Before we begin, there are a few things to remember:
The purpose of bolts is to squish your parts together so that friction holds the parts in place. Bolts should NOT experience shear forces.
Helicoils and keyserts are a great way to add strength to a bolted connection in aluminum or other weaker materials. These devices let you fasten into a steel, which is stronger and reduces the chance of damaging internal threads.
To do the analysis below with a helicoil, simply do the analysis for the bolt to helicoil interface and then repeat it for the helicoil to part interface. Even though helicoils are strong, you can still rip helicoils out or damage them!
Remember: Parts should be held together by FRICTION alone! Bolts provide the normal force and should not experience shear stress.
Determine what the loading conditions are for the bolted connection and, from there, what the maximum expected load is.
Determine what the required clamping force is so that friction holds the two parts together.
Divide the total clamping force by the number of bolts to determine the axial load per bolt.
NOTE: You can also determine the number and size of bolts based on the total clamping force, or iterate between the two methods.
Example: Consider a 15 kg aluminum part bolted to an aluminum vibration table undergoing a 9.47 gs RMS random vibration normal to the bolted connection. The static coefficient of friction is assumed to be 1.05.
The force Fa on the part due to acceleration on the part is:
Fa = ma = 15 [kg] * 9.47g [m/s^2] = 1394 N
The normal force between the part and vibration table, times the static friction coefficient between the two surfaces, must be greater than the acceleration force:
Fa < μFN
Rearranging this, the minimum normal force is:
FN > Fa/μ
FN > 56.5/1.05 = 1327 N
Therefore, the axial load through ALL the bolts must be at least 1327 N. If we mount the part with four bolts, the load per bolt is 1327 N/4 = 332 N
This step is optional, but it's nice to know if you’ll strip a bolt or your part threading first. (It’s easier to replace a bolt than to fix a part!) However, As and An will be used later on. NOTE: These equations are intended for imperial fasteners. They give reasonable results for metric fasteners but for metric fasteners n = 1/(threads per mm)
Calculate the external shear area As using the equation below:
As = π*n*Le*Knmax[1/(2n) + 0.57735*(Esmin-knmax)]
Where n is the number of threads per inch if you are using imperial fasteners or n = 1/(threads per mm) for metric fasteners. Remember, imperial fastener names give the threads per inch as the second number. E.g. a 2-56 fastener has 56 threads per inch. Le is the engagement length between the part and bolt (i.e. the depth of the thread in the part that the bolt actually engages with, see Figure 1.), Knmax is the maximum possible minor diameter of the internal thread, and Esmin is the minimum possible pitch of the external thread.
Figure 1: Length of Engagement
Calculate the internal shear area An using the equation below:
An = πi*n*Le*Dsmin[1/(2n) + 0.57735*(Dsmin-Enmax)]
Where n is the number of threads per inch if you are using imperial fasteners or n = 1/(threads per mm) for metric fasteners, Le is the engagement length between the part and bolt, Dsmin is the minimum possible major diameter of the external thread, and Enmax is the maximum possible pitch of the internal thread.
Calculate the stripping ratio J
J = (As*σs)/(An*σn)
Where σs is the tensile strength of the external thread and σn is the tensile strength of the internal thread.
If J > 1, the internal thread will strip first. You can either multiply the engagement length by J to prevent this or accept it depending on the risk and impact.
Example: Continuing from the previous example, assume we have four 4-40 fasteners to hold the parts together with a 1/8” engagement length. Assume we’re using an 18-8 SS fastener with a yield strength of 70,000 PSI with a UNC3A thread and both the part and vibration table are made of 6061 aluminum with a 8,000 PSI yield strength.
The number of threads per inch for imperial screws is the second number in the fastener name (i.e. 40 for a 4-40 fastener).
You can find Knmax, Esmin, Enmax, and Dsmin from online tables like this one and this one. However, make sure your selected fastener follows the same fastener standard as the table you’re using! Eg the linked table is for the ANSI/ASME B1.1 standard. Make sure your thread class is the same as your selected fastener too! A fastener will have a #A thread fit while the internal thread will have a #B thread fit.
