SOLVE THE TRUSS GIVEN BELLOW WITH THE HELP OF METHODS OF JOINT
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a) REPLACE JOINTS WITH REACTIONS at A and at B
b) Draw FBD of the TRUSS
Applying the conditions of Equilibrium of Coplanar Non-concurrent Force System,
∑ FX = 0; Rb – Rah = 0 ------ (i)
(-) ← ● → (+)
∑ FY = 0; Rav – 10 – 5 – 15 = 0 => Rav = 30 kN ----- (ii)
∑ MA = 0; 10 x 4 + 5 x 4 + 15 x 2 – Rb x 3 = 0 ----- (iii)
Rb = 30 kN
Hence Rah = Rb = 30 kN
Calculation of Angle θ
The angle θ = tan-1(3/2) = 56.3°
All the unknown forces will be taken as Tensile, if their magnitudes is found negative, then they will be treated as compressive forces.
First we shall choose a joint having only two unknown forces, either we shall choose joint D or joint A
Let us choose joint D first.
We shall consider point D first, as it has only two unknown force. FBD of the point D is drawn.
∑ FX = 0; F2 = 0
∑ FY = 0; F1 – 5 = 0
F1 = 5 kN
Our next joint will be point E. FBD of the joint E is drawn. As F1 = 5 kN, hence unknown forces are two. F3 and F4
∑ FX = 0; – F3 – F4 cos 56.3° = 0
∑ FY = 0; – F1 – 10 – F4 sin 56.3° = 0 [ as F1 = 5 kN]
F4 = –15/ sin 56.3° = – 18.02 kN
F3 = – F4 cos 56.3° = 10 kN
Our next joint is C
F5 and F9 are unknown where as F4 = – 18.02 kN
F2 = 0
∑ FX = 0; – F9 + F4 cos 56.3° = 0
F9 = F4 cos 56.3° = 10 kN
∑ FY = 0; F5 + F4 sin 56.3° – 15 = 0
F5 = – F4 sin 56.3° + 15 = 30 kN
F3 = 10 kN; F5 = 30 kN
∑ FX = 0; F3 = F6 + F7 cos 56.3°
∑ FY = 0; – F5 – F7 sin 56.3° =0
F7 = – F5/ sin 56.3° = – 36.05 kN
F6 = F3 – F7 cos 56.3° = 10 + 20 = 30 kN
∑ FX = 0; F6 = Rah = 30 kN
∑ FY = 0; F8 = Rav = 30 kN
Sl no | Link | Force | Magnitude | Nature |
01 | ED | F1 | 5 kN | T |
02 | CD | F2 | 0 | |
03 | FE | F3 | 10 kN | T |
04 | CE | F4 | 18.02 kN | C |
05 | FC | F5 | 30 kN | T |
06 | AF | F6 | 30 kN | T |
07 | BF | F7 | 36.05 kN | C |
08 | AB | F8 | 30 kN | T |
09 | BC | F9 | 10 kN | T |
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