Looking at Toyota's T24A-FTS, open deck design, aluminum block. The head bolt install process is 65 ft-lbf, then 90-deg, and then another 90-deg. Assume 12mm bolts (don't really know, round to 1/2 in bolt), with a nut factor of 0.2, the clamping force, (assuming the torque reaches 100 ft-lbf), is 12000 lbf. did I do the math correctly? if so, this is a lot to ask for the threads in the aluminum block. Obviously everything works, but I am just wanting to know how the threads are formed to be able to hold that force. Thanks.
Question on thread strength in aluminum block
- Thread starter windeye
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Shear strength of aluminum is around 40 ksi, so you need about 0.3 sq in of area in shear for the threads to hold 12kips. With a 1/2" diameter, you get that with 0.2" of thread engagement.
Thanks for your note. 40ksi is actually quite strong. Amazing.
Yeah, 6061-T6 is actually around 30 ksi shear strength so you would need somewhat more engagement. Also, I'm sure the shear area is a bit less than just pi*D*t.
I have heard as a rule of thumb if you have that's engagement of 2-2.5 times the bolt diameter, you will break the bolt before stripping the thread in aluminum.
I have heard as a rule of thumb if you have that's engagement of 2-2.5 times the bolt diameter, you will break the bolt before stripping the thread in aluminum.
Just follow the 2x rule in aluminum. If you have at least 2x the nominal diameter in engagement, you won't strip threads. AN M10 needs >20mm engagement, M12 >24mm, etc.
On critical things like head bolts, typically you'll see more like 2.5x thread engagement to make it more stable against creep.
Im not familiar with this engine. Do they just screw into the block or is there a steel thing like a timesert in there .
Chris, I am wondering about that and Google did not come up with anything. I only found that it’s likely that the threads are roll-formed instead of being cut. This increases the strength of threads. I also checked shear strength at 150 deg C, some reduction, not significant.
100 foot pounds of torque is the same as 100 pounds feet of torque.
100 ft-lb is 1200 inch pounds not 12000 ?
And the post below yours says .2 inch thread engagement? That is .050 less than a 1/4 inch, ask a Northstar mechanic how well that would work out.
So far no one addressed if the aluminum in all the calculations are for is for extruded, forged, cast or what ever, that is important to consider.
I would not want much less than 1 inch of bolt length into cast iron for that torque. Especially for the pressures and heat cycles its working at.
edit here, I saw Hohn's post after I posted this.
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