Finite Element Analysis of Hydraulic Actuator by using CAE tools

DOI : 10.17577/IJERTV9IS030268

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Finite Element Analysis of Hydraulic Actuator by using CAE tools

Manisha. Anil. Gurwani

Student of Mechanical engineering Babasaheb Naik College of Engineering Pusad, Maharashtra

Prof. V. V. Dongaonkar

Assistant Professor

Dept. Of Mechanical engineering Babasaheb Naik College of Engineering Pusad, Maharashtra

Abstract: A hydraulic system is a fluid power system that is commonly used in industries due to its ability to sustain high pressure .So, here in this paper double acting hydraulic actuator is designed based on the force acting on it and the stroke length .Analysis of different parts of actuators is done by using CAE tools checking for the equivalent stresses and deformation. Two different materials are also used to check proper material to be used.

Keywords— Hydraulic ,actuator ,double acting ,analysis


    Hydraulic actuators are the end results of Pascals law. Hydraulic actuator is the device which converts hydraulic energy into mechanical energy it consists of cylinder that transforms the flow of pressurized fluid into a push or pull of piston rod .In double acting actuators the fluid pressure can be exerted from both sides. Hydraulic actuators are rugged and suited for high force applications.


    The load of 21832 N is to be sustained by actuator but if fails suddenly ,so taking into account sudden loading conditions it should be designed for 43664 N ( 21832 * 2).So by using CATIA V5 models are made, and every component is analyzed separately using two different materials that are low carbon steel and E335 steel. Low carbon steel is used as it is lighter in weight with good yield strength, tensile strength, corrosion resistance, ductility whereas E335 steel is practically used in industries as a material. So here we find out which one is better.


    The following assumptions were taken into the consideration of the design of the cylinder, piston, piston rod and seals in the hydraulic cylinder.

    Working fluid is mineral oil

    Available pressured Pa =200bar= 200* 105Pa Atmospheric pressure = 1.0135 *105Pa Stroke length= 1135mm =1.135m

    Cylinder output force = 43664N Factor of safety =3

    End fixing factor = K = 0.7 Properties of materials used

    TABLE 1


    Low carbon steel

    E355 Steel

    Ultimate tensile stress

    430 MPa

    540 – 620 MPa

    Yield tensile stress


    290 – 450 MPa

    Young Modulus (E)

    210 GPa


    1. Design of piston rod

      The rod is more likely to fail by buckling under the compressive load. In this case, the rod behaves like a column and is subjected to buckling. Therefore Eulers formula in the equation below for long column can be used to obtain the piston rod diameter


      Where: P = Buckling load (N) L = the column length (m)

      I = Moment of inertia (m4)

      E = Young's Modulus of Elasticity for the column material (Pa)

      K = the end fixing factor =0.7

      E = Youngs modulus of the material used in this design calculation is 120 GPa

      P = cylinder force * factor of safety = 43664 * 3 = 130992N

      130992 = (2*210*109*I)/(1.1352*0.72) I=39.894*10-9 m4

      *d4/4 =39.894*10-9

      d=15 mm

      from Baym Hydraulics Corporation catalog of metric rod wipers and piston seals the nearest standard rod seal diameter is 20 mm.

    2. Design of the piston

    Let A be the full area of the piston and a be the cross sectional area of the piston rod. Since the design is a double acting double ended hydraulic cylinder, pressure is acts on both sides of the rod, hence the area which the pressure is acting on is given by (A-a). The force produced is given in the equation below.

    A-a= (D4-d4)/4

    200*105= (43664*384)/((D2-0.022) D= 0.0934m=93.48mm

    from Baym Hydraulics Corporation catalog of metric rod wipers and piston seals, the nearest standard rod seal diameter is 100mm

    2 .length of piston = D to 1.5 D=100mm

    3. Thickness of piston head t 1= 0.43D (P/)1/2



    4 .Radial thickness t r = D (3P/)1/2

    =0.1((3*0.03090/ (430/3))1/2

    =2.5 mm

    C . Design of the cylinder

    Let OD = outside diameter of the cylinder. The maximum working stress (m) is given as m = Tensile stress of material/FOS


    =143.3*106Pa OD2=D2( m +P) /( m P)

    =0.12(143.3*106 +200*105)/( 143.3*106 -200*105) OD=115mm

    D. Cylinder Tube thickness

    The wall thickness required for the cylinder can be calculated from the formula in equation

    t =(OD-d)/2 =(115-100)/2=7.5mm

    E . Bursting stress

    The bursting stress can be referred to as the amounts of hoop stress and longitudinal (axial) stress that are produced in the wall of the cylinder when subjected to internal and external pressures that may cause the material which the cylinder is made from to fail. This happens if the hoops stress exceeds the tensile strength of the material.

    2 2 2

    2 2 2

    The hoop stress (H ) of a cylinder can be determined from the Barlow formula as shown in the equation below.

    H =P((do 2+di )/ (do -di )) Where,

    p= oil pressure, 200bar = 200 * 105Pa do= outer diameter of cylinder =55mm di = inner diameter of cylinder,

    = 200*105*((1152+1002)/ (1152+1002))


    2 2 2 2

    2 2 2 2

    Also the longitudinal stress is given by: L= (P1R1 -P2R2 )/( R1 -R2 )

    Where P1 = Internal pressure (200 * 105pa)

    P2 = External pressure (atmospheric pressure = 1.0135

    * 105pa)

    R1 = Internal radius R2 = External radius


    3)2))/((57.5*10-3)2- (50*10-3)2)

    =61.599*106 MPa


    1. Low carbon steel:

      Fig . 1 Stresses in piston rod

      Fig . 2 Deformation in piston rod

      Fig . 3 Stresses in cylinder

      Fig . 4 Deformation in cylinder

      Fig . 5 Stresses in piston

      Fig . 6 Deformation in piston

      Table 2 . Low Carbon Steel




      Stresses (MPa)




      Total deformation (mm)




    2. E355 Steel :

    Fig . 7 Stresses in piston rod

    Fig . 8 Deformation in piston rod

    Fig . 9 Stresses in cylinder

    Fig . 10 Deformation in cylinder

    Fig . 11 Stresses in piston

    Fig . 12 Deformation in piston

    Table 3 . E355 steel




    Stresses (MPa)




    Total deformation (mm)





    Here, design and analysis of different parts of actuator i.e, piston rod, piston ,cylinder is done using CATIA V5 and ANSYS . From the above obtained value , we can conclude that the stresses developed in low carbon steel are more than E355 steel and the total deformation of E355 is more than low carbon steel. Hence from obtained data it is beneficial to use E355 stee.


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