Question

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The oil pressure operating the hydraulic piston given in the figure is 152 bar. The system is loaded as a floating variable. D1 = 100mm and S = 1.5 E360 B=60mm B1=20mm

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OR YOU CAN CHANGE THE WRONG VALUES

a) Find the diameter of Perno by giving all the necessary measurement values ​​yourself. Select perno, arm and fork materials as cementation steel.

b) L = 500 mm and L = 1000 mm values ​​to be carried by the piston rod without buckling
Find for. Select correction steel as the piston rod material.

Answer 1

a) Force transmited to cylinder having diameter D1=100 mm. Ft= P* A=P* * D = 19 *** 100%mm² = 149.2256kN 7T N IT 4. mm2 4. Due to stress concentration the piston side stress increases by a factor 1.5 so stress on piston is = 1.5 * 19 = 28.5MPa Ppiston Ppiston To have the same force getting transmitted to the piston we have Ft Apiston-side 149.2265 * 103 Apiston-side = (100” – d?) = 4 28.5 Solving for d1 we have, d - 58mm T

b) Now let the steel is tempered at 650 degree celsius, Modulus of elasticity of steel is: G 80GPa We have the torsion equation, given as T T GA J L For no torsion is very very small, lets A= 0.001rad G*A*r L for L= 500 mm 80000 * 0.001 * (58/2) = 4.64MPa 500 T = T = 7T Fp = T * Ashear = 4.64 * * 582 12.26kN 4 for L=1000 mm, force on piston will be 12.26/2 = 6.13kN If we considering no torsion means no buckling of the piston rod then load on piston rod will be euler load with both ends 02:38 fixed.

7 F * E * 1 12 e E= 210 GPa T*di I 64 le L/2, Then safe load for L= 500 mm will be 4* 7? * 210000 * 7 * 584 18.42MN 64 * 5002 for L= 1000 mm Fe = 18.42/4= 4.605MN Fe

** Correction for diameter of pin. There are two pins. Let the clearance between piston diameter and pin diameter be 11 mm so

$d_{pin}= \frac{D_1-d_1}{2}-clearance= \frac{100-58}{2}-11=10 mm$