Applied Statics and Strength of Materials (6th Edition)
Applied Statics and Strength of Materials (6th Edition)
6th Edition
ISBN: 9780133840544
Author: George F. Limbrunner, Craig D'Allaird, Leonard Spiegel
Publisher: PEARSON
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Chapter 19, Problem 19.1P
To determine

The allowable tensile loadfor the single shear lap joint.

Expert Solution & Answer
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Answer to Problem 19.1P

  Pg=129.6ksi

Explanation of Solution

Given:

The lap joint is single shear.

Applied Statics and Strength of Materials (6th Edition), Chapter 19, Problem 19.1P

So, n=1

The diameter of the bolt is d=78''

ASTM A36 steel plates- 12''×12''

Calculation:

Now, calculate the area of the bolt

That is,

  AB=π4d2AB=π4( 7 8)2AB=0.60in2

The allowable load for the connection, based on bolt shear

  Ps=n.ABSs(allow)N

Where, N is the number of bolts.

And, from the table of allowable stresses of steel on steel fastener (A325-X)

  Ss(allow)=30ksi

  Ps=n.ABSs( allow)NPs=1×0.6×30×9Ps=162kips.

Now, calculate the allowable load for the connection based on bearing on connected material

  Pb=d.t.Sp(allow)N

Where, Ss(allow)=87ksi from the ASTM A36.

That is,

  Pb=d.t.Ss( allow)NPb=0.875×0.5×87×9Pb=342.56kips

Now, calculating allowable tensile load based on the gross area.

  Pg=Ag×St(all)

Where, Ag is the gross area of the plate.

  Ag=12×0.5=6in2

And, fom the ASTM 36 St(allow)=21.6ksi .

  Pg=6×21.6Pg=129.6ksi

Now, calculating the allowable tensile load based on the net area.

  Pn=An×St(all)

Where, An is the net area.

  An=btNFdHt

  dH=18+ddH=18+78dH=1in

So,

  An=12×0.53×1×0.5An=4.5in2

Therefore, load will be

  Pn=4.5×29Pn=130.5ksi

Conclusion:

By comparing all, the tensile load for the single shear joint is Pg=129.6ksi

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Chapter 19 Solutions

Applied Statics and Strength of Materials (6th Edition)

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