Example

A truss shown in Figure 2.18 is symmetrically loaded, and it is sufficient to solve half the truss by considering the joints 1 to 5. At joint 1, there are two unknown forces. Summation of the vertical components of all forces at joint 1 gives

which in turn gives the force in the member 1-2, F12 = 190 kN (compressive). Similarly, summation of the horizontal components gives

F13 — F12 cos 45° = 0 Substituting for F12 gives the force in the member 1-3 as

Now, joint 2 is cut completely and it is found that there are two unknown forces, F25 and F23. Summation of the vertical components gives

Therefore,

F23 = 135 kN (tensile) Summation of the horizontal components gives

and hence,

F25 = 135 kN (compressive)

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