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Grade 10.9 bolts or capscrews have a minimum tensile strength of 1040N/mm2 and a hardness of 320 - 380Hv10.

Grade 12.9 bolts or capscrews have a minimum tensile strength of 1220N/mm2 and a hardness of 385 - 435Hv10.

The 12.9 fastener is therefore a higher grade (hardness and tensile strength) fastener compared with 10.9 grade

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How do you convert Mpa to Nm?

Depends on what application the Nm is being subjected to. If the torque was on a circular shaft the method is as follows: If the torsional shear stress is 250 Mpa, on a Beryllium Copper wire. Diameter = 1.5mm (d) Length of rod = 40mm (L) Shear Modulus of elasticity (G) = for BCu wire = 48Gpa or 48*109 N/m2 Polar moment of Inertia (J) = (pi *d4) / 32 (for circular cross section) J = (pi * (.00154)) / 32 J = 4.97*10-13 m4 t = torsional shear stress Torque = T where T= ( t max * J ) / c c is distance from the point of interest. In this case c would be from the center of the rod to the edge of the rod. ( c = d/2 ) in m. c = .0015/2 = 7.5*10-4m letting t max be 250 Mpa.. 250 MPa = 250*106 N/m2 T = ((250*106 N/m2) * (4.97*10-13 m4)) / (7.5*10-4m) T = .165Nm Of course the computed Torque seems low, but don't forget how small this wire is. 250MPa is stress Stress = Force / Area - the smaller the Area the greater the stress. SO for such a small wire only .165Nm is needed to create 250MPa of stress. To answer your question, "How do you convert MPa to Nm" You would need the application with the list of materials and dimensions. G was for theta. Im doing that on my homework right now =P theta (angle in rad ) = TL / GJ theta = (.165Nm *.040m) / (48*109 N/m2 * 4.97*10-13m4) theta = .278rad angle = (.278*180)/pi = 15.9o I am pretty sure that is right. I'm studying Strength of Materials right now, best way for me to study is teaching other people the material I learn. PS this is a conversion from a pressure, to a torque....pretty stupid


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