1.4 Statics, Dynamics & Mechanical Equilibrium

Statics, Dynamics & Mechanical Equilibrium

Statics and dynamics form the mechanical bedrock for machine design, structural frame integrity, mechanism kinematics, and vibration analysis. The PE Mechanical exam tests both foundational statics (equilibrium, trusses, friction, centroids) and planar rigid-body dynamics (kinematics, kinetics, work-energy, impulse-momentum).

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1. Free-Body Diagrams & Static Equilibrium

A Free-Body Diagram (FBD) isolates a body or subsystem from its surroundings and depicts all external forces, applied moments, and support reaction forces.

+-----------------------------------------------------------------------------------------+
|                           COMMON 2D SUPPORT REACTIONS                                   |
|                                                                                         |
|   SUPPORT TYPE            DEGREES OF FREEDOM CONSTRAINED     REACTION FORCES/MOMENTS    |
|   --------------------    ------------------------------     -----------------------    |
|   1. Roller / Rocker      Normal translation                 1 Force (perpendicular)    |
|   2. Frictionless Guide   Transverse translation             1 Force (normal to slot)   |
|   3. Pinned Joint / Hinge Horizontal & Vertical translation  2 Forces (R_x, R_y)        |
|   4. Fixed / Built-In     All translations & rotation        2 Forces + 1 Moment (M_R)  |
|   5. Flexible Cable / Rod Line of action along cable         1 Tension Force (axial)    |
+-----------------------------------------------------------------------------------------+

Equations of Static Equilibrium

For a rigid body in static equilibrium in a 2D plane:

$\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M_O = 0 \quad (\text{about any arbitrary point } O)$

For 3D spatial equilibrium:

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

$\sum M_x = 0, \quad \sum M_y = 0, \quad \sum M_z = 0$

Two-Force and Three-Force Members

• Two-Force Members: A structural component with pin joints at both ends and no intermediate loads or moments applied along its length. The reaction forces at both pins must be equal in magnitude, opposite in direction, and collinear along the line connecting the two pins.
• Three-Force Members: A body subjected to forces applied at only three points. For equilibrium, the lines of action of the three forces must be concurrent (intersect at a single point) or parallel.

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2. Truss Analysis: Joints, Sections & Zero-Force Members

Trusses are pin-connected frameworks where all members act as axial two-force members (tension or compression).

+-----------------------------------------------------------------------------------------+
|                              TRUSS ANALYSIS METHODOLOGIES                               |
|                                                                                         |
|   METHOD OF JOINTS                                   METHOD OF SECTIONS                 |
|   - Isolates individual joint pins.                  - Cuts an imaginary plane through  |
|   - Concurrent force system (Sum F_x=0, Sum F_y=0).    at most 3 unknown members.       |
|   - Ideal for finding all member forces.             - Solves directly for interior     |
|   - Solves 2 unknown forces per joint.                 chords using Sum M = 0.          |
+-----------------------------------------------------------------------------------------+

Rapid Visual Inspection Rules for Zero-Force Members

Identifying zero-force members by visual inspection eliminates unnecessary equations:

1. Two-Member Unloaded Joint: If two non-collinear members form an unloaded pin joint, both members carry zero force ($F_1 = 0, F_2 = 0$).
2. Three-Member Joint with Two Collinear: If three members meet at an unloaded joint where two members are collinear, the third (non-collinear) member carries zero force, and the two collinear members carry equal forces.
3. Loaded Two-Member Joint: If two members meet at a joint with an external load aligned collinearly with one member, the other non-collinear member is a zero-force member.

+-----------------------------------------------------------------------------------------+
|                         ZERO-FORCE MEMBER INSPECTION EXAMPLES                           |
|                                                                                         |
|     Rule 1: Unloaded 2-Bar Joint          Rule 2: 3-Bar Joint (2 Collinear, Unloaded)   |
|               (Pin Joint A)                             (Pin Joint B)                   |
|                 /    \                                     | Member 3 = 0               |
|                /      \                                    v                            |
|     Member 1  /        \ Member 2               -----------+-----------                 |
|              v          v                       Member 1       Member 2                 |
|           F1 = 0     F2 = 0                     (F1 = F2, Member 3 is Zero-Force)       |
+-----------------------------------------------------------------------------------------+

