SME
Finite Element Analysis
Isoparametric 2D elements (CST and Q4)
Use this for “A Q4 element has nodal temperatures 0, 0.12, 0.30 and 0.10 (in node order). Find the shape functions and the interpolated value at ξ = 0.5, η = −0.25.”
Diagram
u(ξ, η) = (0.12(1 + ξ)·(1 − η) + 0.3(1 + ξ)·(1 + η) + 0.1(1 − ξ)·(1 + η)) ÷ 4.
At (0.5, −0.25): u = 0.15, ∂u/∂ξ = 0.075 and ∂u/∂η = 0.08.
|∇u|
=
√
0.075
2
+
0.08
2
=
0.1097.
This is an example. Type your own values to start a new problem.
u
=
N₁
×
u₁
+
N₂
×
u₂
+
N₃
×
u₃
+
N₄
×
u₄
The picture’s u(ξ, η) is N₁u₁ + N₂u₂ + N₃u₃ + N₄u₄ with the node values put in.
ξ
Horizontal coordinate
example
η
Vertical coordinate
example
Formulas
N₁
=
(1
−
ξ
)
×
1 −
η
4
0.1563
=
(1
−
0.5)
×
1 − (−0.25)
4
N₂
=
(1
+
ξ
)
×
1 −
η
4
0.4688
=
(1
+
0.5)
×
1 − (−0.25)
4
N₃
=
(1
+
ξ
)
×
1 +
η
4
0.2813
=
(1
+
0.5)
×
1 + (−0.25)
4
N₄
=
(1
−
ξ
)
×
1 +
η
4
0.09375
=
(1
−
0.5)
×
1 + (−0.25)
4
u
=
N₁
×
u₁
+
N₂
×
u₂
+
N₃
×
u₃
+
N₄
×
u₄
0.15 = 0.1563 × 0 + 0.4688 × 0.12 + 0.2813 × 0.3 + 0.09375 × 0.1
Assumptions
•
Nodes 1 to 4 run counterclockwise round the parent square, starting at (−1, −1).
•
Bilinear: each Nᵢ is 1 at its own node, 0 at the others, and linear along each edge.
•
Isoparametric: the same four Nᵢ map x and y as interpolate the field.
Step-by-step
WE KNOW
ξ = 0.5, η = −0.25, u₁ = 0, u₂ = 0.12, u₃ = 0.3, u₄ = 0.1
FIND
shape function 1 (N₁), shape function 2 (N₂), shape function 3 (N₃), shape function 4 (N₄), value at point (u)
Step 1 · Find shape function 1 (N₁)
Use N₁ = (1 − ξ) × (1 − η) ÷ 4
Each
factor
is
0
on
the
far
edge
and
2
at
node
1
at
(−1,
−1),
hence
the
4
.
N₁
=
(1
−
0.5)
×
1 − (−0.25)
4
N₁
=
0.5
×
1.25
4
N₁
=
0.625
4
N₁
=
0.1563
Step 2 · Find shape function 2 (N₂)
Use N₂ = (1 + ξ) × (1 − η) ÷ 4
Each
factor
is
0
on
the
far
edge
and
2
at
node
2
at
(1,
−1),
hence
the
4
.
N₂
=
(1
+
0.5)
×
1 − (−0.25)
4
N₂
=
1.5
×
1.25
4
N₂
=
1.875
4
N₂
=
0.4688
Step 3 · Find shape function 3 (N₃)
Use N₃ = (1 + ξ) × (1 + η) ÷ 4
Each
factor
is
0
on
the
far
edge
and
2
at
node
3
at
(1,
1),
hence
the
4
.
N₃
=
(1
+
0.5)
×
1 + (−0.25)
4
N₃
=
1.5
×
0.75
4
N₃
=
1.125
4
N₃
=
0.2813
Step 4 · Find shape function 4 (N₄)
Use N₄ = (1 − ξ) × (1 + η) ÷ 4
Each
factor
is
0
on
the
far
edge
and
2
at
node
4
at
(−1,
1),
hence
the
4
.
N₄
=
(1
−
0.5)
×
1 + (−0.25)
4
N₄
=
0.5
×
0.75
4
N₄
=
0.375
4
N₄
=
0.09375
Step 5 · Find value at point (u)
Use u = N₁ × u₁ + N₂ × u₂ + N₃ × u₃ + N₄ × u₄
Weight each nodal value by its shape function at the point, then add the four.
u
=
0.1563
×
0
+
0.4688
×
0.12
+
0.2813
×
0.3
+
0.09375
×
0.1
u
=
0.056256
+
0.08439
+
0.009375
u
=
0.15
Check
Put the numbers back in. Both sides should match.
0.1563
=
(1
−
0.5)
×
1 − (−0.25)
4
✓
0.4688
=
(1
+
0.5)
×
1 − (−0.25)
4
✓
0.2813
=
(1
+
0.5)
×
1 + (−0.25)
4
✓
0.09375
=
(1
−
0.5)
×
1 + (−0.25)
4
✓
0.15 = 0.1563 × 0 + 0.4688 × 0.12 + 0.2813 × 0.3 + 0.09375 × 0.1 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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Use this for “A right-angled CST has node 1 at the origin pinned, node 2 at (40, 0) and node 3 at (0, 30) mm. Node 2 moves (12, 4) μm and node 3 (6, 9) μm. Find the strains.”
The Jacobian of a parallelogram Q4
Use this for “A Q4 element is a parallelogram with a 60 mm base, 40 mm height and a 20 mm lean. Find J and det J, and check the area.”