2D Free-Body Diagrams (FBDs)

Thème :Vecteurs 2D (dans le Plan), Diagrammes, Vecteurs

Conceptual Questions: Drawing FBD’s

The image below depicts a beam with external forces F₁ and F₂. The beam is made of members AB and BC with a fixed support at A, an internal hinge at B, and a roller support at C. Use the image to answer the questions below.

Draw the FBD of member ABC. On the FBD of member ABC, how many reaction forces (excluding moments) will there be at supports A and C?

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A
1 at A, 1 at C
B
1 at A, 2 at C
C
2 at A, 1 at C
D
2 at A, 2 at C

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On the FBD of member ABC, how many couple moments will there be at supports A and C?

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A
None at either
B
1 at A, 0 at C
C
0 at A, 1 at C
D
1 at A, 1 at C

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Draw the FBD’s of members AB and BC. When member AB is “freed” from its supports, how many total forces (including F₁) will there be on the FBD?

What external forces will there be at supports A and B?

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A
A_yat A, B_yat B
B
A_x and A_y at A, B_y at B
C
A_x and A_y at A, B_xand B_y at B
D
A_y at A, B_xand B_y at B

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How many moments will there be on the FBD of AB?

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A
none
B
1 at A
C
1 at B
D
1 at A, 1 at B

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When member BC is “freed” from its supports, how many total forces (including F₂) will there be on the FBD?

On the FBD of BC, what external forces will there be at the roller support at C?

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A
none
B
Only C_x
C
Only C_y
D
C_xand C_y

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How many moments will there be on the FBD of BC?

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A
none
B
1 at B
C
1 at C
D
1 at B, 1 at C

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Use the Figure to Analyze the FBD of each Member

In the interactive Figure below, click and drag the red dot in the center of each member to free it from its constraints and see its FBD.

Click and drag the red dot at A to reveal its support reactions on member AB. Why are there two force reactions, in the x and y axes, at the fixed support at A?

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A
It resists translation of the beam along the x and y axes
B
It resists translation of the beam in the x direction, the y-component is due to other external forces
C
It resists translation of the beam in the y direction, the x-component is due to other external forces
D
It allows for translation of the beam along both axes
E
Some other reason not listed here

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Why is there a couple moment at the fixed support at A?

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A
The fixed support allows for rotation of the beam
B
The fixed support does not allow for rotation of the beam (it prevents rotation)
C
The moment is there to balance moments due to forces at B and C
D
Some other reason not listed here

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Click and drag the red dot at C to reveal its support reactions on the beam. Why is there only one vertical force, C_y at the roller support at C?

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A
The roller resists translation of the beam along the x axis
B
The roller resists translation of the beam along the y axis
C
The roller allows translation of the beam along the y axis
D
Some other reason not listed here

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Isolate member BC to see its FBD. As the angle of F₂increases from 0^o to 90^o, what happens to the magnitude of B_x?

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A
It increases
B
It decreases
C
It stays the same

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If its magnitude stays the same, for what angle of F₂ does the magnitudes of B_y and C_yreach a maximum?

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A
0^o
B
45^o
C
90^o
D
180^o

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If F₂ = 400 N at an angle of 90^o, then in terms of F₂, B_y is equal to?

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A
F₂
B
F₂
C
2F₂
D
F₂

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Isolate member AB to see its FBD. For member AB, moment M_a reaches a maximum when force F₁is at which angle?

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A
0^o
B
45^o
C
90^o
D
180^o

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Using FBD’s to write Equations of Equilibrium

Writing equilibrium equations requires defining coordinate axes. We will use the following coordinate system: x is positive to the right

y is positive toward the top of the page

positive moments result in a counterclockwise rotation

Using the above coordinates and with F₁and F₂ at 60^o, which is the correct equilibrium equation in the y direction F_y for member BC?

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A
B_y + C_y - F₂sin(60) = 0
B
B_y + C_y + F₂sin(60) = 0
C
B_y - C_y - F₂sin(60) = 0
D
B_y + C_y - F₂cos(60) = 0

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Using the above coordinates and with F₁and F₂ at 60^o, which is the correct equilibrium equation in the x direction F_x for member BC?

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A
B_x + F₂cos(60) = 0
B
B_x - F₂sin(60) = 0
C
B_x + F₂sin(60) = 0
D
B_x - F₂cos(60) = 0

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Using the above coordinates and with F₁and F₂ at 75^o, which is the correct equilibrium equation in the y direction F_yfor member AB?

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A
A_y + B_y + F₂sin(75) = 0
B
A_y - B_y - F₂sin(75) = 0
C
A_y + B_y - F₂sin(75) = 0
D
A_y + B_y + F₂cos(75) = 0

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Using the above coordinates and with F₁and F₂ at 75^o, which is the correct equilibrium equation in the x direction F_xfor member AB?

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For the FBD of AB and with F₁at 75^o, which expression for the sum of moments about A, M_a is correct?

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A
-M_a - 2F₁sin(75) - 4B_y= 0
B
-M_a - 2F₁cos(75) - 4B_y= 0
C
M_a + 2F₁sin(75) - 4B_y= 0
D
M_a - 2F₁sin(75) - 4B_y= 0

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2D Free-Body Diagrams (FBDs)

Conceptual Questions: Drawing FBD’s

Use the Figure to Analyze the FBD of each Member

Using FBD’s to write Equations of Equilibrium

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