixed bar, total strain = 0.

2.4 Bars with Varying Cross-Section

• General Formula (for any variation):

$$ \delta = \int_0^L \frac{P}{A(x) E} dx $$

• Uniformly Tapered Circular Bar (diameter varies linearly: $$\displaystyle d(x) = d_1 + \frac{(d_2 - d_1)x}{L} $$):

  Area $$\displaystyle A(x) = \frac{\pi}{4} [d(x)]^2 $$.

$$ \delta = \frac{4P}{\pi E (d_2 - d_1)} \left[ \ln \left( \frac{d_2}{d_1} \right) \right] \quad \text{if } d_1 \neq d_2 $$

For **conical bar** ($$\displaystyle d_2 = 0 $$), use limit or integrate directly.

> [!TIP] **Derivation is Key:** You must be able to derive the formula for a tapered bar from the basic integral. This is a **sure 7-mark question**.

3.0 BENDING OF BEAMS

3.1 Pure Bending Theory

• Assumptions:

  1. Plane sections before bending remain plane after bending.

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

  3. Stress is purely longitudinal (no shear).

  4. Radius of curvature is large compared to depth.

  5. Loading is in the plane of bending.

• Neutral Axis (NA): Line of zero stress/strain. Passes through centroid for symmetric sections.

• Bending Stress Formula:

$$ \boxed{\sigma = \frac{My}{I}} $$

*   $M$ = Bending moment at section.

*   $y$ = Distance from NA.

*   $I$ = Second moment of area about NA.

*   **Maximum Stress**: $$\displaystyle \sigma_{\text{max}} = \frac{M}{Z} $$, where $$\displaystyle Z = \frac{I}{y_{\text{max}}} $$ is the **Section Modulus**.

> [!TIP] **Design Formula:** $$\displaystyle \sigma_{\text{max}} \leq \sigma_{\text{allowable}} \Rightarrow M \leq \sigma_{\text{allow}} \cdot Z $$. Section modulus is the key geometric property for bending strength.

3.2 Shear Stress in Beams

• Shear Stress Formula (for horizontal shear):

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

*   $V$ = Internal shear force.

*   $Q$ = First moment of area **above/below** the point about the NA: $$\displaystyle Q = \int y \, dA $$.

*   $I$ = Second moment of area of whole section about NA.

*   $b$ = Width of the section at the point where $\tau$ is calculated.

• Distribution:

  • Rectangular: Parabolic, max at NA ($$\displaystyle \tau_{\text{max}} = \frac{3}{2} \tau_{\text{avg}} $$).

  • Circular: Parabolic, max at NA ($$\displaystyle \tau_{\text{max}} = \frac{4}{3} \tau_{\text{avg}} $$).

  • I-Section: Max in web at NA. Flange carries little shear. Shear flow $$\displaystyle q = \tau \cdot t $$ is constant in web.

  [!TIP] Plotting: Always sketch the shear stress diagram. For I-beams, show parabolic in web, near zero in flanges.

3.3 Deflection of Beams

Standard Cases (Memorize these for cantilever & simply supported with UDL/point load):

3.3.1 Double Integration Method

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

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

• Apply boundary conditions (e.g., $$\displaystyle y=0 $$ at simple support, $$\displaystyle dy/dx=0 $$ at fixed end) to find $$\displaystyle C_1, C_2 $$.

3.3.2 Macaulay's Method (Step-Function Integration)

• Advantage: Single expression for $M(x)$ for discontinuous loads (point loads, UDL starting/ending at a point) using Macaulay brackets $$\displaystyle \langle x-a \rangle^n $$.

  • $$\displaystyle \langle x-a \rangle^0 = 1 $$ for $x \geq a$, $0$ for $$\displaystyle x < a $$.

• Procedure:

  1. Write $M(x)$ with Macaulay brackets for all loads.

  2. Integrate $$\displaystyle \frac{M}{EI} $$ once to get slope, twice to get deflection. Do not break the integration at load points.

