First find the location of the horizontal neutral axis of the cross section (demonstrated in this example problem: th-cam.com/video/VDJCDGRHkzM/w-d-xo.html). It passes through the cross section's centroid. I recommend that you then select your x-axis to be oriented along the longitudinal axis of the beam, in other words, along the length of the beam and located at the beam cross section's centroid. The beam cross section will then be in the y-z plane. Position the z- axis on the cross-section's horizontal neutral axis. The y-axis will be perpendicular to it. The direction of your axes system will be correct if, when using your right hand, curling your fingers from the x-axis to the y-axis your thumb points in the z-axis direction. When using this axes system, you will not need to change the unsymmetric bending equation signs or variables from what is shown in the video. A demonstration of this coordinate system is shown in the following video at approximately 1:10 (th-cam.com/video/KBMij7mNmTc/w-d-xo.html).
Mz causes compression on the cross section on one side of the z-axis and tension on the other side of the z-axis. The negative sign is present so that when you have a positive Mz, the stress caused by Mz on the portion of the cross section above the z-axis (where the y variable is positive) will be in compression (negative), and the stress caused by Mz on the portion of the cross section below the z-axis (where the y variable is negative) will be in tension (positive). The equation is based on the x-, y- and z-axes demonstrated in the video. If the axes system changes, the signs and variables in the equation may also change.
Transverse loads (forces perpendicular to the longitudinal axis of the beam and moments about axes perpendicular to the longitudinal axis of the beam) cause internal bending moments and internal shear forces in the beam. The internal bending moments result in formation of normal stresses (equation: normal stress = (M*y)/I) and the internal shear forces result in the formation of shear stresses (equation: shear stress =(VQ)/(It) ).
so the Mz(y) part of the stress equation is negative because it is in compression in the positive y direction, so a change in sign is needed. And the My(z) part is positive because it is compression in the already negative z direction so no change to sign is needed
Thanks Sir! Your example with showing (y = minus value) then max tension at D is fantastic. Most of the websites I only see Sigma = - Mz.y/Iz + My.z/Iy for the max stress and are very confusing
best video explaining this concept online by far thanks a million!
amazing video; i finally understand the concept!
great! the only video that helped me understand
Great lecture sir!!! Specially the way you visualize the concept.. great .. subscribed 👩🚀
best video. alot of effort
Can somebody tell me why before Mz part negative sign is put in the flexure formula?
That is just the direction of the negative moment vector to left...
It could change
Thank you very much for the amazing explanation..
Could you point me to more videos?
about this subject.
Thank You..really helped with the concept
Unbeatable lecture
For a question that the x,y axis and NA are not given, how do I choose it wisely ? And will I get the same bending stress if I use different datum ?
First find the location of the horizontal neutral axis of the cross section (demonstrated in this example problem: th-cam.com/video/VDJCDGRHkzM/w-d-xo.html). It passes through the cross section's centroid.
I recommend that you then select your x-axis to be oriented along the longitudinal axis of the beam, in other words, along the length of the beam and located at the beam cross section's centroid. The beam cross section will then be in the y-z plane. Position the z- axis on the cross-section's horizontal neutral axis. The y-axis will be perpendicular to it.
The direction of your axes system will be correct if, when using your right hand, curling your fingers from the x-axis to the y-axis your thumb points in the z-axis direction. When using this axes system, you will not need to change the unsymmetric bending equation signs or variables from what is shown in the video. A demonstration of this coordinate system is shown in the following video at approximately 1:10 (th-cam.com/video/KBMij7mNmTc/w-d-xo.html).
Omaigadd thank you very much sir !
best one so far...........
]
Fantastic!!!! Thank you!!
Grate explaination prof.
so helpful thanks alot!
absolute vidio ,many many thanks
why did you have negative sign in your stress equation for Mz ? does it indicate all Mz is in compression ?
Mz causes compression on the cross section on one side of the z-axis and tension on the other side of the z-axis. The negative sign is present so that when you have a positive Mz, the stress caused by Mz on the portion of the cross section above the z-axis (where the y variable is positive) will be in compression (negative), and the stress caused by Mz on the portion of the cross section below the z-axis (where the y variable is negative) will be in tension (positive). The equation is based on the x-, y- and z-axes demonstrated in the video. If the axes system changes, the signs and variables in the equation may also change.
Are Bending moments and Shear Stress caused by the same force?
Transverse loads (forces perpendicular to the longitudinal axis of the beam and moments about axes perpendicular to the longitudinal axis of the beam) cause internal bending moments and internal shear forces in the beam. The internal bending moments result in formation of normal stresses (equation: normal stress = (M*y)/I) and the internal shear forces result in the formation of shear stresses (equation: shear stress =(VQ)/(It) ).
Very helpful!! Thanks a lot, sir.
left from origin should not be negative Z-axis??
To the left of Z is positive. Please check ,"Algebra 11 - Cartesian Coordinates in Three Dimensions
".
so the Mz(y) part of the stress equation is negative because it is in compression in the positive y direction, so a change in sign is needed. And the My(z) part is positive because it is compression in the already negative z direction so no change to sign is needed
hi. this comment seems fairly recent, but isn't Point d in tension from both mz and my? and also how's life man?
Very helpful 👍 Thanks👍👌
what a lecture sir
great video
very good!
amamzing sir
Thank you very much
Thank you!
Thanks Sir! Your example with showing (y = minus value) then max tension at D is fantastic. Most of the websites I only see Sigma = - Mz.y/Iz + My.z/Iy for the max stress and are very confusing
awesome video
very clear.
Thank you !!
Thank you☺️
great video
thank you, you made it so understandable
Great!
👌👌👌👌👌👍👍👍👍
this was unsymmetrical bending of symmetrical parts! thumbs down!