Strength of Material (ME-304) - Unit 4 Short Notes

UNIT 4: STRENGTH OF MATERIALS - EXAM-FOCUSED SHORT NOTES

1. Fundamental Material Properties & Axial Deformation

Elastic Constants & Relationships

• Young's Modulus (E): Ratio of normal stress to normal strain ($$\displaystyle E = \sigma/\epsilon $$).

• Shear Modulus (G): Ratio of shear stress to shear strain ($$\displaystyle G = \tau/\gamma $$).

• Bulk Modulus (K): Ratio of hydrostatic stress to volumetric strain ($$\displaystyle K = p/\epsilon_v $$).

• Poisson's Ratio (ν): Negative ratio of lateral to longitudinal strain ($$\displaystyle \nu = -\epsilon_{lat}/\epsilon_{long} $$).

• Key Relationships:

$$G = \frac{E}{2(1+\nu)}, \quad K = \frac{E}{3(1-2\nu)}$$

[!TIP] For isotropic materials, $$\displaystyle 0 < \nu < 0.5 $$. Typical values: steel $\nu \approx 0.3$, rubber $\nu \approx 0.5$.

Composite Bars (Multi-Material Systems)

• Compatibility: Deformation ($\delta$) is same for all segments in series.

• Equilibrium: Total load $$\displaystyle P = \sum P_i $$.

• Stress in each material: $$\displaystyle \sigma_i = P_i / A_i $$, but $$\displaystyle \delta = \sum (\sigma_i L_i)/(E_i A_i) $$ must be equal.

• Maximum load determination: Check both stress limits ($$\displaystyle \sigma_i \leq \sigma_{allow,i} $$) and deformation limits ($$\displaystyle \delta \leq \delta_{allow} $$). The governing condition gives $$\displaystyle P_{max} $$.

Axial Deformation of Bars with Varying Cross-Section

• Stepped bar: Deformation $$\displaystyle \delta = \sum \frac{P L_i}{E_i A_i} $$ for each step.

• Tapered bar (circular, diameter varies linearly):

  Let diameter $$\displaystyle d(x) = d_1 + \frac{(d_2 - d_1)}{L}x $$. Area $$\displaystyle A(x) = \frac{\pi}{4} d(x)^2 $$.

$$\delta = \int_0^L \frac{P}{E A(x)} dx = \frac{4P}{\pi E} \int_0^L \frac{dx}{[d_1 + kx]^2}, \quad k = \frac{d_2-d_1}{L}$$

$$\boxed{\delta = \frac{4PL}{\pi E (d_1 d_2)} \quad \text{for linear taper}}$$

[!TIP] For a conical bar (tapering from $D$ to $0$), use $$\displaystyle d_2=0 $$ → formula diverges. Actual deformation finite: $$\displaystyle \delta = \frac{4PL}{\pi E D^2} $$.

Thermal Stresses in Composite Systems

• Free expansion: $$\displaystyle \delta_T = \alpha \Delta T L $$.

• If restrained: Stress $$\displaystyle \sigma = E \alpha \Delta T $$ (if fully constrained).

• Composite bar (two materials): Thermal force $P$ develops such that total deformation zero:

$$\delta_{mech} + \delta_{thermal} = 0 \Rightarrow \frac{P L}{E_1 A_1} + \frac{P L}{E_2 A_2} + \alpha_1 \Delta T L + \alpha_2 \Delta T L = 0$$

Solve for $P$, then $$\displaystyle \sigma_1 = P/A_1 $$, $$\displaystyle \sigma_2 = P/A_2 $$.

Strain Energy

• Gradual loading (load applied slowly from 0 to $P$):

$$U = \int_0^P \delta \, dP = \frac{P^2 L}{2 E A} = \frac{1}{2} P \delta$$

• Impact loading (weight $W$ falls height $h$ onto collar):

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

where $$\displaystyle \delta_{static} = \frac{W L}{E A} $$.

[!TIP] Strain energy density $$\displaystyle u = \frac{\sigma^2}{2E} $$ for uniaxial stress.

Bars with Rigid Supports

• Use deformation compatibility and equilibrium.

• Example: Bar fixed between two rigid walls, load applied internally → reactions adjust so net deformation zero.

