SFD and BMD for Beam - Problem 4 - Shear Force and Bending Moment Diagram - Strength of Materials

[@Ekeeda]

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[@atharvapote755]

not working

[@doileeyang2772]

So what about point of contraflexure? Why you have not you calculated it too?

[@sabitakarki4597]

Hello sir i don’t know if i can read this message. But i wanna thank u for your lecture. Because of u i was able to pass my board exam. Thank u so much.

[@anirudhtikariya219]

Board exam? Yeh toh engineering ka hai

[@sabitakarki4597]

@@anirudhtikariya219 engineering ko exam

[@vipintyagi3573]

in FBD of portion of ABCDE, when we find moment. 2.5×6=15 then we get answer Rc=21.25kn

[@arvincaluyo673]

thank you so much sir. engr. professor.

[@pubgmobileindia3639]

How u take the value of RE 2.5

[@wadkarjurez6890]

SF at point E must be 5 KN Since SF at C is 2.5 it will be constant till D cause there is no force on D and hence RE will be added further it will become total SF at E is 5 KN Sir please let me know this plz....?

[@omkarauti]

Thank you!

[@mandlasiddhartha6531]

Nice sir ❤️👍

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[@akshaypatel1568]

How we know point e is internal hinge

[@AsifHashmi1437]

Value of Mf is wrong, it should be -35

[@JibonReja]

Yeah it's wrong. It will be -35

[@kanishk5677]

sir please upload problem based on moment area method

[@shailendrakumar1099]

BM at F should be - 35kNm. U hv written right equation but calculations is wrong

[@vipintyagi3573]

good sir

[@wadkarjurez6890]

Bending moment must be 0.46 KNm.... Cause there is location of zero shear force from E is 0.125m