a) Consider the two nodes’ bar element (no bending), with u(x) being the displac
ID: 1711272 • Letter: A
Question
a) Consider the two nodes’ bar element (no bending), with u(x) being the displacement function. Why is it taken as a linear function of x, and not a polynomial of a higher degree, a more accurate one?
b) Suppose instead of a linear two nodes’ element, we have a 3-node linear “truss” element. What would be the assumed format of u(x) and why?
c) In a 9 node plane stress isoparametric element with “parent” coordinates s, t, what is the value of the shape function N8(s8,t8)?
Same with N7(s8,t8)?
Reasons?
Explanation / Answer
(a) The displacement function for a 2 -noded bar element is expressed as u (x) ( a linear function of x) because of the following reason:
In a 2 noded bar element ,which is considered as a 1-Dimensional element,the displacement( due to external applied force(acting in one direction)) occurs along one direction only. Hence, the displacement function is expressed as linear function of x.
Note: The bar element resists internal axial force along its axial dimensional only.
(b) Suppose, if we consider a 3 -noded truss element, say i,j,k are the three nodes,then the nodal displacements ui , uj and uk are expressed as :
ui = a1 + a2xi
uj = a1 + a2xj
uk = a1 + a2xk
where a1 and a2 are the constants
xi,xj and xk are the nodal coordinates.
so the assumed format of u (x) = u(xi,j,k) = a1 + a2(xi,j,k)
(c) For the 9 noded plane stress isoparametric element with parent coordinates (s,t) ,the value of shape function N8(s8,t8) is calculated by the expression given below:
N8(s8,t8) = (1/2) *(1 - s8) *(1 - t82)
The value of shape function N7(s8,t8) = (1/2) * (1 - s82) * ( 1 + t8)
Reason :Since N5 through N8 equal to 1/2 at s = t = 0,we need to only subtract N9 / 2 from each in order to obtain N5 through N8 of the nine noded plane stress isoparametric element.
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