SME
Finite Element Analysis · Problem type
Load between two fixed ends
Use this for “Springs of 1000 N/mm and 2000 N/mm hold node 2 between two walls, and 3000 N pushes it toward node 3. Find its displacement and both reactions.”
Diagram
f₁ = k₁(u₂ − u₁)
f₁ = 1,000 N/mm × 1 mm = 1,000 N
Spring 1 is in tension.
f₂ = k₂(u₃ − u₂)
f₂ = 2,000 N/mm × −1 mm = −2,000 N
Spring 2 is in compression.
Each free node balances: its load equals the left element’s force less the right one’s.
This is an example. Type your own values to start a new problem.
u₂ (mm)
=
F (N)
k₁ (N/mm)
+
k₂ (N/mm)
R₁
Reaction at node 1
calculated · N
R₃
Reaction at node 3
calculated · N
Formulas
u₂
=
F
k₁
+
k₂
1
=
3,000
1,000 + 2,000
R₁
=
−
k₁
×
u₂
−1,000 = −1,000 × 1
R₃
=
−
k₂
×
u₂
−2,000 = −2,000 × 1
R₁
+
R₃
+
F
=
0
−1,000 + (−2,000) + 3,000 = 0
Assumptions
•
Axial springs or bars, loaded only at the nodes; small displacements, linear elastic.
•
An element’s force is its stiffness times its stretch, k(uⱼ − uᵢ): tension is +.
•
Nodes 1 and 3 are fixed (u₁ = u₃ = 0); F acts at node 2, toward node 3.
Step-by-step
WE KNOW
k₁ = 1,000 N/mm, k₂ = 2,000 N/mm, F = 3,000 N
FIND
displacement of node 2 (u₂), reaction at node 1 (R₁), reaction at node 3 (R₃)
Step 1 · Find displacement of node 2 (u₂)
Use u₂ = F ÷ (k₁ + k₂)
Strike
out
the
rows
of
the
fixed
nodes
1
and
3:
what
is
left
is
(
k₁
+
k₂
)
u₂
=
F
.
u₂
=
3,000
1,000 + 2,000
u₂
=
3,000
3,000
u₂
=
1
mm
Step 2 · Find reaction at node 1 (R₁)
Use R₁ = −k₁ × u₂
Row
1
of
the
full
K:
R₁
=
k₁u₁
−
k₁
u₂
,
with
u₁
=
0.
Spring
1
stretches
and
pulls
the
wall.
R₁
=
−1,000
×
1
R₁
=
−1,000
N
Step 3 · Find reaction at node 3 (R₃)
Use R₃ = −k₂ × u₂
Row
3
of
the
full
K:
R₃
=
−k₂
u₂
+
k₂u₃,
with
u₃
=
0.
Spring
2
is
squeezed
and
pushes
the
wall.
R₃
=
−2,000
×
1
R₃
=
−2,000
N
Check
Put the numbers back in. Both sides should match.
1
=
3,000
1,000 + 2,000
✓
−1,000 = −1,000 × 1 ✓
−2,000 = −2,000 × 1 ✓
−1,000 + (−2,000) + 3,000 = 0 ✓
Topic check
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