Strength of Materials (CE-305) - Unit 5 Short Notes

UNIT 5: STRENGTH OF MATERIALS - EXAM-FOCUSED SHORT NOTES

I. FUNDAMENTAL CONCEPTS AND ELASTIC CONSTANTS

Stress and Strain

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

  • Normal Stress: Tensile (+) or compressive (-), $$\displaystyle \sigma = \frac{P}{A} $$.

  • Shear Stress (τ): Tangential force per unit area.

• Strain (ε): Deformation per unit length.

  • Normal Strain: $$\displaystyle \epsilon = \frac{\delta L}{L} $$.

  • Lateral Strain: $$\displaystyle \epsilon_{lat} = -\nu \epsilon $$ (Poisson's effect).

• Stress-Strain Diagram: Shows elastic limit, yield point, ultimate strength, fracture. Hooke's law valid up to proportional limit: $$\displaystyle \sigma = E \epsilon $$.

Elastic Constants

Relationships between Elastic Constants

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

   Derivation: Consider a cube under shear; relate shear strain to normal strains via Poisson's effect.

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

3. $G$ and $K$: $$\displaystyle G = \frac{3K(1-2\nu)}{2(1+\nu)} $$

[!TIP]

Exam Focus: Derive $$\displaystyle E = 2G(1+\nu) $$ from first principles. Given any two constants, find the third.

II. AXIAL LOADING

Uniform Axial Stress and Deformation

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

• Deformation: $$\displaystyle \delta = \frac{PL}{AE} $$.

Stepped Bars

• Total extension: $$\displaystyle \delta = \sum \frac{P L_i}{A_i E_i} $$ (series).

• Stress same in all sections if $P$ same; deformation adds.

Tapered Bars

• For a rod of length $L$, diameters $d$ (small) and $D$ (large), axial load $P$:

$$ \delta = \frac{4PL}{\pi E (D^2 - d^2)} \ln\left(\frac{D}{d}\right) $$

Derivation: Integrate $$\displaystyle \frac{d\delta}{dx} = \frac{P}{E A(x)} $$, with $A(x)$ linear in $x$.

Composite Bars

• Bars in parallel (no slip): Same deformation $\delta$, total load $$\displaystyle P = P_1 + P_2 $$.

• Stresses: $$\displaystyle \frac{\sigma_1}{E_1} = \frac{\sigma_2}{E_2} = \frac{\delta}{L} $$.

• Example: Steel rod in copper tube → $$\displaystyle \sigma_s = \frac{E_s}{E_s + \frac{A_c}{A_s}E_c} \cdot \frac{P}{A_s} $$.

Thermal Stresses

• Free expansion: $$\displaystyle \delta_{th} = \alpha L \Delta T $$.

• Fixed ends: $$\displaystyle \sigma = E \alpha \Delta T $$ (compressive if heated, tensile if cooled).

• With initial gap $$\displaystyle \delta_0 $$:

  • If $$\displaystyle \delta_{th} < \delta_0 $$: no stress.

  • If $$\displaystyle \delta_{th} > \delta_0 $$: $$\displaystyle \sigma = E(\alpha \Delta T - \frac{\delta_0}{L}) $$.

Impact Loading

• Suddenly applied load $P$:

  $$\displaystyle \sigma_{max} = \frac{P}{A} \left(1 + \sqrt{1 + \frac{2h}{\delta_{static}}}\right) $$, where $$\displaystyle \delta_{static} = \frac{PL}{AE} $$.

• Falling weight $W$ from height $h$:

  $$\displaystyle \sigma_{max} = \frac{W}{A} \left(1 + \sqrt{1 + \frac{2h}{\delta_{static}}}\right) $$.

Self-Weight of Bars

• Vertical bar fixed at top, weight $W$:

  $$\displaystyle \delta = \frac{WL}{2AE} = \frac{\gamma A L^2}{2E} $$ (where $\gamma$ = unit weight).

[!TIP]

Common Pitfall: In composite bars, stresses are NOT equal; deformations are equal. In thermal stress, sign depends on constraint.

III. STRESS TRANSFORMATION AND PRINCIPAL STRESSES

Plane Stress

• State: $$\displaystyle \sigma_x, \sigma_y, \tau_{xy} $$ (no $$\displaystyle \sigma_z $$).

• Transformation Equations (angle $\theta$ from $x$-axis to $n$-axis):

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

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

Principal Stresses and Planes

• Principal planes: $$\displaystyle \tau_{nt} = 0 $$ → $$\displaystyle \tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x - \sigma_y} $$.

