<P> This method can be used when the truss element forces of only a few members are to be found . This method is used by introducing a single straight line cutting through the member whose force wants to be calculated . However this method has a limit in that the cutting line can pass through a maximum of only 3 members of the truss structure . This restriction is because this method uses the force balances in the x and y direction and the moment balance, which gives us a maximum of 3 equations to find a maximum of 3 unknown truss element forces through which this cut is made . Let us try to find the forces FAB, FBD and FCD in the above example </P> <Dl> <Dd> </Dd> <Dd> ∑ M D = 0 = − 5 ∗ 1 + 3 ∗ F A B ⇒ F A B = 5 3 (\ displaystyle \ sum M_ (D) = 0 = - 5 * 1 + (\ sqrt (3)) * F_ (AB) \ Rightarrow F_ (AB) = (\ frac (5) (\ sqrt (3)))) </Dd> <Dd> ∑ F y = 0 = R A y − F B D sin ⁡ (60) − 10 = 5 − F B D 3 2 − 10 ⇒ F B D = − 10 3 (\ displaystyle \ sum F_ (y) = 0 = R_ (Ay) - F_ (BD) \ sin (60) - 10 = 5 - F_ (BD) (\ frac (\ sqrt (3)) (2)) - 10 \ Rightarrow F_ (BD) = - (\ frac (10) (\ sqrt (3)))) </Dd> <Dd> ∑ F x = 0 = F A B + F B D cos ⁡ (60) + F C D = 5 3 − 10 3 1 2 + F C D ⇒ F C D = 0 (\ displaystyle \ sum F_ (x) = 0 = F_ (AB) + F_ (BD) \ cos (60) + F_ (CD) = (\ frac (5) (\ sqrt (3))) - (\ frac (10) (\ sqrt (3))) (\ frac (1) (2)) + F_ (CD) \ Rightarrow F_ (CD) = 0) </Dd> </Dl> <Dd> ∑ M D = 0 = − 5 ∗ 1 + 3 ∗ F A B ⇒ F A B = 5 3 (\ displaystyle \ sum M_ (D) = 0 = - 5 * 1 + (\ sqrt (3)) * F_ (AB) \ Rightarrow F_ (AB) = (\ frac (5) (\ sqrt (3)))) </Dd> <Dd> ∑ F y = 0 = R A y − F B D sin ⁡ (60) − 10 = 5 − F B D 3 2 − 10 ⇒ F B D = − 10 3 (\ displaystyle \ sum F_ (y) = 0 = R_ (Ay) - F_ (BD) \ sin (60) - 10 = 5 - F_ (BD) (\ frac (\ sqrt (3)) (2)) - 10 \ Rightarrow F_ (BD) = - (\ frac (10) (\ sqrt (3)))) </Dd>

Problem 3. what are the two types of solid solution what is the difference between them