Lecture 2, Shear strain (Lecture & examples)

[@victormejia6533]

one of the best videos in mechanics of material. well done.

[@Anime_Saying]

its very good and simple way of explaination...

[@vilivont4436]

Great lesson, thanks a lot😊😊

[@rajeewprasath7879]

It helped to understand shear strain easily. Thank you so much..🙂

[@NishantBagoria]

That is a very very good video, Definitely cleared all my doubts. I suggest the channel owner to change the Microphone to make the voice crystal clear.

[@Tarek1997Graphic]

Thanks for this video! really saved me :)

[@amatoallahouchen5894]

great video man thanks a million

[@Islamic786_26]

very informative video sir nice sir

[@RahulSharma-oc2qd]

when we do have a perpendicular load acting on beam, do we have normal stress acting on the beam too? normal stress and shear stress both acts only if the load is acting at an angle. Am I missing something?

[@MechanicsofMaterialsLibre]

Yes there will be normal and shear stress. In beams subjected to loads perpendicular to its axis, both shear force and bending moment will be developed; bending moment causes normal stress and shear force causes shear stress. Hope that helps.

[@jMoeBeats]

really saved me dude thank you so much

[@ImranKhan-du7hn]

In question 2, we can directly use shear strain=tan'((1/16)/25).then why we r using a long process

[@nicklibre7936]

That equation (strain = atan(delta/L)) is only valid if we can directly measure the change in the angle of the element which is not the case on the second example. If the change in the angle cannot be measured directly, one can use the angle after deformation and subtract it from the initial angle as we did in the 2nd example. Note that the answer of strain = atan(delta/L) = atan(1/16/25) = 2500 micro strains but the correct answer is 2440 as calculate. The difference could be larger in different geometries. Hope that helps.

[@johnroyettedeiparine6419]

Thank you sir for the input

[@zahrinamardina9200]

very helpful! thank you very much!

[@cristianguerrero6569]

very helpful

[@venkateshbuddepu1649]

Very very nice video

[@anaaya3131]

So helpful, thankyou!

[@legistergregory8108]

thanks that was fun!

[@deemaedrees5763]

Thank you so much !

[@babuks82]

Shouldn't the angle for negative shear strain be (pie/2 + gamma_xy). The angle is greater than 90deg

[@foxedmind1950]

Thanks a lot

[@engr.muhammadtahirkhan6729]

thank you Sir.

[@NAYAN-t3e]

4:54 How did you take that shear strain as π/2 - gamma_xy ?? It is greater than π/2.

[@MechanicsofMaterialsLibre]

If you refer to the right image, the shear strain is negative. Therefore the negative sign would cancel out. Hope that explains.

[@AwestrikeFearofGods]

@@MechanicsofMaterialsLibre It's simply a calculation error: Is: -0.002440, Should be: +0.002436

[@danielmatlock4125]

yes very good nice :)

[@Tom-yo3jf]

Thank you sir

[@dailyenglishepisodes]

Thanks

[@behonest4171]

Thank you

[@haydencarlson9700]

WashU kid learning from a Missouri S&T prof because mine is bad 😂

[@akshayshende8201]

@1.25 since there are shear stress as well as complementary shear stress why the strain there is only wrt to y axis? Why there is no strain wrt to x axis? Please reply and also please clarify me whether both shear stress and complementary shear stress are reesponsible for shear strain or only shear stress responsible ? Please reply my all the questions if possible. Please Reply my all the questions.

[@MechanicsofMaterialsLibre]

There is just one shear stress acting on the element because in order to maintain equilibrium, shear stress should exist on all four faces. I'm not sure what you mean by complementary shear stress.

[@akshayshende8201]

@@MechanicsofMaterialsLibre ok no problem but please explain me @1.25 why there is strain angle with respect to only y axis, why there is no strain wrt to x axis as well ? Basically i am asking that the shear stress is acting on all the four faces then how the strain is only wrt to y axis ? Why there is no strain wrt to x axis.?

[@shreyashasane9100]

@@akshayshende8201 Dear, clearly for calculation and formula purposes, Base of the element is assumed fixed/(element is deformed that way so that angle stays complete in either side) ALSO in order for the element to be in equilibrium, equal and opposite shear strain must exist at the bottom face. He has not at all mentioned strain exists wrt to any of the axis. Here Shear Strain is calculated on the point (O) i,e intersection of xy. If you were to mark a fixed point B on the lower-right end of the element and calculate shear strain at Bxy it would be same as at (Oxy) i,e (gamma xy). Further deforming the object, Strain angle would be sum of both angles created wrt y axis as well as x axis.To conclude it completely depends on the way that element is placed and how load is applied. Pl Correct me if I am wrong.