Designing an Eye bolt using materil 4340

[Rob20]

Can anyone tell me the mechanical property of material 4340?

Thanks

Rob

 

[TVP]

It depends on how it is processed, i.e. normalized, quenched and tempered, etc. If it is quenched and tempered, there is still a huge range depending on the tempering temperature used. This is a high hardenability steel typically used at very high strength levels. Go to Matweb or MIL-HDBK-5/MMPDS to find properties.

http://www.matweb.com

http://assist.daps.dla.mil/quicksearch/

type in MIL-HDBK-5

 

[dbooker630]

Rob20,

What are the requirements of the eye bolt you are trying to design? We can give you a better answer based on your needs. TVP has provided some good links on 4340.

 

[Rob20]

The eye bolt is design for 20 tons load and 1 1/4" dia.

 

[salmon2]

Rob20,

I will do the calculation here, if you or anybody else find I am doing incorrectly, please feel free to point out. I will learn too.

Load = 20 tons = 20X2000 = 40,000 lbs
Area = PI*D^2/4 = 1.2265 in.^2

So the shearing stress = Load/Area = 32,611 psi

So the tensile strength needed = 2 X shearing stress = 65 ksi, which is too soft to get for 4340 actually. So you will have a big safety factor.

 

[boo1]

Analysis or actual max working stresses of the lifting pin must be determined based upon working conditions including impact, number of repetitive loading cycles, load stabilty, wind, or other factors pertinent to the application. Actual max working stress must be equal to or less than the allowable stress.

The discussions of the lifting pin
I would consider using ASME B30.20a-2001 Structural and Mechanical Lifting devices as a start.
main section Requires sf=3 (padeye Requires sf=5)

ASME B30.20-2003 20-1.2.2
"The load-bearing structure components of a lifter shall be designed to withstand the stresses imposed by its rated load plus the weight of the lifter, with a minimum design factor of three, based on yield strength of the material, and the stress ranges that do not exceed the values given in ANSI/AWS D14.1 for the applicable condition. ...."

Other specifications to review are ASTM specification A489 for “Carbon Steel Eye Bolts” or ASTM F541 “Standard Specification for Alloy Steel Eyebolts,”

Load case: simple beam with concentrated load at the center
Mmax =?
Fmax = 20 x 2000 = 40,000 (40k)
where: AISI 4340 Steel, normalized, 25 mm round
Fu = 185.9 ksi, Fy = 125 ksi

The allowable stresses:
Fb = Fy /3.0 = 41.6 ksi = Fv = Ft
(per section 20-1.2.2.2, ASME B30.20)
then:
Ixx = Iyy = (? d4) / 64 = 0.1198 in4
Sxx= I ÷ r = 0.1817in3
Apin = (? d2) / 4 = 1.227 in2

fb = Mmax / Sxx = ?in-lbs / 0.1817 in3
= ?ksi < Fb = 41.6 ksi
fv = Ri / Apin= 40,000 lbs/1.227 in2
= 32.6 ksi < Fv = 41.6 ksi

Bearing stresses of the pin hole area:
Spotts gives the elastic analysis equation for max compressive contact stress (Po) for two different radius cylinders (in your case R1 smaller 1.25" pin and R2 larger hole), P1 load per axial inch on cylinder:

Po=0.591*Sqrt[{(P1*E1*E2)/(E1+E2)}*(1/R1-1/R2)]

Note stresses can be very high due to the small contact area. To reduce the stress reduce the clearance between the two bodies.