MIT - 16.001/16.002 Fall, 2008 Unit M1.2 (All About) Mechanical Equilibrium Readings: CDL 1.6 16.001/002 -- “Unified Engineering” Department of Aeronautics and Astronautics Massachusetts Institute of Technology Paul A. Lagace © 2007 MIT - 16.001/16.002 Fall, 2008 LEARNING OBJECTIVES FOR UNIT M1.2 Through participation in the lectures, recitations, and work associated with Unit M1.2, it is intended that you will be able to……… • ….apply the principle/concept of equilibrium to determine the applied and transmitted forces and moments, and related motion, for a particle, set of particles, or body • ….model a body/system and external forces and moments acting on such • ….apply the concept of equipollent force systems to model a set of forces Paul A. Lagace © 2007 Unit M1.2 - p. 2 MIT - 16.001/16.002 Fall, 2008 The concept of equilibrium is the first of the 3 Great Concepts of solid mechanics we will consider. For a structure, need to consider the general case. But let’s build up to this by considering the simplest case. Equilibrium of a particle A particle is a body whose mass is concentrated at a point Figure M1.2-1 Forces acting on a particle produce acceleration (4) _F (1) _F _a (2) _F (3) _F Newton’s law gives: ∑ F(i) = ma i Paul A. Lagace © 2007 Unit M1.2 - p. 3 MIT - 16.001/16.002 Fall, 2008 Rewrite this as: ∑ F(i) − ma = 0 i inertial external forces forces --> “D’Alembert’s Principle” - ma treated as “inertial force” --> In many cases in structures, we have: a = 0 (Statics) So this reduces to: ∑ F(i) = 0 using vector notation: (i) (i ) ∑ F = ∑Fm im = 0 This can be written as: (i) (Static) Equilibrium ∑ Fm = 0 i Paul A. Lagace © 2007 Unit M1.2 - p. 4 MIT - 16.001/16.002 Fall, 2008 which represents three equations: (i) ∑ F1 = 0 sum of forces in x1 = 0 i 3 scalar (i) ∑ F2 = 0 sum of forces in x2 = 0 equations i F(i) = 0 ∑ 3 sum of forces in x3 = 0 i --> Example of static equilibrium versus general case: Elevator…find cable tension Figure M1.2-2 Case of no motion cable 500 kg Model as point mass: box x3 tension in T cable 500 kg m g = W Paul A. Lagace © 2007 Unit M1.2 - p. 5 MIT - 16.001/16.002 Fall, 2008 ∑ F = 0 ⇒ + (give positive direction) m ⇒ T − 500kg (9.8 ) = 0 s 2 ⇒ T = 4900 N (Note: Work/answers not correct without units!) 2 Figure M1.2-3 Case of accelerating upwards at 3 m/s 2 3 m / s x cable 3 T 500 kg box 500 kg m a m g = W 2 inertial force (a = 3 m/s ) in opposite direction from acceleration Same model but now also model acceleration as inertial force Paul A. Lagace © 2007 Unit M1.2 - p. 6 MIT - 16.001/16.002 Fall, 2008 ∑F = 0 m m T − 500kg (9.8 ) − 500 3 = 0 s2 s 2 ⇒ T = 6400 N Let’s move up one level and consider… Equilibrium of a System of Particles Look at… (i) Figure M1.2-4 A system of particles with forces, P , acting on particle i (1) _P represents rigid physical (2) 1 _P connection between particles 2 (3) 3 _P Paul A. Lagace © 2007 Unit M1.2 - p. 7 MIT - 16.001/16.002 Fall, 2008 “Isolate” particle 1… --> What other forces act on particle 1 besides external force P (1) - reaction (internal) forces in connection - inertial forces So for particle 1 draw a “Free Body Diagram” (will discuss this more in Unit M1.5) Figure M1.2-5 Free Body Diagram for Particle 1 (1) _P (12) 1 R (13) _ R_ Notes: - R (ij) represents reaction in connection between particles i and j - Draw reactions initially in tension (+) (convenient convention). If they come out (–), they are in compression, but you “take care of this” through sign Paul A. Lagace © 2007 Unit M1.2 - p. 8 MIT - 16.001/16.002 Fall, 2008 - Newton’s law of action-reaction says R (ij) = -R (ji) (must point in opposite directions but are of equal magnitude) Figure M1.2-6 (ij) (ji) R R_ _ So apply equilibrium: (1) (12) (13) P + R + R = m1 a1 = 0 Statics Similarly for particles 2 and 3: Figure M1.2-7 Free Body Diagram for Particle 2 (2) (21) _P R_ 2 (23) R_ (2) (21) (23) P + R + R = m2 a2 = 0 Paul A. Lagace © 2007 Unit M1.2 - p. 9 MIT - 16.001/16.002 Fall, 2008 Figure M1.2-8 Free Body Diagram for Particle 3 (31) R_ (32) R_ 3 (3) _P (3) (31) (32) P + R + R = m3 a3 = 0 Recalling R (ij) = -R (ji) and summing these gives: (1) (2) (3) P + P + P = ∑mi ai = 0 i Statics For statics, this is simply: ∑ P(i) = 0 (as before) i --> Is this enough? We have dealt with no linear acceleration. What else can a system of particles/body do? ROTATE So there is a second set of equilibrium requirements….. Paul A. Lagace © 2007 Unit M1.2 - p. 10 MIT - 16.001/16.002 Fall, 2008 Moment Equilibrium For there to be no rotation, the moments about a (any) point/axis must be zero --> Why any? Pure moment…magnitude does not change as point changes! --> Consider each particle and the moments acting on it. Take moments about some “convenient” point: Figure M1.2-9 Moment for Particle 1 (1) position vector _P for particle 1 1 (13) (1) R_ r (12) _ R_ See that r (1) × (P(1) + R(12) + R(13)) = 0 Paul A. Lagace © 2007 Unit M1.2 - p. 11 MIT - 16.001/16.002 Fall, 2008 Similarly for particles 2 and 3: (2) (2) (21) (23) r (2) × (P(2) + R(21) + R(23)) = 0 Particle 2 ( ) (3 ) r (3) × (P(3) + R(31) + R(32) ) = 0 Particle 3 --> again use Newton’s Law of Action-Reaction (R(ij) = − R( ji) ) --> and we must further note that: Figure M1.2-10 Pairs Act Along the Same Line of Action (32) same line of action 2 R_ (23) 3 R_ 3 Thus, the perpendicular distance from the point about which we take the moment to the line of action is the same. So using both pieces of info: ( ) ( ) r (1) × R(13) + r 3 +x R 31 = 0 (⇒ pairs give equal but opposite moments!) Paul A. Lagace © 2007 Unit M1.2 - p. 12 MIT - 16.001/16.002 Fall, 2008 So take the three equilibrium equations, add them, and use this general concept: r (i) × R(ij) + r( j) × R( ji) = 0 to get: ∑r (n) × P(n) = 0 (+ any inertial n considerations) or ∑ M (n) = 0 moment on particle (n) due to force on (n) => Equilibrium requires the moments of the external forces to be zero Paul A. Lagace © 2007 Unit M1.2 - p. 13 MIT - 16.001/16.002 Fall, 2008 What about for a general body? Particles vs. Body/Equilibrium of a Body A body can be considered to be a system of particles. In the limit, a body is an infinite set of infinitesimal masses/particles. So can generalize the equations of equilibrium into two sets: General Equations of Equilibrium: (n) ∑ F = 0 Force Equilibrium n (n) (n) ∑r × F = 0 Moment Equilibrium n Paul A. Lagace © 2007 Unit M1.2 - p. 14 MIT - 16.001/16.002 Fall, 2008 Figure M1.2-11 Forces acting on a general body (2) (1) F _F _ (5) (3) _F _F (4) _F --> Deals with external forces only. --> Gives two necessary and sufficient conditions for equilibrium (6 scalar equations total) Let’s now consider these concepts in two-dimensions (2-D) so that we can more easily represent (and comprehend) them. In addition, many force systems can be well-represented as a: Planar Force System: This is a force system where all forces act in one plane. Paul A. Lagace © 2007 Unit M1.2 - p. 15 MIT - 16.001/16.002 Fall, 2008 Example: Airplane in straight and level flight Figure M1.2-12 A model of an airplane x L 3 W L T e x1 D X All forces in T x - x plane d 1 3 a b m g = W = center of gravity = aerodynamic center Forces are: X mg = W Weight Lw Lift at “aerodynamic center” of wing D Drag T Thrust LT Lift on tail (could be downwards for “trim”) This is a “Free Body Diagram” Paul A. Lagace © 2007 Unit M1.2 - p. 16 MIT - 16.001/16.002 Fall, 2008 Important dimensions/distances: a - distance thrust acts off c.g. line (in x3) b - distance drag acts off c.g. line (a-c to c-g in x3) e - distance wing lift acts off c.g. (a-c to c-g in x1) d - distance tail lift acts off c.g. (in x1) Write out equations of equilibrium: + ∑ F1 = 0 ← ⇒ T − D = 0 (A1) show + direction! F 0 0 0 (A2) ∑ 2 = = F 0 + L L - W 0 (A3) ∑ 3 = w + T = M 0 no forces out of x - x plane ∑ 1 = 1 3 ⇒ 0 = 0 (A4) M 0 Ta + Db - L e + L d = 0 (A5) ∑ 2 = w T M 0 0 0 (A6) Paul A. Lagace © 2007 ∑ 3 = = Unit M1.2 - p. 17 (no forces out of x1 - x3 plane) F 0 0 0 (A2) ∑ 2 = = FF3F = = 0 0 0 L0 0w 0 += 0 L0 0T - W = 0 (A2) (A3)(A2) (A2) ∑∑∑ 222== == MF = 0 noL forces+ L out- W of = x0 - x plane (A3) ∑∑F3F31 3= = 0 0 L wLw w+ + L TLT T- -W W = = 0 1 0 3 (A3) (A3) MIT - 16.001/16.002 ∑∑ ⇒ 0 = 0 (A4) Fall, 2008 MMM1 = == 0 0 0 no no no forces forces forces out out out of of of xxx1--x-xx3 plane plane plane ∑∑∑ 1 1 + 1 1 3 3 M 0 Ta + Db - L e + L d = 0 (A5) ∑ 2 = - ⇒⇒⇒ 0 0 0 === 0 0 0 w T (A4) (A4) (A4) use c.g.