6.1 Statics & Dynamics of Rigid Bodies & Engineering Linkages

6.1 Statics & Dynamics of Rigid Bodies & Engineering Linkages

1. Force Systems and Equations of Equilibrium

Engineering mechanics begins with statics—the study of bodies at rest or moving at constant velocity under the action of forces. A rigid body is defined as an ideal body in which the distance between any two given points remains constant under applied loads. In real mechanical engineering applications, structural members undergo small deformations, but statics assumes these deformations do not significantly alter the geometry or force distribution.

1.1 Force Systems and Resultants

Forces are vector quantities characterized by magnitude, direction, sense, and point of application. In a coplanar force system, forces reside in a single plane. The resultant force $\mathbf{R}$ and resultant moment $\mathbf{M}_O$ about an arbitrary reference point $O$ are obtained by vector summation:

$\mathbf{R} = \sum \mathbf{F} = \left(\sum F_x\right)\mathbf{i} + \left(\sum F_y\right)\mathbf{j}$

$M_O = \sum M = \sum (x F_y - y F_x)$

The magnitude and direction angle $\theta$ of the resultant force are:

$R = \sqrt{\left(\sum F_x\right)^2 + \left(\sum F_y\right)^2}, \quad \theta = \tan^{-1}\left| \frac{\sum F_y}{\sum F_x} \right|$

1.2 Equations of Equilibrium and Free-Body Diagrams (FBD)

For a rigid body to remain in static equilibrium, the resultant force and resultant moment about any point must independently vanish. In two-dimensional coplanar analysis, this yields three independent scalar equilibrium equations:

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

Drawing a correct Free-Body Diagram (FBD) is the single most critical step in solving statics problems. An FBD isolates the body or mechanism component from its surroundings and replaces all physical supports and contacts with equivalent reaction forces and moments:

• Pin / Hinge Support: Restrains horizontal and vertical displacement; introduces two unknown reaction force components ($R_x, R_y$).
• Roller / Smooth Surface: Restrains motion perpendicular to the contact plane; introduces one unknown reaction force normal to the surface ($N$).
• Fixed / Built-in Support: Restrains displacement and rotation; introduces two force components ($R_x, R_y$) and one reaction moment ($M_R$).
• Two-Force Member: A light link loaded only at its ends; the line of action of the reaction force MUST pass through both pin centers.

---

2. Dry Friction and Belt Friction Mechanics

Friction is the contact force resisting relative sliding between contacting surfaces.

2.1 Dry (Coulomb) Friction

According to Coulomb's law of dry friction:

1. Impending Motion (Static Region): The static friction force $F_s$ increases up to a maximum limit proportional to the normal force $N$: $F_s \le F_{s,\text{max}} = \mu_s N$ where $\mu_s$ is the coefficient of static friction.
2. Motion Occurring (Kinetic Region): Once sliding initiates, the friction force drops slightly to the kinetic friction force $F_k$: $F_k = \mu_k N$ where $\mu_k < \mu_s$.

2.2 Belt and Band Friction

In mechanical power transmission (V-belts, flat belts, band brakes), friction between a flexible belt or rope wrapped around a cylindrical drum transmits torque. Consider a drum around which a belt is wrapped through an angle of contact $\beta$ (expressed in radians). If slipping is impending in the direction of the larger tension $T_2$ relative to the smaller tension $T_1$, the ratio of tensions obeys the exponential belt friction equation:

$\frac{T_2}{T_1} = e^{\mu \beta} \implies T_2 = T_1 e^{\mu \beta}$

The net torque $M$ transmitted to or from the drum of radius $r$ is:

$M = (T_2 - T_1) r$

---

3. Centroids, Area Moments of Inertia, and Parallel Axis Theorem

3.1 Centroids of Composite Areas

The centroid $(\bar{x}, \bar{y})$ represents the geometric center of a plane cross-section. For a composite shape subdivided into simple geometric elements (rectangles, triangles, circles), the centroid coordinates are evaluated as weighted averages:

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

3.2 Area Moment of Inertia ($I$)

The second moment of area (moment of inertia) quantifies the distribution of cross-sectional area relative to a reference axis, governing flexural rigidity in beam bending:

$I_x = \int y^2 dA, \quad I_y = \int x^2 dA$

For standard shapes about their centroidal axes:

• Rectangle ($b \times h$): $\bar{I}_x = \frac{b h^3}{12}, \quad \bar{I}_y = \frac{h b^3}{12}$
• Solid Circle (diameter $d$, radius $R$): $\bar{I}_x = \bar{I}_y = \frac{\pi R^4}{4} = \frac{\pi d^4}{64}$

3.3 Parallel Axis Theorem

To calculate the moment of inertia $I_x$ about any parallel non-centroidal axis located at a perpendicular distance $d_y$ from the centroidal axis $\bar{x}$:

$I_x = \bar{I}_{x'} + A d_y^2$

where $\bar{I}_{x'}$ is the centroidal moment of inertia, $A$ is the total cross-sectional area, and $d_y$ is the perpendicular distance between the parallel axes.

---

4. Dynamics of Particles and Rigid Linkages

Dynamics divides into kinematics (the study of motion without regard to forces) and kinetics (the study of forces causing motion).

