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Shear modulus 
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#1
Jan712, 04:16 PM

P: 19

I am reading through some calculations in a book that refer to an average shear modulus of steel (19700).....
they give no units and it has completely thrown me off track while following through the calcs can anybody help? I know the unit is imperial and if i convert some of the other units to metric, it should equal approx 350,000 kg/m or something?! 


#2
Jan712, 04:24 PM

P: 418

The unit for modulus is GPa, gigapascals.
Although if its imperial then it probably is pounds per square inch or possibly ksi, thousands of pounds per square inch 


#3
Jan712, 04:28 PM

P: 19

I am reading the stanliforth suspension design he refers to it as The "19,700" is a constant derived from the average modulus of shear for steel in this calculation 19,700 x (OD4  ID4) / Bar Length = Angular Rate in in. lbs. per degree 


#4
Jan712, 04:31 PM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,953

Shear modulus
It has the same units as Young's modulus. Force / area, or stress, or pressure.
For an isotropic material like steel [itex]E/G = 2(1 + \nu)[/itex] where [itex]\nu[/itex] is Poisson's ratio. It should be about 11.5 x 10^{6} psi or 80 GPa. I don't know what units your number is supposed to be. EDIT: I just caught up with post #3. Your 19,700 isn't the shear modulus, it also has some conversion factors from degrees to radians, and probably a factor of 16 or 32 because you are using diameters not radii. Either you just believe it, or take a few steps back to figure out what the formula really is. http://en.wikipedia.org/wiki/Torsion_(mechanics) 


#5
Jan712, 05:12 PM

P: 19

Thanks for the quick response guys ive been looking at this psreadsheet / book all day..
could somebody validate this equation for me? does it look correct? 80GPA x Pi x 9.81 (OD4^  ID^4) / Bar Length = Angular Rate in Nm. kg. per degree 


#6
Jan812, 01:50 PM

Sci Advisor
HW Helper
P: 2,121

Jas1159: (Your equation in post 5 currently looks incorrect.) Let torsional stiffness be called k_{r}. For your round tube, k_{r} = T/phi = G*J/L, where T = applied torque, phi = tube torsional deflection (twist) angle, in radians, and k_{r} is in units of torque per radian. Therefore, converting phi to degrees, we have, k_{r} = [G*(pi^2)/(32*180)](OD^4  ID^4)/L. Simplifying therefore gives,
(eq. 1) k_{r} = (G/583.6100)(OD^4  ID^4)/L,where k_{r} = torsional stiffness (torque/deg), G = shear modulus of elasticity, and L = tube length. Let us assume E = 206.10 GPa, and nu = 0.30. Therefore, G = 79.270 GPa = 79 270 MPa = 11 497 140 psi. Therefore, eq. 1 becomes, (eq. 2) k_{r} = [(11 497 140 psi)/583.6100](OD^4  ID^4)/L,where k_{r} = torsional stiffness (lbf*inch/deg), and OD, ID, and L are in units of inch. Using meters, instead of inch, eq. 2 becomes, (eq. 3) k_{r} = [(79.270e9 Pa)/583.6100](OD^4  ID^4)/L,where k_{r} = torsional stiffness (N*m/deg), and OD, ID, and L are in units of m. Using mm, instead of m, eq. 2 becomes, (eq. 4) k_{r} = [(79 270 MPa)/583.6100](OD^4  ID^4)/L,where k_{r} = torsional stiffness (N*mm/deg), and OD, ID, and L are in mm. 


#7
Jan912, 02:02 PM

P: 19

Thanks guys, thats brilliant I managed to continue following through the book after i realsied where the 19700 came from :D



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