Fastener Theory - Determining the Ideal Torque for a Bolt

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YoshiMoshi
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Homework Statement
Determine the ideal torque for a class 8.8 bolt that is 6 mm in diameter made of steel that is dry (not greased or wet). The bolt has 1 mm pitch threads.
Relevant Equations
T = K * D * P
T = Target tighten torque
K = Coefficient of friction
D = Bolt nominal diameter
P = Bolt's desired tensile load (generally 75 % of yield strength)
Source: http://www.zerofast.com/proper-bolt-torque
So this is for my car. If I have a bolt that I don't know what the ideal torque is supposed to be because the manufacturer of the car doesn't publish a torque, it's possible to calculate it? This can be done just give the specification of the bolt? Regardless of what the bolt is fastening? I'm not entirely sure. Is this correct?

K = 0.12
I don't know what the coefficient of friction is for my bolt, the specifications can be found here: https://www.mcmaster.com/94036A111, it has a "Black Ultra-Corrosion-Resistant-Coated Steel" finish. I'm not using a grease or an anti-seize compound, just he dry bolt. I used this as a reference https://rommet.com/remco/misc/friction.htm, which lists a "Phosphate Steel" coefficient when dry as 0.12-0.18. I should assume worse case of 0.12 to avoid over tightening?

D = 6 mm
I measured with a micrometer.

P = 69,000 psi
Yield strength for a Class 8.8 bolt is 92,000 psi (source: https://www.engineersedge.com/hex_bolt_identification.htm)
T = K * D * P
T = ( 0.12 ) * ( 6 mm ) ( 1/25.4 in/mm ) * ( 69,000 p/in^2 ) = 1,955 p/in
( 1,955 p/in )( 12 in/ft ) = 23,470 p/ft

Looks like I have done something wrong? I'm not sure what though. Also, isn't torque always in foot*pounds or inch*pounds? But the formula I'm given provides units of measurement of a unit of weight per a unit of displacement.

Thanks for any help!
 
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