Calculate beam load and supporting forces
Standard I-Beams Please call for more information. S-Shapes: A B C: Weight Lbs. 20 30 40: S3 x 5.7: 3 2.330.170: 5.7 114 171 228. Properties of HE-B profiled steel beams. Mechanics - Forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more; Beams and Columns - Deflection and stress, moment of inertia, section modulus and technical information of beams and columns. 3/4 The 450-kg uniform I-beam supports the load shown. Determine the reactions at the supports.
Online Beam Support Force Calculator
Beams manufactured from all rolled steel sections and from welded plate, the Section Factor for passive protection system is calculated as: Section factor (m-1) = 1400/t Where t = the thickness (mm) of the lower steel web and applies for beams. The 450-kg uniform I-beam supports the load shown. Determine the reactions at the supports. Three workers are carrying a 4-ft by 8- ft panel in the horizontal position shown.
The calculator below can be used to calculate the support forces - R1 and R2 - for beams with up to 6 asymmetrically loads.
For a beam in balance loaded with weights (or other load forces) the reactions 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, lbf)
R = force from support (N, lbf)
In addition for a beam in balance the algebraic sum of moments equals zero. The moment balance can be expressed as
F1 af1 + F2 af2 + .... + Fn afn = R ar1 + R ar2 (2)
where
a = the distance from the force to a common reference - usually the distance to one of the supports (m, ft)
Example - A beam with two symmetrical loads
A 10 m long beam with two supports is loaded with two equal and symmetrical loads F1 and F2 , each 500 kg. The support forces F3 and F4 can be calculated
(500 kg) (9.81 m/s2) + (500 kg) (9.81 m/s2) = R1 + R2
=>
R1 + R2 = 9810 N
= 9.8 kN
Note! Load due to the weight of a mass - m - is mg 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 = (9810 N) / 2
= 4905 N
= 4.9 kN
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Example - A beam with two not symmetrical loads
A 10 m long beam with two supports is loaded with two loads, 500 kg is located 1 m from the end (R1), and the other load of 1000 kg is located 6 m from the same end. The balance of forces can be expressed as
(500 kg) (9.81 m/s2) + (1000 kg) (9.81 m/s2) = R1 + R2
=>
R1 + R2 = 14715 N
= 14.7 kN
The algebraic sum of moments (2) can be expressed as
(500 kg) (9.81 m/s2) (1 m) + (1000 kg) (9.81 m/s2) (6 m) =?R1 (0 m) + R2 (10 m)
=>
R2 = 6377 (N)
= 6.4 kN
F3can be calculated as:
The 450 Kg Uniform I Beam Deflection
R1= (14715 N) - (6377 N)
= 8338 N
= 8.3 kN
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Related Topics
- Mechanics - Forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more
- Beams and Columns - Deflection and stress, moment of inertia, section modulus and technical information of beams and columns
- Statics - Loads - force and torque, beams and columns
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The 450 Kg Uniform I-beam
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