SME
Finite Element Analysis · Problem type
Truss element stiffness matrix
Use this for “A truss element has AE ÷ L = 50,000 N/mm and lies at 30°. Find the terms of its 4 × 4 global stiffness matrix.”
Diagram
λ is 4 × 1 and λᵀ is 1 × 4, so M is 4 × 4.
Row 1 of λ times column 2 of λᵀ gives entry (1, 2): m₁₂ = (−0.87) × (−1/2) = 0.43.
Tap an entry of M to see its row and column.
This is an example. Type your own values to start a new problem.
k₁₁ (N/mm)
=
k (N/mm)
×
c
2
k₁₂ (N/mm)
=
k (N/mm)
×
c
×
s
k₂₂ (N/mm)
=
k (N/mm)
×
s
2
Each term is k times an entry of M: k₁₂ = k × m₁₂ with m₁₂ = c × s, and likewise m₁₁ = c² and m₂₂ = s².
θ
Angle from x
example · °
λ₁
Entry 1 of λ
worked out (not typed)
λ₂
Entry 2 of λ
worked out (not typed)
Formulas
c
=
cos(
θ
)
0.866 = cos(30°)
s
=
sin(
θ
)
0.5 = sin(30°)
λ₁
=
−1
×
c
−0.866 = −1 × 0.866
λ₂
=
−1
×
s
−0.5 = −1 × 0.5
k₁₁
=
k
×
c
×
c
37,500 = 50,000 × 0.866 × 0.866
k₁₂
=
k
×
c
×
s
+21,650.6351 = 50,000 × 0.866 × 0.5
k₂₂
=
k
×
s
×
s
12,500 = 50,000 × 0.5 × 0.5
Assumptions
•
kₑ = k × λλᵀ with λ = (−c, −s, c, s) and k = AE ÷ L; the picture shows M = λλᵀ.
•
c = cos θ and s = sin θ, with θ from the global x axis, node 1 to node 2.
•
Every entry is ±kc², ±kcs or ±ks²: + in the two diagonal blocks, − in the other two.
Step-by-step
WE KNOW
k = 50,000 N/mm, θ = 30°
FIND
cosine of θ (c), sine of θ (s), entry 1 of λ (λ₁), entry 2 of λ (λ₂), x–x term (k₁₁), x–y term (k₁₂), y–y term (k₂₂)
Step 1 · Find cosine of θ (c)
Use c = cos(θ)
The
bar’
s
x
direction
cosine:
how
much
of
its
length
runs
along
x.
c
=
cos(30°)
c
=
0.86603
Step 2 · Find sine of θ (s)
Use s = sin(θ)
The
bar’
s
y
direction
cosine:
how
much
of
its
length
runs
along
y.
s
=
sin(30°)
s
=
0.5
Step 3 · Find entry 1 of λ (λ₁)
Use λ₁ = −1 × c
Node
1’
s
x
entry
in
λ
is
minus
node
2’
s
,
c
.
λ₁
=
−1
×
0.86603
λ₁
=
−0.866
Step 4 · Find entry 2 of λ (λ₂)
Use λ₂ = −1 × s
Node
1’
s
y
entry
in
λ
is
minus
node
2’
s
,
s
.
λ₂
=
−1
×
0.5
λ₂
=
−0.5
Step 5 · Find x–x term (k₁₁)
Use k₁₁ = k × c × c
Row
1,
column
1
of
kₑ:
the
x
force
at
node
1
for
an
x
move
there,
k
times
c
2
.
k₁₁
=
50,000
×
0.86603
×
0.86603
k₁₁
=
43,301.5
×
0.866
k₁₁
=
37,500
N/mm
Step 6 · Find x–y term (k₁₂)
Use k₁₂ = k × c × s
Row
1,
column
2
of
kₑ:
the
x
force
at
node
1
for
a
y
move
there,
k
times
cs.
k₁₂
=
50,000
×
0.86603
×
0.5
k₁₂
=
43,301.5
×
0.5
k₁₂
=
+21,650.64
N/mm
Step 7 · Find y–y term (k₂₂)
Use k₂₂ = k × s × s
Row
2,
column
2
of
kₑ:
the
y
force
at
node
1
for
a
y
move
there,
k
times
s
2
.
k₂₂
=
50,000
×
0.5
×
0.5
k₂₂
=
25,000
×
0.5
k₂₂
=
12,500
N/mm
Check
Put the numbers back in. Both sides should match.
0.86603 = cos(30°) ✓
0.5 = sin(30°) ✓
−0.866 = −1 × 0.86603 ✓
−0.5 = −1 × 0.5 ✓
37,500 = 50,000 × 0.86603 × 0.86603 ✓
+21,650.64 = 50,000 × 0.86603 × 0.5 ✓
12,500 = 50,000 × 0.5 × 0.5 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.