Answer:
for 1st question the answer is 5th option.
for 2nd question the answer is 2nd option
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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:
180 x 60 inches
Width = 60 inches
Length = 180 inches
Explanation:
Given
Let L = Length
W = Width
P = Perimeter
Length = 3 * Width
L = 3W
Perimeter of Brass = 480 inches
P = 480
Perimeter is given as 2(L + W);
So, 2 (L + W) = 480
L + W = 480/2
L + W = 240
Substitute 3W for L; so,
3W + W = 240
4W = 240
W = 240/4
W = 60 inches
L = 3W
L = 3 * 60
L = 180 inches
Answer:
5984.67N
Explanation:
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?
from continuity equation
v1A1=v2A2
equation of continuity
v1=4ft /s=1.21m/s
d1=14 inch=.35m
d2=14-2=0.304m
A1=pi*d^2/4
0.096m^2
a2=0.0706m^2
from continuity once again
1.21*0.096=v2(0.07)
v2=1.65
force on the pipe
(p1A1- p2A2) + m(v2 – v1)
from bernoulli
p1 + ρv1^2/2 = p2 + ρv2^2/2
difference in pressure or pressure drop
p1-p2=2psi
13.789N/m^2=rho(1.65^2-1.21^2)/2
rho=21.91kg/m^3
since the pipe is cylindrical
pressure is egh
13.789=21.91*9.81*h
length of the pipe is
0.064m
AH=volume of the pipe(area *h)
the mass =rho*A*H
0.064*0.07*21.91
m=0.098kg
(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)
force =5984.67N