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!

Part 1: What is the Needed Clamping Force?

Remember: Parts should be held together by FRICTION alone! Bolts provide the normal force and should not experience shear stress.

1. Determine what the loading conditions are for the bolted connection and, from there, what the maximum expected load is.

2. Determine what the required clamping force is so that friction holds the two parts together.

3. 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 F_a on the part due to acceleration on the part is:

F_a = 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:

F_a < μF_N

Rearranging this, the minimum normal force is:

F_N > F_a/μ

F_N > 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

Part 2: What Will Fail First? Internal vs External Thread

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, A_s and A_n 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)

1. Calculate the external shear area A_s using the equation below:

A_s = π*n*L_e*K_nmax[1/(2n) + 0.57735*(E_smin-k_nmax)]

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. L_e 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.), K_nmax is the maximum possible minor diameter of the internal thread, and E_smin is the minimum possible pitch of the external thread.

2. Calculate the internal shear area A_n using the equation below:

A_n = πi*n*L_e*D_smin[1/(2n) + 0.57735*(D_smin-E_nmax)]

Where n is the number of threads per inch if you are using imperial fasteners or n = 1/(threads per mm) for metric fasteners, L_e is the engagement length between the part and bolt, D_smin is the minimum possible major diameter of the external thread, and E_nmax is the maximum possible pitch of the internal thread.

3. Calculate the stripping ratio J

J = (A_s*σ_s)/(A_n*σ_n)

Where σ_s is the tensile strength of the external thread and σ_n is the tensile strength of the internal thread.

4. 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.

Using these values and the equations in steps 1 and 2, the external shear area A_s is 0.018 in^2 and the internal shear area A_n 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:

1. Threading deeper into the table

2. Using a threaded insert so your internal threads are stronger

3. Using fine thread bolts

4. Using more, smaller fasteners with the same length of engagement

However, for this example, we will stick with the current geometry.

Part 3: Thread Loads

The maximum load you can put onto your threads is a function of their shear area and the material strength:

F_max = σ _* A

1. For the internal threads:

F_nmax = σ_nA_n

2. While the internal thread maximum load will be:

F_smax = σ_sA_s

Your preload axial stress plus any extra loads on the fastener cannot exceed either of these values!

Example: Continuing from our previous example, F_nmax = 8000 [PSI] *0.018 [inches^2] = 235 pounds force while F_smax = 70000 [PSI] *0.029 [inches^2] = 1291 pounds force

Part 4: Fastener Torque and Factor of Safety

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 T_min applied to your fastener is:

T_min = k * d_m * P_p

Where k is the torque coefficient, d_m is the bolt’s minor diameter, and P_p 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 FS_p to this torque to determine what torque to apply:

T = k * d_m * P_p * FS_P

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 FS_F by

FS_F = σ_s / [(P_p * FS_P)/(A_smin)]

Where A_smin is the fastener’s minor cross section (A_smin =0.25 * π * d_m^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:

T_min = 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:

FS_F = 70000 [PSI] / [(74.6 [lbf] * 2)/( 0.25 * π * 0.081^2 )] = 2.4

Part 5: Internal/External Thread Stresses and Factors of Safety

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 P_d for each fastener is the required preload P_p multiplied by our factor of safety FS:

P_d = P_p * FS

Multiplying this design load by our material strength, we find our design stress σ_d for our internal

threads:

σ_dn = P_d/A_n

And external threads:

σ_ds = P_d/A_s

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:

FS_s = σ_s/σ_ds

External Thread Factor of Safety:

FS_n = σ_n/σ_dn

Example: Finishing up our example, the internal thread design stress is:

σ_dn = P_d/A_n = 149 [lbf]/0.029 [in^2] = 5070 PSI

and the internal thread factor of safety is: FS_n = 8000 [PSI] / 5070[PSI] = 1.58

Meanwhile the external thread design stress is:

σ_dn = P_d/A_s = 149 [lbf]/0.018 [in^2] = 8091 PSI

and the external thread factor of safety is: FS_s = 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.