4.1 Kinematics of Particles and Rigid Linkages

For linear translation under constant acceleration $a$:

$v = v_0 + a t, \quad s = s_0 + v_0 t + \frac{1}{2} a t^2, \quad v^2 = v_0^2 + 2 a (s - s_0)$

For rotational motion under constant angular acceleration $\alpha$:

$\omega = \omega_0 + \alpha t, \quad \theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2, \quad \omega^2 = \omega_0^2 + 2 \alpha (\theta - \theta_0)$

In rigid engineering linkages, relative velocity between two points $A$ and $B$ on the same rigid link rotating with angular velocity $\boldsymbol{\omega}$ is:

$\mathbf{v}_B = \mathbf{v}_A + \mathbf{v}_{B/A} = \mathbf{v}_A + (\boldsymbol{\omega} \times \mathbf{r}_{B/A})$

where $|\mathbf{v}{B/A}| = \omega \cdot r{B/A}$.

4.2 Kinetics: Newton's Second Law, Work-Energy, and Impulse-Momentum

For planar rigid body motion, Newton's second law yields three dynamic equations:

$\sum F_x = m a_{Gx}, \quad \sum F_y = m a_{Gy}, \quad \sum M_G = I_G \alpha$

where $G$ is the mass center and $I_G$ is the mass moment of inertia about $G$.

Work-Energy Principle

The total kinetic energy $T$ of a rigid body in planar motion combines translation and rotation:

$T = \frac{1}{2} m v_G^2 + \frac{1}{2} I_G \omega^2$

The Work-Energy Principle states that work done by all external forces $U_{1-2}$ equals the change in kinetic energy:

$T_1 + U_{1-2} = T_2$

Impulse-Momentum Principle

Linear impulse equals change in linear momentum: $\int_{t_1}^{t_2} \mathbf{F} dt = m \mathbf{v}_2 - m \mathbf{v}1$. Angular impulse equals change in angular momentum: $\int{t_1}^{t_2} M_G dt = I_G \omega_2 - I_G \omega_1$.

---

5. Worked Numerical Examples

Step-by-Step Example 1: Belt Friction Torque Capacity

Problem: A band brake holds a rotating drum of radius $r = 0.25\text{ m}$. The belt has a friction coefficient $\mu = 0.30$ and wraps around half the drum surface (contact angle $\beta = 180^\circ = \pi\text{ rad}$). If the maximum allowable belt tension on the tight side is $T_2 = 2500\text{ N}$, determine the loose side tension $T_1$ and the maximum torque $M$ the brake can absorb.

Solution:

1. Convert contact angle to radians: $\beta = 180^\circ \times \frac{\pi}{180^\circ} = 3.14159\text{ rad}$.
2. Apply the belt friction equation: $\frac{T_2}{T_1} = e^{\mu \beta} = e^{(0.30 \times 3.14159)} = e^{0.94248} = 2.5663$
3. Solve for $T_1$: $T_1 = \frac{T_2}{2.5663} = \frac{2500\text{ N}}{2.5663} = 974.16\text{ N}$
4. Calculate maximum torque capacity $M$: $M = (T_2 - T_1) r = (2500\text{ N} - 974.16\text{ N}) \times 0.25\text{ m} = 1525.84\text{ N}\cdot\text{m} \times 0.25 = 381.46\text{ N}\cdot\text{m}$

Step-by-Step Example 2: Centroid and Moment of Inertia of a T-Section

Problem: A structural T-section beam consists of:

• Top Flange: width $b_1 = 150\text{ mm}$, thickness $h_1 = 20\text{ mm}$.
• Vertical Web: width $b_2 = 20\text{ mm}$, height $h_2 = 180\text{ mm}$ attached to the bottom of the flange. Determine the centroidal distance $\bar{y}$ measured from the bottom base of the web, and the centroidal moment of inertia $\bar{I}_x$.

Solution:

1. Calculate sub-areas and centroid locations from bottom base ($y = 0$):
   • Web (Area 1): $A_1 = 20 \times 180 = 3600\text{ mm}^2$; $y_1 = 180 / 2 = 90\text{ mm}$.
   • Flange (Area 2): $A_2 = 150 \times 20 = 3000\text{ mm}^2$; $y_2 = 180 + (20 / 2) = 190\text{ mm}$.
   • Total Area $A = A_1 + A_2 = 3600 + 3000 = 6600\text{ mm}^2$.
2. Calculate overall centroid $\bar{y}$: $\bar{y} = \frac{A_1 y_1 + A_2 y_2}{A} = \frac{(3600 \times 90) + (3000 \times 190)}{6600} = \frac{324000 + 570000}{6600} = \frac{894000}{6600} = 135.45\text{ mm}$
3. Calculate centroidal moments of inertia of individual parts about their own centroids:
   • $\bar{I}_{1} = \frac{20 \times 180^3}{12} = \frac{20 \times 5832000}{12} = 9,720,000\text{ mm}^4$.
   • $\bar{I}_{2} = \frac{150 \times 20^3}{12} = \frac{150 \times 8000}{12} = 100,000\text{ mm}^4$.
4. Determine distance $d_i = |y_i - \bar{y}|$ between sub-part centroids and section centroid:
   • $d_1 = |90 - 135.45| = 45.45\text{ mm} \implies d_1^2 = 2065.70\text{ mm}^2$.
   • $d_2 = |190 - 135.45| = 54.55\text{ mm} \implies d_2^2 = 2975.70\text{ mm}^2$.
5. Apply Parallel Axis Theorem ($I_x = \sum (\bar{I}_i + A_i d_i^2)$):
   • $I_1 = 9,720,000 + (3600 \times 2065.70) = 9,720,000 + 7,436,520 = 17,156,520\text{ mm}^4$.
   • $I_2 = 100,000 + (3000 \times 2975.70) = 100,000 + 8,927,100 = 9,027,100\text{ mm}^4$.
   • Total $\bar{I}_x = 17,156,520 + 9,027,100 = 26,183,620\text{ mm}^4 = 2.618 \times 10^7\text{ mm}^4$.