GPa for (Metric)Steel:f=200 a) Find minimum diameter neededif the allowable shear stress isI40 Mpa and the rotation ofpoint A can't exceed 2.5 degrees.→Shaft made of steel10 kN-m4KN m2KN -m450g00 mm 1goo mmAmm

Solution:- We know that torsion equation is, TJ=τr=GθL here, net torsion is , T=T3-T1+T2 =10-4+2…

Answered: a) Find minimum diameter nceded if the…

a) Find minimum diameter nceded if the allowabe shear stress is 140 Mpa and the otation of point A can't exced 2.5 degreas. 10 kN-m 4KN m QUN -m A 900 mm (l goo mm 450 mm

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter5: Beams

Section: Chapter Questions

Problem 5.8.1P: Compute the nominal shear strength of an M107.5 of A572 Grad 65 steel.

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Question

a) Find minimum diameter needed
if the allowable shear stress is
I40 Mpa and the rotation of
point A can't exceed 2.5 degrees.
→Shaft made of steel
10 kN-m
4KN m
2KN -m
450
g00 mm 1goo mm
A
mm

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