A regular checkup is needed for the railway signal light box because the signal for the loco pilot is directly proportional to the traveler’s safety and railway safety. Sliding down the signal light box is quite difficult process because due some strict rules regarding rail safety. Also it is not possible for the maintenance officials to climb the post and work at that height because it is also banned due to railway safety policies.

Therefore various systems have been employed to achieve this. For example the base of the post is made pivot and is than bolted to make the post stood still and some additional brackets can be added at the lower parts of the post so that lever system can be used to pull down the post. The process of doing this requires 4 to 5 employees and some heavy equipment therefore the task seems to be much tougher.

Our task is to just decrease the labor and minimize the equipment. Portable hydraulic system is employed for that & is explained further in the paper.

Hydraulic Jack:- It is a hydraulic lift with short stroke which is works by hand pump (F., 1881). It is one of the simplest forms of a fluid power system.It can lift a load weighing several tons just by moving handle of device. A small initial force exerted on the handle with very less area is transmitted by a fluid to a much larger area which makes same pressure on both areas and making the force at larger area very large compare to initial force.hydraulic jack works on the principle of Pascal’s law.This states that “whenever fluid is at rest and if pressure is applied at any point in the liquid, the pressure will be transmitted equally at all points in all direction. Two piston-cylinder arrangements of different sizes are suitably connected with a pipe so that pressure in each cylinder will be the same due to Pascal’s law. If ‘p’is pressure and ‘a1,a2’ are the cross sectional area of cylinders, then a force F is applied to the smaller plunger will make available a load ‘W’ is lifted.

Where, p = pressure of the fluid,

 a1 = small cylinder area,

a2 = larger cylinder area,

F = force acting on smaller plunger,

W = load lifted.

For constant volume of liquid. The displacements of large piston and smaller plunger will be proportionall to each other.

Literature Survey

Working of hydraulic jack: – Pascal’s law is the principle on which hydraulic jack works (Malhotra S., 2013). When the handle of jack is used, the plunger reciprocates to and fro and then the oil or any fluid from the reservoir is sucked into the plunger cylinder or smaller cylinder during upward stroke of the plunger piston through the suction valve which is connected to reservoir. The pressurised oil of the plunger cylinder is transferred into the ram cylinder where force is required to lift dduring the downward stroke of the plunger through the delivery valve which is connected to entry valve of ram cylinder. This pressurized oil is used to apply force on the ram piston to lift the load up.when we want to remove the lift of load or lower the ram, the pressurized oil in the ram cylinder is taken out of cylinder into reservoir by unscrewing the lowering screw so the oil is rushed into the reservoir. It consists of plunger cylinder and ram cylinder on each side. Both the cylinders are mounted on base.Base plate is made of mild steel. By operating the handle we can access the plunger and pressurize the oil.Smaller cylinder or plunger cylinder consists of two one sided valves i.e. liquid can go in one direction only, one for suckking the oil from reservoir and other for delivering oil to ram cylinder.larger cylinder or Ram cylinder consists of ram,which lifts the load and two one sided valve.Suction valve of ram cylinder is used to enter pressurized oil from plunger cylinder to ram cylinder and delivery valve of ram cylinder is used to depressurized oil and transfer it back to reservoir. Ram Cylinder also consists of lowering screw which is a hand operated delivery valve used for depressurizing the oil in the ram cylinder to lower the ram.

Finite Element Analysis Of the post:-

Determination of:-

• Deflection under uniform load(Bansal, 1998):-

Let,

 E=modulus of elasticity of the post material (N/m²)= 30 MPa

L = length of the post= 5m

W = uniform distributed load.

Assumptions:-

1. material of post is isotropic and homogeneous
2. Post material is defect free.
3. Self-wt. is neglected.

Potential energy (_p) = strain energy +work potential.

Let, v=a₁y+a₂xy²+a₃xy³ ———polynomialplacement function

Π_p=dyw

Applying boundary condition

At   y=0      =0

.·a₁+2a₂y+3a₃y²

0=a₁+0+0

.·.  a₁=0

.·. v =a₂y²+a₃y³

Π_p=

₂+6a₃y

=₂+6a₃y)²dy-w₂y²+a₃y³)dy

For minimization

Put, 

.·.     (8

         8+12

Again

Put , =0

    Solving equation (2)&(3) we get

Defection = v=

Stress  

Length of post = 5m

Weight of post = 3.6 kg/m=3.65=18.0 kg

F= 189.81=176.58N

E=30 Gpa=30

Deflection due to self-wt. ==5.886

• Deflection/deformation due to self-weight (Elliott, 2017)

U= displacement in x direction

F=self-weight.

Boundary condition at u=0 at x=0

&at x =0   

Potential energy = strain energy +work potential

²dx-

Deflection = u”=0-

     at L= 5m

Deflection u” Answer.

Results: – 

Deflection due to self-wt. ==5.886

Discussion:-. Finite Element Analysis is considered to be the good approach in order to note down the deflection of the post due to some load acting on the post. In FEM, R-R method is adopted to calculate the deflection of the post. As far the labor effort is concerned due to use of hydraulic device only a single person can work easily but for the safety point of view more modifications are needed in future. But till now two labor will be sufficient to carry out the maintenances by using the hydraulic device.

Conclusion: – From the above analysis it is clear that hydraulic jack can be the effective source in order to lift the post.

References

F., C., 1881. Hydraulic, Steam, and Hand Power Lifting and Pressing Machinery. London: Vance Publications.

Malhotra S., B. D., 2013. Modern Certificate Physics. New Delhi: Frank Bros.&co.(Publishers)ltd.