Engineering Mechanics (Hibbeler)

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1. General Principles Newton's First Law: A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this state provided the particle is not subjected to an unbalanced force. Newton's Second Law: A particle acted upon by an unbalanced force $\vec{F}$ experiences an acceleration $\vec{a}$ that has the same direction as the force and a magnitude that is directly proportional to the force. $ \vec{F} = m\vec{a} $ Newton's Third Law: The mutual forces of action and reaction between two particles are equal, opposite, and collinear. Newton's Law of Gravitational Attraction: $ F = G \frac{m_1 m_2}{r^2} $ where $G = 66.73 \times 10^{-12} \text{ m}^3 / (\text{kg} \cdot \text{s}^2)$ Weight: $ W = mg $ 2. Force Vectors 2.1. Scalar and Vector Quantities Scalar: Mass, volume, length, time. Vector: Force, velocity, position, moment. 2.2. Vector Operations Vector Addition (Parallelogram Law): Head-to-tail rule. $\vec{R} = \vec{A} + \vec{B}$ Vector Subtraction: $\vec{R}' = \vec{A} - \vec{B} = \vec{A} + (-\vec{B})$ Resolution of a Vector: Resolving a vector into components. 2.3. Cartesian Vectors Unit Vector: $\vec{u}_A = \frac{\vec{A}}{|\vec{A}|}$ Cartesian Representation: $\vec{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$ Magnitude: $ |\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2} $ Direction Cosines: $ \cos \alpha = \frac{F_x}{F}, \cos \beta = \frac{F_y}{F}, \cos \gamma = \frac{F_z}{F} $ $ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $ Position Vector: $\vec{r} = (x_B - x_A)\hat{i} + (y_B - y_A)\hat{j} + (z_B - z_A)\hat{k}$ Force Vector along a Line: $\vec{F} = F \cdot \vec{u} = F \left( \frac{\vec{r}}{|\vec{r}|} \right)$ 2.4. Dot Product $ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta $ $ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z $ Angle between Vectors: $ \theta = \cos^{-1} \left( \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \right) $ Projection: $ A_B = |\vec{A}| \cos \theta = \vec{A} \cdot \vec{u}_B $ 3. Equilibrium of a Particle Free-Body Diagram (FBD): Essential for solving equilibrium problems. Equations of Equilibrium: 2D: $ \sum F_x = 0, \sum F_y = 0 $ 3D: $ \sum F_x = 0, \sum F_y = 0, \sum F_z = 0 $ Spring Force: $F = ks$ (Hooke's Law), where $k$ is stiffness, $s$ is deformation. 4. Force System Resultants 4.1. Cross Product $ \vec{C} = \vec{A} \times \vec{B} = (A_y B_z - A_z B_y)\hat{i} - (A_x B_z - A_z B_x)\hat{j} + (A_x B_y - A_y B_x)\hat{k} $ Magnitude: $ |\vec{C}| = |\vec{A}| |\vec{B}| \sin \theta $ Direction: Right-hand rule. 4.2. Moment of a Force (Torque) Scalar Formulation (2D): $ M_O = Fd $ ($d$ is perpendicular distance). Vector Formulation: $ \vec{M}_O = \vec{r} \times \vec{F} $ $\vec{r}$ is position vector from point $O$ to any point on line of action of $\vec{F}$. Moment about an Axis: $ M_{ax} = \vec{u}_{ax} \cdot (\vec{r} \times \vec{F}) $ (Mixed triple product) Varignon's Theorem: Moment of a force about a point is equal to the sum of the moments of its components about the same point. $ \vec{M}_O = \vec{r} \times (\vec{F}_1 + \vec{F}_2) = \vec{r} \times \vec{F}_1 + \vec{r} \times \vec{F}_2 $ 4.3. Couple Two parallel forces, equal in magnitude, opposite in direction, separated by a perpendicular distance $d$. Magnitude: $ M = Fd $ Vector: $ \vec{M} = \vec{r} \times \vec{F} $ (where $\vec{r}$ connects points on $F$ and $-F$) 4.4. Reduction of a Simple Distributed Loading Resultant force $F_R$ equals area under the loading curve. Location $\bar{x}$ is centroid of the area. 5. Equilibrium of a Rigid Body Equations of Equilibrium: 2D: $ \sum F_x = 0, \sum F_y = 0, \sum M_O = 0 $ 3D: $ \sum \vec{F} = 0 \Rightarrow \sum F_x = 0, \sum F_y = 0, \sum F_z = 0 $ $ \sum \vec{M}_O = 0 \Rightarrow \sum M_x = 0, \sum M_y = 0, \sum M_z = 0 $ Supports and Their Reactions: Type of Support Number of Unknowns Reactions Cable, Rope, Link 1 Force along cable/link Smooth Surface 1 Normal force perpendicular to surface Roller, Rocker 1 Normal force