00:01
Hello in this question we are given that is arranged okay so let me draw the section i will be cutting the section through b e d e and d c so let me draw the left hand portion of the cutted section so that will come out to be like this and something like this okay so the section will be cut from here okay and this force is fcd this force force is fde, fde and this angle is 30 degree and also this angle is 30 degree and this force is fbe.
01:05
A vertical load of g that is equal to 7 kiloton ejecting here and a vertical upward reaction at c is 6 .61.
01:16
This is 6 .61 and we need to calculate all the three forces okay so for that what we will do is we will balance let me first write name of this b c d e okay something here is b it will be c and d e okay so first of all let me balance moment about point e to be zero so if we write moment sigma m about e to be zero then we will get we will eliminate fbe and f de because both of them pass through the point e.
02:02
So f -c -d into the height is if this is 10 meter and if this is 5 meter then height will be 5 over root 3 will be the height okay because e let us assume this is g.
02:30
Eg over g c is equal to 1030 that is 1 over root 3 so gc is 5 so h will come out to be 5 over root 3 this is 5 over root 3 so f cd into 5 over root 3 this will give a moment of clockwise about e and reaction at c will give anticlockwise into 5 so this will come out to be equal to and f cd will be f cd will come out to be let me calculate the value 6 .61 into root okay so the value is coming out to be 11 .45 kiloton and if this is positive then it will be a tensile okay why tensile because i have drawn the pd in such a way that assuming all the courses are tensile forces okay if the force calculated come out to be positive then it will be tensile and if it come out to be negative it will be compressive okay so now what we will do we will balance the moment about point d okay if we balance the moment about point d then we will be able to eliminate fd and f c d so b will be the only remaining unknown so let me redraw the diagram so like this and like this this will be the point okay so this is fbe, okay, this angle is 30 degree, this height is 5 over root 3 and this distance is 5.
04:41
This is 5.
04:43
This is given by 6 .61.
04:49
Okay, so when balancing the moment about point d, sigma, md to be 0, we will take two components of this force.
04:59
One will be fbe post 30 and another will be fbe sign 30 so both of them will give a clockwise moment sorry anti -clockwise moment and also reaction at c will give anti -clockwise movement so all of them will have same signs so fbe sine 30 into perpendicular distance is 5 okay plus f f f b b cose 30 into 5 over root 3 plus 6 .6 1 into 10 equal to 0 so from here when we calculate the value of f be will come out to be let me calculate this 6 .61 into 10 divided by 5, sine 30 into 5 plus cost 30 into 5 divided by root 3.
06:13
So this is coming out to be 13 .22 minus 13 .22 kilo -neton.
06:19
And the minus sign indicates the force in this member will be compressive.
06:27
Okay.
06:28
So now the only remaining member is f -d -e.
06:32
So let me calculate for this as well.
06:35
So for fde, what we will do is we will balance the force in vertical direction, sigma f vertical must be 0.
06:49
So for that, we will break this into two components, this one and this one.
06:56
And also we will break f, de into two components in this one and this one.
07:01
Okay, so f, de, this is angle.
07:04
If 30 30 then 60 and 60 okay this will be 60 degree so f de course 60 degree is equal to if this angle is 30 then this will be sign 30 so f be b .e sign 30 plus 6 .61 so f de will come out to be fbe we have already calculated fbe what is fbe minus 13 .22...