• Principal stresses:

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

($$\displaystyle \sigma_1 $$ = max, $$\displaystyle \sigma_2 $$ = min).

• Maximum shear stress:

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

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

Stresses on Inclined Plane (Given Two Perpendicular Planes)

• Given stresses on planes AB ($$\displaystyle \sigma_1, \tau_1 $$) and AC ($$\displaystyle \sigma_2, \tau_2 $$), use transformation to find $$\displaystyle \sigma_x, \sigma_y, \tau_{xy} $$ first.

3D Stress State

• Principal stresses $$\displaystyle \sigma_1 \ge \sigma_2 \ge \sigma_3 $$.

• Max shear stress: $$\displaystyle \tau_{max} = \frac{\sigma_1 - \sigma_3}{2} $$.

[!TIP]

Exam Trick: For principal stresses, always compute $$\displaystyle \frac{\sigma_x+\sigma_y}{2} $$ and $$\displaystyle R = \sqrt{...} $$. Then $$\displaystyle \sigma_{1,2} = \text{avg} \pm R $$. Check sign of $$\displaystyle \tau_{xy} $$ to determine $$\displaystyle \theta_p $$ quadrant.

IV. MOHR'S CIRCLE

Construction for 2D Stress

1. Plot points $$\displaystyle X(\sigma_x, \tau_{xy}) $$ and $$\displaystyle Y(\sigma_y, -\tau_{xy}) $$ (shear sign convention: positive $\tau$ causes clockwise rotation of element).

2. Circle center: $$\displaystyle C\left(\frac{\sigma_x+\sigma_y}{2}, 0\right) $$.

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

4. Principal stresses: intersections with $\sigma$-axis.

5. Max shear: top/bottom of circle.

Determination from Mohr's Circle

• $$\displaystyle \sigma_{avg} = \frac{\sigma_x+\sigma_y}{2} $$.

• $$\displaystyle \sigma_{1,2} = \sigma_{avg} \pm R $$.

• $$\displaystyle \tau_{max} = R $$.

• Stresses on plane at $\theta$: move $2\theta$ from $X$ (counterclockwise if $\theta$ CCW on element).

[!TIP]

Key Point: On Mohr's circle, angle is double the physical angle. Shear sign flips between element and circle.

V. THEORIES OF FAILURE

Application to Combined Stresses

• Example: Solid shaft with bending $M$ and torsion $T$:

  $$\displaystyle \sigma_b = \frac{My}{I} $$, $$\displaystyle \tau = \frac{T r}{J} $$.

  Then $$\displaystyle \sigma_1 = \sigma_b + \sqrt{\sigma_b^2 + \tau^2} $$, $$\displaystyle \sigma_2 = \sigma_b - \sqrt{\sigma_b^2 + \tau^2} $$ (plane stress).

  Apply theory: e.g., von Mises: $$\displaystyle \sigma_1^2 + \sigma_2^2 - \sigma_1\sigma_2 \ge \sigma_y^2 $$.

[!TIP]

Remember: Ductile → von Mises or Guest. Brittle → Rankine. For pure shear, von Mises gives $$\displaystyle \tau_{allow} = \sigma_y/\sqrt{3} $$.

VI. BEAM BENDING AND SHEAR

Types of Beams and Loads

• Supports: Simply supported, cantilever, overhanging, fixed.

• Loads: Point load (concentrated), UDL, UVL, moment.

Shear Force (V) and Bending Moment (M)

• Sign Convention:

  • Sagging (concave up): $M$ positive, $V$ positive if left of section.

  • Hogging (concave down): $M$ negative.

• Differential Relations:

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

• Procedure:

  1. Find reactions.

  2. Cut section, equate forces/moments.

  3. Diagram: $V$ constant under point load, linear under UDL; $M$ linear under point load, parabolic under UDL.

Bending Stress

• Assumptions:

  1. Plane sections remain plane.

  2. $E$ constant, $\sigma \propto \epsilon$.

  3. $R \gg$ depth (small curvature).

  4. $$\displaystyle \sigma_x $$ only, $$\displaystyle \tau_{xy}=0 $$.

• Bending Equation: $$\displaystyle \sigma = \frac{My}{I} $$ (y from NA).

• Section Modulus: $$\displaystyle Z = \frac{I}{y_{max}} $$.

  $$\displaystyle \sigma_{max} = \frac{M}{Z} $$.

• Standard Sections:

  • Rectangle: $$\displaystyle Z = \frac{bd^2}{6} $$.

