SME
Structural Analysis
Matrix methods
Use this for “Steel bars of 1000 mm² × 2 m and 500 mm² × 2 m hang in series from a support, with 30 kN at the end. Find the node displacements.”
Diagram
Bar 1: A₁ = 1,000 mm², L₁ = 2 m.
Bar 2: A₂ = 500 mm², L₂ = 2 m.
f₁ = k₁(u₂ − u₁)
f₁ = 100 kN/mm × 0.3 mm = 30 kN
Bar 1 is in tension.
f₂ = k₂(u₃ − u₂)
f₂ = 50 kN/mm × 0.6 mm = 30 kN
Bar 2 is in tension.
Each free node balances: its load equals the left element’s force less the right one’s.
Elastic modulus E = 200 GPa
This is an example. Type your own values to start a new problem.
E
Elastic modulus
example · GPa
A₁
Area of bar 1
example
L₁
Length of bar 1
example
A₂
Area of bar 2
example
L₂
Length of bar 2
example
k₁
Stiffness of bar 1
calculated · kN/mm
k₂
Stiffness of bar 2
calculated · kN/mm
P
Load at node 3
example
u₂
Displacement of node 2
calculated
u₃
Displacement of node 3
calculated
Formulas
k₁
=
A₁
×
E
L₁
100
=
1,000
×
200
2,000
k₂
=
A₂
×
E
L₂
50
=
500
×
200
2,000
u₂
=
P
k₁
0.3
=
30
100
u₃
=
(
k₁
+
k₂
)
×
P
k₁
×
k₂
0.9
=
(100
+
50)
×
30
100 × 50
Formulas are worked in mm. The step-by-step shows the conversions.
Assumptions
•
Node 1 is fixed; the load P pulls along the bars at node 3; axial members only, small displacements.
•
Each bar is a spring of stiffness k = AE ÷ L between its two nodes.
•
Node 1’s row and column struck out leave [k₁ + k₂, −k₂; −k₂, k₂][u₂; u₃] = [0; P].
Step-by-step
WE KNOW
E = 200 GPa, A₁ = 1,000 mm², L₁ = 2 m, A₂ = 500 mm², L₂ = 2 m, P = 30 kN
FIND
stiffness of bar 1 (k₁), stiffness of bar 2 (k₂), displacement of node 2 (u₂), displacement of node 3 (u₃)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to mm).
L₁ = 2 m = 2,000 mm (1 m = 1000 mm)
L₂ = 2 m = 2,000 mm (1 m = 1000 mm)
Step 2 · Find stiffness of bar 1 (k₁)
Use k₁ = A₁ × E ÷ L₁
A
bar
pulls
back
like
a
spring
of
stiffness
AE
L
;
mm²
×
GPa
mm
gives
kN/mm.
k₁
=
1,000
×
200
2,000
k₁
=
200,000
2,000
k₁
=
100
kN/mm
Step 3 · Find stiffness of bar 2 (k₂)
Use k₂ = A₂ × E ÷ L₂
A
bar
pulls
back
like
a
spring
of
stiffness
AE
L
;
mm²
×
GPa
mm
gives
kN/mm.
k₂
=
500
×
200
2,000
k₂
=
100,000
2,000
k₂
=
50
kN/mm
Step 4 · Find displacement of node 2 (u₂)
Use u₂ = P ÷ k₁
Cramer’s
rule
on
K
u
=
F:
u₂
=
D₁
D
,
which
comes
to
k₂
P
k₁k₂
=
P
k₁
.
u₂
=
D₁
D
D = det [[150, −50], [−50, 50]] = 150 × 50 − (−50) × (−50) = 5,000
D₁ = det [[0, −50], [30, 50]] = 0 × 50 − (−50) × 30 = 1,500
u₂
=
D₁
D
=
1,500
5,000
=
0.3
u₂
=
0.3
mm
Step 5 · Find displacement of node 3 (u₃)
Use u₃ = (k₁ + k₂) × P ÷ (k₁ × k₂)
Cramer’s
rule
again:
D₂
has
[0;
P
]
in
the
second
column,
and
D
works
out
to
k₁
k₂
.
u₃
=
D₂
D
D = det [[150, −50], [−50, 50]] = 150 × 50 − (−50) × (−50) = 5,000
D₂ = det [[150, 0], [−50, 30]] = 150 × 30 − 0 × (−50) = 4,500
u₃
=
D₂
D
=
4,500
5,000
=
0.9
u₃
=
0.9
mm
Check
Put the numbers back in. Both sides should match.
100
=
1,000
×
200
2,000
✓
50
=
500
×
200
2,000
✓
0.3
=
30
100
✓
0.9
=
(100
+
50)
×
30
100 × 50
✓
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Spring system by the direct stiffness method
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