Shear Modulus of Steel: Help Understanding Imperial Calcs

[Jas1159]

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?!

 

[rollcast]

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

 

[Jas1159]

The unit for modulus is GPa, gigapascals.

Thanks rollcast, that's exactly why i am confused because for steel i am expecting approximately 80GPA, but to make all the calcs make sense i need an average number equal to 350,000 :/ derived from 19700

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

 

[AlephZero]

It has the same units as Young's modulus. Force / area, or stress, or pressure.

For an isotropic material like steel E/G = 2(1 + \nu) where \nu is Poisson's ratio.

It should be about 11.5 x 10⁶ 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 )

 

[Jas1159]

Thanks for the quick response guys I've 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

 

[nvn]

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,
(eq. 2) k_r = (19 700 psi)(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,
(eq. 3) k_r = (135 827 008 Pa)(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,
(eq. 4) k_r = (135.8270 MPa)(OD^4 - ID^4)/L,​

where k_r = torsional stiffness (N*mm/deg), and OD, ID, and L are in mm.

 

[Jas1159]

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