  • Circle: $$\displaystyle Z = \frac{\pi d^3}{32} $$.

  • I-section: $$\displaystyle Z \approx \frac{I}{d/2} $$ (d = overall depth).

• Strongest Beam from Circular Log:

  $$\displaystyle Z = \frac{\pi}{32} d^3 (1 - k^4) $$, maximize w.r.t. $$\displaystyle k = d/D $$.

  $$\displaystyle \frac{d}{D} = \frac{1}{\sqrt{2}} $$ → depth = $\sqrt{2}$ × width.

Shear Stress

• Derivation: $$\displaystyle \tau = \frac{VQ}{I t} $$, where $$\displaystyle Q = \bar{A} \bar{y} $$ (area above/below point × distance from NA).

• Rectangular Section:

  $$\displaystyle \tau = \frac{3V}{2A} \left(1 - \frac{4y^2}{d^2}\right) $$ → parabolic, $$\displaystyle \tau_{max} = \frac{3V}{2A} = 1.5 \tau_{avg} $$ at NA.

• I-Section:

  • Web carries most shear.

  • Flange: $\tau \approx 0$ (large $t$, small $Q$).

  • Web: $$\displaystyle \tau_{max} \approx \frac{V}{A_{web}} $$ (approx).

• T-Section: Similar; max shear at NA.

[!TIP]

Shear Stress: Always max at NA for symmetric sections. In I-beams, assume web takes all shear for design.

VII. BEAM DEFLECTION

Area Moment Method (Theorems)

1. Slope between A and B: $$\displaystyle \theta_{AB} = \frac{1}{EI} \times $$ (area of $M$ diagram between A and B).

2. Deflection of B relative to A: $$\displaystyle \delta_B = \frac{1}{EI} \times $$ (moment of area of $M$ diagram about B).

• Signs:

  • Area above axis → + slope (concave up).

  • Moment about point → + deflection (upward).

Double Integration Method

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

• Integrate: $$\displaystyle \frac{dy}{dx} = \theta(x) $$, $$\displaystyle y = \delta(x) $$.

• Apply boundary conditions (e.g., $$\displaystyle y=0 $$ at supports, $$\displaystyle \theta=0 $$ at fixed end).

Macaulay's Method

• Represent loads with Macaulay brackets: $$\displaystyle \langle x-a \rangle^n = \begin{cases} 0 & x<a \\ (x-a)^n & x>a \end{cases} $$.

• Write $M(x)$ using brackets, integrate twice:

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

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

• Apply BCs with brackets (treat $\langle x-a \rangle$ as zero when $$\displaystyle x<a $$).

Standard Deflection Cases (Simply Supported, Cantilever)

[!TIP]

Macaulay Tip: For partial UDL from $a$ to $b$, write $$\displaystyle w\langle x-a \rangle^0 - w\langle x-b \rangle^0 $$. Always check BCs at both ends.

VIII. TORSION OF CIRCULAR SHAFTS

Shear Stress Distribution

• Solid shaft: $$\displaystyle \tau = \frac{T r}{J} $$, $$\displaystyle J = \frac{\pi d^4}{32} $$.

• Hollow shaft: $$\displaystyle J = \frac{\pi (D^4 - d^4)}{32} $$, $$\displaystyle \tau_{max} = \frac{T R}{J} $$ (at outer surface).

• Assumptions:

  1. Plane sections remain plane.

  2. $$\displaystyle \gamma = r \frac{d\phi}{dx} $$ (linear variation).

  3. $G$ constant, no warping.

Angle of Twist

• $$\displaystyle \theta = \frac{TL}{GJ} $$ (radians).

• Torsional rigidity: $GJ$.

• Power Transmission: $$\displaystyle P = \frac{2\pi N T}{60} $$ (W), where $N$ = rpm.

Combined Bending and Torsion

• Bending stress: $$\displaystyle \sigma_b = \frac{My}{I} $$ (tension on one side).

• Torsional shear: $$\displaystyle \tau = \frac{T r}{J} $$ (constant on cross-section).

• Principal stresses (at outer fibre, $$\displaystyle y = c $$):

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

• Max shear stress: $$\displaystyle \tau_{max} = \sqrt{\sigma_b^2 + \tau^2} $$.

• Apply failure theories: e.g., von Mises: $$\displaystyle \sigma_1^2 + \sigma_2^2 - \sigma_1\sigma_2 \ge \sigma_y^2 $$.

Hollow vs Solid Shafts

• Same material & weight: Hollow shaft has larger $J$ → higher torque capacity and lower twist.

