Use a Mohr circle to find the maximum shear stress relative to the axial stress.
Here we assume the axial stress is sigma, the transverse axial stress is zero.
So we have a Mohr circle with (0,0) and (0,sigma) as a diameter.
The centre of the circle is therefore (0,sigma/2), and the radius is sigma/2.
From the circle, we determine that the maximum stress is the maximum y-axis values, namely +/- sigma/2, at locations (sigma/2, sigma/2), and (sigma/2, -sigma/2).
Given that the maximum shear stress is 60 MPa, we have
sigma/2=60 MPa, or sigma=120 MPa.
(note: 1 MPa = 1N/mm^2)
Therefore
100 kN/(pi*d^2/4)=100,000 N/(pi*d^2/4)=120 MPa where d is in mm.
Solve for d
d=sqrt(100,000*4/(120*pi))
=32.5735 mm
It is 46 if you divide 34 by 16 you get 2.125 then divide 100 by 2.125 and you get 47.05 the 0 means go down if you are rounding which brings it to 46
Answer:
20x^2 -4x -16
Step-by-step explanation:
20x^2 -4x -16
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Answer:
10 plu 10 plus 100 plus 10 plus 10 plus 10 plus 10 plus 10 plus 5 plus 5
Step-by-step explanation:
