Sir in my exam a question came in which I had to analyse a frame using moment distribution method. But I got confused as, in question it was written that all members have equal flexural rigidity but in diagram one member had different moment of inertia . So what should be done in such situation.
How about if one of the spans has a combination of uniform load and a concentrated load not located at the mid-span. How is the FEM calculated in this scenario?
It is the rotational stiffness(k) for fixed far end in a propped cantilever beam , from M=kθ, where M is the moment, θ is the slope and k is rotational stiffness which is equal to 4EI/L.
Plz help me with this problem a type 3 post tensioned prestressed concrete beam of 10 M span The Beam is post tension is in 3 high tensile bar of 40mm diameter located at an effective depth of 700mm.The effective prestressing force in each bar after all losses is 600 kN. Given f=40 N/mm².f,= 1035 N/mm². E,= 200 kN/mm², E=28 kN/mm², com- pute the width of cracks in the tension zone if the service load moment at mid span is 1040 KN m,
Thank You Sir. It helps very well in my Endsem
why at joint B u make it negative and point C u make it positive for balancing? I need clarification
thank you, nice lesson
thanks the video is useful
Glad it was helpful! All the best
Sir in my exam a question came in which I had to analyse a frame using moment distribution method.
But I got confused as, in question it was written that all members have equal flexural rigidity but in diagram one member had different moment of inertia . So what should be done in such situation.
How about if one of the spans has a combination of uniform load and a concentrated load not located at the mid-span. How is the FEM calculated in this scenario?
Thanks very much sir
Sir is this method can be done in 18schem also?.....gotta clear this backlog subject...plz do reply
Sir why the I is different for all three span?
Sir can we take stiffness for fixed and intermediate support as I/L and for roller ,SS and hinged support as 3I/4l. Is these values are correct
did u get answer for this question
Sir in Distribution factor table why we take D in last video u don't take D
How comes 4EI/ L ???? ANY DERIVATION ??
It is the rotational stiffness(k) for fixed far end in a propped cantilever beam , from M=kθ, where M is the moment, θ is the slope and k is rotational stiffness which is equal to 4EI/L.
Although u could simply find k with I/L for fixed far ends
And 3/4×I/L for any other conditions
Sir kani s method start kab karr rahe hoo sir ?
Next set of videos will be on Kani's method, keep watching and sharing
Thanks
Sir can we get classes for DSMS
From march i will be uploading steel structure videos, keep watching
@@structuralanalysisanddesign i am also very excited to see 🙈 waiting for your DSMS videos ..
4:25
Plz help me with this problem
a type 3 post tensioned prestressed concrete beam of 10 M span The Beam is post tension is in 3 high tensile bar of 40mm diameter located at an effective depth of 700mm.The effective prestressing force in each bar after all losses is 600 kN. Given f=40 N/mm².f,= 1035 N/mm². E,= 200 kN/mm², E=28 kN/mm², com- pute the width of cracks in the tension zone if the service load moment at mid span is 1040 KN m,
Sorry Ramesh, currently I am not in touch with prestressed concrete subject, in future i will definitely try, thanks for watching
Slop and deflection method sir es topick ka video banaye sir
Or dam
Hello Slope deflection method is already uploaded
4:24 should be -9
Thank you so much sir
Thanks