  3. Apply boundary conditions (usually at $$\displaystyle x=0 $$ and $$\displaystyle x=L $$) to find constants.

  4. For deflection at a specific point $$\displaystyle x=a $$, substitute $$\displaystyle x=a $$ after integration.

  [!TIP] Macaulay is King: This is the most efficient method for beams with multiple point loads or partial UDLs. It's a frequent 7-mark question. Practice writing the moment equation correctly.

3.3.3 Area-Moment (Conjugate Beam) Method

• Theorems:

  1. The slope at a point = $$\displaystyle \frac{1}{EI} \times $$ (Area of M/EI diagram between that point and a fixed point of known slope).

  2. The deflection at a point = $$\displaystyle \frac{1}{EI} \times $$ (Moment of area of M/EI diagram about that point).

• Conjugate Beam: A "fictitious" beam with the same length, loaded with the M/EI diagram of the real beam. Its shear = slope, bending moment = deflection of the real beam.

• Advantage: No integration, good for finding deflection at a specific point.

• Limitation: Requires known boundary conditions (fixed end has zero slope & deflection; simply supported has zero deflection but non-zero slope).

  [!TIP] When to use: Area-moment is great for standard cases or when you need deflection at one point. Double integration/Macaulay is better for equations of slope/deflection along entire beam.

3.4 Application Problems

• Point of Contraflexure: Point where bending moment changes sign ($$\displaystyle M=0 $$). Found from SF/BM diagram or $$\displaystyle M(x)=0 $$.

• Curved Beams (Circular section): Neutral axis does NOT pass through centroid. Formula:

$$ \sigma_\theta = \frac{M}{A e} \left( \frac{1}{r} - \frac{1}{r_n} \right) $$

where $$\displaystyle e = r_c - r_n $$, $$\displaystyle r_n $$ is radius of NA. For a circular section, $$\displaystyle r_n $$ is found from:

$$ \frac{1}{r_n} = \frac{\int \frac{dA}{r}}{A} $$

> [!TIP] **Curved Beam Trap:** For a curved beam, **do not use** $$\displaystyle \sigma = My/I $$. Use the curved beam formula. NA is closer to the center of curvature.

4.0 TORSION OF CIRCULAR SHAFTS

4.1 Pure Torsion Theory

• Assumptions: Circular cross-section remains circular, plane sections remain plane, radial lines remain radial, material homogeneous & elastic.

• Shear Stress Distribution: Linear from center (zero) to outer fiber (max).

$$ \tau = \frac{T r}{J} \quad \Rightarrow \quad \tau_{\text{max}} = \frac{T c}{J} $$

where $T$ = Torque, $r$ = radial distance, $c$ = outer radius, $J$ = Polar moment of inertia.

• Angle of Twist:

$$ \boxed{\theta = \frac{TL}{GJ}} \quad \text{(in radians)} $$

*   $L$ = Length, $G$ = Shear modulus.

*   **Torsional Rigidity**: $GJ$ (resistance to twist).

> [!TIP] **Analogy to Axial Loading:** $$\displaystyle \theta \leftrightarrow \delta $$, $$\displaystyle T \leftrightarrow P $$, $$\displaystyle GJ \leftrightarrow AE $$.

4.2 Solid and Hollow Shafts

• Polar Moment of Inertia:

  • Solid: $$\displaystyle J_s = \frac{\pi d^4}{32} $$

  • Hollow: $$\displaystyle J_h = \frac{\pi (D^4 - d^4)}{32} $$

• Comparison for Same Torque & Same Max Stress:

  • Same $T$, same $$\displaystyle \tau_{\text{max}} = \frac{Tc}{J} \Rightarrow J \propto c $$.

  • For hollow shaft with inner dia $$\displaystyle d = kD $$ ($$\displaystyle k<1 $$):

$$ \frac{J_h}{J_s} = \frac{1 - k^4}{1/2} = 2(1 - k^4) \quad \text{(since } c_{solid}=D/2, c_{hollow}=D/2\text{)} $$

*   **Weight ratio** (same length, same material): $$\displaystyle \frac{W_h}{W_s} = \frac{A_h}{A_s} = \frac{1 - k^2}{1/4} = 4(1 - k^2) $$.

