project statics
- 1. Project inrodused to dr.ahmad alshorman Prepared by : Osamah zahi melhem Ahmad bassab alkhazaleh Salah tawalbeh
- 2. Equilibrium of a particle A particle is in equilibrium if the vector sum of the external forces acting on it is zero. Hence a particle is in equilibrium if: 1. It is at rest and remains at rest − Static.Equilibrium 2. It moves with constant velocity − Dynamic.Equilibrium
- 3. What’s the particle The particle is a model of a real body. The word "particle" does not imply that the particle is a small body. Modelling a body as particle is equivalent to the assumption that all forces applied on body act at the same point. This assumption is acceptable in many.practical engineering applications
- 4. Forces affect on Particles
- 5. Statics (MET 2214) Prof. Simin Nasseri Free Body Diagram (FBD) How to draw a Free Body Diagram: Draw outlined shape - Imagine the particle isolated or cut “free” from its surroundings Show all forces and moments - Include “active forces” and “reactive forces”. Place each force and couple at the point that it is applied.
- 6. Statics (MET 2214) Prof. Simin Nasseri FBD
- 7. Statics (MET 2214) Prof. Simin Nasseri FBD F.B.D of the ring A:
- 8. Equilibrium equations in component form 1-in 2-d 2-in 3 -d
- 10. Example of 3-D
- 11. Statics (MET 2214) Prof. Simin Nasseri Example 1 The sphere has a mass of 6 kg and is supported as shown. Draw a free-body diagram of the sphere, cord CE, and the knot at C.
- 12. Statics (MET 2214) Prof. Simin Nasseri Sphere There are two forces acting on the sphere. These are its weight and the force of cord CE. The weight is: W = 6 kg (9.81 m/s2) = 58.9 N.
- 13. Statics (MET 2214) Prof. Simin Nasseri FBD of sphere This is the way we show the FBD of the sphere: FCE 58.9 N
- 14. Statics (MET 2214) Prof. Simin Nasseri Cord CE There are two forces acting on the cord. These are the force of the sphere, and the force of the knot. A cord is a tension only member. Newton’s third law applies.
- 15. Statics (MET 2214) Prof. Simin Nasseri Knot at C There are three forces acting on the knot at C. These are the force of the cord CBA, and the force of the cord CE, and the force of the spring CD.
- 16. Statics (MET 2214) Prof. Simin Nasseri FBD of the knot at C FCE FCBA FCD C 60o
- 17. Statics (MET 2214) Prof. Simin Nasseri Example 2 Draw the FBD diagram of the ring A: W= 2.452 KN
- 18. Statics (MET 2214) Prof. Simin Nasseri FBD of the ring A Is this the FBD of A? No! this is not the free body diagram of A!
- 19. Statics (MET 2214) Prof. Simin Nasseri FBD of the ring A
- 20. Statics (MET 2214) Prof. Simin Nasseri Example 3 Draw the free body diagrams of C and E and the cable CE:
- 21. Statics (MET 2214) Prof. Simin Nasseri FBD of E
- 22. Statics (MET 2214) Prof. Simin Nasseri FBD of C
- 23. Statics (MET 2214) Prof. Simin Nasseri FBD of cable EC
- 24. Statics (MET 2214) Prof. Simin Nasseri Example 4 Draw the FBD of ring A. W=78.5 N
- 25. Statics (MET 2214) Prof. Simin Nasseri FBD of A
- 26. Statics (MET 2214) Prof. Simin Nasseri Part 2 Applying the Equilibrium Equations
- 27. Statics (MET 2214) Prof. Simin Nasseri FBD Draw the free body diagrams: W N W N f Normal force = The force you have when there is a contact between surfaces (the ball is in contact with the ground). Friction force = You have this when the surface in contact is not frictionless and the friction prevents the motion of the object. 30
- 28. Statics (MET 2214) Prof. Simin Nasseri FBD 0 0 x y F F W N W N f 30 0 0 x y F F x y x y
- 29. Statics (MET 2214) Prof. Simin Nasseri y FBD W N N f N = W N - W.cos 30 = 0 f - W.sin 30 = 0 W.cos30 W.sin30 x x y
- 30. Statics (MET 2214) Prof. Simin Nasseri Example 2: Determine the tension in cables AB and AD for equilibrium of the 250 kg engine. FBD of the ring A
- 31. B 0, cos30 0 0, sin30 2.452 0 Solving for T : sin30 2.452 , 4.90 Subsituting into the first equation: 4.25 x B D y B B B D F T T F T kN T kN T kN T kN Solution of Example 2 According to the free body diagram of the ring A, we have three forces acting on the ring. The forces TB and TD have unknown magnitudes but known directions. Cable AC exerts a downward force on A equal to: W = (250kg)(9.81m/s2) = 2452N = 2.245KN TBcos30 TBsin30
- 32. Moment In statics, moments are effects (of a force) that cause rotation. When computing equilibrium, you must be able to calculate a moment for every force on your free-body diagram. To determine a force's moment, you use one of two different calculations, as you can see in the following list.
- 33. calculations Scalar calculation (for two dimensions): To calculate the moment about a Point O in scalar calculations, you need the magnitude of the force and the perpendicular distance from Point O to the line of action of the ForceF. Vector calculation (for two or three dimensions): To compute the moment vector about a Point O in vector calculations, you must determine the Force F in Cartesian vector form and the position vector from Point O to the line of action of the Force F.
- 36. Moment of a couple If two opposite moments act to cause an object to rotate, such as when your two hand are at the 'quarter-past-three' position on a car steering wheel, it is called a couple. The moment of a couple is called the torque. It is quite often said of engines and applys to the ability of the engine to turn the wheels, or wrongly by Jeremy Clarkson from 'Top Gear' as in, "This engine has a lot of torques."
- 37. an example of a moment of a couple