Answer:
c = 18.0569 mm
Explanation:
Strategy
We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.
Given Data
Applied Torque
T = 750 N.m
Length of shaft
L = 1.2 m
Modulus of Rigidity
G = 77.2 GPa
Allowable Stress
г = 90 MPa
Maximum Angle of twist
∅=4°
∅=4*
/180
∅=69.813 *10^-3 rad
Required Diameter based on angle of twist
∅=TL/GJ
∅=TL/G*
/2*c^4
∅=2TL/G*
*c^4
c=
∅
c=18.0869 *10^-3 rad
Required Diameter based on shearing stress
г = T/J*c
г = [T/(J*
/2*c^4)]*c
г =[2T/(J*
*c^4)]*c
c=17.441*10^-3 rad
Minimum Radius Required
We will use larger of the two values
c= 18.0569 x 10^-3 m
c = 18.0569 mm
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Answer:
All of them cause friction
Explanation:
google
Answer:
The strength coefficient is
and the strain-hardening exponent is 
Explanation:
Given the true strain is 0.12 at 250 MPa stress.
Also, at 350 MPa the strain is 0.26.
We need to find
and the
.

We will plug the values in the formula.

We will solve these equation.
plug this value in 

Taking a natural log both sides we get.

Now, we will find value of 

So, the strength coefficient is
and the strain-hardening exponent is
.