Max Load Force on Uniform Pine Beam of 3.6m & 25kg | Safety Factor 5.0

AI Thread Summary
The discussion focuses on calculating the maximum load force that two vertical roof supports can exert on a uniform pine beam without causing shear failure. The beam measures 3.6 m in length and has a cross-section of 9.5 cm x 14 cm, with a total mass of 25 kg. A safety factor of 5.0 is applied, indicating that the beam should be five times stronger than the maximum expected load. The confusion arises around the initial calculation step, specifically why the maximum stress is multiplied by 1/5, which relates to the safety factor. Understanding the safety factor is crucial, as it defines the beam's strength relative to the load it must support.
Neon32
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Homework Statement


A uniform pine beam, 3.6 m long and 9.5 cm x 14 cm in cross section, rests on two supports near its ends, as shown. The beam’s mass is 25 kg and two vertical roof supports rest on it, each one-third of the way from the ends. What maximum load force FL can each of the roof supports exert without shearing the pine beam at its supports? Use a safety factor of 5.0.

Homework Equations

The Attempt at a Solution



Answer sheet:
F= 1/5 x A x Maximum stress

I just don't understand the first step. Why did I multiply it by 1/5? And what does safety factor mean?
 
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Safety factor is how many times stronger the member is than it needs to be.
 
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