00:01
For this problem on the topic of analysis of structures, we are given a frame and loading is shown and we are asked to determine the forces or the components of the forces acting on member d -a -b -c at points b and d.
00:15
So we'll first draw a free body diagram of the entire frame and from here we will take anticlockwise moments to be positive and we know that the sum of the moments of all.
00:32
All the forces about point g must equal to zero since the system is in equilibrium.
00:38
So if we take the sum of these moments, this is force h times a distance of 0 .6 meters minus the force of 12 kiloons at a distance of 1 meter minus 6 kiloons at a distance of 0 .5 meters.
01:06
And the sum of these moments must equal to zero.
01:10
So from here we are left with only one unknown, and we get the magnitude of force h to be 25 kiloons, which means that the vector h is 25 kiloons in magnitude and vertically upward in direction.
01:33
So now we can do the same for the free body diagram of the member beh.
01:41
And from here we can take again anti -clockwise moments to be positive and we'll take the sum of moments about point f and we know the sum of these moments must be equal to zero again from equilibrium.
02:06
So the x component of b bx times 0 .5 meters minus 25 kilo -neutons times 0 .2 meters must equal to 0.
02:24
So immediately we can find the magnitude of the x component of b.
02:30
And this is bx, and we get this to be positive 10 kiloons.
02:40
Which means the vector b in the x direction as a magnitude of 10 kiloons...