What Factors Determine the Maximum Allowable Force on a Wooden Beam?

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Discussion Overview

The discussion centers around determining the maximum allowable force on a wooden beam before failure, focusing on the application of shear and normal stress equations in a homework context.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant describes their approach to resolving the force P into shear and normal components, applying the factor of safety and failure stresses.
  • Another participant requests clarification on how the first participant resolved the force P.
  • A participant proposes that the shear force is represented by Pcos30 and the normal force by Psin30.
  • Another participant challenges this by suggesting that the correct expressions should involve P/2cos30 and P/2sin30 instead.

Areas of Agreement / Disagreement

There is no consensus on the correct resolution of the force P, as participants present differing views on the calculations involved.

Contextual Notes

The discussion does not clarify the role of the distance b in the calculations, which may be a relevant factor in determining the maximum allowable force.

w31ha-
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Homework Statement


Finding the maximum allowable force P before failure

Homework Equations


Shear /normal stress = Force / Area

The Attempt at a Solution


I know that I can resolve the force P to be along and normal to the end of the wooden beam to find the respective forces separately,applying factor of safety, from the failure stresses given and therefore, determine which P is lower which shall be the maximum allowable P. However, I do not see there the distance b come into play. Am i missing something?
 

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Can you show how you "resolved" P?
 
Pcos30 will be my shear force and Psin30 will be my normal force to the end of the beam.
 
Don't you mean P/2cos30 and P/2sin30?
 

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