15- T
16- F
17- T
I think this is correct
Answer:
Option B
1025 psi
Explanation:
In a single shear, the shear area is 
The shear strength=
and in this case 
Shear strength=
hence making load the subject then
Load=Shear area X Shear strength
Load=
Answer:
874 psi
Explanation:
Given a sample mean (x') = 900,
and a standard error (SE) = 10
At 99% confidence, Z(critical) = 2.58
That gives 99% confidence interval as,
x' ± Z(critical) x SE = 900 ± 2.58 x 10
The value of the lower limit is,
900 - 25.8 = 874.2
≈ 874 psi
Answer:
True
Explanation:
This is a true fact because From the time you are born to around the time you turn 30, your muscles grow larger and stronger.
Answer:
Class of fit:
Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).
Here minimum shaft diameter will be greater than the maximum hole diameter.
Medium Drive Force Fits are FN 2 Fits.
As per standard ANSI B4.1 :
Desired Tolerance: FN 2
Tolerance TZone: H7S6
Max Shaft Diameter: 3.0029
Min Shaft Diameter: 3.0022
Max Hole Diameter:3.0012
Min Hole Diameter: 3.0000
Max Interference: 0.0029
Min Interference: 0.0010
Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.
Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.
Explanation:
