I require to find out torsional strength of 35 mm dia steel of grade 1020 and 4140. ? s there

I am required to calculate torsional strength of a 35 mm dia rod of steel grade 1020 and 4140 as we are trying to use higher grade material ….how do i proceed is there any direct tables or formula to calculate. i want it for 1 metre long for each of it. its urgent plz reply. i search google but cudnt find any direct tabular form..

hi neil, had a look at the pdf literature on 2nd page that u sent me the link for. mine is a solid shaft so for 35 mm dia rod i got Ip or polar moment of inertia as 3.14 x 35 ^ 4 /32 i.e., pi x d ^ 4 / 32 or pi x r^ 4 / 2 as is written. Now next step is how do i calculate internal torsion T to use in formula Tr/J for calculating sheer stress ?? And how can i convert mm ^ 4 which i will get from Ip to Nm2 or MPa which is unit of sheer stress. is sheer strength same as torsion strength ??

or am i correct in assuming that i can calculate sheer strength by using 0.6 x UTS of 1020 and 4140 which is 710 Mpa and 1100 MPa respectively and then calculate internal torsion ??? Please reply…whats bottom of page cos 2 theta and sin 2 theta. It says stresses on inclined section – Torsion ? i dont know the angle the shaft is gonna rotate 360 degrees in machine. i am unable to come to a point…first i got to compare torsion strength of these 2 grade of 35 mm rod and then try if 40 mm rod wud be feasible ….

Be sure to properly collect you units. Using the metric system.

Torque = [Newton-meters]

Radius = [meters]

J (moment of inertia = [meters^4]

[Newton-Meters]*[Meters] / [meters^4] = [Newton/meter^2]

which is MPa

As for determining the angle of rotation you will need to know the stiffness and the length of the shaft. The stiffness of both 1020 and 4140 steel is approximately the same.

Niel

If you use the equation for shear stress [1] on page 2, can determine the allowable Torque (T) for the 1020 and 4140 steel by substituting the shear stress for the allowable shear strength for the two materials.

Niel Leon

Community Developer – Engineering.com

Hi Niel,

i did the calculation and got the torque value of max internal torque using that formula. but a small correction Pascal = N/m2 a Mega pascal thus becomes N/mm2 i just verified today. now about length off the shaft it is 680 mm. it has keyway of 9.52 width and 4.76 depp to length of 235 mm from one end. and same end there is hole of 15.88 mm dia drilled to depth of 67 mm …does this effect torque value of shaft ? If so how to calculate that ? And internal torque value is same as torsional strength or do we have to multiply it again with something ? As torque unit is Nm …what is unit of strength ?? i think we are last stage of the problem…i found the following answer : for grade 1020 T or Tmax is 3.58 x 10 e +3 Nm and for grade 4140 its near 5.46 x 10 e +3 Nm taking UTS of 1020 as 710 MPa and 4140 as 1080 Mpa and using same formula…can u verify it for 35 mm dia plz just to make sure i got it right. Thanking you very much in advance……