Question

As shown, the truss is loaded by the forces P1P1P_1 = 425 NN and P2P2P_2 = 865 NN and has the dimension aaa = 1.10 mm .sing the method of joints, determine FABFABF_AB, FBCFBCF_BC, and FBDFBDF_BD, the magnitude of the force in each of the members connected to joint B. Assume for your calculations that each member is in tension, and include in your response the sign of each force that you obtain by applying this assumption

Answer 1

ΣF_x =0

A_x + F_{AB =}0

F_AB = -355 N , negative sign indicates the compression in AB.

Let's consider free body diagram at C.

Sum of the forces along vertical direction is zero. ΣF_y =0

C_y + F_BC Sin45° = 0

615 + F_BC sin45° = 0

F_BC = -869.74 N ,negative sign indicates the compression in BC.

Let's consider free body diagram at B.

Sum of the forces along horizontal direction is zero.

ΣF_x = 0 = - F_AB- F_BD cos45°- Fbc Cos45°=0

= -(-355)- Fbd cos45° +(-869.74cos45°)=0

F_BD = -367.7 N, negative direction indicates the compression in BD.

Lets consider a free body diagram,

Moment about point A,  ΣM_A =0

C_y*2a - P₂ *a - P₁*a = 0

C_y *2 -875-355 =0

C_y = 615 N

Sum of the forces along horizontal direction is zero.

ΣF_r = 0

A_r- P₁ =0

A_r=355N

Free body diagram at A.

Sum of the forces along the horizontal direction is zero. ΣF_x =0

Ax + Fab = 0

F_AB = -355N

All the best.Thank you.