Brass Pin

Hi
A 10 mm dia Brass rod shears at 300 MPa shear stress when loaded
statically.
At what load will it break in case of sudden load. The pin is required
to get broken at a specific transverse force. This force is to be
found.
Thanks
RJ Khan
Reply to
rjkhan
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The shear stress of a pin in double shear is given as:
Ss = F/A = (2F) / (PiD^2)
where F = the force applied, D= the diameter of the pin and Pi = 3.14159+ The 2 is a constant because you have two areas in shear.
However, real life experience dictates the pin will probably fail in bending, not shear.
Sb = (Mc) /
I = (4Fb) / (PiD^3)
Bending Moment, M = Fb/8 where b is the length of the force on the pin created by the rod Moment of Inertia, I = (PiD^4) / 64 where D= the diameter of the pin and Pi = 3.14159+ Centroid, c = D / 2
Be certain to check for bending failure first, then shear. Note: If you use the computer to do your calculation, Pi = 4 times the Arctan of 1 radian = 4 Atan(1)
Jim Y
Reply to
Jim Y
Thanks for your kind advice but here in this case the pin length is proportionally not very longer than the pin dia to regard the pin as a beam. All the beam related formulae are applicable when lengthen is considerably longer than dia.
The pin is actually acting like a hinge-pin (as in a knuckle joint). The other two parts that are joined by this pin are pulled away from each other with a "jerk". Now for this situation, my question is whether the brass pin behaves same as the same jerk load is applied "statically" or gradually. i.e. will the pin shear at 300 MPa (static shear strength) or at some greater shear stress (pertaining to strain hardening effect).
Thanks
Reply to
rjkhan
Mechanical Engineering Design by Shigley from McGraw-Hill has load amplification factors in it. Don't have my copy handy at the moment. ________________________________________________________ Ed Ruf Lifetime AMA# 344007 ( snipped-for-privacy@EdwardG.Ruf.com)
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Reply to
Ed Ruf
Did you look up the maximum shear theory of failure (brittle materials) or the Mises-Hencky theory (ductile materials)? Try:
"Design of Machine Elements", M.F.Spotts, Prentice-Hall, ISBN 0-13-200593-X, Page 99-101 (6th ed)
"Practical Stress Analysis in Engineering Design", A. Blake, Marcel Dekker, ISBN 0-8247-1370-2, Page 390-393 (1982 edition)
Your term "jerk" may be considered as sudden loading which is defined in Roark as approaching twice the static load.
Jim Y
bending, not shear.
Reply to
Jim Y

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