SME
Finite Element Analysis · Problem type
One-element cantilever beam
Use this for “Model a 2 m cantilever, E = 200 GPa, I = 8 × 10⁶ mm⁴, as one beam element with 5 kN at its free end. Solve for v and θ.”
Diagram
2 equations in 2 unknowns, the right-hand sides after the bar.
The solution: v = 25/3, Lθ = 25/2.
This is an example. Type your own values to start a new problem.
P
Tip load
example
L
Length
example
E
Elastic modulus
example
I
Second moment of area
example
q
Scaled load
worked out (not typed) · mm
v
Tip deflection
calculated · mm
θ
Tip rotation
calculated · rad
Lθ
Scaled rotation
worked out (not typed) · mm
Formulas
q
=
P
×
L
3
E
×
I
25
=
5,000
×
2,
000
3
200,000 × 8,000,000
v
=
P
×
L
3
3 ×
E
×
I
8.3333
=
5,000
×
2,
000
3
3 × 200,000 × 8,000,000
θ
=
P
×
L
2
2 ×
E
×
I
0.00625
=
5,000
×
2,
000
2
2 × 200,000 × 8,000,000
Lθ
=
L
×
θ
12.5 = 2,000 × 0.00625
Formulas are worked in N, mm, MPa. The step-by-step shows the conversions.
Assumptions
•
One beam element, node 1 fixed: only node 2’s deflection v and rotation θ are free.
•
Node 2’s rows: (EI ÷ L³)[[12, −6L], [−6L, 4L²]](v, θ) = (P, 0); v is + along P.
•
Written for (v, Lθ), the matrix is [[12, −6], [−6, 4]] and the load column (q, 0), q = PL³ ÷ (EI).
•
One element is exact for a tip load: its cubic shape matches the beam’s real curve.
Step-by-step
WE KNOW
P = 5 kN, L = 2 m, E = 200 GPa, I = 8,000,000 mm⁴
FIND
scaled load (q), tip deflection (v), tip rotation (θ), scaled rotation (Lθ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N, mm, MPa).
P = 5 kN = 5,000 N (1 kN = 1000 N)
L = 2 m = 2,000 mm (1 m = 1000 mm)
E = 200 GPa = 200,000 MPa (1 GPa = 1000 MPa)
Step 2 · Find scaled load (q)
Use q = P × L³ ÷ (E × I)
Multiply
K(
v
,
θ
)
=
(
P
,
0)
by
L
3
EI
:
the
load
column
becomes
(
PL
³
EI
,
0).
q
=
5,000
×
2,
000
3
200,000 × 8,000,000
q
=
5,000
×
2,
000
3
1,600,000,000,000
q
=
5,000
×
8,000,000,000
1,600,000,000,000
q
=
40,000,000,000,000
1,600,000,000,000
q
=
25
mm
Step 3 · Find tip deflection (v)
Use v = P × L³ ÷ (3 × E × I)
Row-reduce
[[12,
−6
|
q
],
[−6,
4
|
0]],
as
the
picture
does:
v
=
q
3,
which
is
PL
³
3
EI
.
v
=
5,000
×
2,
000
3
3 × 200,000 × 8,000,000
v
=
5,000
×
2,
000
3
600,000 × 8,000,000
v
=
5,000
×
2,
000
3
4,800,000,000,000
v
=
5,000
×
8,000,000,000
4,800,000,000,000
v
=
40,000,000,000,000
4,800,000,000,000
v
=
8.3333
mm
Step 4 · Find tip rotation (θ)
Use θ = P × L² ÷ (2 × E × I)
The
same
row
reduction
gives
Lθ
=
q
2,
so
θ
=
PL
²
2
EI
,
in
radians.
θ
=
5,000
×
2,
000
2
2 × 200,000 × 8,000,000
θ
=
5,000
×
2,
000
2
400,000 × 8,000,000
θ
=
5,000
×
2,
000
2
3,200,000,000,000
θ
=
5,000
×
4,000,000
3,200,000,000,000
θ
=
20,000,000,000
3,200,000,000,000
θ
=
0.00625
rad
Step 5 · Find scaled rotation (Lθ)
Use Lθ = L × θ
The rotation times the length: the second unknown of the scaled system, in mm.
Lθ
=
2,000
×
0.00625
Lθ
=
12.5
mm
Check
Put the numbers back in. Both sides should match.
25
=
5,000
×
2,
000
3
200,000 × 8,000,000
✓
8.3333
=
5,000
×
2,
000
3
3 × 200,000 × 8,000,000
✓
0.00625
=
5,000
×
2,
000
2
2 × 200,000 × 8,000,000
✓
12.5 = 2,000 × 0.00625 ✓
Topic check
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