SME
Structural Analysis · Problem type
Spring system by the direct stiffness method
Use this for “Springs of 100, 50 and 50 kN/m join a wall, node 2, node 3 and a second wall. With 5 kN at node 2 and 10 kN at node 3, find the displacements.”
Diagram
D = 150 × 100 − (−50) × (−50)
D = 12,500
x
=
Dₓ
D
=
1,000
12,500
=
2
25
y
=
Dᵧ
D
=
1,750
12,500
=
7
50
Dₓ and Dᵧ put the right sides in that unknown’s column (shaded).
This is an example. Type your own values to start a new problem.
+
−
−
+
=
K₁₁
Entry 1,1 of K
worked out (not typed) · kN/m
K₁₂
Entry 1,2 of K
worked out (not typed) · kN/m
K₂₂
Entry 2,2 of K
worked out (not typed) · kN/m
D
Determinant of K
worked out (not typed)
Formulas
K₁₁
=
k₁
+
k₂
150 = 100 + 50
K₁₂
=
−
k₂
−50 = −50
K₂₂
=
k₂
+
k₃
100 = 50 + 50
D
=
K₁₁
×
K₂₂
−
K₁₂
2
12,500
=
150
×
100
−
(−50)
2
u₂
=
P₂
×
K₂₂
−
K₁₂
×
P₃
D
0.08
=
5 × 100 − (−50) × 10
12,500
u₃
=
K₁₁
×
P₃
−
K₁₂
×
P₂
D
0.14
=
150 × 10 − (−50) × 5
12,500
Assumptions
•
Spring 1 joins the left wall to node 2, spring 2 joins nodes 2 and 3, spring 3 joins node 3 to the right wall.
•
Loads and displacements are + to the right; the springs act along the line only.
•
The assembled K = [k₁ + k₂, −k₂; −k₂, k₂ + k₃] times [u₂; u₃] gives [P₂; P₃].
Step-by-step
WE KNOW
k₁ = 100 kN/m, k₂ = 50 kN/m, k₃ = 50 kN/m, P₂ = 5 kN, P₃ = 10 kN
FIND
entry 1,1 of K (K₁₁), entry 1,2 of K (K₁₂), entry 2,2 of K (K₂₂), determinant of K (D), displacement of node 2 (u₂), displacement of node 3 (u₃)
Step 1 · Find entry 1,1 of K (K₁₁)
Use K₁₁ = k₁ + k₂
Node 2 is held by springs 1 and 2, so both stiffnesses add on the diagonal.
K₁₁
=
100
+
50
K₁₁
=
150
kN/m
Step 2 · Find entry 1,2 of K (K₁₂)
Use K₁₂ = −k₂
Spring 2 alone joins nodes 2 and 3; off the diagonal it enters with a minus sign.
K₁₂
=
−50
kN/m
Step 3 · Find entry 2,2 of K (K₂₂)
Use K₂₂ = k₂ + k₃
Node 3 is held by springs 2 and 3.
K₂₂
=
50
+
50
K₂₂
=
100
kN/m
Step 4 · Find determinant of K (D)
Use D = K₁₁ × K₂₂ − K₁₂²
The determinant of K: down one diagonal, less the other.
D
=
150
×
100
−
(−50)
2
D
=
150
×
100
−
2,500
D
=
15,000
−
2,500
D
=
12,500
Step 5 · Find displacement of node 2 (u₂)
Use u₂ = (P₂ × K₂₂ − K₁₂ × P₃) ÷ D
Cramer’s
rule:
put
the
loads
[
P₂
;
P₃
]
in
the
first
column,
take
that
determinant
D₁,
and
divide
by
D
.
u₂
=
D₁
D
D₁ = det [[5, −50], [10, 100]] = 5 × 100 − (−50) × 10 = 1,000
u₂
=
D₁
D
=
1,000
12,500
=
0.08
u₂
=
0.08
m
Step 6 · Find displacement of node 3 (u₃)
Use u₃ = (K₁₁ × P₃ − K₁₂ × P₂) ÷ D
Cramer’s
rule:
put
the
loads
in
the
second
column,
take
that
determinant
D₂,
and
divide
by
the
same
D
.
u₃
=
D₂
D
D₂ = det [[150, 5], [−50, 10]] = 150 × 10 − 5 × (−50) = 1,750
u₃
=
D₂
D
=
1,750
12,500
=
0.14
u₃
=
0.14
m
Check
Put the numbers back in. Both sides should match.
150 = 100 + 50 ✓
−50 = −50 ✓
100 = 50 + 50 ✓
12,500
=
150
×
100
−
(−50)
2
✓
0.08
=
5 × 100 − (−50) × 10
12,500
✓
0.14
=
150 × 10 − (−50) × 5
12,500
✓
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