00:01
Hello friends as soon in the figure there is a curve member a v is a parabola face vertex at a if the vertical load p of magnitude worth 45 pound is applied at a then calculate internal forces at j for h is equal to 12 inch n is called to 40 inch and a is equal to 24 inch let us start solving it drawing the free body diagram of a b this is b sorry not written fpd of a b at the horizontal force is towards left vertical force is upward at a vertical force is p downward, horizontal force is a.
02:06
This is the point a, this is the point b.
02:32
This is given 12 inch, this is given 40 inch.
02:56
Now, summation of f -py in this direction we are taking positive to be 0.
03:04
So you will get by minus p is called to 0.
03:13
So py is called to p and p is given to us is 450 pounds.
03:25
That is upward.
03:29
Moment of couple about a.
03:33
Sorry, moment of course about a.
03:37
If we're taking clockwise to be positive, so summation of m .a must be 0.
03:44
So we can say it would be vx into 12 vx into this distance minus 450 into this.
03:59
So 450 into 40 inch must be 0.
04:05
From here you will get px to be let me write 1500 pound towards left.
04:27
Now summation of fx is 0.
04:32
So you will get a minus bx to v0.
04:38
So we can write a equal to vx and vx is 1500 and the direction of a force is towards right.
04:53
A .v .r.
04:54
Is a parabola.
04:55
So y is called to kx squared.
05:07
So its equation is y is called to kx squared.
05:12
For x is equal to 40 inch y is 12 inch so we can say 12 is equal to k into 40 squared so value of k you will get 12 y so it becomes 0 .075 so equation of parabola can be written as y equal to 0 .005 so equation of parabola can be written as y equal to 0 .00 now slope of this parabola d -y upon dx differentiating it you will get 015x.
06:21
So at point for point j x is 24 inch so y at j, you may write at j point j point so y .j you will get 0 .075 4 .32 inch.
06:59
And slope at j will be 015 into 24 inch...