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Sunday, February 16, 2014

How to Calculate beams loads and support forces


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 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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