00:01
In this question we are given the diameter capital d is given which is equal to 160 mm small d is given which is equal to 100 mm the value of g of steel is 80 gigapascal and the value of modulus of b is equal to which is the bronze is equal to 40 gigapascal.
00:30
Now here the value of ts which is the torque of steel plus the torque of the bronze is equal to the torque bc which is the bc section which is equal to the torque bc is given in the question which is 120 kilo newton meter.
00:53
Now here in this question the angle of rotation in steel is equal to the angle of rotation in bronze and in equation this can be represented as follows the torque of steel multiplied by the constant k divided by the modulus of steel multiplied by the moment of inertia of steel is equal to the torque of bronze multiplied by the constant k divided by the modulus of bronze multiplied by the moment of inertia of bronze.
01:39
Now here substituting the values ts here the value of k is considered to be 1 divided by gs is 80 multiplied by the moment of inertia will be 5 by 32 multiplied by 160 raise to 4 minus 100 raise to 4 is equal to the torque of bronze divided by 40 into 5 by 32 multiplied by 100 raise to 4.
02:10
Upon solving this we will get the torque of steel is equal to 11 .0 sorry 11 .1072 times the torque of bronze.
02:21
So, this is the relation between torque of steel and torque of bronze here.
02:26
Therefore, tb plus 11 .1072 times the value of torque of bronze will be equal to 120 that is from this relation.
02:47
Now here we can say that upon solving the above equation we will get the value of torque of bronze to be 9 .911 kilo newton meter and here the value of torque of steel will be equal to 110 .088 kilo newton meter.
03:11
Now here we need to calculate the shear stress in bronze.
03:14
The shear stress in bronze is calculated as follows.
03:22
Shear stress in bronze tb is equal to 16 times the torque of bronze divided by 5 into the diameter of bronze its cube...