When the in-plane shear stress is at its highest, it is true that the normal stresses are equal.
Their combined value is σn+σt.
<h3>What is in-plane shear stress?</h3>
- 6.3 ksi is the maximum in-plane shear stress. 10.2 ksi is the maximum out-of-plane shear stress.
- By examining Mohr's circle, we may understand why out of plane shear stress is the highest.
- Maximum in-plane shear stress is indicated by the inner blue circle radius (6.3 ksi).
- Maximum shear stress planes are 45 degrees from the major planes.
- The principle stress is the highest stress that may be created in a plane, and the principal stress refers to the principal plane when shear stress is zero.
The complete question is:
What is true about the normal stresses when the in-plane shear stress is maximum? Select all that apply.
a) They are equal.
b) They are zero.
c) Their sum is equal to σn+σt.
d) They are maximum.
e) They are equal to shear stress.
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Answer:
36.0 kpsi2.
Explanation:
From the question
Sut=110 kpsi
Se’=0.5(110)=55 kpsi
For surface factor ka,
a=2.70, b= -0.265, ka=a(Sut)b=2.70(110)-0.265=0.777
Assuming the worst case for size factor kb, and since 0.11 less than or equal to d less than or equal to 2in,
Therefore:
kb=0.879d-0.107 = 0.879(1.5)-0.107=0.842
Hence, the endurance strength is Se= ka kb Se’ = 36.0 kpsi2.
Answer:
Depending on the amount of torque that is put in the motor, the new velocity is 550 rpm.
Answer:
a) 15.37 mm
b)
c) 5.7186 W/m². K
d) 0.60 m
Explanation:
Given that :
The surface temperature = 130°C = ( 130+ 273 ) K = 473 K
suspended in quiescent air at 25°C = ( 25 + 273 ) K = 298 K
Atmospheric Pressure = 1 atm
The properties obtained from Table A - 4 include :
v =
k = 0.03 W/m K
Pr = 0.700
η = 5
Hence, the boundary layer thickness at a location 0.15 m measured from the lower edge is 15.37 mm
b) The maximum velocity in the boundary layer with f'(n) = 0.275
u = 0.3659 m/s
the maximum velocity in the boundary layer at this location is 0.3659 m/s
the position in the boundary layer where the maximum occur is calculated as:
3.074 mm
c) Using the similarity solution result, , determine the heat transfer coefficient 0.15 m from the lower edge.
we know that:
=
= 28.593
Making the subject from the above formula:
= 5.7186 W/m². K
d) to determine the location on the plate that the boundary layer we become turbulent ; we have the following:
0.60 m