2. Beam Bending, Shear, and Deflection

Shear Force & Bending Moment Diagrams

• Sign convention: Sagging BM (+), clockwise SF (+).

• Relationship:

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

$$\frac{d^2M}{dx^2} = -w(x)$$

• Point of contraflexure: Where $$\displaystyle M=0 $$ (BM changes sign).

• Overhanging beams: SFD may have negative regions; BMD may have positive and negative moments.

Assumptions of Simple Bending Theory

1. Beam initially straight, homogeneous, isotropic.

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

3. Material obeys Hooke's law ($\sigma \propto \epsilon$).

4. Radius of curvature large compared to depth.

5. Pure bending (no shear force) or shear effects negligible on normal stress.

Deflection Methods

• Macaulay's Method (Step-Function Integration):

  1. Write $$\displaystyle EI \frac{d^2y}{dx^2} = M(x) $$ using Macaulay brackets $$\displaystyle \langle x-a \rangle^n $$.

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

  3. Apply boundary conditions (deflection/slope zero at supports) to find constants.

  [!TIP] Advantage: Single expression for $M(x)$ even with discontinuous loads. Use $$\displaystyle \langle x-a \rangle^0 = 1 $$ for $$\displaystyle x>a $$, else 0.

• Area-Moment Method (Theorems):

  • Theorem 1: Slope at $A$ = $$\displaystyle \frac{1}{EI} \times $$ (area of $M/EI$ diagram between $A$ and $B$), taken about $B$.

  • Theorem 2: Deflection at $A$ relative to $B$ = $$\displaystyle \frac{1}{EI} \times $$ (moment of area of $M/EI$ diagram between $A$ and $B$), taken about $A$.

  • Apply to cantilevers and simply supported beams with standard load cases.

Maximum Bending Stress & Section Modulus

• $$\displaystyle \sigma_{max} = \frac{M_{max}}{Z} $$, where $$\displaystyle Z = \frac{I}{y_{max}} $$ (section modulus).

• Rectangular section: $$\displaystyle Z = \frac{b d^2}{6} $$.

• I-section: $$\displaystyle Z \approx \frac{I}{y_{max}} $$ about strong axis.

• Design: Choose $$\displaystyle Z \geq M_{max}/\sigma_{allow} $$.

Shear Stress in Beams

• Formula: $$\displaystyle \tau = \frac{V Q}{I b} $$, where $Q$ = first moment of area above/below point about NA.

• Rectangular section ($b \times d$):

$$\tau_{max} = \frac{3V}{2bd} = 1.5 \tau_{avg}, \quad \text{at NA}$$

Parabolic distribution, zero at top/bottom.

• Circular section ($D$):

$$\tau_{max} = \frac{4V}{3A} = \frac{16V}{3\pi D^2}, \quad \text{at NA}$$

• Quarter-Fourth Rule (Rectangular): Max shear at NA = 1.5×avg; at $$\displaystyle y = d/4 $$ from NA, $$\displaystyle \tau \approx 0.75 \tau_{max} $$.

Composite Beams (Transformed Section Method)

1. Transform one material to equivalent area of other using $$\displaystyle n = E_1/E_2 $$.

2. Find neutral axis from transformed section: $$\displaystyle \bar{y}_{NA} = \frac{\sum (n_i A_i y_i)}{\sum n_i A_i} $$.

3. Calculate bending stress in each material:

$$\sigma_i = \frac{M y_i}{I_{trans}} \cdot \frac{1}{n_i} \quad \text{(if transformed to material 1)}$$

[!TIP] Ensure compatibility of strains: $\epsilon$ same at same $y$, so $$\displaystyle \sigma_1/E_1 = \sigma_2/E_2 $$.

Curved Beams (Circular Section)

• Neutral axis NOT at centroid. Radius of curvature $R$ of NA:

$$R_{NA} = \frac{A}{\int \frac{dA}{r}}$$

• For circular section of radius $$\displaystyle r_s $$, under pure moment $M$:

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

where $$\displaystyle e = R - r_n $$, $$\displaystyle r_n $$ = radius of NA, $y$ measured from NA.

• Maximum stress at inner/outer fibers.

