8.1 Simple Stresses, Strains, Elastic Moduli & Thermal Stresses

1. Engineering vs. True Stress and Strain

In mechanical and structural engineering applications—such as the design of mine hoists, dragline boom trusses, and heavy earthmoving equipment at Coal India Limited (CIL)—characterizing the constitutive response of structural alloys under tensile and compressive loading is fundamental.

Tensile Test of Ductile Materials (Mild Steel)

When a standard standardized specimen (such as IS 2062 or ASTM A36 structural steel) is subjected to uniaxial tensile loading in a Universal Testing Machine (UTM), the resulting load-displacement diagram reveals distinct behavioral regimes:

 Stress (σ)
   ^
   |               E (Ultimate Tensile Strength)
   |              / \
   |     C       /   \  F (Fracture)
   |    / \_ D  /     x
   |   B    \__/ (Strain Hardening)
   |  /   (Yield Plateau)
   | A (Proportional Limit)
   |/ (Linear Elastic Hooke's Region)
   +-------------------------------------> Strain (ε)

1. Proportional Limit ($A$): The upper boundary of linear elasticity where Hooke's Law ($\sigma \propto \epsilon$) strictly applies. Modulus of Elasticity $E$ is the slope of this segment ($E = \tan \theta$).
2. Elastic Limit ($B$): The maximum stress the material can sustain without suffering permanent plastic deformation upon complete unloading. Beyond point $B$, plastic deformation initiates.
3. Upper Yield Point ($C$): The point at which plastic deformation initiates abruptly due to the breakout of interstitial carbon/nitrogen solute atoms from Cottrell atmospheres pinning dislocations.
4. Lower Yield Point ($D$): The steady-state stress level at which plastic yield propagation (Lüders bands) spreads along the specimen gauge length at approximately constant load. In engineering design codes, the lower yield strength ($f_y$ or $\sigma_y$) is used as the nominal yield strength.
5. Strain Hardening Plateau ($D \to E$): Beyond yield propagation, dislocation entanglement increases internal resistance to further deformation. Stress rises continuously with increasing strain up to the Ultimate Tensile Strength (UTS) at point $E$.
6. Necking & Fracture ($E \to F$): At point $E$, uniform plastic elongation ceases, and localized cross-sectional necking initiates. While the actual true stress continues to increase until ductile shear cup-and-cone fracture at point $F$, the nominal engineering stress drops because load is divided by the original (un-necked) cross-sectional area $A_0$.

Mathematical Formulation: Engineering vs. True Metrics

Let $P$ be instantaneous axial load, $L_0$ original gauge length, $A_0$ original cross-sectional area, $L_i$ instantaneous length, and $A_i$ instantaneous area.

$$\sigma_{\text{eng}} = \frac{P}{A_0}, \qquad \epsilon_{\text{eng}} = \frac{\Delta L}{L_0} = \frac{L_i - L_0}{L_0}$$ $$\sigma_{\text{true}} = \frac{P}{A_i}, \qquad \epsilon_{\text{true}} = \int_{L_0}^{L_i} \frac{dL}{L} = \ln\left(\frac{L_i}{L_0}\right)$$

Assuming plastic deformation occurs at constant material volume ($A_0 L_0 = A_i L_i$), the instantaneous area ratio is $\frac{A_0}{A_i} = \frac{L_i}{L_0} = 1 + \epsilon_{\text{eng}}$. Substituting this into the true stress definition yields:

$$\sigma_{\text{true}} = \frac{P}{A_i} = \frac{P}{A_0} \cdot \frac{A_0}{A_i} = \sigma_{\text{eng}}(1 + \epsilon_{\text{eng}})$$ $$\epsilon_{\text{true}} = \ln\left(\frac{L_i}{L_0}\right) = \ln(1 + \epsilon_{\text{eng}})$$

Key Exam Note: In tension ($\epsilon_{\text{eng}} > 0$), $\sigma_{\text{true}} > \sigma_{\text{eng}}$ and $\epsilon_{\text{true}} < \epsilon_{\text{eng}}$. In compression ($\epsilon_{\text{eng}} < 0$), $|\sigma_{\text{true}}| < |\sigma_{\text{eng}}|$ and $|\epsilon_{\text{true}}| > |\epsilon_{\text{eng}}|$.

