Answer:
k = 1.91 × 10^-5 N m rad^-1
Workings and Solution to this question can be viewed in the screenshot below:
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
Answer
For isotropic material plastic yielding depends upon magnitude of the principle stress not on the direction.
Tresca and Von Mises yield criteria are the yield model which is widely used.
The Tresca yield criterion stated that yielding will occur in a material only when the greatest maximum shear stress reaches a critical value.
max{|σ₁ - σ₂|,|σ₂ - σ₃|,|σ₃ - σ₁|} = σ_f
under plane stress condition
|σ₁ - σ₂| = σ_f
The Von mises yielding criteria stated that the yielding will occur when elastic energy of distortion reaches critical value.
σ₁² - σ₁ σ₂ + σ₂² = σ²_f
Answer:
The industrial systems of the future are seen as complex systems, composed of vast numbers of devices, interacting with each other and with enterprise systems continuously.
summary:
they are related because they ARE a system. well, a type.
hope this helps!! :)
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