Presentation is loading. Please wait.

Presentation is loading. Please wait.

5.3 Equations of Equilibrium

Similar presentations


Presentation on theme: "5.3 Equations of Equilibrium"— Presentation transcript:

1 5.3 Equations of Equilibrium
For equilibrium of a rigid body in 2D, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 ∑Fx and ∑Fy represent the algebraic sums of the x and y components of all the forces acting on the body ∑MO represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to x-y plane and passing through arbitrary point O, which may lie on or off the body

2 5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations For coplanar equilibrium problems, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be used Two alternative sets of three independent equilibrium equations may also be used ∑Fa = 0; ∑MA = 0; ∑MB = 0 When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis

3 5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations Consider FBD of an arbitrarily shaped body All the forces on FBD may be replaced by an equivalent resultant force FR = ∑F acting at point A and a resultant moment MRA = ∑MA If ∑MA = 0 is satisfied, MRA = 0

4 5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations If FR satisfies ∑Fa = 0, there is no component along the a axis and its line of axis is perpendicular to the a axis If ∑MB = 0 where B does not lies on the line of action of FR, FR = 0 Since ∑F = 0 and ∑MA = 0, the body is in equilibrium

5 5.3 Equations of Equilibrium
Alternative Sets of Equilibrium Equations A second set of alternative equations is ∑MA = 0; ∑MB = 0; ∑MC = 0 Points A, B and C do not lie on the same line Consider FBD, if If ∑MA = 0, MRA = 0 ∑MA = 0 is satisfied if line of action of FR passes through point B ∑MC = 0 where C does not lie on line AB FR = 0 and the body is in equilibrium

6 5.3 Equations of Equilibrium
Procedure for Analysis Free-Body Diagram Establish the x, y, z coordinates axes in any suitable orientation Draw an outlined shape of the body Show all the forces and couple moments acting on the body Label all the loadings and specify their directions relative to the x, y axes

7 5.3 Equations of Equilibrium
Procedure for Analysis Free-Body Diagram The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed Indicate the dimensions of the body necessary for computing the moments of forces

8 5.3 Equations of Equilibrium
Procedure for Analysis Equations of Equilibrium Apply the moment equation of equilibrium ∑MO = 0 about a point O that lies on the intersection of the lines of action of the two unknown forces The moments of these unknowns are zero about O and a direct solution the third unknown can be obtained

9 5.3 Equations of Equilibrium
Procedure for Analysis Equations of Equilibrium When applying the force equilibrium ∑Fx = 0 and ∑Fy = 0, orient the x and y axes along the lines that will provide the simplest resolution of the forces into their x and y components If the solution yields a negative result scalar, the sense is opposite to that was assumed on the FBD

10 5.3 Equations of Equilibrium
Example 5.6 Determine the horizontal and vertical components of reaction for the beam loaded. Neglect the weight of the beam in the calculations.

11 5.3 Equations of Equilibrium
Solution FBD 600N force is represented by its x and y components 200N force acts on the beam at B and is independent of the force components Bx and By, which represent the effect of the pin on the beam

12 5.3 Equations of Equilibrium
Solution Equations of Equilibrium A direct solution of Ay can be obtained by applying ∑MB = 0 about point B Forces 200N, Bx and By all create zero moment about B

13 5.3 Equations of Equilibrium
Solution

14 5.3 Equations of Equilibrium
Solution Checking,

15 5.3 Equations of Equilibrium
Example 5.7 The cord supports a force of 500N and wraps over the frictionless pulley. Determine the tension in the cord at C and the horizontal and vertical components at pin A.

16 5.3 Equations of Equilibrium
Solution FBD of the cord and pulley Principle of action: equal but opposite reaction observed in the FBD Cord exerts an unknown load distribution p along part of the pulley’s surface Pulley exerts an equal but opposite effect on the cord

17 5.3 Equations of Equilibrium
Solution FBD of the cord and pulley Easier to combine the FBD of the pulley and contracting portion of the cord so that the distributed load becomes internal to the system and is eliminated from the analysis

18 5.3 Equations of Equilibrium
Solution Equations of Equilibrium Tension remains constant as cord passes over the pulley (true for any angle at which the cord is directed and for any radius of the pulley

