00:01
In this question we can write the let theta which is equal to the rotation rotation angle of b n r a b c so we can write theta which is equal to row b d as we can say in this figure divided by l ab so it becomes which is equal to row c divided by l ac as we know that the l -a -b which is equals to l -b -c which is equal to 1 meter and the l -a -c which is equals to 2 meter this both the values we have okay in this question next we can therefore from this equation write the row b -t which is equals to 1 divided by 2 row c e okay this is first equation now we can write the summation of the masses of a which is equal to 0 so it becomes f b d l ab l ab plus f c e l ac which is equal to q l ac from the figure so we get this as q which is equal to f c e plus 1 divided by 2 f b d now assuming that the cable ce is the yielded, then the f, ce, which is equals to a, sigma, y, which is equals to 100 multiplied by 10 to power minus 6, 345 mega, that is 10 to power 6.
01:51
So on simplification, it becomes 34 .5 kilo -newton.
01:56
Then fbd which is equal to 2 multiplied by q minus fce that is 2 multiplied by 50 minus 34 .5 so it becomes a 31 kilo newton now here this fbd which is laysenar equals to a sigma y so the cable bd the cable bd is is a elastic zone when q which is equal to 50 kilo newton.
02:39
Therefore, let's start with the first value.
02:42
We can write the sigma max at bd, which is equals to fbd divided by a.
02:51
So it becomes 31 multiplied by 10 to power 3 divided by 100.
02:56
So it becomes 310 megapascal...