For
a beam in balance loaded with weights (or other load forces) the reaction
forces “R” at the supports equals the
load forces “F”. The force balance can be expressed as:
F1
+ F2 + ........ + Fn = R1 + R2 (1)
where
F
= force from load (N or lbf)
R
= forces from support (N or lbf)
In
addition, for a beam in balance the algebraic sum of moments equals zero. The
moment balance can be expressed as
F1
l1 + F2 l2 + .... + Fn ln = R x1
+ R x2 (2)
where
l
= the distance from the force to a common reference - usually the distance to
one of the supports (m or ft)
x
= the distance from the support force to a common reference - usually the
distance to one of the supports (m or ft)
Example:
A beam with two symmetrical loads
A beam with two symmetrical loads
A
9 m long beam with two supports is loaded with two equal and symmetrical loads
F1 and F2, each 300 kg. The support forces F3
and F4 can be calculated:
(300
kg) (9.81 m/s2) + (300 kg) (9.81 m/s2) = R1 +
R2
Therefore:
R1
+ R2 = 5886 (N)
Note….!
Load
due to the weight of a mass “m” is (F = m.g)
Newton's - where g = 9.81 m/s2.
With
symmetrical and equal loads the support forces also will be symmetrical and
equal. Using
R1
= R2
The
equation above can be simplified to
R1
= R2 = (5886 N) / 2
= 2943 N
Example:
A beam with two not
symmetrical loads
A
9 m long beam with two supports is loaded with two loads, 3 kg is located 3 m
from the end (R1), and the other load of 750kg is located 6 m from
the same end. The balance of forces can be expressed as:
(300
kg) (9.81 m/s2) + (750 kg) (9.81 m/s2) = R1 +
R2
Therefore:
R1
+ R2 = 10300.5 (N)
The
algebraic sum of the moments (2) can be expressed as:
(300
kg) (9.81 m/s2) (3 m) + (750 kg) (9.81 m/s2) (6 m) = R1
(0 m) + R2 (9 m)
So
R2
= 5886 (N)
R1
can be calculated as:
R1 = (10300.5 N) - (5886 N)
= 4414.5 N
References:
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