Summer 2005COE 2001 Statics1 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

Summer 2005COE 2001 Statics2 Particle Equilibrium F1F1 F2F2 F3F3 Finite Body Equilibrium F1F1 F2F2 F3F3 M1M1 where r OA FAFA O A d  F A can act anywhere along “line of action” +θ rotates r OA into F A and

Summer 2005COE 2001 Statics3 Moments 1.Moment of a force about a line (e.g., a hinge) is the projection of M on to the line. 2.Only the component of F perpendicular to contributes to the moment. 3. M due to F is equal to M due to all components of F. 4.To compute Q given P and F i use: 5.Couple: moment developed by equal and opposite forces separated by d; direction is perpendicular to both F and d. Since  F i =0 we see that  M Q =0+M P 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. 1.Resultant is defined as  F and  M 2.For rigid bodies, equipollent systems produce the same results on the body 3.Must satisfy: for any (and all) points, P, so we can pick any point for test and making equivalent Notes:

Summer 2005COE 2001 Statics4 Simplest Resultant Definition: an equipollent system consisting of only a force and NO moment. 1.Concurrent forces (where M P =0) 2.Co-planar forces 3.Parallel forces 4.3D: Screwdriver (wrench) concept Exists for: P Q P Q P Q FRFR MRMR FRFR ? Distributed Loads 1.Forces per unit volume: gravity, acceleration, electromagnetic… 2.Forces per unit area: pressures 3.Forces per unit length: line loads in 2D problems 4.Concentrated forces: defined in the limit as area → 0 Line Loads: dx x q(x) x M r0 FrFr x FrFr r 0x Ex: x q0q0 b x q0bq0b b/2 x q0q0 b x ½q0b½q0b b/3

Summer 2005COE 2001 Statics5 General Equilibrium 1.3D problem: 6 equations 2D problem: 3 equations 2.# unknowns = # equations → determinate (SOLVABLE) # unknowns > # equations → indeterminate (many solutions) # unknowns < # equations → mechanism 3.Free body diagram (FBD) to reveal and isolate forces & moments 4.Equilibrium must apply to ANY and to ALL FBD’s 5.Constraints: create reactions (pin, slot, normal force, rotating disk, rope & pulley, etc) Notes: 1.Idealize loads, constraints and define reactions 2.Determine solvability: - determinate or indeterminate - external vs internal redundancy 3.Create appropriate FBD’s 4.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.) Procedure: Applications: 1.Bars: 1D prismatic members supporting only coaxial forces through end pins - forces & loads are applied only at ends - no lateral (transverse) loads 2.Beams: 1D prismatic members carrying lateral loads - loads applied in a single plane (planar bending) - beam-column combines beam and bar behavior

Summer 2005COE 2001 Statics6 Trusses (application of bars) A B D C Q1Q1 Q2Q2 B F AB F BD F BC Solvability Equations:2J (3D = 3J) Unknowns:M (# members) Reactions:R M+R=2J (=3J) = determinate > indeterminate (redundants exist) < mechanism BarJoint + TensionTension = away -CompressionCompression = towards This eqn. is a necessary but not sufficient condition Method of Joints 1.Label all joints 2.Find external reactions 3.Formulate a joint equilibrium equation 4.Solve for unknown member force 5.Repeat from #3 until done… Tips 1.Force can act anywhere along its line of action (pick location to avoid added moment calculations). 2.To avoid including a bar force, sum forces in direction  to that bar force. 3.Consider using virtual points for 0=  M to eliminate bar forces acting through that point. Method of Sections (use when only one or two member forces are needed) 1.Label all joints 2.Cut to construct a FBD to expose bar forces and eliminate other unknown forces and reactions 3.Solve for bar force (usually using 0=  M)

Summer 2005COE 2001 Statics7 Centroids & Centers of Mass Basic definition of centroid: where Q is either: (a) volume (V), (b) area (A), or (c) line (L) Area: y=f(x) y=g(x) dx x y y2y2 y1y1 y=f(x) y=g(x) x y x2x2 x1x1 y=f(x) y=g(x) dx x y y2y2 y1y1 y* Alternate method for computing y-bar: Composite Areas: x y dy A1A1 A2A2 A3A3

Summer 2005COE 2001 Statics8 Friction Static friction: F m =  s N Model: Friction force depends only on surfaces (m) and normal force (N) Magnitude cannot exceed F s before motion and F k during motion Direction always opposes motion (opposite to relative velocity) Will lock up if force magnitude drops below F k Angle of Friction FsFs FkFk Modes: 1.P=0: no friction force is developed; no motion 2.P=F<F m : no motion but F is whatever is needed to maintain equilibrium (cannot use F=  N to compute F) 3.P=F m =  s N: motion is “impending” (use F=  s N to compute friction force) 4.P>F m : motion is present (problem must now include dynamic forces) P F FmFm N Kinetic friction: F m =  k N W P FmFm N W P W P R ss Use: Replace F m and N with R acting at angle  s from N Knowing  s allows calculation of  s Particle equilibrium problem: R acts on line of action defined by  s Valid at instant of impending motion only 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 F m in magnitude, then motion occurs, otherwise we have a simple static solution. Note that this means F<F m =  s N. Case B: Motion is impending and we find critical values for  s 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.