Statically indeterminate beam problem

[yecko]

Homework Statement

Homework Equations

Beam problem
Deflection
Statically indeterminate

The Attempt at a Solution

Am I correct? Or should the assumed deflection at B be the difference in deflection between B and C? Thanks

 

[Chestermiller]

It looks like you are solving this as the superposition of three separate loadings, but you don't provide enough information (text) about the separate solutions to the three loading problems. Please provide more detailed discussion, rather than just a reference to some table.

 

[yecko]

Sorry that i forgot to embed the reference here. The two photos are the reference i used.

 

[yecko]

Here is the numerical calculation derived from post #1, that i thought you may not need it.
You can just take it as reference.

 

[yecko]

Please provide more detailed discussion,

Indeed, i have no idea with how to derive the formula from the cases in the reference. However, do you mind to tell me if my method of utilising the formulae correct?or should the assumed deflection at B be the difference in deflection between B and C? Thanks

 

[Chestermiller]

What you need to do is express the displacements at B and C in the following form:

$$\delta_B=C_1W+C_2R_B+C_3R_C$$ and $$\delta_C=C_4W+C_5R_B+C_6R_C$$where the subscripted C's are obtained from the solutions to the separate problems in terms of the property- and geometric parameters. You then set $\delta_B$ and $\delta_C$ equal to zero and solve the results pair of linear algebraic equations for the two unknowns, $R_B$ and $R_C$.

 

[yecko]

Am i correct to calculate like this? Thanks

 

[Chestermiller]

View attachment 224589
View attachment 224590
Am i correct to calculate like this? Thanks

I haven't checked it over in detail (because it's hard to read a handwritten submission), but it looks like you now had the right idea.

 

[yecko]

sorry for unable to type all my formulae out.
I have got "δ=δW+δRB+δRC+δM" for B and C, for which each are using the formula from reference list.
thank you very much

 

[Chestermiller]

I get the following: $$\frac{R_B}{24}+5\frac{R_C}{48}=\frac{WL}{128}$$ and
$$5\frac{R_B}{48}+\frac{R_C}{3}=7\frac{WL}{(16)(24)}$$
This simplifies to:
$$R_B+\frac{5}{2}R_C=\frac{3}{16}WL$$and$$\frac{5}{2}R_B+8R_C=\frac{7}{16}WL$$

So, $$R_B=\frac{13}{56}WL$$
and $$R_C=-\frac{WL}{56}$$

 

[yecko]

Well, the sign of my equations are different from yours, the answer is different...