Analysis of 3 Span Beam by Moment Distribution Problem | 1 End Fixed and another End Hinged Beam
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- čas přidán 3. 01. 2023
- This video explains how to analyze a 3 span indeterminate beam by using moment distribution method where the beam is having 1 end fixed and another end hinged, this video explains about modified procedure of moment distribution method where the end span having hinge support is modified by using different stiffness and subsequently made changes in moment distribution table to make end D as zero moment.
The main advantage of modified procedure is to have less iterations in the moment distribution table to get the final moments. Lastly bending moment diagram is drawn for loading cases and than moments are overlapped to get net bending moment.
Moment Distribution Problem 1 : • Moment Distribution Me...
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Thank You Sir. It helps very well in my Endsem
Thanks
thanks the video is useful
Glad it was helpful! All the best
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?
Sir is this method can be done in 18schem also?.....gotta clear this backlog subject...plz do reply
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
Sir why the I is different for all three span?
Sir in Distribution factor table why we take D in last video u don't take D
Sir kani s method start kab karr rahe hoo sir ?
Next set of videos will be on Kani's method, keep watching and sharing
Slop and deflection method sir es topick ka video banaye sir
Or dam
Hello Slope deflection method is already uploaded
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 ..
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
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