Hi everybody in this example I try to find a way to solve it

Hi everybody, in this example, I try to find a way to solve it by the method of matrix.

It seems to me there are in total 9 unknown in the sytem if I want to solve it.

However, I can only write down 7 equations, and not sure if it is correct.

Please help me to complete the system.

How can I write down this problem under the matrix form so that I can solve it numerically.

My equations

EXAMPLE 6.2 Determine the forces acting in all the members of the truss shown in Fig. 6-9a. 3 kN 45° SOLUTION By inspection, there are more than two unknowns at each joint. Consequently, the support reactions on the truss must first be determined Show that they have been correctly calculated on the free-body diagram in Fig. 6-9b. We can now begin the analysis at joint C. Why? 30° Joint C. From the free-body diagram, Fig. 6-9c, 3 kN FCD cos 30° + FCB sin 45° 0 These two equations must be solved simultaneously for each of the two unknowns. Note, however, that a direct solution for one of the unknown forces may be obtained by applying a force summation along an axis that is perpendicular to the direction of the other unknown force. For example, summing forces along the y\' axis, which is perpendicular to the direction of FcD. Fig. 6-9d, yields a direct solution for FcB 3 kN m- 1.5 kN 1.5 kN CB -45 CD 1.5 cos 30° kN-FCB sin 15°-0 FCB = 5.019 kN = 5.02 kN (C) 150 1.5 kN Then CB 15° FcL 5.019 cos l 5°-1.5 sin 30° = 0: FCD4.10 kN (T) Ans. CD Joint D. We can now proceed to analyze joint D. The free-body 30 1.5 kN diagram is shown in Fig, 6-9e. FDA cos 30° + 4.10 cos 300 kN = 0 FDA= 4.10 kN (T) FDB-24.10 sin 300 kN) = 0 FDB 4.10 kN (T) Ans DB 30 30° DA 4.10 kN NOTE: The force in the last member, BA, can be obtained from joint B or joint A. As an exercise, draw the free-body diagram of joint B, sum the forces in the horizontal direction, and show that FBA 0.776 kN (C) Fig. 6-9

Solution

Your approach is fine . At B , you missed F_{y calculation .}

You can get another equation by applying moment of force at C .

It\'s simple .