Two transmission belts pass over a double-sheaved pulley that is attached to an axle supported by bearings at A and D. The radiu
s of the inner sheave is 125 mm and the radius of the outer sheave is 250 mm. Knowing that when the system is at rest, the tension is 100 N in both portions of belt B and 160 N in both portions of belt C, determine the reactions at A and D. Assume that the bearing at D does not exert any axial thrust.
1 answer:
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Answer:
The attached figure shows the hydraulic circuit using one sequence valve to control two simultaneous operations performed in proper sequence in one direction only. In the other direction, both the operations are simultaneous.
When we keep the 4/2 DCV in crossed arrow position, oil under pressure is supplied to the inlet port of the sequence valve. It directly flows to Head end port-1. Hence Cylinder 'C1' extends first.
By the end of the extension of cylinder 'C1', pressure in the line increases and hence poppet of sequence valve is lifted off from its seat and allows oil to flow to port-2 and hence, Cylinder 'C2 extends completing the pressing operation.
In the straight-arrow position of 4/2 DCV the oil under pressure reaches the rod end of both the cylinders C1 and C2 simultaneously through port-3. This causes both the cylinders to retract simultaneously.
Also, a Flow control valve is provided tho control the velocity of clamping
Explanation:
find attached the figure
Answer:
•Estimated density = 39.685Pc/mi/en
•Level of service, LOS frequency = LOSC
Explanation:
We are given:
•Freeway current lane width,B = 12ft
• freeway current shoulder width,b = 6ft
• percentage of heavy vehicle, Ptb = 10℅
• peak hour factor, PHF = 0.9
Let's consider,
•Number of lanes N = 4
• flow of traffic V = 7500vph
• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph
• cars equivalent for recreational purpose Er= 2
•cars to be used for trucks and busses Etb= 2.5
Let's first calculate for the heavy adjustment factor.
We have:
Substituting figures in the equation we have:
= 0.75
Let's now calculate equivalent flow rate of the car using:
= 2777.7 pc/h/en
Calculating for traffic density, we have:
D = 39.685 Pc/mi/en
Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC