Equilibrium  
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The figure shows the free body diagram around joints B and C. The free body diagram cuts through the column at the point below joint B which is marked 2. The forces on either side of joint C have a magnitude of 125 kilo newtons and are directed downwards. The forces on either side of joint B have a magnitude of 175 kilo newtons and are directed downwards. The force F subscript 2 acts vertically downwards along the column at point 2.

Next, draw a free-body diagram (FBD) that cuts through member (2) and includes the free end of the column. Once again, a positive internal axial force will be assumed.

Write the equilibrium equation for the sum of forces in the vertical (x) direction:

∑Fx=-2(125 kN)-2(175 kN)-F2=0

Therefore, the internal force in member (2) is:

F2=-600 kN

Note: The negative sign indicates that the internal force is actually compression.