Question

method of joints

Truss shown below, key parameters are given below: p=2,q=3 R= 5, T = 7. (a) Draw the FBD of the entire structure. (10 pts) (b) Determine the forces in all members (state whether the member is in Tension or Compression and include FBDs for the joints) (20 pts) (Hint: guess vertical component of force at C? Why?) qm pm R KN UN

Answer 1

(a) FBD 3m AJO, 4 - Hр 2 m 3m 5 KN KN Let the be Ve vertical V reactions at CD respectively. Let the Horizontal com reactions at CD be He, Ho respectively, In triangle AEB In triangle BED BE tan ₂ = 3 tano, = AE = 1 o - tan " ( 2 ) = 56.310 O₂ - tani(1) = 45°

Step 1 - Calculation of Reactions at Supports . & Fx = 0 He=Ho - ② {fy = 0 Ve + y = 5+7 VC + VD = 12 - Now, consider [Joint C . Efx=0 FBc = He Bc - ③ FBD of joint c Ve=0 3. Vertical reaction at c = 0. Substituting Ve=0 & in equation , we get Vo = 12 KN

EM = 0 Sum of moments about A is zero. FX2 – Vx(5) - Vx (5) + Hex (3)=0 14 - 1285 - 0 + 3Hc=0 He=15.33 KN. Substituting He = 15:33 bens in equation , we get ABC = 15.33 KN (tension) Lo force in member BC. Consider Joint A FAB vo W TAE 5KN FAB sino = 5 + FAB = 1 n 56.31 FBD of joint A sin 56.31 FAB = 6 KN (tension) Lo force in member AB

(SFx=0 FAE = FAB coso, = 6 cs56-31 IFAE = 3·33 kN (compression / Lo force in member AE Consider Joint E AFBE FAE EL DE 7 KN - FRE = 7 kN (tension) FBD of joint E 4 force in BE [SFx=0 FE=FAE FOE - 3.33 KN (compression 2 Le Force in member DE

Consider Joint D 04 FDE (sfy = 0] FBD of joint D FBD sind = 1 FBD = 12 sin 45° FBD = 12 52 KN (Compression) Lo Force in member BD