largest diameter - beam

[logistic_guy]

The rigid beam is supported by the three posts \(\displaystyle A, B,\) and \(\displaystyle C\). Posts \(\displaystyle A\) and \(\displaystyle C\) have a diameter of \(\displaystyle 60 \ \text{mm}\) and are made of \(\displaystyle \text{aluminum}\), for which \(\displaystyle E_{\text{al}} = 70 \ \text{GPa}\) and \(\displaystyle (\sigma_Y)_{\text{al}} = 20 \ \text{MPa}\). Post \(\displaystyle B\) is made of \(\displaystyle \text{brass}\), for which \(\displaystyle E_{\text{br}} = 100 \ \text{GPa}\) and \(\displaystyle (\sigma_Y)_{\text{br}} = 590 \ \text{MPa}\). If \(\displaystyle P = 130 \ \text{kN}\), determine the largest diameter of post \(\displaystyle B\) so that all the posts yield at the same time.

 

[logistic_guy]

The beam and post \(\displaystyle B\) are exerting on each other the same vertical force at \(\displaystyle x = 4 \ \text{m}\). Let us call this force \(\displaystyle F_{B}\).

This force creates stress in post \(\displaystyle B\) and from previous exercise we know the formula of stress. That is:

\(\displaystyle \sigma = \frac{F}{A}\)

For our pupose, this formula becomes:

\(\displaystyle (\sigma_{Y})_{\text{br}} = \frac{F_{B}}{A_{B}} = \frac{F_{B}}{\pi r^2_{B}} = \frac{4F_{B}}{\pi d^2_{B}}\)

where \(\displaystyle d_{B}\) is the maximum diameter just before any yielding starts.

We cannot find \(\displaystyle d_{B}\) at this moment because we have two unknowns: \(\displaystyle d_{B}\) and \(\displaystyle F_{B}\).

 

[khansaheb]

The rigid beam is supported by the three posts \(\displaystyle A, B,\) and \(\displaystyle C\). Posts \(\displaystyle A\) and \(\displaystyle C\) have a diameter of \(\displaystyle 60 \ \text{mm}\) and are made of \(\displaystyle \text{aluminum}\), for which \(\displaystyle E_{\text{al}} = 70 \ \text{GPa}\) and \(\displaystyle (\sigma_Y)_{\text{al}} = 20 \ \text{MPa}\). Post \(\displaystyle B\) is made of \(\displaystyle \text{brass}\), for which \(\displaystyle E_{\text{br}} = 100 \ \text{GPa}\) and \(\displaystyle (\sigma_Y)_{\text{br}} = 590 \ \text{MPa}\). If \(\displaystyle P = 130 \ \text{kN}\), determine the largest diameter of post \(\displaystyle B\) so that all the posts yield at the same time.

View attachment 39595

The student will have to
(1) Assume no buckling →
(2) Assume the beam to start and remain horizontal →

Equal displacements in all the struts​

​

CONTINUE......

 

[logistic_guy]

The student will have to
(1) Assume no buckling →
(2) Assume the beam to start and remain horizontal →

Equal displacements in all the struts​

​

CONTINUE......

Now you are pretending to be an expert in beams!

 

[khansaheb]

Now you are pretending to be an expert in beams!

I do not need to pretend - I am a student of STRUCTURAL MECHANICS.

 

[logistic_guy]

I do not need to pretend - I am a student of STRUCTURAL MECHANICS.

 

[logistic_guy]

\(\displaystyle (\sigma_{Y})_{\text{br}} = \frac{F_{B}}{A_{B}} = \frac{F_{B}}{\pi r^2_{B}} = \frac{4F_{B}}{\pi d^2_{B}}\)

We can calculate the vertical forces that act on the beam.

\(\displaystyle F_A + F_B + F_C - P - P = 0\)

We know that \(\displaystyle F_A = F_C\), then

\(\displaystyle 2F_A + F_B - 2P = 0\)

Or

\(\displaystyle F_B = 2P - 2F_A\)

This gives:

\(\displaystyle (\sigma_{Y})_{\text{br}} = \frac{4F_{B}}{\pi d^2_{B}} = \frac{8(P - F_A)}{\pi d^2_{B}}\)

We still have two unknowns: \(\displaystyle d_B\) and \(\displaystyle F_A\)!

 

[logistic_guy]

We will write the stress in the aluminum post.

\(\displaystyle (\sigma_Y)_{\text{al}} = \frac{4F_A}{\pi d^2_A}\)

This gives:

\(\displaystyle F_A = \frac{\pi d^2_A(\sigma_Y)_{\text{al}}}{4}\)

Then,

\(\displaystyle (\sigma_{Y})_{\text{br}} = \frac{8(P - F_A)}{\pi d^2_{B}} = \frac{8\left(P - \frac{\pi d^2_A(\sigma_Y)_{\text{al}}}{4}\right)}{\pi d^2_{B}} = \frac{2\left(4P - \pi d^2_A(\sigma_Y)_{\text{al}}\right)}{\pi d^2_{B}}\)


We are readyto find the largest diameter!

 

[logistic_guy]

\(\displaystyle (\sigma_{Y})_{\text{br}} = \frac{8(P - F_A)}{\pi d^2_{B}} = \frac{8\left(P - \frac{\pi d^2_A(\sigma_Y)_{\text{al}}}{4}\right)}{\pi d^2_{B}} = \frac{2\left(4P - \pi d^2_A(\sigma_Y)_{\text{al}}\right)}{\pi d^2_{B}}\)

\(\displaystyle (\sigma_{Y})_{\text{br}} = \frac{2\left(4P - \pi d^2_A(\sigma_Y)_{\text{al}}\right)}{\pi d^2_{B}}\)

Let us plug in numbers.

\(\displaystyle 590000000 = \frac{2\left[4(130000) - \pi (0.06)^2(20000000)\right]}{\pi d^2_{B}}\)

This gives:

determine the largest diameter of post \(\displaystyle B\) so that all the posts yield at the same time.

\(\displaystyle d_B = 0.0178 \ \text{m} = \textcolor{blue}{17.8 \ \text{mm}}\)