00:01
In this question we have to find the internal force that developed in the members, cf, b, and ad.
00:11
From figure a, it is a frivity diagram.
00:15
And if we use equilibrium at sigma fy equal to 0, we have that fad plus fbe plus f cfce, equal to 230 10 to the power of 3 we use it as equation 1 and from sigma m a equal to 0 which point a is this and we have that f be e plus 3 fcf is cf is 460 come 10 to the power of 3 and that is the second equation.
01:05
And from figure b, we have that del be is equal to del ad plus del c .f minus del ad divided by thousand, two hundred times four hundred from figure b.
01:26
So if you have that, del be is 2 over 3, of del ad plus one -third of del c .f.
01:39
Now you have that because l, a, and e of these beams are equals so if b .e is 2 over 3, f.
01:56
Ad plus a third of fcf.
02:02
That is our third equation.
02:05
And from solving those three equations, we have that.
02:11
F.
02:12
Ad is 32 .857 newton of the kilo -newton...