*   **Efficiency** (torque capacity per unit weight):

$$ \text{Efficiency} = \frac{T/W}{} \propto \frac{J/c}{A} = \frac{J}{cA} $$

    For hollow shaft: $$\displaystyle \frac{J_h}{c_h A_h} = \frac{\pi (D^4 - d^4)/32}{(D/2) \cdot \pi (D^2 - d^2)/4} = \frac{D^2 + d^2}{4D^2} $$

    Max when $$\displaystyle d=0 $$ (solid), but hollow transmits more torque for **same weight**.

> [!TIP] **Classic Proof:** "Prove hollow shaft stronger for same weight." Show $$\displaystyle \frac{T_h}{T_s} = \frac{1 + k^2}{1 - k^2} > 1 $$ for $$\displaystyle k>0 $$. For $$\displaystyle d=3D/4 $$ ($$\displaystyle k=0.75 $$), $$\displaystyle \frac{T_h}{T_s} = \frac{1+0.5625}{1-0.5625} = \frac{1.5625}{0.4375} \approx 3.57 $$.

4.3 Stepped Shafts

• Compatibility: Angle of twist at common junctions must be equal (if shafts are connected).

$$ \theta_{AB} = \theta_{BC} = \theta_{CD} $$

$$ \frac{T_{AB} L_{AB}}{G_{AB} J_{AB}} = \frac{T_{BC} L_{BC}}{G_{BC} J_{BC}} = \frac{T_{CD} L_{CD}}{G_{CD} J_{CD}} $$

• Equilibrium: Sum of torques at any section = 0.

• Solution: Use compatibility + equilibrium to find unknown torques, then $$\displaystyle \tau_{\text{max}} = \frac{Tc}{J} $$ in each segment.

  [!TIP] Step Shaft Trap: If torques are given at ends and rotations at junctions are specified, use compatibility of rotations (θ's equal) to find internal torques.

4.4 Combined Bending and Torsion

• Stress State: At a point on the shaft surface (critical point):

  • Bending stress: $$\displaystyle \sigma_b = \frac{M y}{I} = \frac{M}{Z} $$ (tensile on one side, compressive on other).

  • Torsional shear stress: $$\displaystyle \tau_t = \frac{T c}{J} $$ (pure shear on surface).

• Principal Stresses (at surface, $$\displaystyle y=c $$):

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

*   If $$\displaystyle \sigma_b $$ is tensile, $$\displaystyle \sigma_1 $$ is more tensile, $$\displaystyle \sigma_2 $$ less tensile/compressive.

*   **Maximum Shear Stress**: $$\displaystyle \tau_{\text{max}} = \sqrt{ \left( \frac{\sigma_b}{2} \right)^2 + \tau_t^2 } $$

• Design by Failure Theories:

  • Max Principal Stress (Rankine): $$\displaystyle \sigma_1 \leq \frac{\sigma_y}{\text{FOS}} $$

  • Max Shear Stress (Guest): $$\displaystyle \tau_{\text{max}} \leq \frac{\sigma_y}{2 \cdot \text{FOS}} $$ (for ductile, yield in shear ≈ half yield in tension).

  • Max Distortion Energy (Hencky/von Mises):

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

    For pure bending + torsion ($$\displaystyle \sigma_2=0 $$): $$\displaystyle \sigma_{\text{eq}} = \sqrt{\sigma_b^2 + 3\tau_t^2} $$.

> [!TIP] **Key Result:** For combined bending & torsion, von Mises equivalent stress is $$\displaystyle \sqrt{\sigma_b^2 + 3\tau_t^2} $$. This is a **very common formula**.

5.0 STRESS TRANSFORMATION AND MOHR’S CIRCLE

5.1 Plane Stress Transformation

• Stress Components on Inclined Plane (angle θ from x-axis):

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

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

• Proof: Sum of Normal Stresses on Perpendicular Planes is Constant:

$$ \sigma_x + \sigma_y = \sigma_{x'} + \sigma_{y'} = \text{constant} = 2 \times \text{average normal stress} $$

5.2 Principal Stresses and Planes

• Principal Planes: Planes where shear stress is zero.