3. Torsion of Circular Shafts

Solid Circular Shaft

• Shear stress: $$\displaystyle \tau = \frac{T \rho}{J} $$, max at outer surface: $$\displaystyle \tau_{max} = \frac{T}{Z_p} $$, $$\displaystyle Z_p = J/c = \frac{\pi D^3}{16} $$.

• Angle of twist: $$\displaystyle \theta = \frac{T L}{G J} $$, $$\displaystyle J = \frac{\pi D^4}{32} $$.

• Torsional rigidity: $GJ$.

Hollow Circular Shaft

• $$\displaystyle J = \frac{\pi}{32} (D^4 - d^4) $$, $$\displaystyle Z_p = \frac{J}{D/2} = \frac{\pi (D^4 - d^4)}{16 D} $$.

• Comparison with solid shaft (same material, same max stress):

  • Solid: $$\displaystyle T_s = \tau_{allow} \cdot \frac{\pi D_s^3}{16} $$.

  • Hollow: $$\displaystyle T_h = \tau_{allow} \cdot \frac{\pi (D^4 - d^4)}{16 D} $$.

  • For same weight (same volume/length): $$\displaystyle \frac{\pi}{4} D_s^2 = \frac{\pi}{4}(D^2 - d^2) \Rightarrow D_s^2 = D^2 - d^2 $$.

  • Let $$\displaystyle d = kD $$ (e.g., $$\displaystyle k=0.75 $$). Then:

$$\frac{T_h}{T_s} = \frac{(1 - k^4)}{(1 - k^2)^{3/2}} > 1 \quad \text{for } 0<k<1$$

> [!TIP] **Proof**: For $$\displaystyle d = 0.75D $$, $$\displaystyle T_h/T_s \approx 1.18 $$. Hollow shaft carries ~18% more torque for same weight.

Combined Bending & Torsion

• Stresses: Bending gives $$\displaystyle \sigma = \pm M/Z $$, torsion gives $$\displaystyle \tau = T/Z_p $$.

• Principal stresses (2D, $$\displaystyle \sigma_z=0 $$):

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

where $$\displaystyle \sigma = M/Z $$ (tensile/compressive).

• Maximum shear stress:

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

Shaft Design under Combined Loading

• Maximum Shear Stress Theory (Tresca):

$$\tau_{max} \leq \frac{\tau_{allow}}{FOS} \quad \text{or} \quad \frac{\sigma_1 - \sigma_2}{2} \leq \frac{\sigma_y}{2 \cdot FOS}$$

• Maximum Distortion Energy Theory (von Mises):

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

For pure bending + torsion ($$\displaystyle \sigma_1 = \sigma $$, $$\displaystyle \sigma_2 = -\sigma $$ or 0 depending on sign):

$$\sigma_{eq} = \sqrt{\sigma^2 + 3\tau^2} \leq \frac{\sigma_y}{FOS}$$

[!TIP] von Mises is more accurate for ductile materials; Tresca conservative.

Composite Shafts (Stepped, Different Materials)

• Torque equilibrium: $$\displaystyle T_{in} = T_{out} $$ (if no torque between sections).

• Angle of twist compatibility: $$\displaystyle \theta_1 = \theta_2 $$ if connected in series.

$$\theta = \frac{T_1 L_1}{G_1 J_1} = \frac{T_2 L_2}{G_2 J_2}$$

• Solve for unknown torques or angles.

4. Stress Transformation & Mohr's Circle

Stress Transformation Equations (2D)

For plane stress with $$\displaystyle \sigma_x, \sigma_y, \tau_{xy} $$:

• Normal stress on plane at $\theta$:

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

• Shear stress:

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

Mohr's Circle Construction

1. Plot $$\displaystyle X(\sigma_x, \tau_{xy}) $$, $$\displaystyle Y(\sigma_y, -\tau_{xy}) $$ (note sign of $$\displaystyle \tau_{xy} $$).

2. Center $$\displaystyle C = ((\sigma_x+\sigma_y)/2, 0) $$.

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

4. Principal stresses: $$\displaystyle \sigma_{1,2} = \sigma_{avg} \pm R $$.

5. Max shear stress: $$\displaystyle \tau_{max} = R $$ (on planes at $$\displaystyle 45^\circ $$ to principal planes).

6. Angle $2\theta$ measured from $X$-axis to line $CX$.

Principal Planes & Stresses

• Principal planes: Planes where $$\displaystyle \tau_{x'y'}=0 $$, $$\displaystyle \sigma = \sigma_{1,2} $$.

