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In theory of bending of beams, often these terms are used. There are no clear definition present. Most of them contradicting. Which of the property(bending moment and shear force) is zero in which case?

The definition of contraflexure is defined as the point of zero bending moment. Then why have another term, i.e inflection point?

NOTE:The wikipedia link has no references.

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  • $\begingroup$ contraflexure surely? $\endgroup$ – Solar Mike Feb 1 '18 at 13:52
  • $\begingroup$ Instead of wikipedia, why not go to a recognised expert in the field such as Timoshenko : mechzoneblog.files.wordpress.com/2017/08/mos-timoshenko-1.pdf $\endgroup$ – Solar Mike Feb 1 '18 at 13:56
  • $\begingroup$ @Solar Mike, thanks for the book link. But u fortunately Timoshenko doesn't talk about this $\endgroup$ – karthikeyan Feb 1 '18 at 18:26
  • $\begingroup$ Which pages did you check? $\endgroup$ – Solar Mike Feb 1 '18 at 20:38
  • $\begingroup$ I checked the index and also chapter 3. $\endgroup$ – karthikeyan Feb 2 '18 at 2:44
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A point of inflection is defined as the point where a function changes from convex to concave or vice versa.

For a function $f(x)$, this is frequently mathematically defined as the point where $f''(x) = 0$, since $f''(x)$ describes $f(x)$'s curvature[1]. Being more pedantic about it, this is a necessary but not sufficient condition: a function might have no curvature $\left(f''(x) = 0\right)$ but not actually change between convex and concave (this is strictly called an undulation point). That being said, in common parlance any point which satisfies $f''(x) = 0$ is considered an inflection point.

Note that the above has nothing to do with bending moment, it is a characteristic of any mathematical function.

Contraflexure, on the other hand, is as you've said: the point where a beam suffers no bending moment.

For example, here's the bending moment of a cantilever beam with a linear load which generates a point of contraflexure and a point of inflection (notice that to the left of the point, the curve has negative curvature, and to the right of the point, it has positive curvature). The point of contraflexure is easily identifiable on a diagram (it's where the bending moment diagram equals zero). The point of inflection is harder to identify precisely; you can quickly tell that there is a point of inflection and that is it "around here", but you can't easily point to its exact location (without doing the math, I can tell it's somewhere in the circled region).

enter image description here

However, when speaking of beams, it is possible to state that the point of contraflexure is the same as the point of inflection. However, one will actually be talking about two different things at the same time.

As previously stated, the point of contraflexure is that point where bending moment is equal to zero. However, according to the fundamental beam equation, we know that bending moment is the second derivative of deflection (times the stiffness). So, since the point of inflection is that point where the second derivative of a function is equal to zero[2], the point of contraflexure is equal to the deflection's inflection point.


[1] As mentioned by @Mohan's answer, there is a possible exception to this rule: continuous functions $f(x)$ with continuous $f'(x)$ but discontinuous $f''(x)$. If $f''(x)$ is discontinuous at $x=a$ and $\text{sign}\left(\lim\limits_{x\rightarrow a^+}f''(x)\right) \ne \text{sign}\left(\lim\limits_{x\rightarrow a^-}f''(x)\right)$ (in English: the sign of $f''(x)$ to the left of the discontinuity is different than to its right), then $x=a$ is $f(x)$'s inflection point.

This only works in cases where $f'(x)$ is continuous. This is because for a point to serve as an inflection point for $f(x)$, there must be a single valid tangent at that point. So if the derivative is continuous, that condition is met even if $f''(x)$ (the curvature) is discontinuous).

[2] The exception described above in [1] applies to beams. If a beam suffers a concentrated bending moment, that leads to a discontinuous bending (or curvature) diagram, but the deflection's derivative (tangent angle diagram) will still be continuous. So if that concentrated bending moment leads to a change in the bending diagram's sign at that point, then that will be an inflection point in the beam's deflection.

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    $\begingroup$ contraflexure could also mean, the point where the moment changes from positive to negative. This impacts the design process of structural members. $\endgroup$ – Jem Eripol Feb 3 '18 at 0:26
  • $\begingroup$ @Wasabi I think the bending moment need not be zero at point of contraflexure.Its just the nature of bending(hoging/sagging) is different before and after the point $\endgroup$ – user17332 Jan 31 '19 at 13:10
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From what my teacher taught me, both are the same point.It is called contraflexure in the bending moment diagram (where the bending moment changes sign, i.e hoging to saging or the other way) .Here bending moment neednot be zero, it just changes sign(it neednot change gradually, it can also make a sudden jump).It is called the point of inflection in the deflection curve also called elastic curve (the point where the deformation changes from concave to convex or the other way) enter image description here

Refer this question I asked in Math stack exchange https://math.stackexchange.com/questions/3095917/can-point-of-inflection-occur-at-a-point-where-second-derivative-doesnot-exist , for better understanding of point of inflection. For furthur reference use these links https://math.stackexchange.com/questions/1021582/definition-of-point-of-inflection https://math.stackexchange.com/questions/402459/an-inflection-point-where-the-second-derivative-doesnt-exist https://math.stackexchange.com/questions/688503/what-is-inflection-point

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  • $\begingroup$ No definition of inflection is no deformation/displacement. You mixed up the definition of contraflexture with inflection. Please do look up the accepted answer which gives better explanation $\endgroup$ – karthikeyan Jan 31 '19 at 5:41
  • $\begingroup$ @karthikeyan In the context of beams, the function f(x) stated in the accepted answer is the deflection curve itself (vertical deflection of the points on the beam as a function of horizontal position) $\endgroup$ – user17332 Jan 31 '19 at 13:12
  • $\begingroup$ @karthikeyan I think the user edited his answer today(the last passage was included today).Please read it once again . $\endgroup$ – user17332 Jan 31 '19 at 14:01
  • $\begingroup$ @karthikeyan Try and understand the basics of functions and calculus.Then you will be able to understand and relate the function and its derivatives with the proper curves. $\endgroup$ – user17332 Feb 5 '19 at 1:32
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Point of inflection is the point where a curve changes its curvature, that is f''(x) = 0 and f"'(x) not equal to 0. Contraflexure is a point in Bending Moment diagram where Bending Moment changes its sign and since it changes sign, local value is taken as zero (Bending Moment=0 only will not give Contraflexure point, its sign should also be different before and after the point). Now let k be the second derivative of the elastic curve ( curve followed by beam under Bending ) wrt beam length. We know the second derivative gives curvature. We know that EIk = M (E - Elastic Modulus, I = Area moment of cross-section of Beam, M - Bending Moment Applied), Now as we know that Bending Moment is equal to the second derivative of the elastic curve. So, at the point of contraflexure (M=0), the second derivative of elastic curve = 0. Also when bending moment changes sign at contraflexure curvature of the elastic curve also changes. That is, the point of contraflexure is for Bending Moment Diagram while Point of inflection is the corresponding point in Elastic curve of the beam and here beam changes curvature.

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    $\begingroup$ Please try to format your text somehow better. Try to use also latex formatting (so: $\frac{E}{k}$ will be: $\frac{E}{k}$). $\endgroup$ – peterh Jan 30 '19 at 14:40

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