00:01
Hello students, for part a we are going to find out the normal stress and shear stress at point h.
00:08
Now we are going to consider sigma h is equals to rho a plus m x divided by i x into y.
00:27
As per this we are going to substitute the values pi into 10 to the power of 3 divided by pi by 4 into 95 square minus 85 square plus 2 .1 into 10 to the power of 6 divided by pi by 64 into 95 to the power of 4 minus 85 to the power of 4 into 47 .5.
01:06
After simplification we get sigma h is equals to 73 .01 mega pascal.
01:20
Next we are going to find out the tau h is equals to t divided by z p plus v a into y bar divided by i b.
01:45
As per this we are going to substitute the so 3 .96 into 10 to the power of 6 divided by pi by 16 into 95 to the power of 4 minus 85 to the power of 4 divided by 95 plus 18 into 10 to the power of 3 into 706 .5 into 20 .1 divided by 1 .4 into 10 to the power of 6 into 95.
02:34
After simplification we get tau h is equals to 66 .3 mega pascal.
02:50
For part b we are going to find the same thing at k.
02:57
For first we are going to find out the ratio between sigma 1 divided by sigma 2 is equals to sigma by 2 plus or minus square root of sigma by 2 the whole square plus tau square.
03:14
Now we can substitute the values 73 .01 divided by 2 plus or minus sigma plus or minus 73 .01 whole square divided by 2 plus 66 .34 whole square...