Shear Diagram - Beam with 3 supports

[teeth]

Homework Statement

Draw the shear diagram for the compound supported beam.

Homework Equations

Statics... so sum of moments at any point or forces in any direction are 0

The Attempt at a Solution

I summed the forces in the y direction, as well as the moments at A and moments at B, but using these 3 equations causes the variables to cancel out when solving for the reactions. I get 0 = 0...
I guess I cannot use two moment equations, but then how am I to solve this? I'm in first year statics.

My equations are...
$$0 = \sum F_y = A_y + B_y + D_y -3*6 - 5$$
$$0 = \sum M_A = B_y*3 + D_y*9 -3*6*3 - 5*7.5$$
$$0 = \sum M_B = -A_y*3 + D_y*6 - 5*4.5$$
Counter-clockwise moment is positive

 

[SteamKing]

You have a pinned joint at point C. Try splitting the single beam into two parts: A-B-C and C-D. Write separate equations of static equilibrium for each piece of the beam. What's the bending moment at point C?

 

[teeth]

Sorry, is the bending moment at C zero?
I tried getting isolating for $$C_y$$ but I keep getting $$C_y = 0$$
Is the upward force at pin C the same as the "shear force" at C?

 

[SteamKing]

Sorry, is the bending moment at C zero?

Yes. The pin cannot develop a moment.

I tried getting isolating for $$C_y$$ but I keep getting $$C_y = 0$$
Is the upward force at pin C the same as the "shear force" at C?

If you write equilibrium equations for beam C-D, you'll see that Cy cannot be zero.

 

[Aman Ratan]

Homework Statement

Draw the shear diagram for the compound supported beam.

Homework Equations

Statics... so sum of moments at any point or forces in any direction are 0

The Attempt at a Solution

I summed the forces in the y direction, as well as the moments at A and moments at B, but using these 3 equations causes the variables to cancel out when solving for the reactions. I get 0 = 0...
I guess I cannot use two moment equations, but then how am I to solve this? I'm in first year statics.

My equations are...
$$0 = \sum F_y = A_y + B_y + D_y -3*6 - 5$$
$$0 = \sum M_A = B_y*3 + D_y*9 -3*6*3 - 5*7.5$$
$$0 = \sum M_B = -A_y*3 + D_y*6 - 5*4.5$$
Counter-clockwise moment is positive

Hi Teeth, I had a query.

By taking the moments about Point A and B, You have formulated the equations 2 and 3.
Is this way of generating equations right ?

Looking forward to a reply from you soon.

Aman Ratan
Thanks

 

[SteamKing]

Hi Teeth, I had a query.

By taking the moments about Point A and B, You have formulated the equations 2 and 3.
Is this way of generating equations right ?

Looking forward to a reply from you soon.

Aman Ratan
Thanks

Teeth hasn't been around PF since posting this question a year and a half ago, so I hope you are not looking for a reply any time soon.

PF generally frowns on resurrecting old posts like this.

It is OK to link to an old post if you have a particular question about it, but you should start your own thread in that case.

 

[Aman Ratan]

Teeth hasn't been around PF since posting this question a year and a half ago, so I hope you are not looking for a reply any time soon.

PF generally frowns on resurrecting old posts like this.

It is OK to link to an old post if you have a particular question about it, but you should start your own thread in that case.

Thanks SteamKing