Access to website nexote.net/nexote/SALN/ is not restricted. You should be able to view all the videos there. Send me a note if you are still having difficulty accessing the site or viewing any of the videos.

Yes, it is correct. The phrase concentrated moment is generally used for applied load/moment, not an internal bending moment in a beam. If you think about it, internal bending moment in the beam is not concentrated, it is distributed throughout the member.

If the elastic curve (the displacement function) has a point of inflection, then yes, that would be a point of maximum/minimum moment. But, consider the case of a simply supported beam, say subjected to a uniformly distributed load. The elastic curve is a concave up function without any point of inflection, yet maximum moment takes place at mid-span where displacement is maximum and slope is zero. So, determining maxima and minima of bending moment using the elastic curve is a bit tricky. Why is the moment at a point of inflection a (local) maximum? If we refer to the beam's displacement function as w, then w' (first derivative of w, with respect to x, is slope of the curve), and w'' is the first derivative of the slope function (also the second derivative of the displacement function). At a point of inflection of w, the derivative of w'' must be zero. So, when w'''(x) = 0, then x is a point of inflection. What is the relationship between bending moment and w? w'' = M/EI where E is modulus of elasticity of the material, and I is moment of inertial of the cross-section of the beam. The above equation states that the derivative of the slope of the elastic curve equals bending moment over some constant. So, we can rewrite the above equation as: M = EI w'' Where is moment maximum? when dM/dx = 0. So, dM/dx = d (EI w'')/dx = 0 or, dM/dx = EI w''' = 0 That is, the derivative of M is zero when w''' = 0. And since w''' is zero at a point of inflection, then moment is maximum at such a point.

Cut the beam at some distance x past the distributed load where 5 < x < 10. The moment of the left support reaction about the cut point is: 375 x The moment of the distributed load about the cut point is: 100(5)(x-2.5). The distributed load can be replaced with its equivalent concentrated load. The magnitude of that load is 100(5) = 500. The location of that load is (x - 2.5). This is the midpoint of the rectangular load measured from the left end of the cut point. So. the total moment about the cut point is: 375 x - 100(5)(x-2.5) = 1250 - 125 x.

There is going to be a jump/drop in the moment at the beam’s midpoint where the concentrated moment is located. Suppose the the beam’s length is 10m, the magnitude of the downward concentrated load is 2 kN, and the concentrated bending moment is acting in the clockwise direction at the midpoint of the beam. The moment has a magnitude of 5 kN-m. The reaction bending moment at the fixed end is: (10m)(2 kN) + 5 kN-m = 25 kN-m. This would be a counterclockwise moment. And the reaction at the support is 2 kN. Bending moment diagram starts at -25 kN-m. It reaches -25 + (2 kN)(5 m) = -15 kN-m just to the left of the concentrated moment at the midpoint of the beam. The clockwise moment at the midpoint of the beam causes a jump of 5 kN-m (the magnitude of the applied moment) in the moment value. So, moment goes from -15 just to the left of the midpoint to -10 just to the right of the midpoint. Then the moment changes linearly until it reaches zero at the free end of the beam.

@@DrStructure I resolved your example and it's correct you are right. Thank you for your time 🙏🙏🙏🙏 My real issue is in an example, where the bending moment in the midpoint of the segment act conter clockwise. If I consider a conter clockwise direction as a positif direction the result will totally change ! So, the sign convention of the positif/negatif bending moment is it always true ? I mean we can't consider a moment where the beam's segment tend to deform concave down as a positif moment right ?

@@seddiksoundous9883 The same example as above, but with a change in the concentrated moment direction. Now the 5 kN-m moment is acting the in the counterclockwise direction. The reactions at the fixed support are: Bending moment: (2 kN)(10 m) - 5 kN.m = 15 kN.m (counterclockwise) Vertical reaction: 2 kN (upward). The bending moment diagram starts at -15 and decreases to -5 just to the left of the beam's midpoint. This time, the applied bending moment causes a drop (not a jump) in the moment value. The moment goes from -5 at the left end of the midpoint to -10 at the right end of the midpoint. The moment changes linearly from -10 at the right end of the midpoint to 0 at the free end of the beam. Both in this case and in the pervious case, the beam bends concave down. And that is what we want to keep in mind. That is, of significance is to know which fiber (top/bottom) of the beam is being compressed and which fiber is being stretched. This distinction becomes important when we are designing the beam as the design of the parts under compression differ from those under tension. In these lectures, we have referred to the bending moment that causes the beam to bend concave up as positive. So, by that convention, bending moment in the cantilever beam is negative. But this sign convention is arbitrary. We could have called this a positive moment. That is, we could have taken a concave down deflection as positive when writing numbers. As long as we keep in mind, under our assumed sign convention, which beam fiber (top/bottom) is being compressed and which is being stretched, everything will work out fine at the design stage. In our traditional sign convention, a positive moment means the top fiber of the beam is being compressed and the bottom fiber is being stretched. In the alternative sign convention, a positive moment means the top fiber is being stretched and the bottom fiber is being compressed. For design purposes, we really don't care above positive vs negative labels; we only care about which side of the beam is being compressed and which side is being stretched.

Thank you @@DrStructure 🙏🙏🙏 it's clear now

Regardless of the nature of the boundary conditions (supports), deflection of a beam segment is either concave up or concave down. If the segment carries a positive moment, the deformed shape is concave up. If the segment carries a negative moment, the deformation assumes a concave down configuration.

No, generally speaking we cannot make such a conclusion. It is best to keep the mathematical relationship between shear, moment, and deflection in mind, when thinking about such questions. Moment is the integral of shear, and deflection is the double integral of moment divided by EI. To determine the maximum moment, we can take the derivative of the moment equation and set it to zero. This translates into: moment is maximum when shear is zero. To determine the maximum deflection, we can take the derivative of the deflection equation. That resulting equation represent the slope of the deflection . So, when the slope of the deflection is zero, the deflection is maximum. But the slope equation is not the same as the moment equation or the shear equation. So we can make general statements like: deflection is max when moment is max. If we take the derivative of slope of, we get the moment equation. So, we can say slope is maximum/minimum when moment is zero. But that is different than saying moment is maximum when slope is zero. In some special cases, we may have the max deflection correspond to max moment (zero shear), but that is not true in most cases.

Interesting question. I am not aware of any method for solving the problem in reverse, but in most cases, we can infer a lot of information from the moment diagram. For example, from the moment values at the boundaries of the beam, we can infer the beam's supports. Also, if we can locate the point at which the slope of the bending moment is zero (if such a point exists), then we know where shear is zero. Also, If we know the nature of the curve for the moment diagram (i.e., quadratic, linear,...) then we know the shape of the shear diagram, and the type of load that is placed on the beam. For example, a quadratic moment diagram means a linear shear diagram, both associated with a uniformly distributed load. But if the exact (mathematical) shape of the moment diagram is not available to us, it would not be possible to accurately determine the nature of the loads placed on the beam.

@@DrStructure that was helpful. Thank you so much dr structure 🙏🙏🙏🙏

Why should it be 375 + 50x instead of 375 - 100x?

He use derivative .

the derivative is to be of M, and its equation is M= 375x-50x^2,so dM/dx should be 375-50x

Not quite. Your M equation is correct. Your dM/dx equation however is incorrect. More specifically, the derivative of -50x^2 is -100x, not -50x.