• Same outer diameter: Hollow has less $J$? No, $J$ decreases if $d$ increases? Actually for same $D$, $J$ decreases as $d$ increases? Wait: $$\displaystyle J = \frac{\pi}{32}(D^4 - d^4) $$, so as $d$ increases, $J$ decreases. But for same weight, $$\displaystyle A_{solid} = A_{hollow} $$ → $$\displaystyle \frac{\pi D_s^2}{4} = \frac{\pi (D^2 - d^2)}{4} $$ → $$\displaystyle D_s^2 = D^2 - d^2 $$. Then $$\displaystyle J_{hollow} = \frac{\pi}{32}(D^4 - d^4) = \frac{\pi}{32}(D^2 - d^2)(D^2 + d^2) = \frac{\pi}{32} D_s^2 (D^2 + d^2) > \frac{\pi}{32} D_s^4 = J_{solid} $$. So hollow stronger/stiffer for same weight.

[!TIP]

Design: For combined loading, use $$\displaystyle \sigma_{max} $$ from principal stresses. Check both normal and shear criteria.

IX. COLUMNS AND BUCKLING

Definitions

• Strut: Member under compression, any length.

• Column: Strut with $$\displaystyle L_e/k > $$ limit (slenderness important).

• Slenderness ratio: $$\displaystyle \lambda = \frac{L_e}{k} $$, where $$\displaystyle k = \sqrt{I/A} $$ (least radius of gyration), $$\displaystyle L_e $$ = effective length.

Euler's Buckling Load

• For pinned ends: $$\displaystyle P_{cr} = \frac{\pi^2 EI}{L^2} $$.

• Effective Length $$\displaystyle L_e $$:

  | End Condition | $$\displaystyle L_e $$ | |---------------|-------| | Both pinned | $L$ | | Both fixed | $L/2$ | | One fixed, one free | $2L$ | | One fixed, one pinned | $L/\sqrt{2}$ |

• Euler's formula: $$\displaystyle P_{cr} = \frac{\pi^2 EI}{L_e^2} $$.

Rankine's Formula (Intermediate Columns)

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

• $$\displaystyle P_e = \frac{\pi^2 EI}{L_e^2} $$ (Euler load).

• $$\displaystyle P_0 = \sigma_c A $$ (crushing load, $$\displaystyle \sigma_c $$ = ultimate compressive stress).

• For ductile materials, $$\displaystyle \sigma_c \approx \sigma_y $$.

• Constant $a$: For cast iron, $$\displaystyle a = \frac{1}{1600} $$ (in Rankine-Gordon: $$\displaystyle P_{cr} = \frac{\sigma_c A}{1 + a \lambda^2} $$).

Eccentric Loading

• Axial load $P$ with eccentricity $e$:

  $$\displaystyle \sigma_{max} = \frac{P}{A} + \frac{Pe}{Z} $$,

  $$\displaystyle \sigma_{min} = \frac{P}{A} - \frac{Pe}{Z} $$.

• Middle Third Rule (rectangular section):

  No tensile stress if $$\displaystyle e \le \frac{b}{6} $$ (for $b$ = width in plane of eccentricity).

[!TIP]

Euler vs Rankine: Euler for long slender columns ($$\displaystyle \lambda > \lambda_{cr} $$). Rankine for all lengths. Always use least $I$ for $k$.

X. SPECIAL TOPICS AND APPLICATIONS

Thin Cylindrical Shells

• Hoop stress (circumferential): $$\displaystyle \sigma_h = \frac{p d}{2t} $$.

• Longitudinal stress: $$\displaystyle \sigma_l = \frac{p d}{4t} $$.

• Change in dimensions:

  • $$\displaystyle \Delta d = \frac{p d^2}{2tE}(1 - \nu) $$.

  • $$\displaystyle \Delta L = \frac{p d L}{4tE}(1 - 2\nu) $$.

• Fluid volume change:

  Additional fluid volume = original volume × $$\displaystyle \frac{p d}{2tE}(1 - \nu/2) $$? Actually:

  $$\displaystyle \Delta V_{fluid} = V_0 \left( \frac{p d}{2tE}(1 - \nu) + \frac{p d}{4tE}(1 - 2\nu) \right) = V_0 \frac{p d}{4tE}(3 - 4\nu) $$?

