Finding Normal Compressive & Tensile Stress: a Example

[Precursor]

Homework Statement

[PLAIN]http://img12.imageshack.us/img12/6109/17007711.jpg


The attempt at a solution

For part (a), I can only find a normal compressive stress. It is as follows:

\sigma_{c} = \frac{F}{A}

\sigma_{c} = \frac{100*9.81}{\pi(0.013)^{2}}

\sigma_{c} = 1847704 Pa

\sigma_{c} = 1.848 MPa

Is this correct? How do I find the normal tensile stress?

 

[nvn]

Precursor: No, you did not include the bending stress. You would need to read a few pages in your textbook. We are not allowed to tell you how to solve your homework. You would need to post a valid attempt; and then someone might check your math.

 

[Precursor]

The reason I did not include the bending stress is because I thought it only wanted the normal stress. But if I do include the bending stress, then it is as follows:

\sigma_{b} = - \frac{My}{I}

\sigma_{b} = - \frac{(-5000*0.04)(0.013)}{\frac{\pi}{4}(0.013)^{4}}

\sigma_{b} = 115907104.7 Pa

\sigma_{b} = 115.9 MPa


Now if I add both of them I will get:

\sigma_{total} = 1.848 + 115.9

\sigma_{total} = 117.7 MPa

Now is this correct?

 

[nvn]

Precursor: Nice work. The only thing is, compressive stress is negative. So you need to take into account the correct signs on the stresses.

 

[Precursor]

So is the maximum tensile stress 115.9 MPa and the maximum compressive stress 1.848 MPa?

How would I indicate their location on the given coordinate system? Would they simply be located at the origin of the x-y plane?

 

[nvn]

No, put the correct signs on the stresses, before you add them together. The location would be on the two extreme fibers.

 

[Precursor]

Alright, so the total stress will be 115.9 - 1.848 = 114 MPa in tension.

Will it be in tension at 13 mm to the left of the origin, and in compression 13 mm to the right of the origin?

 

[nvn]

Yes, very good. However, if you mean the stress at x = 13 mm is the same as at x = -13 mm, then you have not taken into account the correct signs on the stresses before adding them at x = 13 mm.

 

[Precursor]

For part (b), if I were to take a square element on the xy plane, would \tau_{xy} = 114 MPa?

How would I determine \sigma_{x} and \sigma_{y}?

 

[Precursor]

So is the stress element something like this?

[PLAIN]http://img191.imageshack.us/img191/1921/48140514.jpg


How would I find \sigma_{x} and \sigma_{y} ?

 

[nvn]

No, 114.1 MPa is a normal tensile stress, sigma_y, at x = -13 mm. The shear stress at x = +/-13 mm is, in reality, zero. But check how your textbook wants you to do it. I do not know if they want you to use actual shear stress, or average shear stress.

 

[Precursor]

Is it possible that I can use the transverse shear stress equation to determine the shear force caused by F_{2} = 5000 N?

 

[nvn]

Maybe (except I think you meant to say "stress"). Give it a try and let's see what you mean.

 

[Precursor]

Ok, so this is what I was thinking:

\tau = \frac{VQ}{It}

\tau = \frac{(5000)(0.5\pi(0.013)^{2}(\frac{4(0.013)}{3\pi}))}{(\frac{\pi(0.013^{4})}{4})(0.013)}

\tau = 25114311.8 Pa

\tau = 25.1 MPa

So 25.1 MPa is the shear force acting on the element at the extreme fibres. From here I can calculate the principal stresses and the max shear stress, by simply using the formulas.

Did I go wrong anywhere?

 

[nvn]

Precursor: That would be the shear stress at the neutral axis, not the extreme fibre (except you divided by the wrong t).

 

[Precursor]

Oh so t should be 26 mm, making the shear stress I calculated 12.6 MPa.

Yes, I just realized that 12.6 MPa acts at the N.A. So this value is not relevant to the problem.

If shear doesn't act at the extreme fibres, then the stress element will only have \sigma_{y}?

 

[nvn]

Correct.

 

[Precursor]

From here I simply use the principal stress and maximum in-plane shear stress formulas to calculate the required values?

Thanks for all the help by the way.

 

[nvn]

Yes, but review posts 8 and 11.

 

[Precursor]

Yes, very good. However, if you mean the stress at x = 13 mm is the same as at x = -13 mm, then you have not taken into account the correct signs on the stresses before adding them at that location.

So at x = -13 mm, the normal stress will be 115.9 - 1.848 = 114.1 MPa (tension).
At x = +13 mm, the normal stress will be -115.9 - 1.848 = -117.7 MPa (compression).

And for part (b), I will use -117.7 MPa as \sigma_{y}. There is infact an average shear stress formula, \tau_{ave} = \frac{V}{A}. And average shear stress is maximum at the extreme fibres and 0 at the neutral axis.

So,

\tau_{ave} = \frac{5000}{\pi(0.013)^{2}}
\tau_{ave} = 9417452.3 Pa
\tau_{ave} = 9.42 MPa

However, it is clear from the image below that transverse shear will not have an affect on the bending stress distribution on the cross section.

[PLAIN]http://www.roymech.co.uk/images23/shear_7.gif
(http://www.roymech.co.uk/Useful_Tables/Beams/Shear_stress.html)


So all in all, my stress element will have \sigma_{y} = -117.7 MPa and \tau_{xy} = 9.42 MPa.

Is this correct?

 

[nvn]

Precursor: Excellent work. Average shear stress is uniform over the entire cross section; it is an approximation. A more accurate shear stress approximation is supposedly V*Q/(I*t). You can decide which one you wish to use. As a footnote, the diagrams in post 20 are for a rectangular cross section, not a circular cross section; but point well taken.

By the way, generally always maintain four (or five) significant digits throughout all your intermediate calculations. Then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits.

 

[Precursor]

I have looked through textbook examples, and it seems that they don't consider average shear stress, only transverse shear stress(VQ/It). And since the transverse shear stress is 0 at the extreme fibres, the only relevant stress on the element is the normal stress -117.7 MPa.

 

[nvn]

Excellent work, Precursor.