19 5.3 Equations of Equilibrium
Solution

20 5.3 Equations of Equilibrium
Example 5.8 The link is pin-connected at a and rest a smooth support at B. Compute the horizontal and vertical components of reactions at pin A

21 5.3 Equations of Equilibrium
Solution FBD Reaction NB is perpendicular to the link at B Horizontal and vertical components of reaction are represented at A

22 5.3 Equations of Equilibrium
Solution Equations of Equilibrium

23 5.3 Equations of Equilibrium
Solution

24 5.3 Equations of Equilibrium
Example 5.9 The box wrench is used to tighten the bolt at A. If the wrench does not turn when the load is applied to the handle, determine the torque or moment applied to the bolt and the force of the wrench on the bolt.

25 5.3 Equations of Equilibrium
Solution FBD Bolt acts as a “fixed support” it exerts force components Ax and Ay and a torque MA on the wrench at A

26 5.3 Equations of Equilibrium
Solution Equations of Equilibrium

27 5.3 Equations of Equilibrium
Solution

28 5.3 Equations of Equilibrium
Solution Point A was chosen for summing the moments as the lines of action of the unknown forces Ax and Ay pass through this point and these forces are not included in the moment summation MA must be included Couple moment MA is a free vector and represents the twisting resistance of the bolt on the wrench

29 5.3 Equations of Equilibrium
Solution By Newton’s third law, the wrench exerts an equal but opposite moment or torque on the bolt For resultant force on the wrench, For directional sense, FA acts in the opposite direction on the bolt

30 5.3 Equations of Equilibrium
Solution Checking,

31 5.3 Equations of Equilibrium
Example 5.10 Placement of concrete from the truck is accomplished using the chute. Determine the force that the hydraulic cylinder and the truck frame exert on the chute to hold it in position. The chute and the wet concrete contained along its length have a uniform weight of 560N/m.

32 5.3 Equations of Equilibrium
Solution Idealized model of the chute Assume chute is pin connected to the frame at A and the hydraulic cylinder BC acts as a short link

33 5.3 Equations of Equilibrium
Solution FBD Since chute has a length of 4m, total supported weight is (560N/m)(4m) = 2240N, which is assumed to act at its midpoint, G The hydraulic cylinder exerts a horizontal force FBC on the chute Equations of Equilibrium A direct solution of FBC is obtained by the summation about the pin at A

34 5.3 Equations of Equilibrium
Solution

35 5.3 Equations of Equilibrium
Solution Checking,

36 5.3 Equations of Equilibrium
Example 5.11 The uniform smooth rod is subjected to a force and couple moment. If the rod is supported at A by a smooth wall and at B and C either at the top or bottom by rollers, determine the reactions at these supports. Neglect the weight of the rod.

37 5.3 Equations of Equilibrium
Solution FBD All the support reactions act normal to the surface of contact since the contracting surfaces are smooth Reactions at B and C are acting in the positive y’ direction Assume only the rollers located on the bottom of the rod are used for support

38 5.3 Equations of Equilibrium
Solution Equations of Equilibrium

39 5.3 Equations of Equilibrium
Solution Note that the line of action of the force component passes through point A and this force is not included in the moment equation Since By’ is negative scalar, the sense of By’ is opposite to shown in the FBD

40 5.3 Equations of Equilibrium
Solution Top roller at B serves as the support rather than the bottom one

41 5.3 Equations of Equilibrium
Example 5.12 The uniform truck ramp has a weight of 1600N ( ≈ 160kg ) and is pinned to the body of the truck at each end and held in position by two side cables. Determine the tension in the cables.

42 5.3 Equations of Equilibrium
View Free Body Diagram Solution Idealized model of the ramp Center of gravity located at the midpoint since the ramp is approximately uniform FBD of the Ramp

43 5.3 Equations of Equilibrium
Solution Equations of Equilibrium By the principle of transmissibility, locate T at C

44 5.3 Equations of Equilibrium
Solution Since there are two cables supporting the ramp, T’ = T/2 = N

45 5.4 Two- and Three-Force Members
Simplify some equilibrium problems by recognizing embers that are subjected top only 2 or 3 forces Two-Force Members When a member is subject to no couple moments and forces are applied at only two points on a member, the member is called a two-force member


Download ppt "5.3 Equations of Equilibrium"

Similar presentations


Ads by Google