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A A’ dr  F The work of a force F corresponding to the small displacement dr is defined as dU = F dr This scalar product can be written as dU = F ds cos.

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Presentation on theme: "A A’ dr  F The work of a force F corresponding to the small displacement dr is defined as dU = F dr This scalar product can be written as dU = F ds cos."— Presentation transcript:

1 A A’ dr  F The work of a force F corresponding to the small displacement dr is defined as dU = F dr This scalar product can be written as dU = F ds cos  where ds is the magnitude of the displacement. The work dU is positive if  90 o. Bölüm 10 Virtüel İşler Metodu The work of a couple of moment M acting on a rigid body is dU = M d  where d  is the small angle in radians through which the body rotates.

2 A A’ d r F1F1 F2F2 FnFn Considering a particle located at A acted upon by several forces F 1, F 2,... F n, we imagine the particle moves to a new position A’. Since this displacement does not actually take place, it is referred to as a virtual displacement and is denoted as  r.

3 A A’ d r F1F1 F2F2 FnFn The corresponding work of the forces is called the virtual work, and is denoted as  U, and written as dU = F 1  r + F 2  r +... + F n  r The principle of virtual work states that if a particle is in equilibrium, the total virtual work  U of the forces acting on the particle is zero for any virtual displacement of the particle. The principle of virtual work can be extended to rigid bodies and systems of rigid bodies.

4 A1A1 O ds s1s1  A A2A2 s2s2 s F The work of a force corresponding to a finite displacement of its point of application is designated as U 1 2. U 1 2 = F dr A1A1 A2A2 = (F cos  ) ds s1s1 s2s2 Similarly, the work of a couple of moment M corresponding to a finite rotation from  1 to  2 of a rigid body is expressed as U 1 2 = M d  11 22

5 U 1 2 = - Wdy y1y1 y2y2 y dy W A1A1 A2A2 The work of the weight W of a body as its center of gravity moves from an elevation y 1 to y 2 is y1y1 y2y2 = Wy 1 - Wy 2 The work is negative when the elevation increases, and positive when the elevation decreases. A

6 U 1 2 = - k x dx x1x1 x2x2 = kx 1 - kx 2 B B x1x1 A1A1 A2A2 A AOAO F B x2x2 x The work of the force F exerted by a spring on a body A as the spring is stretched from x 1 to x 2 is obtained by making F = kx 2 2 1212 1212 The work is positive when the spring is returning to its undeformed position. spring undeformed

7 U 1 2 = V 1 - V 2 When the work of a force F is independent of the path followed between points a A 1 and A 2, the force is a conservative force. The work is expressed as where V is the potential energy associated with F, and V 1 and V 2 represent the values of V at A 1 and A 2.The potential energies associated with the force of gravity W and the elastic force F are V g = Wy V e = kx 2 1212

8 When the position of a mechanical system depends upon a single independent variable , the potential energy is a function V (  ) of that variable. It follows that  U = -  V = -( dV / d  ) . The condition  U = 0 required for equilibrium of the system can be replaced by the condition dV d  = 0 The second derivative of V provides useful information regarding the system d 2 V/d  2 > 0, V is minimum and the equilibrium is stable; d 2 V/d  2 < 0, V is maximum and the equilibrium is unstable; d 2 V/d  2 = 0, higher order derivatives must be examined.


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