SME
Finite Element Analysis · Problem type
Quadratic bar element
Use this for “A three-node element has nodal displacements 2, 30 and 50 μm at ξ = −1, 0 and 1. Find u at ξ = 0.5.”
Chart
u(0.5) = 41
Ringed: the nodes (−1, 2), (0, 30), (1, 50)
This is an example. Type your own values to start a new problem.
u (μm)
=
N₁
×
u₁ (μm)
+
N₂
×
u₂ (μm)
+
N₃
×
u₃ (μm)
ξ
Natural coordinate
example
Formulas
N₁
=
ξ
×
ξ
− 1
2
−0.125
=
0.5
×
0.5 − 1
2
N₂
=
(1
−
ξ
)
×
(1
+
ξ
)
0.75 = (1 − 0.5) × (1 + 0.5)
N₃
=
ξ
×
ξ
+ 1
2
0.375
=
0.5
×
0.5 + 1
2
u
=
N₁
×
u₁
+
N₂
×
u₂
+
N₃
×
u₃
41 = −0.125 × 2 + 0.75 × 30 + 0.375 × 50
Assumptions
•
Nodes 1, 2 and 3 sit at ξ = −1, 0 and 1; node 2 is at the middle of the element.
•
Each Nᵢ is 1 at its own node and 0 at the other two; together they add to 1.
•
u(ξ) is the parabola through the three nodal displacements.
Step-by-step
WE KNOW
u₁ = 2 μm, u₂ = 30 μm, u₃ = 50 μm, ξ = 0.5
FIND
shape function 1 (N₁), shape function 2 (N₂), shape function 3 (N₃), displacement at ξ (u)
Step 1 · Find shape function 1 (N₁)
Use N₁ = ξ × (ξ − 1) ÷ 2
N₁
is
0
at
ξ
=
0
and
ξ
=
1
(its
factors)
and
1
at
ξ
=
−1,
hence
the
2
.
N₁
=
0.5
×
0.5 − 1
2
N₁
=
0.5
×
−0.5
2
N₁
=
−
0.25
2
N₁
=
−0.125
Step 2 · Find shape function 2 (N₂)
Use N₂ = (1 − ξ) × (1 + ξ)
N₂
is
0
at
both
end
nodes
and
1
at
the
middle
node:
1
−
ξ
2
.
N₂
=
(1
−
0.5)
×
(1
+
0.5)
N₂
=
0.5
×
1.5
N₂
=
0.75
Step 3 · Find shape function 3 (N₃)
Use N₃ = ξ × (ξ + 1) ÷ 2
N₃
is
0
at
ξ
=
−1
and
ξ
=
0
and
1
at
ξ
=
1,
hence
the
2
.
N₃
=
0.5
×
0.5 + 1
2
N₃
=
0.5
×
1.5
2
N₃
=
0.75
2
N₃
=
0.375
Step 4 · Find displacement at ξ (u)
Use u = N₁ × u₁ + N₂ × u₂ + N₃ × u₃
Each
nodal
displacement
is
weighted
by
its
shape
function
at
ξ
,
then
the
three
are
added.
u
=
−0.125
×
2
+
0.75
×
30
+
0.375
×
50
u
=
−0.25
+
22.5
+
18.75
u
=
41
μm
Check
Put the numbers back in. Both sides should match.
−0.125
=
0.5
×
0.5 − 1
2
✓
0.75 = (1 − 0.5) × (1 + 0.5) ✓
0.375
=
0.5
×
0.5 + 1
2
✓
41 = −0.125 × 2 + 0.75 × 30 + 0.375 × 50 ✓
Topic check
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