DAEDALUS wrote:You didn't mention the grade of the hardware. I calculate over 100ksi stress on those bolts just from bending alone.
Good point. Below is my best guess calculation of the bolt strength:
Did I do the math right? The numbers seem too good to be true. Just one bolt, tightly torqued to the VH45DE head bosses, with a steel plate at 90 degrees could hold up more than 15 Q45 engines without hitting the yield strength.
However, I'm not sure if this calculation is in a line parallel to the threads, or perpendicular to the threads (see photo attached); so I repeat my calculations here for others to critique if necessary.
I'm given the three indications below:a) Local dealership mechanics do NOT use front engine slingersb) Scottsdale dealership mechanics do not use front engine slingersc) Scottsdale has sold less than half a handful in 14 years of businessMy interpretation is that that nearly EVERYONE is using some other method of VH45DE support.
Perhaps, if they simply bolt a steel plate onto the engine, I can too.
Assuming we attempt the suggested overhead support mode, it's a good idea, as you suggest, to THINK about the calculated stresses & then use parts vastly able to handle them. I don't have any formal training in engineering sheer, stress, torsion, tension, pressure, etc. (i.e., see
http://www.esabna.com/EUWeb/OX...1.htm ), so, I need to rely on published enforced standards even more than you do. I knew nothing about bolt grade markings yesterday; but, they were simple to look up (GOOGLE: "Bolt Grade Marking"). (It's kinda' humbling; more than I could ever remember was found in that single web search.)
Since I could get squished if I get this math wrong (the true difference between mass and force ), I'll need to go slowly in the MKS system (using conversion charts at
http://www.engnetglobal.com/ti...id=15):
Assuming:A) Maximum engine weight = 700 pounds = 318 kilogramsB) Apparently, 1 kilogram = 9.18 Newtons, so 318 kg ~= 3,000 NewtonsC) And given the nominal bolt diameter = 10 mm (so the radius is 5 mm)It appears:i) Cross-sectional area = pi x r^2 = 22/7 x 5^2 mm ~= 80 square mmii) That gives me a force per unit area (i.e., pressure) on one bolt of:iii) Newtons / square mm of bolt = 3,000 Newtons / 80 mm^2 ~= 40 N/mm^2
The metric bolts I purchased were marked 8,8, which (according to
http://home.jtan.com/~joe/KIAT/kiat_2.htm ) have the break-point properties of: a) Tensile Strength === the stress level where the bolt breaks.b) Yield Strength === the stress level where the bolt bends.Where an 8,8 indicates:a) Tensile Strength = 8 x 100 = 800 Newtons/mm2 (116,000 psi)b) Yield Strength = 8 x 8 x 10 = 640 N/mm2 (93,000 pounds per square inch)
If my math is correct, the Q45 engine (700 pounds, 3,118 Newtons) if held up by a single 10 mm diameter bolted metal plate similar to an engine slinger, would only put a pressure of 40 newtons per square millimeter on the bolt (which, according to
http://www.engnetglobal.com/ti...id=17, is about 6,000 pounds of force per square inch); while that bolt can hold more than fifteen times that pressure without yielding.
One problem with this calculation is that I'm not sure if I just calculated the stresses tearing the bolt apart perpendicular or parallel to the threads.
I'm assuming it's parallel (which is what a force from a thin steel plate similar to the engine slinger is where the engine support J-hook is attached would be).
Also, I'm not assuming any force multiplication (torque) due to a cantilevered arm.
If I assume the engine bolt is not attaching a steel plate, but is cantilevered with a chain slung around the head 80 mm from the attachment point, the force x distance equation becomes (how much greater)?
