| Author | Message | ||
     rajbeer malhotra (rajbeer) New member Username: rajbeer Post Number: 1 Registered: 06-2004 |
Consider a spring subjected to a load say "P", undergoing a displacement "d" , then as given in various texts the expression for external work done is given by: External work done = P x d/2 My question is that what is the reason for this divisibility by 2???? Plz help me , urgent!!!!!! | ||
| Fred Sachs (sachs) Member Username: sachs Post Number: 8 Registered: 01-2004 |
It is the average force of the distance x | ||
| Mongi Mansouri (mongim_pde) Member Username: mongim_pde Post Number: 23 Registered: 11-2003 |
Rajbeer, when u integrate the differential work done on linear spring, dW = k*x*dx over x = 0..d, you get W = 1/2*k*d^2 = k*d * d/2 = P*d/2, where P=kd is the spring force at deformation x = d. Hope this helps. | ||
| Patricio A. Greco (pagreco) Member Username: pagreco Post Number: 25 Registered: 08-2003 |
The spring force is: F= K* (X-Xo) Where K : Elastic constant X : Extreme displacement Xo: Initial position If we fix one extreme and take a coordenate axis that Xo=0 then F= K*X The work is : W= Int(F dx , {L,0,d}) Calculating W= (K*d^2) / 2 But K*d= Final force P W= P*d / 2 | ||
| Patricio A. Greco (pagreco) Member Username: pagreco Post Number: 26 Registered: 08-2003 |
In rigor : F=-K*(X-Xo) and the external work , lets We = -W . | ||
| rajbeer malhotra (rajbeer) New member Username: rajbeer Post Number: 2 Registered: 06-2004 |
ya,the above replies give a mathematical reasoning.That's ok.Thanks a million. what's the physical reason??In texts it says " -- as the load is gradually applied , hence we divide by 2" Whats this "gradually applied load"?? | ||
| Patricio A. Greco (pagreco) Member Username: pagreco Post Number: 27 Registered: 08-2003 |
The spring force is F= -k*(x-xo) then a displacement from the equilibrium point generates a proportional force applied to the mass if you calculate the resultant of both forces are: Fr= P-k*(x-xo) The mass stops when there are a force equilibrium then the diferential external work is "bigger"at the start of movement and 0 at the end then the integral calculates the average. I hope you understand the idea.  | ||
| rajbeer malhotra (rajbeer) Junior Member Username: rajbeer Post Number: 3 Registered: 06-2004 |
Can i say like this: Initially the full value of externAL FORCE "F" acts.Then, the internal force gradually develops(opposing the external force "P") . When the full opposing value of internal force is developed displacement stops.Hence the value of "P" goes on decreasing from start and finally becomes zero when the full value of internal force is developed. Since, "P" keeps on decreasing from start to end we divide by 2? Is thT OK? | ||
| Patricio A. Greco (pagreco) Member Username: pagreco Post Number: 28 Registered: 08-2003 |
Yes , you understood the point. Imagine a graph with the resultant force versus the displacement. With the coordinate axis you will form a triangle , the area of this triangle is the external work. | ||
| rajbeer malhotra (rajbeer) Member Username: rajbeer Post Number: 4 Registered: 06-2004 |
Thanx, another question: Now, suppose a load P acts on a beam at a point C producing a displacement at a point say "delta_c". Now , the load P1 is applied at a point other than C, producing a displacement "sigma_c" at C. Then the work done is: =(P x delta_c/2)+(P x sigma_c) If you go by my above post(last one where u mentioned i got it correct) the value of "P" has become zero after displacement delta_c took place then only P1 is applied then why have an expression P x sigma_c?????(P has already become zero since it is already statically equilibriated).Plz help | ||
| rajbeer malhotra (rajbeer) Member Username: rajbeer Post Number: 5 Registered: 06-2004 |
Eagerly awaiting response!!! | ||
| Patricio A. Greco (pagreco) Member Username: pagreco Post Number: 29 Registered: 08-2003 |
Dear Rajbeer: Please see the original formula F= -K*(x-xo) Xo is the initial position of the spring , this is independent of the forces that are actuating.For this readon and applying the superposition principle the external work is We= 0.5*delta_c*P + 0.5*sigma_c*P1 Let me a couple of hours and I'll recomend a book to study these basic pinciples . | ||
| rajbeer malhotra (rajbeer) Member Username: rajbeer Post Number: 6 Registered: 06-2004 |
I said above "P1" is not acting at "C" .It is just that when P1 is applied at a point other than C it produces a displacement "sigma_c" at "C". The component "sigma_c" is due to "P1" but under "P". The work done by "P" is given as "P * sigma_c" (virtual work) Again , my question, ----the value of "P" has become zero after displacement delta_c took place then only P1 is applied then why have an expression P x sigma_c????? | ||
| rajbeer malhotra (rajbeer) Member Username: rajbeer Post Number: 7 Registered: 06-2004 |
I have finished my masters recently just getting some gaps filled---- | ||
| Patricio A. Greco (pagreco) Member Username: pagreco Post Number: 30 Registered: 08-2003 |
The problem with this supositions is that we must to know how the think happens. If you apply a force to a spring you must to define how you generate this. If you use a mass you will have a oscilatory movement , if you add an ammortiguator at the end of the oscilation the formula you first used are valid but one part of the energy was gone (the friction in the ammortiguator transforms the energy on heat (for example)) .Then if you add a new weight a new ammortiguated process appears , when it stops you loose another part of the energy. This happens in electrostatics with the charged capacitors interconnection , the charge is the same but the total energy is less (one part was gone?) If you imagine a spring without mass and an ideal force the movement will be instantaneous and this is impossible. Imagine a real scheme and solve the ecuations , then calculate the energy balance and the you will find the answer to your questions. | ||
| rajbeer malhotra (rajbeer) Member Username: rajbeer Post Number: 8 Registered: 06-2004 |
Thank you, but what you said above is not understandable.I am not well familiar with electrostatics but am from a civil engg. background Can u be specific and answer the post no. 6 above!!!Thanks again!! | ||
| rajbeer malhotra (rajbeer) Member Username: rajbeer Post Number: 9 Registered: 06-2004 |
I shall be extremely garateful if u answer that, its been giving me sleepness nights. ----"the value of "P" has become zero after displacement delta_c took place then only P1 is applied, then why have an expression P x sigma_c?????[P has already been equilibriated] Rajbeer | ||
| rajbeer malhotra (rajbeer) Member Username: rajbeer Post Number: 10 Registered: 06-2004 |
Plz help!! | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 159 Registered: 06-2003 |
Rajbeer - We would like to help, but it appears that what you need is tutoring in science, and we at PDE Solutions do not have the staff to engage in internet science tutoring. If you have a question about FlexPDE, we would be glad to answer. | ||
| rajbeer malhotra (rajbeer) Member Username: rajbeer Post Number: 11 Registered: 06-2004 |
Anyway, thanx very much, but whats Flexpde exactly? Actually,though, i have been a good student academically i had this doubt from the beginning .Shall be still glad if anyone could just explain about it if possible. |