  Standard: $$\displaystyle \Delta V = \frac{p d}{4tE} (5 - 4\nu) V $$? Let's derive:

  $$\displaystyle \frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta L}{L} = 2\left[\frac{pd}{2tE}(1-\nu)\right] + \frac{pd}{4tE}(1-2\nu) = \frac{pd}{tE}\left(1-\nu + \frac{1-2\nu}{4}\right) = \frac{pd}{tE}\left(\frac{4(1-\nu) + (1-2\nu)}{4}\right) = \frac{pd}{tE}\left(\frac{5 - 6\nu}{4}\right) $$.

  So $$\displaystyle \Delta V = \frac{pd V}{4tE}(5 - 6\nu) $$. But many texts use $(5-4\nu)$. Check: Actually for thin cylinder, $$\displaystyle \Delta V \approx \frac{\pi d^3 p}{4tE}(1 - \nu/2) $$? I'll use standard formula:

  $$\displaystyle \Delta V = \frac{\pi d^3 p}{4tE}(1 - \frac{\nu}{2}) $$? No. Better to state:

  $$\displaystyle \frac{\Delta V}{V} = \frac{pd}{tE}\left(\frac{1}{2} - \nu\right) $$? That's for spherical?

  Correct: For thin cylinder:

  $$\displaystyle \epsilon_h = \frac{\sigma_h}{E} - \nu \frac{\sigma_l}{E} = \frac{pd}{2tE} - \nu \frac{pd}{4tE} = \frac{pd}{4tE}(2 - \nu) $$.

  $$\displaystyle \epsilon_l = \frac{\sigma_l}{E} - \nu \frac{\sigma_h}{E} = \frac{pd}{4tE} - \nu \frac{pd}{2tE} = \frac{pd}{4tE}(1 - 2\nu) $$.

  Then $$\displaystyle \Delta V/V = 2\epsilon_h + \epsilon_l = 2\cdot\frac{pd}{4tE}(2-\nu) + \frac{pd}{4tE}(1-2\nu) = \frac{pd}{4tE}(4 - 2\nu + 1 - 2\nu) = \frac{pd}{4tE}(5 - 4\nu) $$.

  So: $$\displaystyle \Delta V = \frac{pd V}{4tE}(5 - 4\nu) $$.

Thin Spherical Shells

• Stress: $$\displaystyle \sigma = \frac{p d}{4t} $$ (same in all directions).

• Change in diameter: $$\displaystyle \Delta d = \frac{p d^2}{4tE}(1 - \nu) $$.

Compound Cylinders (Shrink Fit)

• Radial pressure $p$ at interface after shrinking.

• Stresses in each cylinder ( Lamé equations for thick cylinders):

  • Hoop: $$\displaystyle \sigma_\theta = \frac{p r_i^2}{r_o^2 - r_i^2} \left(1 + \frac{r_o^2}{r^2}\right) $$.

  • Radial: $$\displaystyle \sigma_r = \frac{p r_i^2}{r_o^2 - r_i^2} \left(1 - \frac{r_o^2}{r^2}\right) $$.

• Final stresses with internal pressure $$\displaystyle p_i $$: Superpose pressure $p$ at interface and $$\displaystyle p_i $$ inside.

Point of Contraflexure

• Point where $$\displaystyle M = 0 $$ (changes sign).

• In overhanging/cantilever beams, often at support or between loads.

St. Venant's Principle

• Stress distribution far from load application point becomes independent of exact load distribution.

• Use: Replace complex load with statically equivalent system for stress analysis away from load.

Section Modulus Recap

• $$\displaystyle Z = I / y_{max} $$. Key for bending strength: $$\displaystyle \sigma_{max} = M/Z $$.

Middle Third Rule

• For rectangular column under axial load with eccentricity $e$:

  No tensile stress if $$\displaystyle e \le \frac{b}{6} $$ (where $b$ = width in plane of eccentricity).

  Stress distribution linear: $$\displaystyle \sigma = \frac{P}{A} \pm \frac{Pe}{Z} $$.

Stress due to Initial Strain

• Example: Bolt tightened by quarter turn → initial strain $$\displaystyle \epsilon_0 = \frac{pitch}{2\pi r} $$ (threads).

  Stress $$\displaystyle \sigma = E \epsilon_0 $$ if no slip.

[!TIP]

Thin Shells: Hoop stress = 2 × longitudinal stress for cylinders. For spheres, stress = half of cylinder hoop stress for same $p,d,t$.

Final Note: This compilation covers all high-frequency topics from past RGPV papers. Focus on derivations (Euler, shear stress, bending), numerical problems (principal stresses, beam deflection, torsion, columns), and theory comparisons (failure theories). Use Mohr's circle for quick stress transformation.