• Principal Stresses ($$\displaystyle \sigma_1, \sigma_2 $$): Normal stresses on principal planes (max & min).

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

• Orientation of Principal Planes:

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

• Maximum Shear Stress:

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

Occurs on planes inclined at $$\displaystyle 45^\circ $$ to principal planes.

5.3 Mohr’s Circle of Stresses

• Construction:

  1. Plot point $$\displaystyle X(\sigma_x, \tau_{xy}) $$ and $$\displaystyle Y(\sigma_y, -\tau_{xy}) $$ on σ-τ plane.

  2. Line XY is a diameter. Center $$\displaystyle C = ((\sigma_x+\sigma_y)/2, 0) $$.

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

  4. Circle: $$\displaystyle (\sigma - \sigma_{avg})^2 + \tau^2 = R^2 $$.

• Readings:

  • Principal Stresses: Points on σ-axis (τ=0) → $$\displaystyle \sigma_1, \sigma_2 $$.

  • Max Shear Stress: Top/bottom of circle → $$\displaystyle \tau_{\text{max}} = R $$.

  • Stress on plane at angle θ: Move 2θ from X-axis (counterclockwise for +θ).

  [!TIP] Mohr's Circle Rule: On Mohr's circle, angles are doubled. Physical plane angle θ corresponds to 2θ on circle. Sign convention: +τ on +x face causes counterclockwise rotation of element → plot +τ upward.

5.4 Special Cases

• Pure Shear: $$\displaystyle \sigma_x = \sigma_y = 0 $$, $$\displaystyle \tau_{xy} = \tau $$.

  • Principal Stresses: $$\displaystyle \sigma_1 = +\tau $$, $$\displaystyle \sigma_2 = -\tau $$.

  • Conclusion: Pure shear is equivalent to equal tensile and compressive stresses on planes at 45°.

• Two Perpendicular Tensile Stresses ($$\displaystyle \sigma_x, \sigma_y > 0 $$, $$\displaystyle \tau_{xy}=0 $$):

  • Principal Stresses: $$\displaystyle \sigma_1 = \max(\sigma_x,\sigma_y) $$, $$\displaystyle \sigma_2 = \min(\sigma_x,\sigma_y) $$.

  • Plane of Maximum Obliquity: Angle where angle between resultant stress and normal is maximum.

$$ \tan \phi_{\text{max}} = \frac{\tau_{\text{max}}}{\sigma_{\text{avg}}} = \frac{|\sigma_x - \sigma_y|/2}{(\sigma_x+\sigma_y)/2} = \left| \frac{\sigma_x - \sigma_y}{\sigma_x + \sigma_y} \right| $$

    Occurs on planes at $$\displaystyle 45^\circ $$ to principal planes.

6.0 THEORIES OF FAILURE

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

  • Guest: $$\displaystyle \frac{\sigma_1 - \sigma_2}{2} \leq \frac{\sigma_y}{2} $$

  • Hencky: $$\displaystyle \sigma_{\text{eq}} = \sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2} \leq \sigma_y $$

• Comparison:

  • For pure shear: Guest gives $$\displaystyle \tau_{\text{allow}} = \sigma_y/2 $$, Hencky gives $$\displaystyle \tau_{\text{allow}} = \sigma_y/\sqrt{3} \approx 0.577\sigma_y $$.

  • For hydrostatic stress: All theories agree (no failure).

  [!TIP] Decision Tree:

  1. Brittle material? → Use Rankine.

  2. Ductile material? → Use Hencky (von Mises) for accuracy. Guest is conservative.

  3. Pressure vessel (thin-walled)? → Rankine (hoop & axial stresses are principal).

7.0 COLUMNS AND BUCKLING

7.1 Euler’s Buckling Theory

• Assumptions:

  1. Column is perfectly straight, homogeneous, isotropic.

  2. Load is axial, centroidal.

  3. Material obeys Hooke's law (stress < proportional limit).

  4. Column is long (slenderness ratio high).

  5. Buckling occurs in the plane of least stiffness.

• Euler's Crippling Load:

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

where $$\displaystyle L_e $$ = **Effective Length**.

