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Problem 5

The document describes a problem to determine the optimal outer diameter of brass spacers used between 16mm diameter steel bolts that hold two steel plates together. It is given that the average normal stress in the bolts must not exceed 200 MPa and in the spacers must not exceed 130 MPa. The solution shows the equations used to relate the tensile force in the bolts to the compressive force in the spacers when in equilibrium. Solving these equations determines that the optimal outer diameter of the spacers that provides the most economical and safe design is 25.2 mm.

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Bob Karl
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1K views1 page

Problem 5

The document describes a problem to determine the optimal outer diameter of brass spacers used between 16mm diameter steel bolts that hold two steel plates together. It is given that the average normal stress in the bolts must not exceed 200 MPa and in the spacers must not exceed 130 MPa. The solution shows the equations used to relate the tensile force in the bolts to the compressive force in the spacers when in equilibrium. Solving these equations determines that the optimal outer diameter of the spacers that provides the most economical and safe design is 25.2 mm.

Uploaded by

Bob Karl
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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PROBLEM 1.

Two steel plates are to be held together by means of 16-mm-


diameter high-strength steel bolts fitting snugly inside cylindrical
brass spacers. Knowing that the average normal stress must not
exceed 200 MPa in the bolts and 130 MPa in the spacers,
determine the outer diameter of the spacers that yields the most
economical and safe design.

SOLUTION

At each bolt location the upper plate is pulled down by the tensile force Pb of the bolt. At the same time, the
spacer pushes that plate upward with a compressive force Ps in order to maintain equilibrium.

Pb = Ps

Fb 4Pb π
For the bolt, σb = = or Pb = σ bdb2
Ab π db2 4

Ps 4Ps π
For the spacer, σs = = or Ps = σ s (d s2 − db2 )
As π (d s2 − db2 ) 4

Equating Pb and Ps ,

π π
σ bdb2 = σ s (d s2 − db2 )
4 4
σb 200
ds = 1+ d = 1+ (16) d s = 25.2 mm
σs b 130

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