Question

Mechanics Help Please.
A tube B, length b, is fixed at both ends. Tube B is bisected by a beam A. A force F is applied at the end of Beam A. The distance between the force and the center of the tube is a. Calculate the deflection due to bending of the tube.

The Elastic Modulus is equal to E
The Second Moment of are of the tube is equal to I.

Answer 1

I understand that I need to use \[Ymax = \frac{ 1 }{ 192 }\frac{ W L^3}{EI }\] but I am unsure if the value for W is affected by the distance a.

Thank you

Answer 2

Oh boy, I want to apologise that I haven't studied this. But still since you asked for qualified help I'll try tagging @dan815 @ganeshie8

Answer 3

Thank you @ParthKohli

Answer 4

@pooja195 @nincompoop @mayankdevnani @imqwerty

Answer 5

@Michele_Laino

Answer 6

here we have a compound stress, namely we have an inflection plus a twist. So we have to compute the equivalent twisting moment or the equivalent bending moment, by means of the formula of \(Poncelet\):
\[\Large {M_{ET}} = \frac{3}{8}{M_B} + \frac{5}{8}\sqrt {{{\left( {{M_B}} \right)}^2} + {{\left( {{M_T}} \right)}^2}} \]
where:
\[\Large \begin{gathered}
{M_T} = F \cdot {d_{AB}} \hfill \\
\hfill \\
{M_B} = P \cdot \frac{L}{8} \hfill \\
\end{gathered} \]
are the twisting moment, and Bending moment, and \(P\) is the weight of the driven shaft

Answer 7

@Michele_Laino
If I were to calculate the bending moment alone, (I have been instructed to ignore self weight of the parts) should I just use P=F and then use Mb as the Force for the fixed end beam formula please?

Answer 8

if we can neglect the weight \(P\) of the driven shaft, then we have, from the \(Poncelet's\) formula:
\[\Large \begin{gathered}
{M_{ET}} = \frac{3}{8}{M_B} + \frac{5}{8}\sqrt {{{\left( {{M_B}} \right)}^2} + {{\left( {{M_T}} \right)}^2}} = \hfill \\
\hfill \\
= \frac{5}{8}{M_T} = \frac{5}{8} \cdot F \cdot {d_{AB}} \hfill \\
\end{gathered} \]:

Answer 9

Ok thank you. How do I use the Equivalent twisting moment to calculate the actual deflection of the tube please?

Answer 10

I'm thinking...

Answer 11

I think that we can use such moment, at numerator, namely:
we have to do this substitution:
\[\Large P{L^3} \to {M_{ET}}{L^2}\]

Answer 12

even if I'm not so sure

Answer 13

oops.. I have made a typo:
\[\Large F{L^3} \to {M_{ET}}{L^2}\]

Answer 14

Ah ok how do I get that into the formula that I was given please?

Answer 15

this one \[Ymax = \frac{ 1 }{ 192 }\frac{ W L^3 }{ E I }\]

Answer 16

I think, that we can write this:
\[\Large Ymax = \frac{1}{{192}}\frac{{{M_{ET}}{L^2}}}{{EI}}\]

Answer 17

Thank you that makes sense!

Answer 18

:)