SME
Structural Analysis
Indeterminate structures
Use this for “A 6 m beam, fixed at A and propped at B, carries 20 kN/m. Find the reactions and the wall moment.”
Diagram
ΣF = 0: RA + RB = 75 + 45 = 120 kN, the total load.
The wall at A holds MA = 90 kN·m, counterclockwise.
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.
RB (kN)
=
3
×
w (kN/m)
×
L (m)
÷
8
RA (kN)
=
5
×
w (kN/m)
×
L (m)
÷
8
MA (kN·m)
=
w (kN/m)
×
(
L (m)
)
2
÷
8
Formulas
RB
=
3
×
w
×
L
8
45
=
3
×
20
×
6
8
RA
=
5
×
w
×
L
8
75
=
5
×
20
×
6
8
MA
=
w
×
L
2
8
90
=
20
×
6
2
8
Assumptions
•
Fixed at A, a roller (the prop) at B; EI, the bending stiffness, is the same along the beam.
•
Compatibility: without the prop, B sags wL⁴ ÷ (8EI); the prop lifts it back RBL³ ÷ (3EI).
•
w covers the whole span; reactions are + up and the wall moment MA bends the beam hogging.
Step-by-step
WE KNOW
w = 20 kN/m, L = 6 m
FIND
prop reaction (RB), wall reaction (RA), wall moment (MA)
Step 1 · Find prop reaction (RB)
Use RB = 3 × w × L ÷ 8
Set
the
prop’s
lift
equal
to
the
sag,
wL
⁴
8EI
=
RBL
3
3EI
,
and
divide
both
sides
by
L
3
3EI
.
RB
=
3
×
20
×
6
8
RB
=
60
×
6
8
RB
=
360
8
RB = 45 kN
Step 2 · Find wall reaction (RA)
Use RA = 5 × w × L ÷ 8
The
two
supports
carry
wL
together:
RA
=
wL
−
3
w
L
8
.
RA
=
5
×
20
×
6
8
RA
=
100
×
6
8
RA
=
600
8
RA = 75 kN
Step 3 · Find wall moment (MA)
Use MA = w × L² ÷ 8
Moments
about
A:
the
load
wL
acts
at
L
2
and
the
prop
at
L
,
so
MA
=
wL
²
2
−
RBL
=
wL
²
8
.
MA
=
20
×
6
2
8
MA
=
20
×
36
8
MA
=
720
8
MA = 90 kN·m
Check
Put the numbers back in. Both sides should match.
45
=
3
×
20
×
6
8
✓
75
=
5
×
20
×
6
8
✓
90
=
20
×
6
2
8
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.