MECH419/928 : Simulation Assignment

MECH419/928 1
Name
Institution

MECH419/928 2
Simulation assignment
Part a
The stresses and strains on common edges, including the ‘cut’ edges are continuous and uniform
across the boundaries (L18 and L34). The plate is transversely homogeneous.Forces acting
perpendicular to the plane are zero or may be neglected, as the ratio of thickness to a
fundamental x or y of any section is < 0.1. The plate is, assumed transversely homogeneous.The
plate may be modeled using a quarter sections because of reflection symmetry on the x- and y-
axes.
o Each edge and shared node experiences equal displacement (e.g., L28, L36, L18, or
L34).
o The shape functions, N, must sum to 1[Nh + Ni + Nj = 1].
The plate is free floating. The distributed load on the annulus affects it uniformly.Each edge and
shared node experiences equal displacement.A06 = A01, A02 = A04, and A03=A05 by symmetry. If L
denotes the line between two nodes: L18 = L27, L12 = L34 = L78, L26 = L35, and L23 = L56 by
symmetry.
Displacement Matrix, {d}, and Displacement function, 
For each node h, i, or j on the triangular element, the displacement vector, {d} is { d }=
{dh
di
d j
}
, where {dh }= {uh
vh }, {di }= {ui
vi },∧ {d j }= {u j
v j }.

MECH419/928 3
The linear displacement function for triangular element n, n, is
Ψ n = {u (x , y )
v (x , y ) }=
{a1
a2
a1 xh
a2 xi
a1 yh
a2 y j
a3 a3 x j a3 y j
a4
a5
a6
a4 xh
a5 xi
a6 x j
a4 yh
a5 yi
a6 y j
}. This gives {u} and {v}:, where ( xh , y h ), ( xi , yi ), and ( x j . y j ) are
the coordinates of node h, i, or j. Thus {u}= {uh
ui
u j
}=
{1 xh yh
1 xi yi
1 x j y j
}{a1
a2
a3
} and {v} =
{vh
vi
v j
}=
{1 xh yh
1 xi yi
1 x j y j
}{a4
a5
a6
}, where { Χ }=
{1 xh yh
1 xi yi
1 x j y j
}= {1 X Y }. If we define { a }=
{a1
a2
a3
}∧ {a ' }=
{a4
a5
a6
},
then { u } = { X } { a } and { v }= { X } {a' }. Transposing, { a }= { Χ }−1 { u } and { a ' }= { Χ }−1 { v }.
The inverse of {X} is { Χ }−1= 1
2 A {α h α i α j
βh βi β j
γ h γ i γ j
}, where

MECH419/928 4
{ α }=
{α h
α i
α j
}=
{ xi y j − yi x j
xh y j − yh x j
xh yi − yh xi
}, { β }=
{βh
βi
β j
}=
{ yi− y j
y j− yh
yh− yi
},∧ { γ }=
{γ h
γ i
γ j
}=
{ x j−xi
xh−x j
xi−xh
}, and
2 A=xh ( yi− y j ) + xi ( y j− yh ) + x j ( yh− yi ), where A is area of the triangular element and the (xm, ym)
represent the Cartesian coordinates of node “m”.
Substituting, the expressions for u(x,y) and v(x,y) become: u ( x , y ) = { Nh uh+ N i ui + N j u j }
and v ( x , y ) = { Nh vh + Ni vi + N j v j }, where Nh, Ni, and Nj are the shape functions.
Nh= 1
2 A ( αh + βh x + γh y ), Ni= 1
2 A ( αi + βi x +γ i y ), and N j= 1
2 A ( α j + β j x+ γ j y ). This gives i in
terms of N and d: {i}= [N]{d}, [ N ]= [N h 0 Ni 0 N j 0
0 Nh 0 Ni 0 N j ]. Determining equations for
{σ x
σ y
τ xy
}, using the definition of the components of [N}, We now write:
ε x= ∂ u
∂ x = ∂ Nh
∂ x uh+ ∂ Ni
∂ x ui + ∂ N j
∂ x u j,
ε y= ∂ v
∂ y = ∂ Nh
∂ y vh+ ∂ Ni
∂ y vi+ ∂ N j
∂ y v j, and
γ xy= ∂u
∂ y + ∂ v
∂ x = ( ∂ N h
∂ y uh + ∂ Ni
∂ y ui+ ∂ N j
∂ y uj )+( ∂ Nh
∂ x vh + ∂ Ni
∂ x vi + ∂ N j
∂ x v j ). The
derivatives of the Ns for node
m are given by: ∂ N m
∂ x =
βm
2 A ∧∂ Nm
∂ y = γm
2 A
which can