00:01
Hello students, here a steel bar ab of the frame is assumed to be pin connected at its ends for yy axis buckling.
00:08
In this case given that p is 20 kilo newton and the modulus of elasticity of steel is 200 gigapascal, sigma y equal to 300 megapascal and the total length l equal to 6 meter, l2 is 3 meter and l1 is 4 meter.
00:42
In this case we have to find the factor of safety.
00:47
So let's see the body sketch.
00:50
Here a is 50 millimeter or we can write 0 .05 meter and b is 100 millimeter which is equal to 0 .1 meter.
01:06
So first we have to determine the area.
01:13
So area equal to a dot b which is 0 .05 into 0 .1.
01:24
This gives the area of 0 .005 meter square and next we have to find the moment of inertia about the y -axis.
01:35
So the moment of inertia iy equal to 1 by 12 a cube b gives the moment of inertia as 1 .0417 into 10 power minus 6 meter power 4 and by using pythagoras theorem, by pythagoras theorem we have to find the length lac.
02:09
Here it is given that this is a, b and c.
02:15
This is 3 meter, this is 4 meter.
02:18
So lac is found out by using pythagoras theorem as lac is equal to square root of 3 square plus 4 square.
02:27
So lac equal to 5 meter.
02:32
Now we need to draw the forces that is the free body diagram for this case.
02:38
The force p is acting and this point is a.
02:44
In this straight line force ab is acting and another force is acting in this direction which is 4fac and here the length is 5, 4 and this is 3.
03:04
So now we can determine the force fac in the rod by using the equilibrium equation of joint a with respect to x -axis.
03:16
So this is given by here we are taking in the positive x direction.
03:24
So summation of forces in the x direction is 0.
03:28
So p minus 3 by 5 fac equal to 0.
03:34
From this we can able to calculate fac.
03:37
So fac equal to 5 by 3 into 20 kilo newton...