SME
Finite Element Analysis · Problem type
Mapping and the Jacobian
Use this for “A bar element runs from x = 20 to 60 mm. Where is ξ = 0.5, and what is the Jacobian?”
Chart
x(0.5) = 50
Ringed: the nodes (−1, 20), (1, 60)
This is an example. Type your own values to start a new problem.
x (mm)
=
N₁
×
x₁ (mm)
+
N₂
×
x₂ (mm)
J (mm)
=
(
x₂ (mm)
−
x₁ (mm)
)
÷
2
ξ
Natural coordinate
example
Formulas
N₁
=
1 −
ξ
2
0.25
=
1 − 0.5
2
N₂
=
1 +
ξ
2
0.75
=
1 + 0.5
2
x
=
N₁
×
x₁
+
N₂
×
x₂
50 = 0.25 × 20 + 0.75 × 60
ξ
=
2 ×
x
−
x₁
−
x₂
x₂
−
x₁
0.5
=
2 × 50 − 20 − 60
60 − 20
J
=
x₂
−
x₁
2
20
=
60 − 20
2
Assumptions
•
A linear two-node element: ξ = −1 at node 1 and ξ = 1 at node 2.
•
Isoparametric: N₁ = (1 − ξ) ÷ 2 and N₂ = (1 + ξ) ÷ 2 place x as they interpolate u.
•
J = dx/dξ = L ÷ 2, the same all along the element; dx = J dξ in integrals.
Step-by-step
WE KNOW
x₁ = 20 mm, x₂ = 60 mm, ξ = 0.5
FIND
shape function 1 (N₁), shape function 2 (N₂), point position (x), Jacobian (J)
Step 1 · Find shape function 1 (N₁)
Use N₁ = (1 − ξ) ÷ 2
N₁
is
1
at
ξ
=
−1
and
0
at
ξ
=
1.
N₁
=
1 − 0.5
2
N₁
=
0.5
2
N₁
=
0.25
Step 2 · Find shape function 2 (N₂)
Use N₂ = (1 + ξ) ÷ 2
N₂
is
0
at
ξ
=
−1
and
1
at
ξ
=
1.
N₂
=
1 + 0.5
2
N₂
=
1.5
2
N₂
=
0.75
Step 3 · Find point position (x)
Use x = N₁ × x₁ + N₂ × x₂
The nodal positions weighted by the shape functions place the point.
x
=
0.25
×
20
+
0.75
×
60
x
=
5
+
45
x
=
50
mm
Step 4 · Find Jacobian (J)
Use J = (x₂ − x₁) ÷ 2
dx/dξ
=
x₁
dN₁/dξ
+
x₂
dN₂/dξ
=
x₂
−
x₁
2
:
ξ
’s
2
units
cover
L.
J
=
60 − 20
2
J
=
40
2
J
=
20
mm
Check
Put the numbers back in. Both sides should match.
0.25
=
1 − 0.5
2
✓
0.75
=
1 + 0.5
2
✓
50 = 0.25 × 20 + 0.75 × 60 ✓
0.5
=
2 × 50 − 20 − 60
60 − 20
✓
20
=
60 − 20
2
✓
Topic check
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