• Maximum shear stress planes: At $$\displaystyle 45^\circ $$ to principal planes; $$\displaystyle \tau_{max} = (\sigma_1 - \sigma_2)/2 $$.

• Pure shear state: $$\displaystyle \sigma_x = \sigma_y = 0 $$, $$\displaystyle \tau_{xy} = \tau $$ → $$\displaystyle \sigma_1 = \tau $$, $$\displaystyle \sigma_2 = -\tau $$ (tension & compression at $$\displaystyle 45^\circ $$).

Applications of Mohr's Circle

• Stress on any inclined plane: Read $\sigma$, $\tau$ at angle $2\theta$ from $X$-axis.

• Maximum obliquity: $$\displaystyle \tan \phi_{max} = \tau_{max}/\sigma_{avg} $$.

• Graphical solution: Avoids trigonometry.

[!TIP] Common Pitfall: On Mohr's circle, $\theta$ is double the physical angle. Shear sign: $$\displaystyle \tau_{xy} $$ positive causes clockwise rotation of element → plot $Y$ with $$\displaystyle -\tau_{xy} $$.

5. Theories of Failure & Factor of Safety

Ductile vs. Brittle Materials

• Ductile (steel, Al): Significant plastic deformation before failure. Use Tresca or von Mises.

• Brittle (cast iron, concrete): Little plastic deformation, fails by cleavage. Use Rankine (max principal stress).

Maximum Principal Stress Theory (Rankine)

• Statement: Failure occurs when max principal stress reaches ultimate stress in simple tension.

• Formula:

$$\sigma_1 \leq \frac{\sigma_{ult}}{FOS} \quad \text{(tension)}$$

$$\sigma_3 \geq -\frac{\sigma_{ult,c}}{FOS} \quad \text{(compression)}$$

• Application: Brittle materials, pressure vessels (thick), concrete.

• Factor of Safety: $$\displaystyle FOS = \sigma_{ult}/\sigma_1 $$.

Maximum Shear Stress Theory (Tresca)

• Statement: Failure when max shear stress equals shear stress at yield in simple tension.

• Formula:

$$\tau_{max} = \frac{|\sigma_1 - \sigma_2|}{2} \leq \frac{\sigma_y}{2 \cdot FOS}$$

• Application: Ductile materials, conservative.

• For pure shear: $$\displaystyle \sigma_y = 2\tau_y $$.

Maximum Distortion Energy Theory (von Mises)

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

• Derivation: $$\displaystyle U_d = \frac{1}{12G}(\sigma_1-\sigma_2)^2 + (\sigma_2-\sigma_3)^2 + (\sigma_3-\sigma_1)^2) $$.

• Formula:

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

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

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

• Application: Ductile materials, more accurate than Tresca.

Comparison of Theories

Application to Thin-Walled Pressure Vessels

• Hoop stress: $$\displaystyle \sigma_h = \frac{p r}{t} $$ (longitudinal $$\displaystyle \sigma_l = \frac{p r}{2t} $$).

• Design thickness:

  • Rankine: $$\displaystyle \frac{p r}{t} \leq \frac{\sigma_y}{FOS} \Rightarrow t \geq \frac{p r FOS}{\sigma_y} $$.

  • Tresca: $$\displaystyle \tau_{max} = \frac{\sigma_h - \sigma_l}{2} = \frac{p r}{4t} \leq \frac{\sigma_y}{2 FOS} \Rightarrow t \geq \frac{p r FOS}{2 \sigma_y} $$.

  • von Mises: $$\displaystyle \sigma_{eq} = \sqrt{\sigma_h^2 - \sigma_h \sigma_l + \sigma_l^2} = \frac{p r}{t} \sqrt{\frac{3}{4}} \leq \frac{\sigma_y}{FOS} \Rightarrow t \geq \frac{\sqrt{3} p r FOS}{2 \sigma_y} $$.

  [!TIP] von Mises gives thinnest wall, Tresca thickest for same FOS.