• Effective Length ($$\displaystyle L_e $$):

  | End Condition | $$\displaystyle L_e $$ | $$\displaystyle L_e / L $$ | | :--- | :--- | :--- | | Both ends pinned (hinged) | $L$ | 1.0 | | Both ends fixed | $L/2$ | 0.5 | | One end fixed, other free | $2L$ | 2.0 | | One end fixed, other pinned | $L/\sqrt{2}$ | 0.707 |

7.2 Crippling Load & Slenderness Ratio

• Crippling Load ($$\displaystyle P_{cr} $$): Minimum axial load causing buckling.

• Slenderness Ratio ($\lambda$):

$$ \lambda = \frac{L_e}{r} \quad \text{where } r = \sqrt{\frac{I}{A}} \text{ (radius of gyration)} $$

*   High $\lambda$ → Euler's formula valid (elastic buckling).

*   Low $\lambda$ → Material yields before buckling (use Rankine).

7.3 Rankine’s Formula (for Intermediate Columns)

• Formula:

$$ \frac{1}{P_{cr}} = \frac{1}{P_e} + \frac{1}{P_y} \quad \Rightarrow \quad P_{cr} = \frac{P_y}{1 + a (L_e/r)^2} $$

where:

*   $$\displaystyle P_e = \pi^2 E / (L_e/r)^2 $$ (Euler load)

*   $$\displaystyle P_y = \sigma_y A $$ (Yield load)

*   $a$ = Rankine constant ($$\displaystyle a = \sigma_y / \pi^2 E $$ for ideal column).

• Transition Length: Length where $$\displaystyle P_e = P_y $$.

$$ \left( \frac{L_e}{r} \right)_{\text{crit}} = \sqrt{\frac{\pi^2 E}{\sigma_y}} $$

> [!TIP] **When to use what:**

> *   If $$\displaystyle L_e/r > \left( L_e/r \right)_{\text{crit}} $$ → **Euler** (long column).

> *   If $$\displaystyle L_e/r < \left( L_e/r \right)_{\text{crit}} $$ → **Rankine** (short/intermediate column).

> *   **Shortest length for Euler applicability**: $$\displaystyle L_{\text{min}} = \left( L_e/r \right)_{\text{crit}} \times r $$.

7.4 Application Problems

• Safe Load: $$\displaystyle P_{\text{safe}} = \frac{P_{cr}}{\text{FOS}} $$ (Euler) or $$\displaystyle \frac{P_{\text{Rankine}}}{\text{FOS}} $$.

• Required Dimensions: Given $P$, find $I$ or $r$ from buckling formula, then relate to cross-section geometry.

• Columns with Different End Conditions about Different Axes (e.g., rectangular section):

  • Compute $$\displaystyle I_{xx} $$, $$\displaystyle I_{yy} $$ → $$\displaystyle r_{xx} $$, $$\displaystyle r_{yy} $$.

  • Compute $$\displaystyle \lambda_x = L_{e,x}/r_{xx} $$, $$\displaystyle \lambda_y = L_{e,y}/r_{yy} $$.

  • Buckling occurs about the axis with higher slenderness ratio (lower $r$ or higher $$\displaystyle L_e $$).

  • Use that $\lambda$ in Euler/Rankine formula.

  [!TIP] Rectangular Column Trap: For a $b \times d$ column ($$\displaystyle d > b $$), $$\displaystyle I_{min} = bd^3/12 $$ about the axis parallel to $b$. Buckling will be about the weaker axis (smaller $I$). Always check both axes.

8.0 SPECIAL TOPICS (Recurring in Exams)

8.1 Strain Energy Relationships

• Strain Energy due to Gradual Loading: $$\displaystyle U = \frac{1}{2} \times \text{load} \times \text{deformation} $$.

