Answer:
When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.
Explanation:
When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.
The engine knock problem can be caused due to the following reason
a) When the octane rating of the fuel used is low.
b) The deposition of the carbon around the cylinder walls takes place.
c) The spark plug used in the vehicle is not correct.
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:
Answer:
Absolute pressure=70.72 KPa
Explanation:
Given that Vacuum gauge pressure= 30 KPa
Barometer reading =755 mm Hg
We know that barometer always reads atmospheric pressure at given situation.So atmospheric pressure is equal to 755 mm Hg.
We know that P= ρ g h
Density of 
So P=13600 x 9.81 x 0.755
P=100.72 KPa
We know that
Absolute pressure=atmospheric pressure + gauge pressure
But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.
Absolute pressure=atmospheric pressure + gauge pressure
Absolute pressure=100.72 - 30 KPa
So
Absolute pressure=70.72 KPa
The question is incomplete. The complete question is :
The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?
Solution :
Given data :
Diameter of the rod : 46 mm
Torque, T = 85 Nm
The polar moment of inertia of the shaft is given by :


J = 207.6 
So the shear stress at point A is :



Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.