0.2% Offset Proof Stress

For metals without a well-defined yield plateau (e.g., cold-rolled steel, aluminium alloys, brass, titanium), the yield strength is established via the 0.2% offset method (offset strain $\epsilon = 0.002$). A line is drawn parallel to the initial linear elastic slope starting at $\epsilon = 0.2\%$ on the strain axis; the intersection with the stress-strain curve defines the 0.2% proof stress ($\sigma_{0.2}$).

2. Generalized Hooke's Law, Poisson's Ratio & Volumetric Strain

Poisson's Ratio ($\mu$ or $\nu$)

When a prismatic bar is subjected to longitudinal tensile stress, it undergoes axial elongation accompanied by lateral (transverse) contraction. Poisson's Ratio is defined as the absolute ratio of lateral strain to longitudinal strain under uniaxial stress:

$$\mu = -\frac{\epsilon_{\text{lateral}}}{\epsilon_{\text{longitudinal}}} = -\frac{\delta d / d_0}{\delta L / L_0}$$

• Thermodynamic theoretical bounds: $-1.0 \le \mu \le 0.5$
• Practical engineering material range: $0.0 \le \mu \le 0.5$
  • Cork: $\mu \approx 0.0$ (ideal for bottle stoppers because axial compression causes negligible lateral bulging)
  • Concrete & Glass: $\mu \approx 0.10 - 0.20$
  • Cast Iron: $\mu \approx 0.23 - 0.27$
  • Structural Mild Steel: $\mu \approx 0.28 - 0.33$
  • Copper & Aluminium Alloys: $\mu \approx 0.33 - 0.36$
  • Rubber & Saturated Clay: $\mu \approx 0.48 - 0.50$ (incompressible materials)
  • Auxetic Materials (special polymer foams, cellular honeycombs): $\mu < 0$ (expand laterally when stretched axially)

3D Generalized Hooke's Law

For a homogeneous, isotropic, linearly elastic material subjected to a general triaxial normal stress state ($\sigma_x, \sigma_y, \sigma_z$), applying the principle of superposition yields the generalized strain equations:

$$\epsilon_x = \frac{\sigma_x}{E} - \frac{\mu}{E}(\sigma_y + \sigma_z)$$ $$\epsilon_y = \frac{\sigma_y}{E} - \frac{\mu}{E}(\sigma_x + \sigma_z)$$ $$\epsilon_z = \frac{\sigma_z}{E} - \frac{\mu}{E}(\sigma_x + \sigma_y)$$

Volumetric Strain ($\epsilon_v$)

Volumetric strain is defined as the fractional change in volume $\epsilon_v = \frac{\Delta V}{V_0}$. For small elastic deformations:

$$\epsilon_v = \epsilon_x + \epsilon_y + \epsilon_z = \frac{\sigma_x + \sigma_y + \sigma_z}{E}(1 - 2\mu)$$

For a uniform hydrostatic stress state where $\sigma_x = \sigma_y = \sigma_z = \sigma$:

$$\epsilon_v = \frac{3\sigma(1 - 2\mu)}{E} = \frac{\sigma}{K}$$

Where $K$ is the Bulk Modulus of Elasticity. Notice that if $\mu = 0.5$, $\epsilon_v = 0$, signifying that the material is perfectly incompressible ($K \to \infty$).

3. The Four Elastic Moduli & Number of Independent Constants

An isotropic elastic material possesses four interrelated elastic moduli:

1. Young's Modulus ($E$): Ratio of tensile/compressive normal stress to longitudinal normal strain in uniaxial loading ($E = \sigma / \epsilon$).
2. Shear Modulus or Modulus of Rigidity ($G$ or $C$): Ratio of shear stress to shear strain ($G = \tau / \gamma$).
3. Bulk Modulus ($K$): Ratio of uniform hydrostatic pressure to volumetric strain ($K = -p / \epsilon_v$).
4. Poisson's Ratio ($\mu$ or $\nu$): Ratio of lateral transverse strain to axial longitudinal strain.

Fundamental Interrelationships

Every candidate preparing for CIL MT Paper-II must master these four algebraic identities:

$$\mathbf{E = 2G(1 + \mu)}$$ $$\mathbf{E = 3K(1 - 2\mu)}$$ $$\mathbf{E = \frac{9KG}{3K + G}}$$ $$\mathbf{\mu = \frac{3K - 2G}{6K + 2G}}$$

Another convenient form frequently tested in PSU examinations is:

$$\frac{9}{E} = \frac{3}{G} + \frac{1}{K} \quad \Longleftrightarrow \quad \frac{1}{E} = \frac{1}{3G} + \frac{1}{9K}$$

Elastic Constants by Material Anisotropy

4. Axial Deformations: Prismatic, Tapered, and Self-Weight Bars

1. Prismatic Bar Under Axial Force

For a bar of length $L$, constant cross-sectional area $A$, and elastic modulus $E$ subjected to axial load $P$:

$$\Delta L = \frac{PL}{AE}$$

2. Linearly Tapered Circular Bar

Consider a solid circular bar of length $L$ whose diameter varies linearly from $d_1$ at one end to $d_2$ at the other. Under axial tensile load $P$, the diameter at distance $x$ from the $d_1$ end is $d(x) = d_1 + \left(\frac{d_2 - d_1}{L}\right)x$. The total elongation is derived by integrating the deformation of a differential element $dx$:

$$\Delta L = \int_0^L \frac{P dx}{A(x) E} = \int_0^L \frac{P dx}{\frac{\pi}{4}\left[d_1 + \left(\frac{d_2 - d_1}{L}\right)x\right]^2 E} = \mathbf{\frac{4PL}{\pi E d_1 d_2}}$$

(Note: If $d_1 = d_2 = d$, this reduces directly to $\frac{4PL}{\pi E d^2} = \frac{PL}{AE}$).

3. Flat Rectangular Bar with Linearly Tapered Width

For a flat plate of uniform thickness $t$ and length $L$, with width tapering linearly from $b_1$ to $b_2$ under axial load $P$:

$$\Delta L = \int_0^L \frac{P dx}{t \cdot b(x) E} = \mathbf{\frac{PL}{Et(b_2 - b_1)} \ln\left(\frac{b_2}{b_1}\right)}$$

4. Elongation Under Self-Weight (Prismatic Bar)

Consider a vertical bar of length $L$, uniform cross-sectional area $A$, mass density $\rho$, and specific weight $\gamma = \rho g$ suspended from its upper end. The total weight of the bar is $W = \gamma A L$. The tensile force supported by a cross-section at distance $x$ from the free lower end is simply the weight of the segment below it: $P(x) = \gamma A x$.

$$\Delta L_{\text{self}} = \int_0^L \frac{P(x) dx}{AE} = \int_0^L \frac{\gamma A x dx}{AE} = \frac{\gamma L^2}{2E} = \mathbf{\frac{WL}{2AE}}$$

  Fixed Support ///////////////
        |          Stress σ_max = γL = W/A
        | 
        | |        Cross-section A
        | | 
        | | Length L, Specific weight γ
        | | 
        |          Stress σ = 0 at free bottom
        v

• Maximum tensile stress: Occurs at the fixed support ($x = L$): $\sigma_{\max} = \frac{W}{A} = \gamma L$.
• Minimum tensile stress: Occurs at the free bottom tip ($x = 0$): $\sigma_{\min} = 0$.
• Elongation comparison: Elongation under self-weight is exactly half (50%) of the elongation produced if the entire weight $W$ were applied as an external concentrated load at the bottom tip: $\Delta L_{\text{self}} = \frac{1}{2}\Delta L_{\text{end-load}}$.

5. Elongation Under Self-Weight (Conical Bar)

For a conical bar hanging vertically with its circular base (area $A_{\text{base}}$) fixed at the top and its vertex at the bottom:

$$\Delta L_{\text{cone}} = \mathbf{\frac{\gamma L^2}{6E}} = \frac{WL}{2 A_{\text{base}} E} = \mathbf{\frac{1}{3} \Delta L_{\text{prismatic-self}}}$$

(Here $W = \frac{1}{3}\gamma A_{\text{base}} L$ is the total weight of the cone). Maximum stress at the support is $\sigma_{\max} = \frac{W}{A_{\text{base}}} = \frac{1}{3}\gamma L$.

