Question

B 5m sm/ sm sm/ 5 m C5 5 m Calculate the support reactions for the pin at E and cable at D, then use the method of ioints to calculate the force in each member of the cantilever truss shown in the diagram. Given that F, = 30 kN and F2 = 20 kN. Enter your results using the convention that Tension is positive. Please include the unit ("kN") within all your answers and round to 3 significant figures (e.g. if the calculated answer was 55.562 kilo Newtons please input: 55.6 kN).

Answer 1

Calculations for finding all the internal force in the members and tension in cable at D and Reaction forces at E are done below.

Note:

• Let the tension in the cable at D is T
• Let the horizontal reaction at E is F_{​​​​​​EX}​​​​
• Let the vertical reaction at E is F_{​​​​​​EY}
• _​​​Use Equilibrium equations wherever required.
• Equilibrium equations are
  • $\Sigma$F_X=0
  • $\Sigma$F_Y=0
  • $\Sigma$M_Z=0

GIVEN: 6= 50 KN F = 20 KN 5m 5m 5m Fi SOLUTION: Dnaw Free body diagram of the given conti kwer truss : STEP 1: 60 C FEX 20 AN 30 KN STEP 2: Apply Equilibrium condition, EMat E =0. = 30 (10) + 20 (5) – T(5) =0 - 5T =0 400 Magnitude of tinsion in cable ot D = 80KN T 80 KN STEP 3: dpply Equilibrium condition, Ef, =D0 + T CO 30' - FEX =0 80 Cod 30' - FEx =0 Fex= 69.3 KN Magnitude of horizont al reaction at E = 69.3 KN

STEP 4 : nply Equi libnium condition, EF,yco E +Tsin 30 - 30 - 20 =0 - 50 FEy + 80 sin 30 Fey = 10 KN vertical reaction at E = 10KN Magnitude Free bedy dicaquam at joint A: STEP 5 Apply Equilibrium condition, SF, =0 FAC + Fe sin b0 - 30 =0 30 KN FAR = 34-6 KN Intornal foTce within member AB, FAR= 34.6 KN Arply Equililrium condition, Efx=0 » FAR Co 60 t FAC =0 FAC = - 17.3 KN Internal force within member AC, FAC= -17.3 KN STEP 6: Free bedy diagram at joint B: 4rply Tquilitnium condition, FBD 60 FAB FBC FBC = - FAR FBc = - 34.6 KN Internal force within membar BC, FBc = - 34.6 KN Arply Equlilrium condition , E F =0 FBD + FBr cos 60' FAB CS60 FBD + 34.6) co6o - (34.6).coo60 =0 FBD= 34.6 KN Internal force oithin member BD, FED= 34.6 EN

STEP 7: Free body dioqam at joint c: Feg FOE FAC 20 KN Arply equilibrium condition, EF =0 = Fec sin60 + FeD sin 60 20 E34.6) sin 60 + FeD sin6o - 20 =0 FCD = 57.7 KN Dnternal force within member CD, Fep 57.7 KN rpey Equililnium condition, EFx=0 FAC =0 FBC Ce60' cos 6o + FCE 57.7 cos 60 t For - E34.6) cos60 -G17.3) = 0 FCE =(17.3 + 34.6.c60 +57. 7 cod6o) FE = -63. 45 KN %3D FCE = -63.5 KN Internal force within menmber CE, FE =-63-5 KN

Fre erody diagram at jointE: FEN SVEP 8: =69.3 KN FEX У FOE Fy=10KN Aryply equilibnium condition, EF, =0. FDE sinbo + Ey DE sinbo' + FDE = -11.6KN %3D Internol force within member DE , FDE - -1.6 KN