Chapt 9.ppt

ENGR 107: Engineering Fundamentals Lecture 20: Chapter 9 Free Body Diagrams (FBD) Department of Electrical and Computer Engineering George Mason University http://eta.physics.uoguelph.ca/tutorials/fbd/intro.html http://www.physicsclassroom.com/Class/newtlaws/u2l2c.html References Free Body Diagrams Free Body Diagrams - one of the most important tools in structural engineering FBD: drawing of a ?body,? a structure or portion - showing all the forces acting on it. First step in any structural analysis Drawing FBDs Isolate the body of interest and draw the outline of the structure Place all external force vectors (or loads) for that body on the FBD At the location of each support, draw and label the appropriate reactions Draw the x-y coordinate system Indicate all relevant dimensions Free-Body Diagrams A Free-body diagram (FBD) is a sketch of the body and all forces acting on the body. The body is cut away from all supports and connectors. Only the forces acting on the body are shown. Inclined plane, no friction A force normal to the plane through the center of gravity of the body Cable, chain or rope A tension force along the cable away from or pulling the body. Free-Body Diagrams Examples Roller or ball A force perpendicular to the surface on which the roller could roll. A reaction at some angle, usually unknown; components are therefore indicated on FBD; directions of the force components are assumed and may need correction after solving. Mass/earth Mass multiplied by the earth?s gravitational constant produces a force directed toward the center of the earth Hinge or pin Equilibrium ? Fx = 0 ? Fy = 0 ? M = 0 Conventions for this class Fx is positive to the right (?). Fy is positive upward ( ). M is positive CCW ( ). Class Example 220 pound mass block sitting on the ground. Find the Reaction force on the block in newtons. 220 lbm Convert lbm to kg 1 kg = 2.2 lbm therefore, F = mgL/gC F = 980.7 N F = 980.7 N RA ? Fx = 0 ? Fy = 0 ? M = 0 RA = 980.7 N = -980.7 + RA F = 100 kg * 9.807 m/s2 1 kg·m/N·s2 x y FBD = 100 kg 1 kg 2.2 lbm 220 lbm Example Problem 9.3 Construct a free-body diagram (FBD) for object A, shown in Fig. 9-14a. The surface is smooth (no friction). Object A is a homogeneous cylinder weighing 4.00x102 lbm. 45? A 4x102 lbf 45? 60? RN RT FBD A x y 60? Assume gL = 32.174 ft/s2 gC = 32.174 lbm·ft lbf·s2 F = mgL/gC F = 400 lbm (32.174 ft/s2) 32.174 lbm·ft/lbf·s2 = 400 lbf Equilibrium Newton?s first law states that if the resultant force acting upon a particle is zero, then the particle will remain at rest or move with a constant velocity. A body will be in equilibrium when the sum of all external forces and moments acting upon the body is zero. This requires the body to be at rest or moving with a constant velocity. Example Problem 9.4 A beam, assumed weightless, is subject to the load as shown in Fig. Below. Determine the reaction on the beam at supports A and B for the equilibrium condition. 3.0 m 6.0 m 12 kN 24 kN A 3.0 m B 12 kN 24 kN Ay By 3.0 m 6.0 m 3.0 m y x FBD A B Ax Example Problem 9.4 3.0 m 6 m 24 kN A 3.0 m B Ay Ax By 12 kN x y Ax = 0 = Ay + By ? 12k ? 24k ? MA= -12k (3.0) ?24k (6.0) + By( 12.0) =0 FBD Fx = 0 ? Fy = 0 ? Mpt = 0 Example 9.4 ? Fy = 0 = Ay + By ? 12 ? 24 = 0 M? = 0 = -12k(3.0) ?24k(6.0) + By( 12.0) 12 By = 36k + 144k and By = 15 kN Substituting into the forces summed in the Y-direction eqn. Ay + 15k ? 12k ? 24k = 0 Ay = 21 kN Example 9.5 Solve Example prob. 9.3 for the cable tension and reaction of the inclined surface on the cylinder. A 45? A 4x102 lbf 45? 60? x y RN RT FBD 60? F = 400 lbf Example Problem 9.5 ? Fx = 0 A 4x102 lbf 45? 60? x y RN RT RT cos 60° -RN cos 45° 293 lbf 60° 1.414 (207 lbf) = RT = 1.414 RN = & = 0 1.932 RN = 4x102 or RN = 207 lbf 45? 1.414 RN (sin 60?) + RN Sin 45? = 4.00x102 lbf Substituting for RT into the Forces in y-direction, ? Fy = 0 = RT sin 60 ? + RN Sin 45? -4.00x102 RT = RN Cos 45?/cos 60? =1.414 RN Example Problem 9.6 For the crane system shown in Fig 916a, determine the reaction on the crane at pin A and roller B. A B 6.5 m 60? 94 kN 4.0 m 2.0 m 60? 94 kN FBD A 2.0 m 6.5 m 4.0 m By B Ay Ax x y Example Problem 9.6 Ax y 94k cos 60? B 6.5 m 60? 4.0 m 2.0 m Ay x 94k sin 60? By FBD A or, Ay = -54 kN = 54 kN = +Ay +135k ? 94k sin 60? ?Fy = 0 By = 135 kN + By ( 6.5) ?(94k cos 60?)(4.0) = -(94k sin 60?)(8.5) ? MA = 0 or Ax = 47 kN ? & Ax = -47 kN = 0 + 94k cos 60? +Ax ? Fx = 0 Example Problem 9.7 6.00 m cable 20.0 kg 10.00 m 1 1 For the structure shown above, determine the pin reaction at G and the tension T in the cable. G ? 2.00 m 90° I ? Example Problem 9.7 6.00 m cable 20.0 kg 10.00 m H ? ? Gy ? G? = 8.0 m ( 3,4,5 Right triangle) 1 1 45? G 2.00 m FBD Gx x y = 8.13? ? = 45? -36.87? ? = tan-1 .75 = 36.87? ? = tan-1 6/8 I Example Problem 9.7 MG = 0 6.00 m cable (20.0x9.807) N 10.00 m 1 1 Gx H 2.00 m ? ? ? 45? G 8 m Gy FBD = 9.84 N or Gy -20(9.807) + |H| sin 45? ? Fy= 0 = 206 N? Gx =291.3 cos 45? Gx = |H| cos 45? -|H| cos 45?= 0 + Gx ? Fx = 0 H = 291.3 N 45 ? +(8.00)H {(cos 8.13?)(12.0)} = {-(20.0x9.807)} = -206+196 = +Gy = -9.84 N Example Problem 9.7 Gx =291.3 cos 45? = 206 N? Gy = -206+196 = 9.84 N |G| = Gx2 + Gy2 = 206 N = 2062 + (-9.84) 2 G = 206 N -2.73? = -2.73º = tan-1 (-.004776) = tan-1 -9.84/206 ? = tan-1 Gy/Gx Homework Week 13- Homework: Problems:9.25, 9.26 and 9.27