When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero.

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2 When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero k M j M i M M k R j R i R F R z y x z y x

3 Forces and moments acting on a rigid body could be external forces/moments or internal forces/moments. Forces acting from one body to another by direct physical contact or from the Earth are examples of external forces. Fluid pressure acting to the wall of a water tank or a force exerted by the tire of a truck to the road is all external forces.

4 The weight of a body is also an external force. Internal forces, on the other hand, keep the particles which constitute the body intact. Since internal forces occur in pairs that are equal in magnitude opposite in direction, they are not considered in the equilibrium of rigid bodies. The first step in the analysis of the equilibrium of rigid bodies must be to draw the free body diagram of the body in question.

5 Common Support / Connection Element Types in Two Dimensional Analysis In rigid bodies subjected to two dimensional force systems, the forces exerted from supports and connection elements are shown in the free body diagram as follows: It should be kept in mind that reaction will occur along the direction in which the motion of the body is restricted.

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10 Equations of Equilibrium in Two Dimensional Case If all the forces acting on the rigid body are planar and all the couples are perpendicular to the plane of the body, equations of equilibrium become two dimensional k M M F R F R j R i R F R z y y x x y x or in scalar form, O y x M F F At most three unknowns can be determined.

11 Alternative Equations of Equilibrium In two dimensional problems, in alternative to the above set of equations, two more sets of equations can be employed in the solution of problems. Points A, B and C in the latter set cannot lie along the same line, if they do, trivial equations will be obtained C B A B A x M M M M M F

12 Two-Force Member Members which are subjected to only two forces are named as two force members. Forces acting on these members are equal in magnitude, opposite in direction and are directed along the line joining the two points where the forces are applied. A B P A x A y B y B x F A F B F A =F B Weight is neglected. If weight is considered, the member will not be a two force member! P P Examples of two force members P P P Hydraulic cylinder P

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14 Three-Force Member In rigid bodies acted on by only three forces, the lines of action of the forces must be concurrent; otherwise the body will rotate about the intersection point of the two forces due to the third force which is not concurrent. If the forces acting on the body are parallel, then the point of concurrency is assumed to be in infinity. F A P F B A P B

15 Free Body Diagram (FBD) The procedure for drawing a free body diagram which isolates a body or system consists of the following steps: 1) If there exists, identify the two force members in the problem. 2) Decide which system to isolate. 3) Isolate the chosen system by drawing a diagram which represents its complete external boundary. 4) If not given with the problem, select a coordinate system which appropriately suits with the given forces and/or dimensions. 5) Identify all forces which act on the isolated system applied by removing the contacting or attracting bodies, and represent them in their proper positions on the diagram. 6) Write the equations of equilibrium and solve for the unknowns.

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18 A y A x M O O y O x

19 B x A y A x

20 F f F f

21 B y B x A x A y M A A x

22 T W=mg F f W=mg N N

23 T A x A x N L A y A y mg m O g

24 mg T A y B y A x B x F f N

25 B x B y T T A y A x A x A y mg L

26 1. Determine the magnitude P of the vertical force required to lift the wheelbarrow free of the ground at point B. The combined mass of the wheelbarrow and its load is 110 kg with center of mass at G.

27 2) A 35 N axial force at B is required to open the spring loaded plunger of the water nozzle. Determine the required force F applied to the handle at A and the magnitude of the pin reaction at O. Note that the plunger passes through a vertically-elongated hole in the handle at B, so that negligible vertical force is transmitted there.

28 3. The small crane is mounted on one side of the bed of a pickup truck. For the position q=40 o, determine magnitude of the force supported by the pin at O and the oil pressure p against the 50-mm diameter piston of the hydraulic cylinder BC.

29 4. The pin A, which connects the 200-kg steel beam with center of gravity at G to the vertical column, is welded both to the beam and to the column. To test the weld, the 80-kg man loads the beam by exerting a 300 N force on the rope which passes through a hole in the beam as shown. Calculate the torque (couple) M supported by the pin.

30 5. The pipe bender consists of two grooved pulleys mounted and free to turn on a fixed frame. The pipe is bent into the shape shown by a force P = 300 N. Calculate the forces supported by the bearings of the pulleys.

31 6. The mass center of 15-N link OC is located at G, and the spring constant of k=25 N/m is unstretched length when q=0. Calculate the tension T and the reactions at O for q=45 o.

32 7. Pulley A delivers a steady torque of 100 Nm to a pump through its shaft at C. The tension in the lower side of the belt is 600 N. The driving motor B has a mass of 100 kg and rotates clockwise. As a design consideration, determine the magnitude of the force on the supporting pin at O.

33 8. Plate AB contains a smooth parabolic slot. Fixed pins B and C are located at the positions shown in the figure. The equation of the parabolic slot is given as y = x 2 /160, where x and y are in mm. If it is known that the force input P = 4 N, determine the forces applied to the plate by the pins B and C and also the force output Q. Q 120 mm P A y B C 46 mm 20 mm x 140 mm 60 mm 40 mm

34 b Q 120 mm P=4 kn B F F M 60B A 0. 6C Q y x y 0. 8C Q y 0 C Q 4 90 y B y 140 mm 60 mm 40 mm B 0 P B b C a a C P C sina Q sin b 0 y C cos a Q cos b 0 Tangent to the parabolic slot 46 mm 20 mm x 2 x y 160 dy 2x y dx 160 y x tan b B 60 Q sin b46 40tan b y b o mm 160 tana 3 4 C N Q N By N