Using these values and the equations in steps 1 and 2, the external shear area As is 0.018 in^2 and the internal shear area An is 0.029 in^2.
J = (0.018 [in^2] * 70000 [PSI])/(0.029 [in^2] * 8000 [PSI] ) = 5.5
J > 1, so the internal threads in the vibration table will fail first. A vibration table is probably more expensive than the bolts you’re using, so consider either:
Threading deeper into the table
Using a threaded insert so your internal threads are stronger
Using fine thread bolts
Using more, smaller fasteners with the same length of engagement
However, for this example, we will stick with the current geometry.
The maximum load you can put onto your threads is a function of their shear area and the material strength:
Fmax = σ * A
For the internal threads:
Fnmax = σnAn
While the internal thread maximum load will be:
Fsmax = σsAs
Your preload axial stress plus any extra loads on the fastener cannot exceed either of these values!
Example: Continuing from our previous example, Fnmax = 8000 [PSI] *0.018 [inches^2] = 235 pounds force while Fsmax = 70000 [PSI] *0.029 [inches^2] = 1291 pounds force
To apply the necessary clamping force (i.e. preload), you must torque the fasteners to some required value. There are maximum torque charts available for your fasteners, but make sure they are for the same standard as the fasteners you’re using! Also remember the given torques are the MAXIMUM values the fasteners can be torqued to before failing and will approximately equal their failure point.
The minimum torque Tmin applied to your fastener is:
Tmin = k * dm * Pp
Where k is the torque coefficient, dm is the bolt’s minor diameter, and Pp is the preload force. The torque coefficient for your load case can be found in various tables like this one, but generally a dry fit is around 0.2 and loctite is 0.14.
To ensure that we have enough clamping force, we will need to add a load factor of safety FSp to this torque to determine what torque to apply:
T = k * dm * Pp * FSP
The calculated torque must not exceed the maximum recommended torque for your fastener type and standard.
Next, we can calculate the axial load factor of safety for the fastener FSF by
FSF = σs / [(Pp * FSP)/(Asmin)]
Where Asmin is the fastener’s minor cross section (Asmin =0.25 * π * dm^2)
Example: For our 332 N (74.6 pounds force) preload calculated in Part 1, assuming k = 0.14 for a loctited connection, a load factor of safety of 2, and a minor bolt diameter of 0.081 inches, the minimum required torque is:
Tmin = 0.14 * 0.081 [inches] * 74.6 [lbf] * 2 = 1.7 inch pounds
This torque is below the maximum 4.9 inch pounds for an ASTM A574 stainless steel fastener, which is a good sanity check.
The associated axial load factor of safety is:
FSF = 70000 [PSI] / [(74.6 [lbf] * 2)/( 0.25 * π * 0.081^2 )] = 2.4
Finally, we must make sure that the preload stress on our internal and external threads does not exceed the maximum allowable loads for either.
The design-load Pd for each fastener is the required preload Pp multiplied by our factor of safety FS:
Pd = Pp * FS
Multiplying this design load by our material strength, we find our design stress σd for our internal
threads:
σdn = Pd/An
And external threads:
σds = Pd/As
And by dividing the yield strength of both materials by the design stresses, we can find the internal and external thread factors of safety:
Internal Thread Factor of Safety:
FSs = σs/σds
External Thread Factor of Safety:
FSn = σn/σdn
Example: Finishing up our example, the internal thread design stress is:
σdn = Pd/An = 149 [lbf]/0.029 [in^2] = 5070 PSI
and the internal thread factor of safety is: FSn = 8000 [PSI] / 5070[PSI] = 1.58
Meanwhile the external thread design stress is:
σdn = Pd/As = 149 [lbf]/0.018 [in^2] = 8091 PSI
and the external thread factor of safety is: FSs = 70000 [PSI] / 8091[PSI] = 8.65
The bolt’s axial load factor of safety as well as the internal and external factors of safety are above 1, At 1.58, the internal thread factor of safety is the lowest and where we should expect failure to occur.