SME
Finite Element Analysis · Problem type
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.”
Diagram
J₁₁ = 30 mm, J₁₂ = 0 mm, J₂₁ = +10 mm, J₂₂ = 20 mm
det J = J₁₁J₂₂ − J₁₂J₂₁ = 30 × 20 − 0 × 10 = 600 − 0 = 600 mm²
Down the main diagonal, take away the other diagonal
This is an example. Type your own values to start a new problem.
a
Base length
example
h
Height
example
s
Top lean
example
J₁₁
Entry ∂x/∂ξ
calculated · mm
J₂₁
Entry ∂x/∂η
calculated · mm
J₂₂
Entry ∂y/∂η
calculated · mm
θ
Skew angle
calculated · °
det J
Jacobian determinant
calculated · mm²
A
Element area
calculated
Formulas
J₁₁
=
a
2
30
=
60
2
J₂₁
=
s
2
+10
=
20
2
J₂₂
=
h
2
20
=
40
2
θ
=
tan⁻¹(
J₂₁
J₂₂
)
+26.5651
=
tan⁻¹(
10
20
)
det
J
=
J₁₁
×
J₂₂
600 = 30 × 20
A
=
4
×
(det
J)
2,400 = 4 × 600
Assumptions
•
Nodes (0, 0), (a, 0), (a + s, h) and (s, h), counterclockwise from node 1.
•
A parallelogram has the same J at every point; J₁₂ = ∂y/∂ξ = 0 here.
•
det J > 0 means the nodes run counterclockwise and the element is not folded.
Step-by-step
WE KNOW
a = 60 mm, h = 40 mm, s = +20 mm
FIND
entry ∂x/∂ξ (J₁₁), entry ∂x/∂η (J₂₁), entry ∂y/∂η (J₂₂), skew angle (θ), Jacobian determinant (det J), element area (A)
Step 1 · Find entry ∂x/∂ξ (J₁₁)
Use J₁₁ = a ÷ 2
Along
ξ
the
bottom
and
top
edges
both
run
a
;
ξ
covers
them
in
2
units,
so
∂x/∂ξ
=
a
2
.
J₁₁
=
60
2
J₁₁
=
30
mm
Step 2 · Find entry ∂x/∂η (J₂₁)
Use J₂₁ = s ÷ 2
Going
up
η,
x
shifts
by
the
lean
s
over
2
units,
so
∂x/∂η
=
s
2
.
J₂₁
=
20
2
J₂₁
=
+10
mm
Step 3 · Find entry ∂y/∂η (J₂₂)
Use J₂₂ = h ÷ 2
Going
up
η,
y
rises
the
height
h
over
2
units,
so
∂y/∂η
=
h
2
.
J₂₂
=
40
2
J₂₂
=
20
mm
Step 4 · Find skew angle (θ)
Use θ = tan⁻¹(J₂₁ ÷ J₂₂)
The
sides
lean
s
across
for
h
up,
so
the
skew
from
square
is
tan⁻¹(
s
h
),
the
same
as
J₂₁
over
J₂₂
.
θ
=
tan⁻¹(
10
20
)
θ
=
+26.5651°
Step 5 · Find Jacobian determinant (det J)
Use det J = J₁₁ × J₂₂
det
J
=
J₁₁
J₂₂
−
J₁₂
J₂₁
,
and
J₁₂
=
0,
so
only
the
main
diagonal
is
left.
det J = 30 × 20
det J = 600 mm²
Step 6 · Find element area (A)
Use A = 4 × (det J)
dA
=
det
J
dξ
dη
and
the
parent
square’
s
area
is
2
×
2
=
4;
the
result
equals
base
times
height.
A
=
4
×
600
A
=
2,400
mm²
Check
Put the numbers back in. Both sides should match.
30
=
60
2
✓
+10
=
20
2
✓
20
=
40
2
✓
+26.5651
=
tan⁻¹(
10
20
)
✓
600 = 30 × 20 ✓
2,400 = 4 × 600 ✓
Topic check
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