SME
Statics · Problem type
Reactions of a simply supported beam
Use this for “A 6 m beam on a pin and a roller carries 12 kN 2 m from the pin. Find both reactions.”
Diagram
ΣF = 0: RA + RB = 8 + 4 = 12 kN, the total load.
This is an example. Type your own values to start a new problem.
RA (kN)
×
L (m)
=
P (kN)
×
(
L (m)
−
a (m)
)
RB (kN)
×
L (m)
=
P (kN)
×
a (m)
Formulas
RA
×
L
=
P
×
(
L
−
a
)
8 × 6 = 12 × (6 − 2)
RB
×
L
=
P
×
a
4 × 6 = 12 × 2
Assumptions
•
A pin at A and a roller at B; the beam’s own weight is ignored.
•
The load is vertical, so both reactions are vertical; + up, and counterclockwise moments +.
•
Moments about B give RA, and moments about A give RB; then RA + RB = P checks them.
Step-by-step
WE KNOW
L = 6 m, P = 12 kN, a = 2 m
FIND
reaction at A (RA), reaction at B (RB)
Step 1 · Find reaction at A (RA)
Use RA × L = P × (L − a)
Take
moments
about
B:
RA
turns
the
beam
one
way
with
arm
L
,
the
load
the
other
way
with
arm
L
−
a
.
RA
=
P
×
L
−
a
L
RA
=
12
×
6 − 2
6
RA
=
12
×
4
6
RA
=
48
6
RA = 8 kN
Step 2 · Find reaction at B (RB)
Use RB × L = P × a
Take
moments
about
A:
RB
with
arm
L
balances
the
load
with
arm
a
.
RB
=
P
×
a
L
RB
=
12
×
2
6
RB
=
24
6
RB = 4 kN
Check
Put the numbers back in. Both sides should match.
8 × 6 = 12 × (6 − 2) ✓
4 × 6 = 12 × 2 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Force vectors and equilibrium
Main lesson
Resultant of three forces by components
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Moment of a force about a point
Use this for “Find the moment of F = 30i + 40j N, acting at (2, 1) m, about the origin.”