A problem about virtual work principle

1 reply · 2K views · last reply
athosanian
Messages
66
Reaction score
8
hello, I want to apply the virtual work principle to a continuous mechanical system to derive equilibrium equation, naemly, a long thin bar with one end fixed in the wall and other end applied with pressure P. but I can not derive the correct equilibrium equations. I hope some expert could give some advice.
my thhinking is shown below:
Typed page titled Virtual work principle applied to a continuous system, with section 1 introduction giving equilibrium equation dT/dx = 0, boundary condition T = P at the right end, constitutive law T = ES and strain S = du/dx.


Diagram of a long bar fixed at the left wall with pressure P applied at the right end, coordinate x along the bar, and an inset of a small element showing forces T and T + dT/dx dx and displacements δu and δu + dδu/dx dx.


Handwritten-style typed derivation using the virtual work principle: equations (5) to (7) integrate T du/dx and dT/dx terms by parts to give delta W = T delta u at the endpoints, with delta u = 0 at x=0, leading to T at x=L equal to 0.
 
Attachments
  • Typed page titled 'Virtual work principle applied to a continuous system' with section 1 introduction, showing equations (1) dT/dx = 0, (2) T = ES and (3) S = du/dx with boundary condition T at x = L equals P.
    virtual work-1.png
    54.1 KB · Views: 602
  • Diagram of a long bar fixed at the left wall with pressure P applied at the right end, with a free-body sketch of a small element showing forces T and T + dT/dx dx and displacements δu and δu + dδu/dx dx, plus a page of text on the virtual work principle.
    virtual work-2.png
    55.1 KB · Views: 616
  • Handwritten-style typeset derivation of the virtual work principle for a bar: equations (5) to (7) show the variation dW as an integral from 0 to L of T times du/dx plus dT/dx times du, then integrated by parts to T du evaluated at L and 0, followed by a closing question in red.
    virtual work-3.png
    11 KB · Views: 589
Obviously, if the two terms from (4) result in ## T \delta u |_0^L = 0 ##, which gives you no condition on ##T## because ## \delta u(0) = \delta u(L) = 0 ##, it is the neglected third term that you must take into account.