00:01
Hello students, in this question we have given value of b is 100, h is 90, t is 4 and b is 5.
00:16
Now we have three parts.
00:19
First in a part we have to find the distance e to the shear centre.
00:24
Now using the formula for this we have e is equal to 3b square t by 2ht plus 6bt.
00:35
Now substituting the values here we get 3 into 100 square into 4 divided by 2 into h90 into 4 plus 6 into 100 into 4.
00:54
After solving this we get the distance will be 38 .4615 mm.
01:03
Now for part b we have to find the shearing stress at d.
01:09
So for centroid we have y will be 102 multiply 4 into 2 plus 176 into 4 into 92 plus 102 into 4 into 182 divided by this whole value will be divided as 102 into 4 plus 176 into 4 plus 102 into 4.
01:47
Now after solving this we get y will be 92 mm.
01:56
Now for moment of inertia using relation that is bd cube divided by 12 plus a y minus y1 square.
02:13
Now substituting the values here we get first 102 into 4 cube divided by 12 plus 102 into 4 into 92 minus 2 square.
02:29
Now for l2 this will be y minus y2 and for l3 this will be y3 minus 1.
02:36
Now calculating this we get 3304800 so plus 544 so we get i1 is 3305344 power 4.
02:57
Now i2 this will be 4 into 176 cube divided by 12 plus 176 into 4 into 92 minus 92 square...