• Strain Energy in Shafts (torsion):

$$ U = \frac{T^2 L}{2GJ} = \frac{\tau^2}{2G} \times \text{Volume} $$

*   **Maximum strain energy stored** in solid shaft: $$\displaystyle U_{\text{max}} = \frac{\tau_{\text{allow}}^2}{2G} \cdot \frac{\pi d^2}{4} L $$.

8.2 Shear Stress in Circular Sections (Derivation)

• Consider a circular shaft of radius $c$, torque $T$.

• Shear stress at radius $r$: $$\displaystyle \tau = \frac{T r}{J} $$ (linear distribution).

• Derivation from first principles: Assume $\tau \propto r$ (from Hooke's law & geometry). Equate total torque:

$$ T = \int_0^c \tau \cdot 2\pi r dr \cdot r = 2\pi \int_0^c \tau r^2 dr $$

With $$\displaystyle \tau = k r $$, solve for $k$, get $$\displaystyle \tau = \frac{T r}{J} $$.

> [!TIP] **Plot:** Shear stress vs. $r$ is a straight line from 0 at center to $$\displaystyle \tau_{\text{max}} $$ at $$\displaystyle r=c $$.

8.3 Area-Moment Method: Advantages & Limitations

• Advantages:

  1. No integration/differentiation needed after M/EI diagram is drawn.

  2. Quick for finding slope/deflection at specific points.

  3. Geometrical interpretation (areas & centroids).

• Limitations:

  1. Requires known slope/deflection at one point (usually fixed end).

  2. Difficult for discontinuous M/EI diagrams (needs separate areas).

  3. Less straightforward for obtaining equations of deflection curve along entire beam (compared to Macaulay).

8.4 Quarter-Fourth Rule

• Used to approximate moment of inertia of built-up sections (like I-beams with flanges & web).

• Rule: Calculate $I$ about NA by summing $I$ of each rectangle about its own centroid + $$\displaystyle A d^2 $$ (parallel axis), but neglect the flange/web junctions (i.e., assume no material in the quarter-fourth regions near NA).

• Purpose: Simplifies calculation; gives slightly conservative (lower) $I$ value.

8.5 Curved Beams: Neutral Axis Position

• For a curved beam of circular section, NA does not pass through centroid.

• Location from center of curvature:

$$ \frac{1}{r_n} = \frac{\int \frac{dA}{r}}{A} = \frac{A_c}{A \cdot r_c} \quad \text{(for circular section)} $$

where $$\displaystyle A_c $$ is area of cross-section, $$\displaystyle r_c $$ is distance from center of curvature to centroid.

• Bending Stress Formula:

$$ \sigma_\theta = \frac{M}{A e} \left( \frac{1}{r} - \frac{1}{r_n} \right) $$

where $$\displaystyle e = r_c - r_n $$.

> [!TIP] **Key Difference:** In straight beams, $\sigma \propto y$. In curved beams, $$\displaystyle \sigma \propto (1/r - 1/r_n) $$. **NA is closer to the center of curvature** than the centroid.

SUMMARY OF HIGH-FREQUENCY TOPICS FROM PAST PAPERS:

1. Composite Bars (stress & deformation compatibility) – Every paper.

2. Thermal Stresses (especially composite bars with temp change) – Very frequent.

3. Tapered Bar Elongation (derivation) – Every paper.

4. Beam Deflection (Macaulay's method, area-moment, standard cases) – Every paper.

5. Solid vs Hollow Shaft (comparison for same torque/weight) – Every paper.

6. Stepped Shafts (torque & angle of twist) – Frequent.

7. Combined Bending & Torsion (principal stresses, failure theories) – Every paper.

8. Mohr's Circle (construction, principal stresses) – Every paper.

9. Failure Theories (Rankine, Guest, Hencky – comparison & application) – Every paper.

10. Euler's Buckling (derivation for different end conditions, effective length) – Every paper.

11. Rankine's Formula (for intermediate columns) – Frequent.

12. Curved Beams (NA position) – Frequent.

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