Question

The structure is in static equilibrium. It is constructed using two-force members, joined at their ends. 1) Label every membe
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Answer #1

Solution:

First of all, let's redraw the diagram to number the joints.

30N Lyo F3 F8 х 600 60° F5 F1 F11 F9 F7 F10 F6 © 60 60° 60° 20N 4 F2 non

Now, Assuming compressive forces as negative and tensile forces as positive .

For Joint 1,

Joint 1 F1 6. 0 F2 10N

\sum F_{y} =0

F1Sin 60 +10 = 0

F1= 11.547 (c) # Sin 60= 1.547

\sum F_{x} =0

F2 + F1cos 60 = 0

F2= 5.7735 (T)

-------------------------------------------

For Joint 2:

Joint 2 F3. 60° F1 F4

\sum F_{y} =0

F4 + F1Sin 60 = 0

F4= 10 N (T)

\sum F_{x} =0

F3 = F1cos 60

F3= 5.7735 (C)

--------------------------------

For Joint 3:

Joint 3 F5 F4 60 F6 F2

\sum F_{y} =0

F4 + F5Sin 60 = 0

F5= 11.547  (C)

\sum F_{x} =0

F6 = F2- F5cos 60

F6= 11.547 (T)

---------------------------------

For joint 4:

\sum F_{y} =0

F7=0 (z) (zero force member)

\sum F_{x} =0

F10= F6

F10= 11.547 (T)

--------------------------------------------------------

For Joint 6:

\sum F_{y} =0  

F11 = 0, Zero force member

\sum F_{x} =0

F8 = 0 , Zero force member

----------------------------------------

For Joint 7:

\sum F_{y} =0

Fg Sin 60 +F11 +20 =0

Fg = 23.094 (c)

-----------------------------------

Now, there we can illustrate the representation of Forces:

Here, tensile forces are away from the joints

Compressive forces are towards the joint.

30N F3 F8 Z 600 60° F5 F1 F11 Z T F7 F10 F6 60 60 60° T T T 20N 2 hion

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