COE2001 Review Material

1. COE2001 Review Material • Basic equilibrium equations are from Physics I • Reinforce fundamental understanding of force & moments • Techniques for applying to mechanical problems • Applications • Rigid body problems • Trusses • Multi-force members: axial, shear forces & bending moments • Machines • Problems with friction (more complex equilibrium conditions) • Calculation of centroids • 1D (wires), 2D (areas) and 3D (volumes) • Distributed forces and their resultants COE 2001 Statics

2. Particle Equilibrium F2 F1 F3 Finite Body Equilibrium F2 F1 M1 F3 where O FA rOA • FA can act anywhere along “line of action” • +θ rotates rOA into FA A q d and COE 2001 Statics

3. Moments • Moment of a force about a line (e.g., a hinge) is the projection of M on to the line. • Only the component of F perpendicular to contributes to the moment. • M due to F is equal to M due to all components of F. • To compute M @ Q given M @ P and Fi use: • Couple: moment developed by equal and opposite forces separated by d; direction is perpendicular to both F and d. Since SFi=0 we see that SMQ=0+MP so a couple produces a moment that is constant Equipollent Systems Two force & moment systems are equipollent (“equal power”) if they have the same resultant at any point. Notes: • Resultant is defined as SF and SM • For rigid bodies, equipollent systems produce the same results on the body • Must satisfy: for any (and all) points, P, so we can pick any point for test and making equivalent COE 2001 Statics

4. Simplest Resultant FR Definition: an equipollent system consisting of only a force and NO moment. FR ? Q Q Q Exists for: MR P P P • Concurrent forces (where MP=0) • Co-planar forces • Parallel forces • 3D: Screwdriver (wrench) concept Distributed Loads • Forces per unit volume: gravity, acceleration, electromagnetic… • Forces per unit area: pressures • Forces per unit length: line loads in 2D problems • Concentrated forces: defined in the limit as area → 0 Line Loads: q(x) r0x Fr Mr0 Fr x x x dx Ex: b/2 q0 q0b x x b q0 b/3 ½q0b x x b COE 2001 Statics

5. Notes: General Equilibrium • 3D problem: 6 equations2D problem: 3 equations • # unknowns = # equations → determinate (SOLVABLE)# unknowns > # equations → indeterminate (many solutions)# unknowns < # equations → mechanism • Free body diagram (FBD) to reveal and isolate forces & moments • Equilibrium must apply to ANY and to ALL FBD’s • Constraints: create reactions (pin, slot, normal force, rotating disk, rope & pulley, etc) Procedure: • Idealize loads, constraints and define reactions • Determine solvability:- determinate or indeterminate- external vs internal redundancy • Create appropriate FBD’s • Solve equilibrium equations (Note: FBD’s simply help you decouple and solve equilibrium equations in the simplest way. This is not a unique process and many ways are possible. Use all of the above techniques for working with forces and moments.) Applications: • Bars: 1D prismatic members supporting only coaxial forces through end pins- forces & loads are applied only at ends- no lateral (transverse) loads • Beams: 1D prismatic members carrying lateral loads- loads applied in a single plane (planar bending)- beam-column combines beam and bar behavior COE 2001 Statics

6. Trusses (application of bars) FAB B A B This eqn. is a necessary but not sufficient condition FBD FBC SolvabilityEquations: 2J (3D = 3J)Unknowns: M (# members)Reactions: R M+R=2J (=3J)= determinate> indeterminate (redundants exist)< mechanism Q1 C D Q2 Method of Joints Method of Sections (use when only one or two member forces are needed) • Label all joints • Find external reactions • Formulate a joint equilibrium equation • Solve for unknown member force • Repeat from #3 until done… • Label all joints • Cut to construct a FBD to expose bar forces and eliminate other unknown forces and reactions • Solve for bar force (usually using 0=SM) Tips • Force can act anywhere along its line of action (pick location to avoid added moment calculations). • To avoid including a bar force, sum forces in direction  to that bar force. • Consider using virtual points for 0=SM to eliminate bar forces acting through that point. COE 2001 Statics

7. y x Centroids & Centers of Mass where Q is either: (a) volume (V), (b) area (A), or (c) line (L) Basic definition of centroid: Area: y=f(x) y=f(x) y2 y x1 y y=g(x) y=g(x) dy x2 y1 x x dx Alternate method for computing y-bar: y=f(x) y2 y y=g(x) y* y1 x dx Composite Areas: A2 A3 A1 COE 2001 Statics

8. Friction • Model: • Friction force depends only on surfaces (m) and normal force (N) • Magnitude cannot exceed Fs before motion and Fk during motion • Direction always opposes motion (opposite to relative velocity) • Will lock up if force magnitude drops below Fk F W Fs Static friction: Fm=msN Fk P Fm Kinetic friction: Fm=mkN P N • Modes: • P=0: no friction force is developed; no motion • P=F<Fm: no motion but F is whatever is needed to maintain equilibrium (cannot use F=mN to compute F) • P=Fm=msN: motion is “impending” (use F=msN to compute friction force) • P>Fm: motion is present (problem must now include dynamic forces) Angle of Friction • Use: • Replace Fm and N with R acting at angle fs from N • Knowing ms allows calculation of fs • Particle equilibrium problem: R acts on line of action defined by fs • Valid at instant of impending motion only W W P P R fs Fm N General Solution Approach: Case A: All forces are specified and all reactions & internal forces to maintain equilibrium are computed. If any forces that must develop from friction are above Fm in magnitude, then motion occurs, otherwise we have a simple static solution. Note that this means F<Fm=msN. Case B: Motion is impending and we find critical values for ms or for the geometry of the problem (angles, lengths) Case C: Determine valuse of a load necessary to cause motion. For this case, we must choose directions for friction forces (opposing impending motion) and at least one friction force is at the limit value. COE 2001 Statics