4.2 Engineering Mechanics & Strength of Materials

4.2 Engineering Mechanics & Strength of Materials

1. Statics of Rigid Bodies

Statics investigates bodies at rest or moving at constant velocity under the action of balanced force systems.

Conditions for Static Equilibrium

For a two-dimensional coplanar force system acting on a rigid body, static equilibrium requires three scalar equations:

$$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_z = 0$$

where $\sum M_z$ is the summation of moments about any arbitrary axis perpendicular to the plane.

Moment of a Force & Couples

The moment $\vec{M}_O$ of a force $\vec{F}$ about a point $O$ is a vector cross product:

$$\vec{M}_O = \vec{r} \times \vec{F} = F \cdot d$$

where $\vec{r}$ is the position vector from point $O$ to any point on the line of action of $\vec{F}$, and $d$ is the perpendicular distance (moment arm).

Friction (Dry / Coulomb Friction)

When two contacting surfaces tend to slide relative to each other, a tangential friction force $f$ develops:

• Static friction: $f_s \le \mu_s N$ (Maximum static friction $f_{s,\text{max}} = \mu_s N$ occurs at impending motion)
• Kinetic friction: $f_k = \mu_k N$ (where $\mu_k < \mu_s$)
• Angle of static friction: $\tan\phi_s = \mu_s$

Centroids & Moment of Inertia

The centroid $(\bar{x}, \bar{y})$ of a composite area is determined by dividing it into simple geometric shapes:

$$\bar{x} = \frac{\sum A_i \bar{x}_i}{\sum A_i}, \quad \bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i}$$

The area moment of inertia $I_x$ quantifies the resistance of an area to bending about an axis:

$$I_x = \int y^2 dA$$

Parallel Axis Theorem: The moment of inertia $I_x$ about any axis parallel to a centroidal axis $I_{\bar{x}}$ at distance $d$ is:

$$I_x = I_{\bar{x}} + A d^2$$

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2. Dynamics of Moving Bodies

Kinematics of Rectilinear Motion (Constant Acceleration $a$)

1. $v = v_0 + at$
2. $s = s_0 + v_0 t + \frac{1}{2} a t^2$
3. $v^2 = v_0^2 + 2a(s - s_0)$
4. $s - s_0 = \left(\frac{v_0 + v}{2}\right) t$

Kinetics & Newton's Second Law

When an unbalanced force system acts on a particle of mass $m$, it accelerates in the direction of the resultant force:

$$\sum \vec{F} = m \vec{a}$$

Work-Energy Principle

The total work $U_{1-2}$ performed by all forces on a body equals the change in kinetic energy $\Delta T$:

$$U_{1-2} = T_2 - T_1 = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2$$

Impulse-Momentum Principle

The linear impulse $\vec{J}$ of a force acting over time interval $\Delta t$ equals the change in linear momentum $\Delta \vec{p}$:

$$\vec{J} = \int_{t_1}^{t_2} \vec{F} dt = m\vec{v}_2 - m\vec{v}_1$$

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3. Strength of Materials & Stress-Strain Analysis

Strength of Materials analyzes internal stress distributions and structural deformations under external loading.

Normal Stress & Strain

• Normal Stress $\sigma$: Force per unit area acting perpendicular to the cross section: $$\sigma = \frac{P}{A}$$
• Normal Strain $\epsilon$: Fractional change in length: $$\epsilon = \frac{\delta}{L_0}$$
• Hooke's Law: Within the elastic limit of a material: $$\sigma = E \epsilon$$ where $E$ is Young's Modulus of Elasticity (e.g., $E_{\text{steel}} \approx 200 \text{ GPa}$).

Axial Deformation Formula

Combining $\sigma = \frac{P}{A}$, $\epsilon = \frac{\delta}{L}$, and $\sigma = E\epsilon$ yields the total elongation $\delta$ of an axially loaded bar:

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

Thermal Stress & Strain

When a constrained bar experiences a temperature change $\Delta T$, internal thermal stress arises:

$$\delta_{\text{thermal}} = \alpha L \Delta T \implies \sigma_{\text{thermal}} = E \alpha \Delta T$$

where $\alpha$ is the coefficient of thermal expansion.

Poisson's Ratio & Shear Modulus

• Poisson's Ratio $\nu$: Ratio of lateral strain to axial strain under uniaxial loading: $$\nu = -\frac{\epsilon_{\text{lateral}}}{\epsilon_{\text{axial}}}$$
• Shear Modulus $G$ (Modulus of Rigidity): Relates shear stress $\tau$ to shear strain $\gamma$: $$\tau = G \gamma, \quad G = \frac{E}{2(1 + \nu)}$$

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4. Shaft Torsion & Beam Flexure

Torsion of Circular Shafts

When a torque $T$ is applied to a circular shaft of radius $r = d/2$ and length $L$, the maximum shear stress $\tau_{\text{max}}$ occurs at the outer boundary:

$$\tau_{\text{max}} = \frac{T r}{J} = \frac{16 T}{\pi d^3}$$

where $J = \frac{\pi d^4}{32}$ is the polar moment of inertia for a solid circular cross section.

The angle of twist $\phi$ (in radians) over length $L$ is:

$$\phi = \frac{T L}{G J}$$

Flexural Stress in Beams (Flexure Formula)

When a beam is subjected to a bending moment $M$, internal normal stresses $\sigma$ vary linearly with distance $y$ from the neutral axis:

$$\sigma = -\frac{M y}{I}$$

The maximum flexural stress $\sigma_{\text{max}}$ at the extreme fiber ($y = c$) is:

$$\sigma_{\text{max}} = \frac{M c}{I} = \frac{M}{S}$$

where $S = \frac{I}{c}$ is the Section Modulus of the beam.