A college course may have one or more scheduled sections, or may not have a scheduled section. Attributes of Course include cour
1 answer:
You might be interested in
Answer:
Pressure = 11.38 psi
Force = 13.981 Ibf
Explanation:
Step by step solution is in the attached document.
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