SME
Structural Analysis · Problem type
Fixed-end moments
Use this for “A 6 m beam fixed at both ends carries 30 kN at midspan and 10 kN/m. Find the fixed-end moments.”
Diagram
ΣF = 0: RA + RB = 45 + 45 = 90 kN, the total load.
The wall at A holds MA = 52.5 kN·m, counterclockwise.
The wall at B holds MB = 52.5 kN·m, clockwise.
M peaks where V crosses 0: M = 37.5 kN·m at x = 3 m.
The area under V between two points is the change in M.
The bent shape is drawn bigger than life; it meets every support.
This is an example. Type your own values to start a new problem.
FEMP (kN·m)
=
P (kN)
×
L (m)
÷
8
FEMw (kN·m)
=
w (kN/m)
×
(
L (m)
)
2
÷
12
M (kN·m)
=
FEMP (kN·m)
+
FEMw (kN·m)
Formulas
FEMP
=
P
×
L
8
22.5
=
30
×
6
8
FEMw
=
w
×
L
2
12
30
=
10
×
6
2
12
M
=
FEMP
+
FEMw
52.5 = 22.5 + 30
Assumptions
•
Both ends fixed, so neither end turns; EI, the bending stiffness, is the same along the span.
•
P stands at midspan and w covers the whole span; their end moments add (superposition).
•
Each wall holds the same moment, bending the beam hogging at the ends.
Step-by-step
WE KNOW
P = 30 kN, w = 10 kN/m, L = 6 m
FIND
moment from P (FEMP), moment from w (FEMw), fixed-end moment (M)
Step 1 · Find moment from P (FEMP)
Use FEMP = P × L ÷ 8
Holding
both
end
slopes
at
0
under
a
load
at
midspan
takes
PL
8
at
each
wall.
FEMP
=
30
×
6
8
FEMP
=
180
8
FEMP = 22.5 kN·m
Step 2 · Find moment from w (FEMw)
Use FEMw = w × L² ÷ 12
Holding
both
end
slopes
at
0
under
a
uniform
load
takes
wL
²
12
at
each
wall.
FEMw
=
10
×
6
2
12
FEMw
=
10
×
36
12
FEMw
=
360
12
FEMw = 30 kN·m
Step 3 · Find fixed-end moment (M)
Use M = FEMP + FEMw
Both loads act together, so their end moments add.
M
=
22.5
+
30
M
=
52.5
kN·m
Check
Put the numbers back in. Both sides should match.
22.5
=
30
×
6
8
✓
30
=
10
×
6
2
12
✓
52.5 = 22.5 + 30 ✓
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Indeterminate structures
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Two-span continuous beam by moment distribution
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