Finding Equilibrium Forces in a Two-Member Frame

In summary, the frame supports a 200-lb cylinder and 500 lb–ft couple moment. The forces acting on the pins in the frame determine the equations for the pins. The roller does not contact the pins.
  • #1
rico22
51
0

Homework Statement



The two-member frame (see picture attached) supports the 200-lb cylinder and 500
lb–ft couple moment. Determine the force of the roller at B on member
AC and the horizontal and vertical components of force which the pin at
C exerts on member CB and the pin at A exerts on member AC. The
roller C does not contact member CB.


Homework Equations





The Attempt at a Solution


I split the frame into members AC and CB. The I found the forces acting on D(pulley) which gave me Dx = 200 lb and Dy = 200 lb. I plugged this into member CB; giving me the next equations:
ƩFx = 0: Cx +Bx -200 = 0
∴ Cx + Bx = 200

ƩFy = 0: Cy - 200 = 0
∴ Cy = 200

ƩMc = 0: 200(4) - 4Bx = 0
∴ Bx = 200 which also give me Cx = 0 from the summation of Forces in the x direction.
Then I plug the relevant values of Cx, Cy and Bx into the AC member; this gives me the following equations:
ƩFx = 0: Ax - Bx + 200(from cable) = 0
∴ Ax = 0

ƩFy = 0: Ay - 200 = 0
∴ Ay = 200
The issue that I am having is that this last answer is wrong according to the book. Its supposed to be Ay = 175 lbs. I know I didn't use the couple moment at all so, maybe this is the reason? But I really don't see how it could be 175 lbs. Any help would be greatly appreciated.
 

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  • #2
rico22 said:

Homework Statement



The two-member frame (see picture attached) supports the 200-lb cylinder and 500
lb–ft couple moment. Determine the force of the roller at B on member
AC and the horizontal and vertical components of force which the pin at
C exerts on member CB and the pin at A exerts on member AC. The
roller C does not contact member CB.


Homework Equations





The Attempt at a Solution


I split the frame into members AC and CB. The I found the forces acting on D(pulley) which gave me Dx = 200 lb and Dy = 200 lb.
this is ok
I plugged this into member CB; giving me the next equations:
ƩFx = 0: Cx +Bx -200 = 0
∴ Cx + Bx = 200

ƩFy = 0: Cy - 200 = 0
∴ Cy = 200
No you cannot just isolate CB without including the pulley forces and torque.
The issue that I am having is that this last answer is wrong according to the book. Its supposed to be Ay = 175 lbs. I know I didn't use the couple moment at all so, maybe this is the reason?
well it has to factor in somewhere, right?
But I really don't see how it could be 175 lbs.
It is.
Any help would be greatly appreciated.
Before pulling the frame apart, try summing external forces = 0 and summing moments of external forces about A = 0. Only the cylinder weight and couple and reactions at A and C are external to the system.
 
  • #3
Thank you for the reply. So if I take the entire frame would C then become a roller and not a pin?
 
  • #4
ok I got it now... I was supposed to plug in the roller support reaction into the FBD of AC on top of Cy and Cx...that was throwing the whole thing off... thanks again.
 
  • #5
You are missing the normal reaction because of the roller.
Link to the solution.
https://faculty.uca.edu/njaustin/ENGR2311/Quizzes/quiz07.pdf

Good job guyz. Enjoy.
 

1. What is a two-member frame in statics?

A two-member frame in statics is a type of structure composed of two connected members that are subject to external forces. These frames are commonly used in engineering and construction to support loads and maintain stability.

2. How do you determine the reactions at the supports of a two-member frame?

The reactions at the supports of a two-member frame can be determined by applying the equations of static equilibrium. This involves summing the forces and moments acting on the frame and setting them equal to zero. Solving these equations will give you the reactions at the supports.

3. What is the difference between a statically determinate and indeterminate two-member frame?

A statically determinate two-member frame is one in which the reactions at the supports can be determined using static equilibrium equations alone. In contrast, a statically indeterminate two-member frame requires additional equations, such as compatibility equations, to solve for the reactions at the supports.

4. How do you calculate the internal forces and moments in a two-member frame?

The internal forces and moments in a two-member frame can be calculated using the equations of static equilibrium along with equations of equilibrium for each member. These equations will allow you to determine the internal forces and moments at any point along the members.

5. What factors can affect the stability of a two-member frame?

The stability of a two-member frame can be affected by various factors, including the magnitude and direction of external forces, the geometry and material properties of the members, and the support conditions. Changing any of these factors can alter the reactions at the supports and potentially affect the stability of the frame.

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