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3. Coulomb Friction & Belt/Capstan Friction

Coulomb Dry Friction & Impending Slip

For two dry surfaces in contact with normal force $N$:

• Static Equilibrium (No slip): $F_f \le \mu_s N$
• Impending Motion / Maximum Static Friction: $F_{f,\max} = \mu_s N$
• Kinetic (Dynamic) Motion: $F_k = \mu_k N \quad (\text{where } \mu_k < \mu_s)$

$\text{Angle of Static Friction: } \tan \phi_s = \mu_s$

Belt and Capstan Friction (Eytelwein's Formula)

For a flexible belt, rope, or band wrapped around a curved drum with friction coefficient $\mu$ and total wrap angle $\beta$:

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

Where:

• $T_2 = \text{tension on the tight side (side resisting or pulling motion)}$
• $T_1 = \text{tension on the slack side } (T_2 > T_1)$
• $\mu = \text{coefficient of friction between belt and drum}$
• $\beta = \text{total contact angle in } \mathbf{radians} \quad (\beta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180})$

$\text{Braking Torque: } \tau_{\text{brake}} = (T_2 - T_1) R$

$\text{Power Transmitted: } P = (T_2 - T_1) v = (T_2 - T_1) (\omega R)$

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4. Centroids, Area & Mass Moments of Inertia

Composite Centroid Calculation

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

Parallel Axis Theorem

• Area Moment of Inertia (Bending Rigidity):

$I_x = \bar{I}_{x'} + A d_y^2, \qquad I_y = \bar{I}_{y'} + A d_x^2$

• Mass Moment of Inertia (Rotational Inertia):

$I_{\text{axis}} = I_G + m d^2$

Where $I_G$ is the centroidal mass moment of inertia, $m$ is body mass, and $d$ is the perpendicular distance between the centroidal axis and the parallel axis of rotation.

+-----------------------------------------------------------------------------------------+
|                        COMMON MASS MOMENTS OF INERTIA (I_G)                             |
|                                                                                         |
|   SHAPE GEOMETRY                    AXIS OF ROTATION          FORMULA                   |
|   ------------------------------    ---------------------     -----------------------   |
|   Slender Rod (Length L)            Through Center (G)        I_G = (1/12) * m * L^2    |
|   Slender Rod (Length L)            Through End Pivot         I_end = (1/3) * m * L^2   |
|   Solid Cylinder / Disk (Radius R)  Central Polar Axis (G)    I_G = (1/2) * m * R^2     |
|   Thin Cylindrical Shell (Radius R) Central Polar Axis (G)    I_G = m * R^2             |
|   Solid Sphere (Radius R)           Through Center (G)        I_G = (2/5) * m * R^2     |
|   Thin Spherical Shell (Radius R)   Through Center (G)        I_G = (2/3) * m * R^2     |
+-----------------------------------------------------------------------------------------+

$\text{Radius of Gyration: } k = \sqrt{\frac{I}{m}} \implies I = m k^2$

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5. Kinematics of Particles and Rigid Bodies

+-----------------------------------------------------------------------------------------+
|                            PLANAR KINEMATICS EQUATIONS                                  |
|                                                                                         |
|   RECTILINEAR (Constant a)             ANGULAR / ROTATIONAL (Constant alpha)            |
|   v = v_0 + a * t                      omega = omega_0 + alpha * t                      |
|   s = s_0 + v_0*t + 0.5*a*t^2          theta = theta_0 + omega_0*t + 0.5*alpha*t^2      |
|   v^2 = v_0^2 + 2*a*(s - s_0)          omega^2 = omega_0^2 + 2*alpha*(theta - theta_0)  |
|                                                                                         |
|   CURVILINEAR NORMAL-TANGENTIAL ACCELERATION:                                           |
|   Tangential: a_t = dv/dt = r * alpha  (changes speed)                                  |
|   Normal:     a_n = v^2 / rho = r * omega^2  (changes direction toward center)          |
|   Total Acceleration: a = sqrt(a_t^2 + a_n^2)                                           |
+-----------------------------------------------------------------------------------------+

Relative Velocity in Rigid Bodies

For points $A$ and $B$ located on the same rigid body rotating with angular velocity $\boldsymbol{\omega}$:

