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Summer 2005COE 2001 Statics1 COE2001 Review Material Basic equilibrium equations are from Physics I –Reinforce fundamental understanding of force & moments.

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Presentation on theme: "Summer 2005COE 2001 Statics1 COE2001 Review Material Basic equilibrium equations are from Physics I –Reinforce fundamental understanding of force & moments."— Presentation transcript:

1 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

2 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

3 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 M @ Q given M @ 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:

4 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

5 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

6 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)

7 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

8 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.


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