SME
Structural Analysis · Problem type
Two-span continuous beam by moment distribution
Use this for “Two spans of 6 m and 4 m, pinned at the far ends, carry 12 kN/m. Find the moment over the middle support.”
Diagram
ΣF = 0: RA + RB + RC = 29 + 77.5 + 13.5 = 120 kN, the total load.
Moments clockwise + on each member end; 1 cycle balance every joint.
This is an example. Type your own values to start a new problem.
L₁
Span AB
example
L₂
Span BC
example
w
Uniform load
example · kN/m
k₁
Stiffness of AB
calculated · m⁻¹
k₂
Stiffness of BC
calculated · m⁻¹
DF₁
Distribution factor BA
calculated
DF₂
Distribution factor BC
calculated
FEM₁
Fixed-end moment BA
calculated · kN·m
FEM₂
Fixed-end moment BC
calculated · kN·m
MB
Moment over B
calculated · kN·m
Formulas
k₁
=
3
L₁
0.5
=
3
6
k₂
=
3
L₂
0.75
=
3
4
DF₁
=
k₁
k₁
+
k₂
0.4
=
0.5
0.5 + 0.75
DF₂ = 1 − DF₁
0.6 = 1 − 0.4
FEM₁
=
w
×
L₁
2
8
54
=
12
×
6
2
8
FEM₂
=
w
×
L₂
2
8
24
=
12
×
4
2
8
MB = FEM₁ − DF₁ × (FEM₁ − FEM₂)
42 = 54 − 0.4 × (54 − 24)
Assumptions
•
Far ends A and C pinned: each span is 3EI ÷ L stiff; EI is the same, so k = 3 ÷ L.
•
Nothing carries over to a pinned end, so one balance at B settles the moment there.
•
w covers both spans; moments are sizes, MB bending the beam hogging over B.
Step-by-step
WE KNOW
L₁ = 6 m, L₂ = 4 m, w = 12 kN/m
FIND
stiffness of AB (k₁), stiffness of BC (k₂), distribution factor BA (DF₁), distribution factor BC (DF₂), fixed-end moment BA (FEM₁), fixed-end moment BC (FEM₂), moment over B (MB)
Step 1 · Find stiffness of AB (k₁)
Use k₁ = 3 ÷ L₁
A
span
pinned
at
its
far
end
is
3E
I
L
stiff;
EI
is
shared,
so
keep
3
L
.
k₁
=
3
6
k₁
=
0.5
m⁻¹
Step 2 · Find stiffness of BC (k₂)
Use k₂ = 3 ÷ L₂
The same for span BC.
k₂
=
3
4
k₂
=
0.75
m⁻¹
Step 3 · Find distribution factor BA (DF₁)
Use DF₁ = k₁ ÷ (k₁ + k₂)
Each member at B takes its share of the joint’s stiffness.
DF₁
=
0.5
0.5 + 0.75
DF₁
=
0.5
1.25
DF₁ = 0.4
Step 4 · Find distribution factor BC (DF₂)
Use DF₂ = 1 − DF₁
The factors at a joint add to 1.
DF₂ = 1 − 0.4
DF₂ = 0.6
Step 5 · Find fixed-end moment BA (FEM₁)
Use FEM₁ = w × L₁² ÷ 8
With
its
far
end
pinned,
a
uniform
load
needs
wL²
8
to
hold
the
end
at
B
still.
FEM₁
=
12
×
6
2
8
FEM₁
=
12
×
36
8
FEM₁
=
432
8
FEM₁ = 54 kN·m
Step 6 · Find fixed-end moment BC (FEM₂)
Use FEM₂ = w × L₂² ÷ 8
The same for span BC.
FEM₂
=
12
×
4
2
8
FEM₂
=
12
×
16
8
FEM₂
=
192
8
FEM₂ = 24 kN·m
Step 7 · Find moment over B (MB)
Use MB = FEM₁ − DF₁ × (FEM₁ − FEM₂)
Balance B: the unbalance FEM₁ − FEM₂ is shared by the factors, and BA keeps FEM₁ less its share.
MB = 54 − 0.4 × (54 − 24)
MB = 54 − 0.4 × 30
MB = 54 − 12
MB = 42 kN·m
Check
Put the numbers back in. Both sides should match.
0.5
=
3
6
✓
0.75
=
3
4
✓
0.4
=
0.5
0.5 + 0.75
✓
0.6 = 1 − 0.4 ✓
54
=
12
×
6
2
8
✓
24
=
12
×
4
2
8
✓
42 = 54 − 0.4 × (54 − 24) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.