5. Thermal Stresses, Boundary Restraints & Composite Bars

When a structural component experiences a temperature variation $\Delta T$, it undergoes thermal expansion or contraction. The free thermal strain is $\epsilon_{\text{th}} = \alpha \Delta T$, where $\alpha$ is the material's Coefficient of Linear Thermal Expansion ($/^{\circ}\text{C}$ or $/\text{K}$).

Thermal Restraint Conditions

1. Unrestrained (Free Expansion):

   $$\Delta L_{\text{free}} = L \alpha \Delta T, \qquad \sigma_{\text{th}} = 0$$

   No thermal stress is generated as long as deformation is unconstrained.

2. Fully Restrained Between Rigid Supports: If rigid immovable supports prevent expansion completely, a compressive mechanical strain equal to the prevented thermal strain is induced:

   $$\epsilon_{\text{mech}} = -\epsilon_{\text{th}} = -\alpha \Delta T$$ $$\mathbf{\sigma_{\text{th}} = E \alpha \Delta T \quad (\text{Compressive for } \Delta T > 0)}$$

   The thrust force exerted on the rigid walls is $P_{\text{th}} = \sigma_{\text{th}} A = AE\alpha\Delta T$. Note that $\sigma_{\text{th}}$ is completely independent of the bar length $L$.

3. Partially Restrained with Yielding / Expansion Gap ($\delta$): If the supports permit a free axial gap $\delta$ before preventing further movement (where $L\alpha\Delta T > \delta$):

   $$\Delta L_{\text{prevented}} = L\alpha\Delta T - \delta$$ $$\mathbf{\sigma_{\text{th}} = \frac{(L\alpha\Delta T - \delta)E}{L} = E\left(\alpha\Delta T - \frac{\delta}{L}\right)}$$

   Rigid Wall                                Rigid Wall
   |==========[ Prismatic Bar (L, E, α) ]====| ======|
   |<------------------- L ----------------->|<--δ-->|

Composite / Bimetallic Bars Under Temperature Change

Consider a bimetallic bar composed of two materials—e.g., Copper ($\alpha_c, E_c, A_c$) and Steel ($\alpha_s, E_s, A_s$)—firmly bonded or fastened between rigid end plates such that they deform together.

Since $\alpha_{\text{copper}} > \alpha_{\text{steel}}$ (typically $\alpha_c \approx 17 \times 10^{-6}/^{\circ}\text{C}$ vs $\alpha_s \approx 12 \times 10^{-6}/^{\circ}\text{C}$), upon heating ($\Delta T > 0$):

• Copper attempts to expand more than steel, but the common rigid end connection forces a compromise final length.
• Therefore, Copper is subjected to Compressive stress ($\sigma_c$), while Steel is subjected to Tensile stress ($\sigma_s$).

To solve for stresses, apply equilibrium and compatibility:

1. Equilibrium Equation (Internal axial forces balance):

   $$P_{\text{steel}} = P_{\text{copper}} \implies \mathbf{\sigma_s A_s = \sigma_c A_c}$$

2. Compatibility Equation (Equal final extension $\Delta L$):

   $$L\alpha_s \Delta T + \frac{\sigma_s L}{E_s} = L\alpha_c \Delta T - \frac{\sigma_c L}{E_c}$$ $$\mathbf{(\alpha_c - \alpha_s)\Delta T = \frac{\sigma_s}{E_s} + \frac{\sigma_c}{E_c}}$$

Combining these two equations yields the exact stresses in both materials.

