A circular steel tube of length L=1.0 m is loaded in torsion...

A circular steel tube of length $L=1.0 \mathrm{m}$ is loaded in torsion by torques $T$ (see figure). (a) If the inner radius of the tube is $r_{1}=45 \mathrm{mm}$ and the measured angle of twist between the ends is $0.5^{\circ},$ what is the shear strain $\gamma_{1}$ (in radians) at the inner surface? (b) If the maximum allowable shear strain is 0.0004 rad and the angle of twist is to be kept at $0.45^{\circ}$ by adjusting the torque $T,$ what is the maximum permissible outer radius $\left(r_{2}\right)_{\max } ?$

Transcript

00:01 In this question we are given that there is this steel tube that is followed.

00:05 It has a length of 1 meter and applied by a torque t.

00:10 The inner radius of the tube is 45 mm.

00:15 If this is some tube, this inner radius is 45 mm and the angle of twist is 0 .5 degrees.

00:35 What is the searing strain at inner surface? all right that is the searing stain at the inner surface as we know that t over ip is equal to tau by y is called to g theta y this is known to us now we are given that the angle of rotation is 0 .5 degree all right angle of twist is 0 .5 degrees this value is 0 .g g into 0 .5 into 3 .14 by 180 this is this much radian your length is one meter all let us convert it into millimeters so thousand millimeter so tau over y y is 45 so from here we will get our value of tau so tau would be is it the value of g given to us let me check it is a steel road so and anyways we can calculate the value of g also let me calculate the value of g it is 80 into 10 to the power 9 g is 80 into 10 to the power 9 newton per millimeter square newton per meter square actually right if you want to write it in the form of newton 1 millimeter scale then 80 ,000 newton per millimeter square would be the answer so from here if we solve it it would be 80 thousand into 45 into 0 .5 into 3 .14 divided by 1000 divided by 180.

02:32 Tau would be this one.

02:33 So, 30 cancels out.

02:35 One zero also cancels out.

02:37 So this would be equal to 45 into 8 into 0 .5 into 3 .14 divided by 18.

02:48 31 .4 newton per millimeter square will be the shear strain at the inner surface.

02:55 This is the solution for the first part.

02:58 The second part is if the maximum allowable shear strain is 0 .04 radiant and the angle of twist is to be kept at 45 degree to adjust the torque, but is the maximum permissible outer radius.

03:14 So angle of twist now is limited to 0.

03:20 0 .45 degree.

03:23 Theta is 0 .45 degrees.

03:26 Right.

03:27 And maximum shear strain is 0 .004 radian.

03:34 That means tau would be g into theta, g into 0 .00, 04.

03:44 Did so? yeah, one more 0.

03:48 So, tau would be 80 ,000 into 4.

03:53 Let me calculate it, 80 ,000 into 0 .004...

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