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

$\mathbf{a}_B = \mathbf{a}_A + \boldsymbol{\alpha} \times \mathbf{r}_{B/A} - \omega^2 \mathbf{r}_{B/A}$

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6. Planar Kinetics: Force-Acceleration, Work-Energy & Impulse

+-----------------------------------------------------------------------------------------+
|                          THREE KINETIC FORMULATION APPROACHES                           |
|                                                                                         |
|   1. NEWTON'S 2ND LAW (Forces & Accelerations at an instant)                            |
|      Sum F_x = m * a_Gx,    Sum F_y = m * a_Gy,    Sum M_G = I_G * alpha                |
|      (About fixed pivot O: Sum M_O = I_O * alpha)                                       |
|                                                                                         |
|   2. WORK-ENERGY THEOREM (Relates Forces to Velocity over a Distance)                   |
|      T_1 + Sum U_{1->2} = T_2                                                           |
|      Kinetic Energy: T = 0.5 * m * v_G^2 + 0.5 * I_G * omega^2                          |
|      Spring Work: U_spring = -0.5 * k * (x_2^2 - x_1^2)                                 |
|      Gravity Work: U_grav = -m * g * (y_2 - y_1)                                        |
|                                                                                         |
|   3. IMPULSE-MOMENTUM (Relates Forces to Velocity over a Time Interval)                 |
|      Linear:  m * v_1 + Integral(Sum F dt) = m * v_2                                    |
|      Angular: I_G * omega_1 + Integral(Sum M_G dt) = I_G * omega_2                      |
|      (About fixed pivot O: I_O * omega_1 + Integral(Sum M_O dt) = I_O * omega_2)        |
+-----------------------------------------------------------------------------------------+

Direct Central Impact and Coefficient of Restitution ($e$)

For colliding bodies $A$ and $B$ along their line of impact:

$\text{Conservation of Linear Momentum: } m_A v_{A1} + m_B v_{B1} = m_A v_{A2} + m_B v_{B2}$

$\text{Coefficient of Restitution: } e = \frac{v_{B2} - v_{A2}}{v_{A1} - v_{B1}}$

• Elastic Impact ($e = 1.0$): Total kinetic energy is conserved.
• Plastic / Inelastic Impact ($e = 0$): Bodies stick together ($v_{A2} = v_{B2}$), maximizing energy dissipation.

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7. Step-by-Step Worked Problem: Industrial Band Brake

Problem: A band brake holds a rotating drum of diameter $D = 24 \text{ in.}$ ($R = 1.0 \text{ ft}$). The coefficient of friction between the band and drum is $\mu = 0.35$. The band wraps around the drum by an angle of $\theta = 225^\circ$. The drum is rotating counterclockwise and transmitting a torque of $\tau = 450 \text{ ft}\cdot\text{lbf}$.

Calculate the required slack side tension $T_1$ and tight side tension $T_2$ to hold the load.

+-----------------------------------------------------------------------------------------+
|                        BAND BRAKE STEP-BY-STEP SOLUTION                                 |
|                                                                                         |
|   STEP 1: Convert Wrap Angle to Radians                                                 |
|           beta = 225 deg * (pi / 180 deg) = 3.9270 radians                              |
|                                                                                         |
|   STEP 2: Compute Capstan Friction Ratio (T2 / T1)                                      |
|           T2 / T1 = e^(mu * beta) = e^(0.35 * 3.9270) = e^(1.37445) = 3.9529            |
|           T2 = 3.9529 * T1                                                              |
|                                                                                         |
|   STEP 3: Relate Torque to Belt Tensions                                                |
|           Torque: tau = (T2 - T1) * R                                                   |
|           450 ft*lbf = (3.9529 * T1 - T1) * (1.0 ft)                                    |
|           450 = 2.9529 * T1                                                             |
|                                                                                         |
|   STEP 4: Solve for Tensions                                                            |
|           Slack Side Tension T1 = 450 / 2.9529 = 152.39 lbf                             |
|           Tight Side Tension T2 = 3.9529 * 152.39 = 602.39 lbf                          |
|                                                                                         |
|   CHECK: tau = (602.39 - 152.39) * 1.0 = 450.0 ft*lbf (Verified).                      |
+-----------------------------------------------------------------------------------------+

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