Question

Question regarding simple Solid Mechanics problem

Answer 1

`A 4-m long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when the rod is subjected to a 10-kN axial laod. Knowing that E = 200 GPa, determine the required diameter.`

\[\huge \mathbf{My~Work:}\]\[\large \delta = \frac{PL}{AE}\]Where:
\(\delta\) is the deflection of the rod
P is the applied load
L is the length of the rod
A is the cross sectional area of the rod
E is the modulus of elasticity of the Steel rod

Thus,\[0.0003\text{m}=\frac{(10,000\text{N})(4.0\text{m})}{A*(200 \times 10^9\text{N}/\text{m}^2)}~~ \rightarrow~~ A=6.67 \times 10^{-5}\text{m}^2\]By using the equation of the area of a circle, I concluded that \(\mathbf{d=0.0092m}\)

Answer 2

However, I noticed that I can receive the same exact answer simply by using the equation of stress and using the max stress constraint that they gave me:\[\large \sigma_{max}=\frac{F}{A_c}\]\[150 \times 10^6\text{N}/\text{m}^2=\frac{10,000\text{N}}{A}~~ \rightarrow ~~ A=6.67 \times 10^{-5}\text{m}^2\]Which is the same exact area that I got before.

My question is that will this always happen?
Can I always make this shortcut? Or was this just coincidence?
And if I get two different areas, would I just use the larger area?

Answer 3

@Michele_Laino @IrishBoy123 :)

Answer 4

First of all 3mm is not 0.0003, it is 0.003 m, I hope that was just a typo

Secondly, with a constraint on stress and on strain, you would have to calculate for both and then check, which answer will satisfy both of them

In our example, if the question was, the stress should not exceed 300Mpa, then by doing max stress calculation you would end up with half the area, thus lesser radius, but that would not satisfy the strain constraint right?

Answer 5

To supplement mashy's answer, no, it will not always work out like that. Many times you will obtain two separate answers, however in this case it just happened to work out as such.

Answer 6

I think that such problem is overdetermined, namely there are too many initial data. We can accept the stretching and contemporarily to neglect the unitary stress, or vice versa. Please keep in mind that unitary stress \(\sigma\) is a function of the unitary stretching \(\epsilon\), by means of the Hooke law:
\[\Large \sigma = \epsilon E\]
if we are inside the proportionality region

Answer 7

@Mashy Yes, it was just a typo but the resultant area is still correct. I am still a bit confused about what you are describing, particularly with referencing the strain constraint.

@Michele_Laino In similar problems, we have been using the specific equation for deflection rather than Hooke's Law. In fact, we haven't exactly been utilizing strain to solve the problems. I'm afraid that later down the road I would be expected to apply the deflection-equation-method instead.

Answer 8

I am fairly confident that I have the correct path(s) for solving this problem. The only thing I am unsure about is what to do should the two areas that I solved for are different and how to proceed accordingly.

Answer 9

I meant that both of your solutions are correct, since, there is the coincidence that the unitary stress is equal to the accepted maximum limit
@CShrix

Answer 10

@Michele_Laino Oooh, I see. So if I end up with two different areas, then they are both correct? In the real world when doing such calculations, which would be the best area (larger or smaller)?

Answer 11

I think that the best area is the one for which the subsequent condition holds:
\[\Large \frac{F}{A} \leqslant {\sigma _{adm}}\]
where the \(admissible\) unitary stress \(\sigma_{adm}\) is \(150\; MPa\)

Answer 12

Alright, that makes sense! Thank you!