6. Engineering Reference Tables & Physical Properties

7. Worked Step-by-Step Engineering Calculations

Example 1: Determining Elastic Moduli from Tensile Test Data

Problem: A mild steel tie rod of diameter $d = 20\text{ mm}$ and gauge length $L = 200\text{ mm}$ is subjected to an axial tensile load of $P = 50\text{ kN}$. The measured extension is $\Delta L = 0.1516\text{ mm}$ and the lateral diameter contraction is $\Delta d = 0.00455\text{ mm}$. Calculate:

1. Young's Modulus $E$
2. Poisson's Ratio $\mu$
3. Shear Modulus $G$
4. Bulk Modulus $K$

Solution:

• Cross-sectional area $A = \frac{\pi}{4}(20)^2 = 314.16\text{ mm}^2 = 314.16 \times 10^{-6}\text{ m}^2$.
• Longitudinal strain $\epsilon_{\text{long}} = \frac{\Delta L}{L} = \frac{0.1516}{200} = 7.58 \times 10^{-4}$.
• Lateral strain $\epsilon_{\text{lat}} = \frac{\Delta d}{d} = \frac{0.00455}{20} = 2.275 \times 10^{-4}$.
• 1. Young's Modulus: $$E = \frac{\sigma}{\epsilon_{\text{long}}} = \frac{P}{A \cdot \epsilon_{\text{long}}} = \frac{50 \times 10^3}{(314.16 \times 10^{-6})(7.58 \times 10^{-4})} = 209.97 \times 10^9\text{ Pa} = \mathbf{210.0\text{ GPa}}$$
• 2. Poisson's Ratio: $$\mu = \frac{\epsilon_{\text{lat}}}{\epsilon_{\text{long}}} = \frac{2.275 \times 10^{-4}}{7.58 \times 10^{-4}} = \mathbf{0.300}$$
• 3. Shear Modulus: $$G = \frac{E}{2(1 + \mu)} = \frac{210.0}{2(1 + 0.30)} = \frac{210.0}{2.6} = \mathbf{80.77\text{ GPa}}$$
• 4. Bulk Modulus: $$K = \frac{E}{3(1 - 2\mu)} = \frac{210.0}{3(1 - 2(0.30))} = \frac{210.0}{3(0.40)} = \frac{210.0}{1.2} = \mathbf{175.0\text{ GPa}}$$

---

Example 2: Thermal Stress in Stepped Bar with Expansion Gap

Problem: A stepped bar fixed at one end consists of a steel segment ($L_s = 300\text{ mm}, A_s = 400\text{ mm}^2, E_s = 200\text{ GPa}, \alpha_s = 12 \times 10^{-6}/^{\circ}\text{C}$) and an aluminium segment ($L_a = 200\text{ mm}, A_a = 600\text{ mm}^2, E_a = 70\text{ GPa}, \alpha_a = 24 \times 10^{-6}/^{\circ}\text{C}$). The initial gap between the free end and a rigid right wall is $\delta = 0.20\text{ mm}$. If the temperature rises by $\Delta T = 50^{\circ}\text{C}$, determine the compressive force developed.

Solution:

• Free thermal expansion of both segments: $$\Delta L_{\text{free}} = L_s \alpha_s \Delta T + L_a \alpha_a \Delta T$$ $$\Delta L_{\text{free}} = (300)(12 \times 10^{-6})(50) + (200)(24 \times 10^{-6})(50) = 0.180\text{ mm} + 0.240\text{ mm} = 0.420\text{ mm}$$
• Since $\Delta L_{\text{free}} = 0.420\text{ mm} > \delta = 0.200\text{ mm}$, the bar contacts the wall and undergoes restrained compression.
• Net prevented expansion $\Delta L_{\text{prevented}} = 0.420 - 0.200 = 0.220\text{ mm}$.
• The compressive force $P$ is common to both segments (in series): $$\Delta L_{\text{prevented}} = \frac{P L_s}{A_s E_s} + \frac{P L_a}{A_a E_a}$$ $$0.220 = P \left[\frac{300}{(400)(200 \times 10^3)} + \frac{200}{(600)(70 \times 10^3)}\right] = P \left[3.75 \times 10^{-6} + 4.762 \times 10^{-6}\right] = P(8.512 \times 10^{-6})$$ $$P = \frac{0.220}{8.512 \times 10^{-6}} = 25846\text{ N} = \mathbf{25.85\text{ kN}}$$