zxzxz

1. 4.8.5.2PART DESIGN OF HYDRAULIC CYLINDERS
 General specifications from geometrical analysis & assumptions ;
 Cylinder retracted height =900 mm
 Cylinder extracted height=1500 mm
A) Designof piston rod
- It is the column that pushes the boom in ordered to lift the engine ;
- Subjected to higher compression stress
- Length of cylinder =600 mm<from geometrical analysis>
 Material selection; gray cast iron
-𝜎𝑢𝑡=400 𝑀𝑝𝑎, 𝜎𝑦=250 𝑀𝑝𝑎 , 𝜏 𝑦=145 𝑀𝑝𝑎 ,𝐸=200 𝐺𝑝𝑎, 𝐺=77 𝐺𝑝𝑎
F=design load =load * factor of safety
=19600*2
=39200 N,
𝑑 𝑝𝑟=𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑝𝑖𝑠𝑡𝑜𝑛 𝑟𝑜𝑎𝑑 ,
𝜎 𝑎𝑙𝑙 =
𝐹
𝐴
= 𝐹
3.14
4𝑑 𝑝𝑟∗2 ,
⁄⁄
𝜎𝑎𝑙𝑙 =
𝜎𝑦
4
=
250
4
= 62.5 𝑀𝑝𝑎
𝑑 𝑝𝑟 = √
4 × 19600
3.14 × 62.5 × 106
= √4.01× 10−4
𝒅 𝒑𝒓 ≈ 𝟐𝟎 𝒎𝒎
- Cheeked for buckling of the piston road ;
 Apply Rankin’s formula for all types of columns;
𝑊𝑐𝑟 =
𝜎 𝑦×𝐴
1+⁄
𝑎( 𝐿
𝑘)⁄
2
=
250 × 106
× 3.14 × 0.032
1 + 1.266 × 10−4(
1.2
7.5 × 10−3 )
𝑾 𝒄𝒓 = 𝟒𝟑𝟔.𝟑 𝑲𝑵
Design factor /nd /=
𝑾 𝒄𝒓=𝟒𝟑𝟔.𝟑 𝑲𝑵
63080
= 𝟔. 𝟓so the design of piston road is safe.

2. 𝐁) Designof cylinder
- The cylinder is used to store the fluid or oil under pressure.
 Material selection ;
Steel standard < structural ASTM a 36>
𝜎𝑢𝑡=400 Mpa, 𝜎𝑦=250 𝑀𝑝𝑎 , 𝜏 𝑦=145 𝑀𝑝𝑎 , 𝐸=200 𝐺𝑝𝑎, 𝐺=77 𝐺𝑝𝑎
 To find the inner diameter of the cylinder/𝑑 𝑐𝑦,𝑜=𝑎𝑝𝑝𝑙𝑦 𝑝𝑎𝑠𝑐𝑎𝑙𝑠 𝑙𝑎𝑤 𝑃1 =𝑃2
 R/A1 = F2/A2
𝑑 𝑐𝑦,𝑖𝑛= 𝑑1√ 𝐹
𝑅⁄ And let Fh=240 N
𝑡𝑎𝑘𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝑜; -Fh*(x+L) +R(Xcos 𝜃) = 0
R=
240 ∗0.5
0.04 cos30
= 3464N
SO, 𝑑 𝑐𝑦,𝑖𝑛=0.02√
19.6∗103
3464
= 0.3989
𝒅 𝒄𝒚,𝒊𝒏= 𝟒𝟎 𝒎𝒎
 The outer diameter of the cylinder will be;
𝑑 𝑐𝑦,𝑜𝑢 = 𝑑 𝑐𝑦,𝑖𝑛 +2𝑡
= 40+2*3
𝑑 𝑐𝑦,𝑜𝑢= 46 mm
𝑓𝑜𝑟 𝑠𝑎𝑓𝑒𝑡𝑦 𝑡𝑎𝑘𝑒 𝒅 𝒄𝒚,𝒐𝒖= 𝟓𝟎𝒎𝒎
Cheek either the cylinder is thin or thick; Pin<<1/6(𝜎𝑎𝑙𝑙 )
Pin = 19600*4/3.14*0.052
= 9.98 Mpa 9.98 Mpa <<< 1/6(250 Mpa) =41.67 Mpa so it is thin cylinder.
This thin cylinder is subjected to tensile stresses;
1. Circumferential /hoop stress
𝜎
𝑡1 =
𝑝 𝑖∗𝑑 𝑐𝑦
2𝑡
⁄
So t =
𝑝 𝑖∗𝑑 𝑐𝑦
2𝜎𝑡1
=
9.98∗106
2∗125 ∗106 × 0.05
=2.5mm
Take t ≈ 3mm
2. Longitudinal stress/
𝜎
𝑡2=
9.98∗0.05∗106
4∗3∗10−3
𝜎𝑡2=40.5*106
pa
 According to the maximum shear stress theory, the maximum shear stress is one half of the
algebraic different of the maximum & minimum principal stresses.s

3. 𝜏
𝑚𝑎𝑥 =
𝜎𝑡2 −𝜎 𝑡1
2
=
125 − 40.5
2
𝜏 𝑚𝑎𝑥=42.25 Mpa
𝑠𝑖𝑛𝑒 𝛕 𝐦𝐚𝐱≪ 𝛕 𝐲 where the design is safe.
C) Designof pipe
- Pre used to transport working fluid under pressure
- use flexible material for the pipe-rubber
 material selection; Rubber (grade Amg)
𝜎𝑢𝑡=200 Mpa, 𝜎𝑢𝑐=240 𝑀𝑝𝑎 , 𝜏 𝑢𝑡=350 𝑀𝑝𝑎 ,
Assumptions; internal diameter of the pipe, d=5mm, L=500 mm
Let us assume thin cylinder; this applied when
 The stress across the pipe section is uniform
 The internal diameter of the pipe is more than twenty times its wall thickness/t d/t > 20.
𝜎𝑡=
𝑝𝑑
2𝑡
, pressure is the same in the closed direction so p=9.8 Mpa
𝑡 =
𝑝𝑑
2𝜎𝑎𝑙𝑙
𝜎 𝑎𝑙𝑙=
𝜎𝑡
𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑡𝑦
= 20/1.5=13.33 𝑁
𝑚𝑚2⁄
𝑡 =
𝑝𝑑
2𝜎 𝑎 𝑙𝑙
=
9.8∗5
2∗13.33
𝒕 = 𝟐 𝒎𝒎
Cheek for thin;
 D/t>20
 5/2 >20 so it is not correct.
Therefore the pipe is thick and use lame’s equation;
𝑡 𝑝 = 𝑟
𝑝(√
𝜎 𝑎𝑙𝑙 +𝑝
𝜎 𝑎𝑙𝑙 −𝑝
−1)
= 2.5(√6.55-1)
𝒕 𝒑 ≈ 𝟒 𝒎𝒎 Thickness of the pipe