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I know that the neutral axis occurs where the sum of stresses (bending + axial) is equal to 0. I have found the stresses and drawn a diagram to represent forces. How do I determine the neutral axis?

Question

What ive done so far

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  • $\begingroup$ Am I the only one who has never seen this use of the term neutral axis? I've always seen the neutral axis defined as the point where stresses are null under bending (not axial). Considering axial, then the axis with no stresses is just... "the axis with no stresses". $\endgroup$ – Wasabi Nov 3 '15 at 2:16
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In order to calculate the neutral axis under this definition (see my comment under the OP), you need to figure out the stress distribution for each of the applied loads and then superpose them.

So, for this we need to calculate the properties: $$\begin{align} A &= 80\cdot80 - 60\cdot60 = 2800 \\ I &= \dfrac{80\cdot80^3 - 60\cdot60^3}{12} = 2333333 \end{align}$$

So, under the axial force, the stress distribution is uniform and equal to (positive for tension) $$\sigma = \dfrac{N}{A} = \dfrac{72000}{2800} = +25.71$$

Now, under the moment, the stress distribution is linear, and the maximum stresses are equal to $$\sigma = \dfrac{My}{I} = \dfrac{(72000\cdot(60+40))\cdot(\pm40)}{2333333} = \pm123.43$$ where the stresses on the left side are negative (compression) and on the right side are positive (tension). The stress profile due to bending can therefore be described by the equation $$\sigma = -123.43 + \dfrac{2\cdot123.43}{80}y$$ where $y=0$ is on the left side.

Now, adding the effect of the axial load's stress, this equation becomes $$\sigma = -123.43 +25.71 + \dfrac{2\cdot123.43}{80}y$$

So, to find the neutral axis, you just need to find the zero of this equation: $$\begin{gather} \sigma = 0 = -97.72 + \dfrac{2\cdot123.43}{80}y \\ y = \dfrac{97.72}{\left(\dfrac{2\cdot123.43}{80}\right)} = 31.7 \end{gather}$$

So there you have it, the neutral line is 31.7 mm from the left side.

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  • $\begingroup$ Thank you very much!! If u dont mind me asking tho, why is it 31.7 from the left side , why do u assume left hand side y=0? $\endgroup$ – CcoderBeginner Nov 4 '15 at 18:26
  • $\begingroup$ I didn't assume $y=0$ was the left-hand side. I just defined it as such in order to create the stress equation. I could just as easily have defined $y=0$ on the right-hand side, in which case the equation would have changed and the result would come out as $y=48.3$. Also, I could define $y=0$ in the midpoint, changing the equation again and then the result becomes $y=-8.3$ (defining positive values as to the right of the center). Hell, you could theoretically define $y=0$ anywhere, though those three positions are the most obvious. $\endgroup$ – Wasabi Nov 4 '15 at 18:47
  • $\begingroup$ im just confused on how u derived the expression (2*123.43*y/80) $\endgroup$ – CcoderBeginner Nov 4 '15 at 22:23
  • $\begingroup$ 25.71 is the axial stress,-123.43 is the bending, now whats that 3rd stress and why does it carry these values $\endgroup$ – CcoderBeginner Nov 4 '15 at 22:24
  • $\begingroup$ The stress distribution is linear and can therefore be described by an equation of the form $\sigma(y) = a + by$. Looking at just the bending stresses and defining $y=0$ on the left, then we know that $\sigma(0) = -123.43 = a$. And we know that $\sigma(80) = +123.43$ (on the right side). The slope of a line ($b$) is equal to $\dfrac{\sigma_1-\sigma_0}{y_1-y_0} = \dfrac{123.43+123.43}{80-0} = \dfrac{2\cdot123.43}{80}$. Therefore the bending stress distribution is given by $\sigma = -123.43 + \dfrac{2\cdot123.43}{80}$. With axial, just add the uniform stress. $\endgroup$ – Wasabi Nov 4 '15 at 23:40

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