perpendicular to surface Pin (2D) 2 Two force components Fixed Support (2D) 3 Two force components, one moment Ball-and-Socket (3D) 3 Three force components Fixed Support (3D) 6 Three force components, three moments 6. Trusses, Frames, and Machines 6.1. Trusses Members are two-force members (only axial forces). Method of Joints: Apply particle equilibrium ($\sum F_x=0, \sum F_y=0$) at each pin joint. Method of Sections: Cut through members, apply rigid body equilibrium ($\sum F_x=0, \sum F_y=0, \sum M=0$) to a section. Zero-Force Members: If only two non-collinear members form a truss joint and no external load or reaction is applied to the joint, the members are zero-force members. If three members form a truss joint for which two of the members are collinear, the third member is a zero-force member, provided no external force or reaction is applied. 6.2. Frames and Machines Contain at least one multi-force member. Procedure: Draw FBD of the entire structure (if possible) to find external reactions. Disassemble the frame/machine into its component parts. Draw FBD for each part, showing all internal forces (equal and opposite on connecting members). Apply equilibrium equations to each part. 7. Internal Forces Axial Force (N): Perpendicular to cross-section. Shear Force (V): Tangent to cross-section. Bending Moment (M): Causes bending. Sign Convention: Axial: Tension (+), Compression (-). Shear: Up on right face (+), Down on right face (-). Moment: Causes compression in top fibers (+), Tension in top fibers (-). Relations between Load, Shear, and Moment: $ \frac{dV}{dx} = w(x) $ (load per unit length) $ \frac{dM}{dx} = V(x) $ 8. Friction Static Friction ($F_s$): Opposes impending motion. $ F_s \le \mu_s N $ Kinetic Friction ($F_k$): Opposes relative motion. $ F_k = \mu_k N $ Typically, $ \mu_s > \mu_k $. Angle of Static Friction ($\phi_s$): $ \tan \phi_s = \mu_s = F_s/N $ Angle of Repose: Maximum angle of inclination before an object slides. Wedges: Used to increase applied force or slightly adjust position. Analyze using FBDs and friction equations. Journal Bearings: Create a friction couple if the shaft slips. Thrust Bearings & Disks: Friction moment depends on pressure distribution. Uniform pressure: $ M = \frac{2}{3} \mu P \frac{R_2^3 - R_1^3}{R_2^2 - R_1^2} $ Uniform wear: $ M = \frac{1}{2} \mu P (R_1 + R_2) $ Belt Friction: $ T_2 = T_1 e^{\mu \beta} $ (where $\beta$ is contact angle in radians) 9. Center of Gravity & Centroid Center of Gravity (CG): Point where entire weight of body acts. $ \bar{x} = \frac{\sum \tilde{x}W}{\sum W} $, etc. Centroid: Geometric center of an area or volume. $ \bar{x} = \frac{\sum \tilde{x}A}{\sum A} $ for area, etc. Composite Bodies: Divide into simple shapes, find centroid/CG of each, then use summation formulas. Theorems of Pappus and Guldinus: Area of Surface of Revolution: $ A = \theta \bar{r} L $ (where $\theta$ is angle in radians) Volume of Revolution: $ V = \theta \bar{r} A $ 10. Moments of Inertia Area Moment of Inertia: $ I_x = \int y^2 dA $ $ I_y = \int x^2 dA $ Polar Moment of Inertia: $ J_O = \int r^2 dA = I_x + I_y $ Parallel-Axis Theorem: $ I = \bar{I} + Ad^2 $ (where $\bar{I}$ is moment of inertia about centroidal axis, $d$ is distance between parallel axes). Radius of Gyration: $ k = \sqrt{I/A} $ Composite Areas: Sum $I$ for each part using parallel-axis theorem. Mass Moment of Inertia: $ I = \int r^2 dm $ Parallel-Axis Theorem: $ I = \bar{I} + md^2 $ 11. Virtual Work Principle of Virtual Work: If a body is in equilibrium, the total virtual work done by all the forces acting on the body is zero for any virtual displacement. $ \delta U = 0 $ $ \sum \delta U = \sum F \delta s \cos \theta + \sum M \delta \theta = 0 $ Conservative Forces: Work done is independent of path (e.g., gravity, spring force). Gravitational Potential Energy: $ V_g = Wy $ Elastic Potential Energy: $ V_e = \frac{1}{2} ks^2 $ Potential Energy Criterion for Equilibrium: Stable: $ \frac{d^2V}{ds^2} > 0 $ Unstable: $ \frac{d^2V}{ds^2} Neutral: $ \frac{d^2V}{ds^2} = 0 $