Shoulder Bolt Loading Question. 1

[JasonEm]

Thanks for Reading

I have a question on the stress analysis that has to do with a design that will be using a shoulder bolt. The bolts will be the primary loading points at either end of a vertically loaded swivel.

The mounting hole for the bolt will be called out at 0.3750" (+.0015/-.0010) as to have a close fit with the bolt.

Bolt Specs:
Nominal Shoulder Diameter: 3/8"
Length of Shoulder: 1.25"
Bolts spans a gap of 0.65"
Minimum tensile strength: 126,000 psi (416 stainless hardened to 26-32 RC)

A 3,000 lb maximum load will be centrally located on the bolt shoulder and will be considered to be acting at a single point as this would be the scenario producing the greatest stress on the bolt shoulder.

My question is this: When doing my stress analysis for the loading conditions I have looked at the double shear stress that will occur, but is the span (0.65") small enough in relation to the bolt diameter (0.375") that the instance of stress due to bending would or would not have to be looked at? When I do consider the stress due to bending the fact that I am loading the bolt to 3,000 lbs (not to mention the 11,250 lb test loading that will occur!) creates a Moment of around 500 lb*in - combined with a very small Section Modulus (1/1000 degree) makes the stress go through the roof.

I feel as though I am not looking at this problem properly and would like some direction or an example of how to go about this analysis.

Any direction would be greatly appreciated.

 

[vonlueke]

JasonEm: Yes, pin bending stress should be checked when performing a lug and clevis design or analysis. (1) What is the thickness, t1, of the lug (i.e., the thickness of the part held between the two clevis plates)? (2) What factor of safety against yield, FSy, and/or what factor of safety against rupture, FSu, is required for your clevis pin (shoulder bolt)? (3) Will the load be applied only very slowly during actual use?

For now, I'll arbitrarily assume (1) t1 = 15.50 mm, (2) that your project requires the pin must have a factor of safety against yield of FSy = 4.0 and a factor of safety against rupture of FSu = 4.7, and (3) that the loading will always be applied only very slowly (static loading). (Note: The above is not a recommendation to use FSy = 4.0. You must use the correct factor of safety required by your program or design code for this lifting application, considering the loading uncertainty, loading rate, required life span, consequences of failure, etc. See question 2, above.) You can consider the pin as a simply supported beam (conservative).

Gap on each side of centered lug, gap = 0.50 mm. Clevis plate thickness, t2 = 7.62 mm. Pin diameter, D. Moment of inertia, I = (pi)(D^4)/64. Applied load, P = 13350 N. Tensile ultimate strength, Stu = 870 MPa. Tensile yield strength, Sty = approximately 730 MPa. For stainless steels, shear ultimate strength, Ssu = approximately 0.55(Stu), and for alloy or carbon steels, shear ultimate strength, Ssu = approximately 0.62(Stu). Hence, Ssu = 480 MPa.

Compute moment arm from midpoint of clevis plate to quarter point of lug thickness (with lug centered between clevis plates). I.e., moment arm, b = 0.5*t2 + gap + 0.25*gamma*t1, where gamma = bearing stress peaking factor. You can conservatively assume gamma = 1.0. Therefore, moment arm, b = 0.5*7.62 + 0.50 + 0.25*1*15.50 = 8.19 mm (conservative). Moment, M = (b)(0.5*P). Pin bending stress, sigma = M*c/I. Substituting the above gives sigma = 16*b*P/(pi*D^3).

Stress level, R = FSy*sigma/Sty. R must not exceed 100%. Letting R = 1, and solving for D, gives D = [(FSy)(16*b*P)/(pi*Sty)]^0.3333. Therefore, D = [4.0*16*8.19*13350/(3.14159*730)]^0.3333 = 14.50 mm. Therefore, the analysis indicates the pin would need to have a diameter of 14.50 mm to avoid being overstressed in bending during operation.

Now let's say your project requires a factor of safety against yield of 1.1 during testing. Applied load, P_test = 50040 N. Pin bending stress, sigma = 16*b*P_test/(pi*D^3) = 16*8.19*50040/(pi*14.50^3) = 684.6 MPa. Stress level, R = FSy*sigma/Sty = 1.1*684.6/730 = 103.2% > 100%. Not OK. Pin bending stress exceeds allowable bending stress during testing. Try D = 14.70 mm.

Also check pin ultimate average shear stress. Average shear stress, tau = (0.5*P)/A = (0.5*13350)/(0.25*pi*14.70^2) = 39.33 MPa. Stress level, R = FSu*tau/Ssu = 4.7*39.33/480 = 38.5% < 100%. OK. Pin shear stress does not exceed allowable shear stress during operation.

 

[vonlueke]

JasonEm: Also, (4) what is the thickness, t2, of each of your two clevis plates? Notice the above example assumes each of the two clevis plates is t2 = 7.62 mm (or 7.625 mm) thick, corresponding to the shoulder bolt shank length of 31.75 mm.