00:01
So, given in this question that tau equals to 55 mpa, x dash equals to 50 mm.
00:15
So, we can find the value for y dash equals to a1y1 plus a2y2 plus a3y3 divided by a1 plus a2 plus a3 that is equals to 150t multiplied with 75 plus 150t multiplied with 75 plus 100t multiplied with 0 divided by 150t plus 150t plus 100t.
00:59
So, solving down this we get our final value of y dash is equals to 56 .25 mm and a is given by a1 plus a2 plus a3 which is equals to 400t mm square.
01:25
So, from this we can calculate the primary shear stress which is given by tau1 equals to p divided by a that is equals to 25 multiplied with 10 to the power 3 divided by 400t.
01:51
So, we get the value equals to 62 .5 divided by t mpa and here r is given by ga which is equals to square root of gb square plus va square and now we have to just substitute the individual values.
02:17
So, this is equals to 150 minus 56 .25 square plus 50 square.
02:30
So, solving open the values for the square root we get r equals to 106 .25 mm.
02:41
Now, let us find the angle.
02:44
So, tan theta is given by 150 minus 56 .25 divided by 50 which is equals to 1 .875 and we know that from here we can find the theta by the inverse function of tan.
03:03
So, tan inverse 1 .875 which is equals to 61 .92.
03:10
This is the very important angle that we need to proceed with the question.
03:15
So, phi equals to 90 minus theta as we are subtracting it from the perpendicular.
03:26
So, 90 minus 61 .92 which is equals to 28 .07.
03:34
So, substituting these values we can calculate r1 equals to g1g which is equals to square root of 2851 .56 and this is equals to 53 .4 mm.
03:52
R2 equals to g2g equals to g1g equals to 53 .4 mm and finally r3 equals to g3g equals to 56 .25 mm...