Finding the Force P from Normal Stress

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SUMMARY

The discussion focuses on determining the maximum load P that can be applied at point A in a steel structure using a 6-mm-diameter pin at C and 10-mm-diameter pins at B and D. The ultimate shearing stress is established at 150 MPa, while the ultimate normal stress in link BD is 400 MPa, with a desired factor of safety of 3.0. The analysis reveals that link BD is in tension, and the diameter of the pin is excluded from the width of member BD when calculating stress, as that material is not present to support the load. The necessity of evaluating shear area and potential buckling of the link is also emphasized.

PREREQUISITES
  • Understanding of ultimate shearing stress and normal stress concepts
  • Familiarity with Free Body Diagrams (FBD) in structural analysis
  • Knowledge of stress calculations using σ = F/A
  • Basic principles of factors of safety in engineering
NEXT STEPS
  • Study the application of the factor of safety in structural engineering
  • Learn about the mechanics of materials, focusing on tension and compression analysis
  • Investigate shear area calculations in pin connections
  • Explore buckling analysis techniques for structural members
USEFUL FOR

Structural engineers, mechanical engineers, and students studying mechanics of materials who are involved in load analysis and design of pin connections in steel structures.

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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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