10.1 Stress Types, Axial Stress-Strain, and Thermal Deformation

10.1 Stress Types, Axial Stress-Strain, and Thermal Deformation

Core Engineering Principle: Mechanics of Materials (Strength of Materials) relates external forces and moments acting on a deformable body to internal stress distributions, elastic strain deformations, and mechanical failure limits. Understanding internal force-displacement relationships is central to structural and mechanical engineering design on the NCEES FE exam.

Uniaxial Stress-Strain, Hooke's Law, and Thermal Deformation

When a structural member of uniform cross-sectional area $A$ and original length $L$ is subjected to an axial force $P$, the internal normal stress $\sigma$ and normal strain $\epsilon$ are defined as:

$\sigma = \frac{P}{A}, \quad \epsilon = \frac{\delta}{L}$

where $\delta$ is the total axial elongation or contraction.

Tensile Test and Material Mechanical Properties

In the linear elastic region of a material's stress-strain diagram (below the proportional limit), stress is directly proportional to strain according to Hooke's Law:

$\sigma = E \epsilon$

where $E$ is the Young's Modulus (Modulus of Elasticity) of the material. Combining Hooke's Law with the definitions of stress and strain yields the fundamental axial displacement formula:

$\delta = \frac{P L}{A E}$

If the axial load or cross-sectional area varies along the length of the member, total displacement is found by integrating along the longitudinal axis $x$:

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

Poisson's Ratio and Shear Modulus

Axial elongation induces lateral contraction. Poisson's ratio $\nu$ is the negative ratio of lateral strain to longitudinal axial strain:

$\nu = - \frac{\epsilon_{lateral}}{\epsilon_{axial}}$

For isotropic elastic materials, the Elastic Modulus $E$, Shear Modulus $G$, and Poisson's ratio $\nu$ are inter-related by:

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

Thermal Strain and Statically Indeterminate Axial Members

A temperature change $\Delta T = T_{final} - T_{initial}$ causes a free thermal strain $\epsilon_{th}$ and unconstrained thermal deformation $\delta_{th}$:

$\epsilon_{th} = \alpha \Delta T, \quad \delta_{th} = \alpha L \Delta T$

If a member is fully or partially constrained by rigid supports, internal thermal stresses develop. For statically indeterminate systems, solution requires combining:

1. Equilibrium Equations: $\sum F_x = 0$
2. Compatibility Equations: Geometric constraints on total displacement (e.g., $\delta_{total} = \delta_{mechanic} + \delta_{thermal} = 0$)
3. Force-Displacement Relations: $\delta_{mechanic} = \frac{P L}{A E}$

The Four Stress Types NCEES Names

The specification opens with "Stress types (e.g., normal, shear)", and the exam expects you to identify which type a given loading produces before reaching for a formula.

Single vs. double shear is the most common bolt-sizing trap. A bolt in a lap joint carries the full load across one plane (single shear, $\tau = V/A$); the same bolt in a butt joint with two cover plates has two shear planes, so each carries half the load ($\tau = V/2A$) and its capacity doubles. Count the shear planes by counting how many gaps the bolt crosses.

Stress, Strain, and Hooke's Law

$\sigma = \frac{P}{A}, \qquad \varepsilon = \frac{\delta}{L}, \qquad \sigma = E\varepsilon$

Combining these gives the single most-used equation in the chapter:

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

Read the structure: elongation rises with load and length, and falls with area and stiffness. The group $AE/L$ is the member's axial stiffness — the spring constant of a bar — which is what lets you treat a truss member as a spring in indeterminate problems.

Worked Example: Stepped Bar Elongation

A steel bar ($E = 200$ GPa) carries a 60 kN tensile load. Segment 1 is 800 mm long with 400 mm² area; segment 2 is 500 mm long with 250 mm² area. Find the total elongation and the maximum stress.

Both segments carry the full 60 kN (they are in series), so:

$\delta_1 = \frac{60{,}000(800)}{400(200{,}000)} = \frac{4.8\times10^7}{8.0\times10^7} = 0.600\ \text{mm}$

$\delta_2 = \frac{60{,}000(500)}{250(200{,}000)} = \frac{3.0\times10^7}{5.0\times10^7} = 0.600\ \text{mm}$

$\delta_{\text{total}} = 0.600 + 0.600 = \boxed{1.20\ \text{mm}}$

$\sigma_{\max} = \frac{60{,}000}{250} = \boxed{240\ \text{MPa}} \quad\text{(in the smaller segment)}$

The two segments happen to elongate equally: segment 2 is shorter but proportionally thinner. Maximum stress always occurs at the minimum area, which is why stress concentrations at holes, notches, and fillets govern failure even when the average stress is modest.

Thermal Deformation

$\delta_T = \alpha L\,\Delta T, \qquad \varepsilon_T = \alpha\,\Delta T$

The recurring insight: stress arises from restrained deformation, never from temperature change itself. A bar heated on a frictionless surface develops no stress at all, no matter how hot. And $\sigma = E\alpha\Delta T$ contains no length term — a 200 mm and a 20 m restrained bar reach identical stress for the same temperature rise.

For a bar restrained between rigid supports with both load and temperature change, superpose the two effects and enforce the geometric constraint:

$\delta_{\text{total}} = \frac{PL}{AE} + \alpha L\Delta T = 0 \;\Rightarrow\; P = -\alpha\,\Delta T\,A E$

which is the compatibility equation used in the statically indeterminate worked problem below.

Worked Engineering Problems

Problem 1: Statically Indeterminate Thermal & Mechanical Axial Loading

Scenario: A solid aluminum alloy rod ($E = 70\text{ GPa}$, $\alpha = 23 \times 10^{-6}\text{ /}^\circ\text{C}$, cross-sectional area $A = 400\text{ mm}^2 = 4.0 \times 10^{-4}\text{ m}^2$) is rigidly fixed between two unyielding walls separated by length $L = 1.2\text{ m}$. Initially, the rod is stress-free at temperature $T_1 = 20^\circ\text{C}$. The temperature is then raised to $T_2 = 70^\circ\text{C}$ ($\Delta T = 50^\circ\text{C}$). Determine the compressive normal stress $\sigma$ induced in the rod and the force $P$ exerted against the rigid end supports.

Solution:

1. Formulate the compatibility condition: Since the rigid end supports do not move, the total displacement must equal zero: $\delta_{total} = \delta_{thermal} + \delta_{mechanical} = 0$

2. Express thermal elongation and mechanical compression: $\delta_{thermal} = \alpha L \Delta T$ $\delta_{mechanical} = - \frac{P L}{A E} = - \frac{\sigma L}{E}$

3. Substitute into compatibility equation: $\alpha L \Delta T - \frac{\sigma L}{E} = 0 \implies \sigma = E \alpha \Delta T$

4. Calculate numerical compressive stress $\sigma$: $\sigma = (70 \times 10^9\text{ Pa}) \times (23 \times 10^{-6}\text{ /}^\circ\text{C}) \times (50^\circ\text{C})$ $\sigma = 70 \times 10^9 \times 0.00115 = 8.05 \times 10^7\text{ Pa} = 80.5\text{ MPa (compressive)}$

5. Calculate reaction force $P$: $P = \sigma A = (80.5 \times 10^6\text{ N/m}^2) \times (4.0 \times 10^{-4}\text{ m}^2) = 32,200\text{ N} = 32.2\text{ kN}$

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