Pin diameter in frame with peg and slot connection under shear

[papasmurf]

Could someone please check this for me? I attached a rough free body diagram of the whole frame. Any help is appreciated.

Homework Statement

The member ACF of the frame loaded as shown is connected to member BCD by means of a smooth peg and slot C with force P = 930 N at point D. Determine the diameter of pin C in mm if the allowable shear stress is 150 MPa.

Homework Equations

ƩM=0
ƩF_Y=0
ƩF_X=0
τ=F/A
A=(pi/4)*d²

The Attempt at a Solution

Taking ƩM_E=0, I have F_Y(4.8 m) - 930 N(3.6 m), F_Y=697.5 N.
Next, looking at member ACF, take ƩM_A=0, where I have 697.5 N*(4.8 m) - C*cos(θ)*(2.4 m) - C*sin(θ)*(2.7 m), where θ=tan^-1(5.4/4.8)=48.36°. Solving for C, I get C=926.79 N.
Shear is given by τ=F/A where F is the force acting on the pin C and A is the cross-sectional area of pin C. Solving for diameter I get 3.62 mm.

 

[PhanthomJay]

Could someone please check this for me? I attached a rough free body diagram of the whole frame. Any help is appreciated.

Homework Statement

The member ACF of the frame loaded as shown is connected to member BCD by means of a smooth peg and slot C with force P = 930 N at point D. Determine the diameter of pin C in mm if the allowable shear stress is 150 MPa.

Homework Equations

ƩM=0
ƩF_Y=0
ƩF_X=0
τ=F/A
A=(pi/4)*d²

The Attempt at a Solution

Taking ƩM_E=0, I have F_Y(4.8 m) - 930 N(3.6 m), F_Y=697.5 N.

This is OK.

Next, looking at member ACF, take ƩM_A=0, where I have 697.5 N*(4.8 m) - C*cos(θ)*(2.4 m) - C*sin(θ)*(2.7 m), where θ=tan^-1(5.4/4.8)=48.36°. Solving for C, I get C=926.79 N.

You are assuming that there is no x component of the force at F. Is there? Try looking at member DCB instead, to solve for C directly.

 

[nvn]

papasmurf: We do not yet know the answer with certainty, because you did not show boundary conditions (constraints) on your free-body diagram (FBD). Also, you did not state the static coefficient of friction (COF), mu. Therefore, the given problem is not fully defined yet.

If there is a horizontal roller at point F, and if mu = 0, then everything you did is correct, except for your pin diameter.

 

[papasmurf]

Sorry for no clarifying, both "legs" if you will are connected by pins so there are vertical and horizontal force components. Taking that into account I was able to figure out the pin length to be about 4.2 mm. Thanks guys.

 

[nvn]

papasmurf: That is doubtful. Because you did not yet draw clear constraint symbols on your FBD, your given question is not yet defined. We probably cannot help you until you draw clear constraint symbols on your FBD. If you currently do not realize there is a difference between, e.g., a pin and a roller, then you currently might be missing a fundamental concept. Also, "legs, if you will" is unclear. When you talk about point locations, specify the point label, such as point A, point E, etc.

mu = 0 is conceivable (although we cannot be certain if you do not state the given problem). However, with no clearly-shown constraints in the FBD, this question is anyone's guess. Also, your current pin diameter looks wrong, as far as I can tell. Keep trying.

 

[PhanthomJay]

You have correctly noted (perhaps by chance?) that because of the slot, no axial force can be transferred to ACF, and thus the force on the pin at C must be perpendicular to ACF, and hence Cy = Ccostheta. If you draw a FBD of DCB, you can solve for Cy, and then C, and then the area of the bolt, and then it's diameter. I also do not get your answer.