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Dear Miroslav, I am always amazed at how a seemingly simple question doesn't have a simple answer. Given my limited resources, I found very little information regarding bolts smaller than ¼". I found some torque charts for bolts above that size, but most companies that produced these tables were quick to point out that these were recommendations only. An unreferenced torque spreadsheet that shows reasonable numbers is shown at http://www.raskcycle.com/webd oc14.html However, it doesn't include the sizes you need. So we will have to use an equation to help solve your question. Pre-tensioning of bolts is what keeps your connection together. A much better explanation of the importance of pretensioning is given at the following websight from Unified Engineering Inc. http://www.uni fied-eng.com/scitech/bolt/clamping.html Bolts are pretensioned by turning them until they are snug, then using a twisting force (torque) from a specialized torque wrench or adjustable impact hammer to further turn the bolt. Since the head of the bolt is snug against the washer or surface, turning the bolt actually elongates it. The elongated bolt acts as a stretched spring, pulling the two materials together. This is the pretension force. Most designers want to maximize the pretension force used, since using larger bolts at less than maximum force is wasteful and more expensive (unless the engineer determines a necessary degree of safety requires larger bolts). This is how most tables are set up, for the maximum recommended pretensioning for a certain bolt. If you do have some tables for the larger bolts, you will see there are two major factors when determining the maximum torque for a bolt. Grade and lubrication. Grade is the type of steel your bolts are made out of. The higher the number, the stronger the bolt. Grades 1 and 2 are the lowest, cheapest, and are frequently what you buy from the hardware store. I would HIGHLY recommend not using these for high stress applications as they also tend to have a wider range of strengths. The next major grade is grade 5, which is medium strength and used when grade 8, a high strength bolt, is not needed. A good strength table is shown at the following websight: http://www.bsn.com/Cycling/T orques.html This sight also gives a shortened example of the equation I will talk about. The other factor is lubrication between the bolt and the threads. If there is no lubrication, some of the torque force is needed to overcome the friction between the inner and outer threads. This friction then dissipates, meaning the net pretensioning is less. By lubricating the bolt, the torque needed to get the same pretension force is less. I won't go through the derivation of the necessary torque equation, since I barely understood it as well. But here it is: (From "Mechanical Engineering Design" Shigley & Mischke. 5th edition, McGraw-Hill Pg. 346). T = K*Fi*d Where: T = Torque K = Torque Factor Fi = Preload Force d = Major bolt diameter (ex. .250" for ¼" bolt) The bolt diameter is the easy one to find, it is the preload force and torque factor that are a little more subjective. The torque factor generally depends on lubrication. Here is a quick table extracted from the above reference. Bolt Condition K Factor Nonplated, black .30 Zinc-plated .20 Lubricated .18 There are other lubrications that can reduce that number even further, but for general purposes a K value of .20 is used. Then what is the max preload force? Again from Shigley and Mischke, who referenced the following info from Russell, Burdall, & Ward Corp's book "Helpful Hints for Fastener Design and Application". "For reused connections, pretensioned load (Fi) should be 75% of the proof load (Fp) of the bolt." The proof load is obtained from the equation Fp = At*Sp. Where: Fp = Proof load of bolt At = Tensile Stress Area Sp= Minimum Proof Strength So before I confuse you any further, lets do an example using a ¼-20 grade 5 bolt. If you can't find a table listing proof strength, it is generally 90% of the yield strength. (Not tensile strength!) (See the values at the bicycle websight). Proof strength for a grade 5 bolt of this size is 85 kpsi (1 kpsi = 1,000 psi or pounds per square inch). Tensile stress area is .0318 in^2. (The following website has tables that give the stress area for small bolts. Look under fasteners). http://www.srl.gatech.edu/education/ME3110/design-reports/RSVP/DR4/ catalog/fast_bas.html Therefore, Fp = 85,000 psi * .0318 in^2 = 2703 lbs. And we wand Fi to be 75% of this, so 0.75*2703 lbs = 2027 lbs. And from our earlier equation, T = K*Fi*d. We will chose the base K of .20 in this case, and d of ¼ bolt is .250 (d is also found at the last websight). Then we get… T = .20*2027 lbs*.25in = 101 in.lbs, or divide by 12 and we get 8.5 ft.lbs. The table from the first websight gives a torque for an unlubricated ¼-20 bolt of 8 ft.lbs, so our equation seems quite close. It is likely the table is a little on the safe side. So now you have a way to figure out approximate maximum recommended torques for your machine screws. Please note that this is only a recommendation and there can be other factors involved. Also, the manufacturer of your bolts may have better strength information on their product, and be VERY sure about what grade of bolt you have! It should be obvious that a low grade bolt could snap off at the recommended torque of a grade 8. Best of luck and I hope this helps. BK

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