Question

The structure is in static equilibrium. It is constructed using two-force members, joined at their ends. 1) Label every member as tension T, compression C, or zero-force Z. Draw arrows to indicate the direction of the tension or compression forces on every non- zero member. 2) Use the Method of Joints to calculate the forces F1, F2, and F3. Show all your work. Draw free-body- diagrams and write Newton's 2nd law for every joint. 30N у WTT Lx X '60° 600 הט IL 1609 602 60 NTI 20N 10N

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

ΣF, = 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)

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For Joint 2:

Joint 2 F3. 60° F1 F4

ΣF, = 0

F₄ + F₁Sin 60 = 0

F4= 10 N (T)

$\sum F_{x} =0$

F3 = F₁cos 60

F₃= 5.7735 (C)

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For Joint 3:

Joint 3 F5 F4 60 F6 F2

ΣF, = 0

F₄ + F5Sin 60 = 0

F5= 11.547  (C)

$\sum F_{x} =0$

F₆ = F₂- F5cos 60

F₆= 11.547 (T)

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For joint 4:

Joint 4 F7 F10 F6

ΣF, = 0

F₇=0 (z) (zero force member)

$\sum F_{x} =0$

F10= F₆

F10= 11.547 (T)

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For Joint 6:

Joint 6 F8 F11

  

ΣF, = 0

F₁₁ = 0, Zero force member

$\sum F_{x} =0$

F₈ = 0 , Zero force member

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For Joint 7:

ΣF, = 0

F_g Sin 60 +F₁₁ +20 =0

F_g = 23.094 (c)

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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