Finding the Force P from Normal Stress

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Discussion Overview

The discussion revolves around determining the largest load P that can be applied at point A in a steel structure involving pins and links, focusing on the concepts of normal stress and shear stress in structural members. The context includes homework-related inquiries about the mechanics of materials and structural analysis.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asserts that link BD is in tension based on the free body diagram (FBD) analysis.
  • Another participant explains that the diameter of the pin is excluded from the width of member BD when calculating stress because that material is absent and cannot support any load.
  • A question is raised about whether the same exclusion of the pin diameter would apply if link BD were in compression instead of tension.
  • A response indicates that if link BD were in compression, the shear area available would need to be analyzed to prevent the pin from tearing out of the pinhole, suggesting a different approach to evaluating the structural integrity.
  • There is a mention of checking for buckling of the link itself if sufficient shear area is present and the pin is deemed acceptable.

Areas of Agreement / Disagreement

Participants generally agree that link BD is in tension, but there is a discussion about the implications of this on stress calculations and whether the same principles apply in a compression scenario. The discussion remains unresolved regarding the treatment of the pin diameter in different loading conditions.

Contextual Notes

Limitations include assumptions about the material properties and the specific loading conditions that may affect the analysis. The discussion does not resolve the mathematical steps involved in determining the largest load P.

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Homework Statement



In the steel structure shown, a 6‐mm‐diameter pin is used at C and 10‐mm‐diameter pins are
used at B and D. The ultimate shearing stress is 150 MPa at all connections, and the ultimate normal stress is 400 MPa in link BD. Knowing that a factor of safety of 3.0 is desired, determine the largest load P that can be applied at A. Note that link BD is not reinforced around the pin holes.

Homework Equations



F.S = σultallow

σ = F/A

The Attempt at a Solution



I understand the parts concernging the ultimate shearing stress, my confusion is with the normal stress.

When I draw a FBD of beam AB, I assign the force at B, By, to point upwards and the force at C, Cy, to point downwards.

ƩMc = -By(.120) + P(0.280)

By = 2.33P

My question is, does this this result mean that member BD is in compression or tension? Having drawn a FBD of AC I understand that the forces are acting on beam AC, thus while By points in the +y-direction, if I were to draw a FBD of BD, the force, By, would point in the opposite direction thereby indicating that member BD is in tension.

Hence when I use the normal stress equation

σ = F/A

where F is the internal force and A is the cross sectional area, why is the diameter of the pin removed form the width of member BD?
 

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1. Link BD is in tension.

2. The diameter of the pin is removed from link BD width because that material is not present and cannot be considered when calculating stress. If you look at the pin at point B or D and the link, draw a FBD of the link and make a cut horizontally thru the diameter of the pin. Only the material outside the pin hole can support any load.
 
If link BD had been in compression instead of tension, would you still need to remove the diameter of the pin?
 
If link BD were in compression, then you would have to look at the shear area available to keep pin B from tearing out of the pinhole. In this case, a vertical cut on either side of the pin thru the link would be analyzed to determine how much area is able to resist the tendency of the pin to shear thru the end of the link.

If there is sufficient shear area and the pin is OK, then buckling of the link itself should be checked.
 

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