Factor of Safety for Combined Stresses

• Given axial stress $$\displaystyle \sigma_a $$ and shear $\tau$:

  • Rankine: $$\displaystyle \sigma_{max} = \sigma_a + \sqrt{\tau^2} \leq \sigma_y/FOS $$? No—Rankine uses principal stresses. Compute $$\displaystyle \sigma_{1,2} $$ first.

  • Tresca: $$\displaystyle \tau_{max} = \sqrt{(\sigma_a/2)^2 + \tau^2} \leq \sigma_y/(2FOS) $$.

  • von Mises: $$\displaystyle \sigma_{eq} = \sqrt{\sigma_a^2 + 3\tau^2} \leq \sigma_y/FOS $$.

• Example: Bolt under axial $P$ and shear $V$ → $$\displaystyle \sigma = P/A $$, $$\displaystyle \tau = V/A $$. Apply theories.

6. Column Buckling & Stability

Euler's Buckling Load

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

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

• Boundary conditions determine $$\displaystyle P_{cr} $$.

Effective Length $$\displaystyle (L_e) $$ & End Conditions

Slenderness Ratio

$$\lambda = \frac{L_e}{r}, \quad r = \sqrt{\frac{I}{A}}$$

• Euler valid for $$\displaystyle \lambda > \lambda_{critical} $$ (typically 80-100 for steel).

• Short columns: Use inelastic formulas (Rankine, Johnson).

Rankine's Formula (Intermediate Columns)

$$\frac{1}{P_{cr}} = \frac{1}{P_e} + \frac{1}{P_0}$$

where $$\displaystyle P_e = \pi^2 EI / L_e^2 $$ (Euler), $$\displaystyle P_0 = \sigma_0 A $$ (crushing load, $$\displaystyle \sigma_0 $$ = yield/proportional limit).

• For long columns ($$\displaystyle L_e \to \infty $$): $$\displaystyle P_{cr} \to P_e $$.

• For short columns ($$\displaystyle L_e \to 0 $$): $$\displaystyle P_{cr} \to P_0 $$.

• Given $$\displaystyle a = 1/7500 $$: $$\displaystyle P_{cr} = \frac{\sigma_0 A}{1 + a (L_e/r)^2} $$.

Crippling Load & Limitations of Euler

• Crippling load = buckling load.

• Euler limitations:

  1. Material must be linearly elastic (stress < proportional limit).

  2. Column perfectly straight, load axial.

  3. Valid only for long columns ($$\displaystyle \lambda > \lambda_{critical} $$).

• Shortest length for Euler: $$\displaystyle L_{min} $$ such that $$\displaystyle \sigma_{cr} = P_{cr}/A \leq \sigma_{proportional} $$.

Application Problems

• Safe load: $$\displaystyle P_{safe} = P_{cr} / FOS $$.

• Critical length: $$\displaystyle L_c $$ where Euler stress = proportional limit: $$\displaystyle L_c = \pi \sqrt{\frac{E}{\sigma_p}} r $$.

• Columns with composite sections or different end conditions:

  • Compute $I$ about axis of buckling.

  • Use appropriate $$\displaystyle L_e $$ for each axis if different end conditions (e.g., fixed about one axis, hinged about other → use smaller $$\displaystyle L_e $$ for design).

[!TIP] For rectangular section, $$\displaystyle r_{min} = \frac{b}{\sqrt{12}} $$ (buckling about weaker axis). Always check both axes if end conditions differ.

Final Exam Strategy:

• Composite bars & thermal stress: Always write compatibility equation $$\displaystyle \delta_{total}=0 $$ or given $\delta$.

• Beam deflection: Macaulay's method for point loads, area-moment for standard cases. Check boundary conditions carefully.

• Torsion: For hollow vs. solid comparison, derive $$\displaystyle T_h/T_s $$ with same weight constraint.

• Mohr's circle: Draw circle, label principal stresses, read values. Remember $2\theta$ convention.

• Failure theories: Identify material (ductile/brittle). For combined stresses, compute principal stresses first.

• Buckling: Identify end conditions → $$\displaystyle L_e $$ → check slenderness → choose Euler or Rankine.

\boxed{\text{END OF UNIT 4 NOTES}}