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levacccp [35]
3 years ago
13

A 50-kN hydraulic press performs pressing and clamping actions. The clamping cylinder force is 4 kN. The pressing cylinder strok

e is 30 cm and its extension speed is 8 cm/s. Design the hydraulic circuits, perform the preliminary calculations, and select the needed hydraulic elements. Then, calculate the different operating modes of the system.

Engineering
1 answer:
galben [10]3 years ago
4 0

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

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A small family home in Tucson, Arizona has a rooftop area of 2667 square feet, and it is possible to capture rain falling on abo
Kazeer [188]

Answer:

volume  = 53.747 m3 = 14198.138 gal

weight = 526652 N = 118396.08 lbf

Explanation:

We know that volume of water

volume  =  A'\times H

where A' = 61% of A

              = 0.61\times 2667 = 1626.87 sq ft

volume  =  1626.87 \times (\frac{14}{12} ft)

               =1898.015 ft^3

in\ m^3 = \frac{ 1898.015}{35.315} =   53.7457 m^3

in\ gallon = 1898.015 \times 7.481 = 14198.138 gallon

weight = \rho Vg

       = 1000\times 53.74\times 9.8

             =526652 N

In\ lbf =  \frac{526652}{4.448} = 118396.08 lbf

7 0
3 years ago
If link AB of the four-bar linkage has a constant counterclockwise angular velocity of 58 rad/s during an interval which include
katrin2010 [14]

Answer:Vb=-6i-(-0.1ωab+8)j m/s

Explanation:

Va=V0+Va0

Va=V0+(ra0 x ωao)

ω=Angular velocity of link A0

Using r0a=0.1m;

Va=V0+(0.1i x ω0a K)

Va=0

ixk=j

Va=0+0.1ω0aj

Calculating te velocity of using te equation below

Vb=Va+Vba

Vb=Va+ωab x rba

ωab=40rad/s

rab=-0.21i+0.15j

Va=0.1ω0aj

Vb=Va+ωabxrba

Vb=0.1ω0aj+40k x -(0.21i+0.15j)

Vb=0.1ω0aj-8j-6i

Vb=-6i-(-0.1ωab+8)j m/s

5 0
3 years ago
Problem 3: Soil Classification using the AASHTO and USCS Systems
nataly862011 [7]

<u>Solution:</u>

Given\\                   \(\quad W=3000 Ib , \quad m=\frac{W}{g}=\frac{3000}{322} \ slug =93.1677 slug\)\\K_{e q}=2160 lbs / wp =2100 \frac{ lbs }{10} \frac{ x 12}{1 ft }=(2160 \times 12) lb / ft$$

a) The natural frequency

\begin{aligned}&\left(\omega_{n}\right)=\sqrt{\frac{K_{e q}}{m}}\\&=\sqrt{\frac{2160 \times 12}{93.1677}}\\&\omega_{n}=16.68 \text { rad } | s\\&\omega_{n}=\frac{2 \pi}{T}\\&16.68=\frac{2 \pi}{T}\\&T=0.3767 s\end{aligned}

b)

Given, \(t=10 s , \quad y(t)=6 in = A\)\\\(y(t)=A \cos \left(\omega_{n} t+\phi\right) \rightarrow 0\)\\\(6=6 \cos (16.68 \times 10+\phi)\)\\\(1=\cos (166.8+\phi)\)\\\(166.8+\phi=0\)\\\phi=-166.8\)\\At \(t=0, \quad y(0)=6 \cos (16.68 \times 0-166.8)\) {y(0)}=-5.74 in

5 0
3 years ago
How old are legos? Who created them? Why did they create them?
Sunny_sXe [5.5K]
“How old are the legos” 89 years old.


“Who created them?” Ole Kirk Christiansen.


“Why did they created them?” As a you man, Christiansen turned his love of whittling and playing with wood into a Business and, in 1916, he opened his own shop. Since times were so hard, Christiansen made the hard decision to use his wood to create inexpensive goods that might actually sell. Among them were Cheap toys.





Hope this helps•


Bye~



•Kate•
4 0
2 years ago
A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th
Helga [31]

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at  

Replacing by the values already known for vo, vf and a, and solving for t, we get:

t = vf-vo /a = (29 m/s – 11.2 m/s) / 2.2 m/s = 8 sec

5 0
2 years ago
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