Determine the force in member BC.ya 4.50 kN(C)c 8.00 kN(C)b 6.00 kN(C)d 14.00 kN(C)XВDetermine the force in member CD.а 0b 7.50 kN(T)3 mc 10.0 kN (T)d 3.60 kN(C)Determine the force in member AB.Ab 4.80 kN(C)a 7.50 kN(C)c 10.0 kN(C)d 6.00 kN(C)6kNDetermine the force in member BD.a 6.00 kN(T)4 m3 mb 10.0 kN(T)d 12.0 kN(T)

The free-body diagram of the given truss can be drawn as below, Here RC is the horizontal reaction at support C, RAV is the vertical reaction at support A and RAH is the horizontal reaction at support A.Using the equilibrium concept, the summation of the moment due to all forces about point A can be equated to zero, ∑MA=0RC3 m-6 kN4 m=0RC=8 kN The equation of equilibrium of forces in the vertical direction at point C is given as, RCDcos45°=0RCD=0 Here RCD is the force in member CD. The equation of equilibrium of forces in the horizontal direction at point C can be written as, RBC+RCDsin45°+RC=0 Here RBC is the force in member BC. Substitute 0 for RCD and 8 kN for RC in the above expression. RBC+0sin45°+8 kN=0RBC=-8 kNThe equation of equilibrium of forces in the horizontal direction at point B can be written as, RBC-RABsin53.13°=0 Here RAB is the force in member AB. Substitute -8 kN for RBC in the above expression. -8 kN-RABsin53.13°=0RAB=-10 kN The equation of equilibrium of forces in the vertical direction at point B can be written as, RBD+RABcos53.13°=0 Here RBD is the force in member BD. Substitute -10 kN for RAB in the above expression. RBD+-10 kNcos53.13°=0RBD=6 kNTherefore, (1) The force in member BC is 8 kN and it is compressive in nature, hence option (c) is correct. (2) The force in member CD is zero, hence option (a) is correct. (3) The force in member AB is 10 kN and it is compressive in nature, hence option (c) is correct. (4) The force in member BD is 6 kN and it is tensile in nature, hence option (a) is correct.