Equilibrium of a Rigid bodies Part 1-Basic Mecanical Engineering-Lecture Slides, Slides of Mechanical Engineering

Download Equilibrium of a Rigid bodies Part 1-Basic Mecanical Engineering-Lecture Slides and more Slides Mechanical Engineering in PDF only on Docsity! In contrast to the forces on a particle, the forces on a rigid-body are not usually concurrent and may cause rotation of the body (due to the moments created by the forces). Equilibrium of a Rigid bodies 2 Couple moment docsity.com An obvious condition for equilibrium is that the net force acting must be zero. Although this is the only requirement necessary for a point mass to be in equilibrium it is not sufficient to guarantee the equilibrium of a rigid body Equilibrium of a Rigid body 3 In all three cases the net force on the stick is zero. For above fig in the last case the stick does not remain at rest. It rotates clockwise. We are therefore forced to conclude that while a necessary condition for equilibrium is that the net force is zero, this is not a sufficient condition for a rigid body. A second condition is required to insure rotational equilibrium. There must be no net turning effect of forces rotating an object about a pivot point if the rigid object is to be in equilibrium. docsity.com Procedure For Drawing A Free Body Diagram 6 Idealized model Free body diagram 1. Draw an outlined shape. Imagine the body to be isolated or cut “free” from its constraints and draw its outlined shape. 2. Show all the external forces and couple moments. These typically include: a) Applied loads b) Support reactions c) Weight of the body docsity.com Procedure For Drawing A Free Body Diagram 7 Idealized model Free body diagram 3. Label loads and dimensions  All known forces and couple moments should be labeled with their magnitudes and directions.  For the unknown forces and couple moments, use letters like Ax, Ay, MA, etc..  Indicate any necessary dimensions. docsity.com The three common types of connections which join a built structure to its foundation are; 1. Roller Support 2. Pinned Support 3. Fixed Support These supports can be located anywhere along a structural element. Commonly used Supports 8 They are found at the ends or at the midpoints, or at any other intermediate points. Pin Support Roller SupportFixed Support docsity.com A pinned support can resist both vertical and horizontal forces but not a moment. They will allow the structural member to rotate, but not to translate in any direction Commonly used Supports-Pinned Support 11 Pin Support docsity.com Reactions of the Supports nt . fixed support CT pin F y The joint can not move in vertical and horizontal . directions. ® docsity.com Reactions of the Supports 13 As a general rule, if a support prevents translation of a body in a given direction, A force is developed on the body in the opposite direction. Similarly, if rotation is prevented, a couple moment is exerted on the body. docsity.com Reactions of the Supports (ty Oee gnknown. The reicice i adenge whieh pete a ferpendkeular bo the surface atthe poi at contact F rocker (i a = — . Chee anknomn. The teaction mn force which acts E Pad : perpendicular to the surtace at the point of eomact. eich on acting arate F or a — rats Che enknown. “Che reacion a force which arts. a as ay perpendicular te the rod. r rTiet p-oOnecied fo collar an nomi by pod 16 ® docsity.com Reactions of the Supports {Ha Fr. i rae A Tun unkncems The reactions ape mo eqmpomesia of i f is * feree, or the muigenuce and direction deed the resulinnt Fr Ge foto, Wale [at d aenl © ame nit Hoocesarily equal [asially . not, tnless the nod shown is a bik as in-(2)]. Selo Hn or hits ay Zp f Peo unknowns, The reactions are ihe couple mone uM and Ue force ahich acts porpeneticuios tt lhe. rod Ineritect Rego Gurnee ich call ae atin artes fri) {len F. ii # Three unknowns. The reactions are the couple monnent = (t es (1 ft wad he own fore: compodents, or theconple einiens ind lhe ma vitiude and direction ol the reeatiant bore, docsity.com ‘Ona cable One unknown. The reaction is a tension foree which acts away from the member in the direction of the cable. The cable exerts a force on the bracket in the direction of the cable. (1) This concrete girder rests on the ledge that is assumed to act as a smooth contacting surface. (6) smooth contacting surface One unknown, The reaction is a force which acts perpendicular to the surface at the point of contact. docsity.com Example 21 A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. docsity.com Example 22 • Determine B by solving the equation for the sum of the moments of all forces about A. ( ) ( ) ( ) 0m6kN5.23 m2kN81.9m5.1:0 =− −∑ += BM A kN1.107+=B docsity.com • Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces. 0:0 =+=∑ BAF Example 23 xx kN1.107−=xA 0kN5.23kN81.9:0 =−−=∑ yy AF kN 3.33+=yA docsity.com Example 5.6 600 sin 45° N Equations of Equilibrium. Summing forces in the x direction yields SSF, = 0; 600 cos 45° N — B, = 0 B, = 424N | docsity.com Example 5.6 600 sin 45° N A direct solution for A, can be obtained by applying the moment equation > M, = Oabout point B. C+2Mp=0; 100N(2m) + (600 sin 45° N)(5m) — (600 cos 45° N)(0.2m) — A,(7m) = 0 Ay = 319N Ans. 27 ® docsity.com Example 5.6 “s 600 sin 45° N 100 N Summing forces in the y direction, using this result, gives +TZF, = 0; 319.N — 600 sin 45° N — 100 N — 200N + B, = 0 B, = 403N Ans. 28 ® docsity.com Example 5.8 31 docsity.com Example 5.8 Equations of Equilibrium. Summing moments about A, we obtain a direct solution for Ng. G+iM, =0; —90N-m— 60N(1 m) + N,(0.75 m) = 0 Ng = 200N Using this result, 4 IF, = 0; A, — 200 sin 30° N = A, = 100N +TZF, = 0; Ay — 200 cos 30° N — 60N = 0 Ay = 233.N 32 ® docsity.com The platform assembly has weight W1 of 250 N and center of gravity at G1. If it is intended to support a maximum load W2 placed at point G2, Determine the smallest counterweight W that G2 Problem 5.10 33 should be placed at B in order to prevent the platform from tipping over. docsity.com 900 N Free-Body Diagram. The free-body diagram of the member is shown in Fig. 5-19b. The collar exerts a horizontal foree A, and moment M, on the member. The reaction Ng of the roller on the member is vertical, 56 docsity.com Equations of Equilibrium. The forces A, and Ny can be determined directly from the force equations of equilibrium. SF, = 0; Ar=0 Ans, +125, = 0; Ny — 900N = 0 Naz — 900.N Ans. 37 ® docsity.com The moment M, can be determined by summing moments either about point A or point B. C+ EM, = 0; M, — 900 N(1.5 m) — 500 N-m + 900 N [3m + (J m) cos 45°] = 0 M, = —1486 N«m = 149kN-m) Ans. or C+3Mp = 0; M, + 900N [15m + (1 m) cos 45°] — SOON-m = 0 M, = —1486 N+«m = 1.49kN+m) Ans, The negative sign indicates that M,, has the opposite sense of rotation to that shown on the free-body diagram. docsity.com Spring Formula m01046.0 10*15 1570 s m07848.0 10*15 2.1177 s 3B 3A == == 41 o 1 5.10 ) 0.1 0785.0105.0 (tan = += −θ 1 m SA+SB docsity.com The device is used to hold an elevator door open If the spring has the stiffness of k=40N/m and it is compressed 0.2m Find out the horizontal component and vertical components of the reaction at pin “A” And the resultant force at the wheel bearing “B” Problem 5.11 42 A B k docsity.com A F=ks Ax Problem 5.11 43 Ay FB docsity.com Problem 5.7 46 Determine the magnitude of the reactions on the beam at “A” and at “B” docsity.com 600 N 15o 400 N B A A Problem 5.7 47 ByAx y docsity.com Problem 5.7 0F 0F N586B 04*60012*15cos40012B 0M y x y o y A = = = =−− = ∑ ∑ ∑ 48 Ax=103.5 N Ay=400 N FA=413 N docsity.com 0F =∑ N11.105B 0B2)81.9(6039.378 0F y y y = =+− =∑ Problem 5.13 51 N105~F 0B B x x = docsity.com The boom is intended to support two vertical loads F1 and F2. If the cable CB can sustain a maximum load of 1500 N before it fails, determine the critical loads if Problem 5.20 52 F1 = 2F2. docsity.com 4 3 5 1500 N Problem 5.20 53 30oAX A y F2 2F2 docsity.com Problem 5.11, 5.12, 5.13, 5.14, 5.17 56 docsity.com The solution to some equilibrium problems can be simplified if we recognize members that are subjected to forces at only two points (e.g., at points A and B). 57 If we apply the equations of equilibrium to such a member, we can quickly determine that the resultant forces at A and B must be equal in magnitude and act in the opposite directions along the line joining points A and B docsity.com A B BA 58 In the cases above, members AB can be considered as two- force members, provided that their weight is neglected. This fact simplifies the equilibrium analysis of some rigid bodies since the directions of the resultant forces at A and B are thus known (along the line joining points A and B). docsity.com