00:01
Hello students, according to the data which is given in the question, we have made a free body diagram.
00:07
So, now let us consider r a is the reaction at point a, then the bending moment will be equal to the r a x minus 60 x minus 1 minus 50 x minus 3 equation number.
00:31
Now, we will take the moment about the point b, moment about point b.
00:49
So, we can write r a into 4, this will be equal to the 60 x 3 plus 50 x 1.
01:04
So, r a will be equal to the 57 .5 kilo newton.
01:12
Now, we can write again the equation number 1 in the form of deflection d 2 y upon dx square, it will be equal to the value of r a is given as 57 .5, which we have calculated above into x minus 60 x minus 1 minus 50 x minus 3.
01:40
Now, after integrating both side, we will get y is equal to any constant c 2 plus another constant c 1 x plus 28 .75 x cube by 3 minus 330 x minus 1 whole square by 3 and minus 25 by 3 x minus 3 whole square.
02:17
Sorry, there will be the cube.
02:18
Now, we will apply the boundary condition.
02:24
So, at x is equal to 0, the deflection will be equal to the 0.
02:31
Now, c 2 will be equal to 0 and another boundary condition at x is equal to 4, deflection again will be equal to the 0.
02:44
Now, another constant c 1 is equal to minus 83 .75...