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Ivahew [28]
2 years ago
11

A double-threaded Acme stub screw of 2-in. major diameter is used in a jack having a plain thrust collar of 2.5-in. mean diamete

r. Coefficients of running friction are estimated as 0.10 for the collar and 0.11 for the screw.
a. Determine the pitch, lead, thread depth, mean pitch diameter, and lead angle of the screw

b. Estimate the starting torque for raising and for lowering a 5000-lb load.

c. What is the efficiency of the jack when raising the load?

d. Would the screw overhaul if a ball thrust bearing of negligible friction were used in place of the plain thrust washer?
Engineering
1 answer:
Temka [501]2 years ago
6 0
This is the answer for the question

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An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
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Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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