A semi-eliptical laminated vehicle spring to carry a load 8 kN is to consist of seven leaves 5,1 mm wide, two of the leaves extending the full length of the spring. The spring is to be 1100 mm in length and attached to the axle by two U-bolts 90 mm apart. These bolt hold the central portion of the spring so rigidly that they may be considered equivalent to a band having a width equal to the distance between bolts. The leaves are to be silicon manganese steel. Assuming an allowable stress 3500 N/mm^{2}, determine :

- Thickness of the leaves
- Deflection of the spring
- Diameter of eye
- Length of the leaves
- Radius to which leaves should be initially bent, assume modulus of elasticity as 2,1×10
^{6}N/mm^{2}

**1. Effective length :**

2L = 1100 – 2/3 x 90 = 1040 mm

L = 520 mm

**Load at the center :**

2W = 8000

W = 4000 N

**Thickness of the leaves :**

**2. Deflection of the spring :**

**3. Diameter of eye**

**From bearing consideration**

**The length of pin between links**

Maximum bending moment

By equating bending moment to the resistance of the pin with

The required pin diameter is greater from bending consideration. Hence a pin of diameter 20 mm is suggested. For bush wall thickness of 3 mm, **the eye diameter will be**

**4**. **Length of the leaves :**

Smallest leaves :

** 5. Radius**

**SUMBER:**

*http://books.google.co.id/books?id=08FkBSjG2bMC&pg=PA322&lpg=PA322&dq=a+semi-eliptical+laminated&source=bl&ots=MaKD5SYyqv&sig=qVxu1ttFKj5ar18K6sV7aZM1Fs4&hl=id&sa=X&ei=AiBoUKv9E8vNrQfg_YDICQ&ved=0CDEQ6AEwAA#v=onepage&q=a%20semi-eliptical%20laminated&f=false*

You can leave a response, or trackback from your own site.