SME
Finite Element Analysis
Direct stiffness method
Use this for “Springs of 1000 N/mm and 500 N/mm are joined end to end, fixed at node 1, and 2000 N pulls node 3. Find the node displacements, the spring forces and the reaction.”
Diagram
f₁ = 2,000 N.
Spring 1 is in tension.
f₂ = 2,000 N.
Spring 2 is in tension.
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.
k₁
Stiffness of spring 1
example
k₂
Stiffness of spring 2
example
F
Load at node 3
example
u₂
Displacement of node 2
calculated
u₃
Displacement of node 3
calculated
f₁
Force in spring 1
calculated
f₂
Force in spring 2
calculated
R₁
Reaction at node 1
calculated
Formulas
u₂
=
F
k₁
2
=
2,000
1,000
u₃
=
u₂
+
F
k₂
6
=
2
+
2,000
500
f₁
=
k₁
×
u₂
2,000 = 1,000 × 2
f₂
=
k₂
×
(
u₃
−
u₂
)
2,000 = 500 × (6 − 2)
f₂
=
F
2,000 = 2,000
f₁
=
f₂
2,000 = 2,000
R₁
=
−
f₁
−2,000 = −2,000
R₁
=
−
k₁
×
u₂
−2,000 = −1,000 × 2
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 +.
•
Node 1 is fixed (u₁ = 0); the load F pulls node 3 away from the wall.
Step-by-step
WE KNOW
k₁ = 1,000 N/mm, k₂ = 500 N/mm, F = 2,000 N
FIND
displacement of node 2 (u₂), displacement of node 3 (u₃), force in spring 1 (f₁), force in spring 2 (f₂), reaction at node 1 (R₁)
Step 1 · Find displacement of node 2 (u₂)
Use u₂ = F ÷ k₁
Write
K
u
=
(0,
F
)
as
an
augmented
matrix
and
add
row
2
to
row
1:
the
k₂
terms
cancel,
leaving
k₁
u₂
=
F
.
u₂
=
2,000
1,000
Start: [[1500, −500 | 0], [−500, 500 | 2,000]]
R₁
→
R₁
+
R₂:
[[1000,
0
|
2,000],
[−500,
500
|
2,000]]
u₂
=
2
mm
Step 2 · Find displacement of node 3 (u₃)
Use u₃ = u₂ + F ÷ k₂
Row
2
of
K
u
=
(0,
F
)
reads
k₂
(
u₃
−
u₂
)
=
F
:
node
3
moves
u₂
plus
spring
2’s
stretch.
u₃
=
2
+
2,000
500
u₃
=
2
+
4
u₃
=
6
mm
Step 3 · Find force in spring 1 (f₁)
Use f₁ = k₁ × u₂
Spring
1
stretches
by
u₂
−
u₁,
and
u₁
=
0
at
the
wall.
f₁
=
1,000
×
2
f₁
=
2,000
N
Step 4 · Find force in spring 2 (f₂)
Use f₂ = k₂ × (u₃ − u₂)
Spring 2 stretches by the difference of its two nodes’ displacements.
f₂
=
500
×
(6
−
2)
f₂
=
500
×
4
f₂
=
2,000
N
Step 5 · Find reaction at node 1 (R₁)
Use R₁ = −f₁
Node 1 balances: the wall holds it back against spring 1.
R₁
=
−2,000
N
Check
Put the numbers back in. Both sides should match.
2
=
2,000
1,000
✓
6
=
2
+
2,000
500
✓
2,000 = 1,000 × 2 ✓
2,000 = 500 × (6 − 2) ✓
2,000 = 2,000 ✓
2,000 = 2,000 ✓
−2,000 = −2,000 ✓
−2,000 = −1,000 × 2 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
One bar element: stiffness, force and stress
Use this for “A steel bar element 1,000 mm long has an area of 100 mm² and E = 200 GPa. Its nodes move 0.02 mm and 0.12 mm. Find k, the element force and the stress.”
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.”