Discussion :: RCC Structures Design - Section 1 (Q.No.8)
|Sushmitha said: (May 1, 2015)|
|Please give the explanation.|
|Piyush Singh said: (Sep 3, 2015)|
w | w
Reaction Ra = W.
Reaction Rb = W.
Bending Moment at Center of the Beam = [W * (L/3 + L/6) ]-[W* (L/6) ].
Now Moment due to prestressing force = P*h (h= maximum dip).
Dip required => P*h = WL/3.
Therefore h = WL/3P.
|Arun said: (Oct 4, 2016)|
|Lokesh said: (Dec 5, 2016)|
|Good and thanks @Piyush Singh.|
|Mamun Biswal said: (Apr 2, 2017)|
|Chandru M said: (Apr 16, 2017)|
|Thank you @Piyush Singh.|
|Rajesh said: (May 11, 2017)|
Done very cleverly....
|Manohar said: (Aug 18, 2017)|
|Somnath Gorain said: (Sep 14, 2017)|
|Mriganka Roy said: (Sep 22, 2017)|
|Thank you @Piyush.|
|Subhra said: (Oct 25, 2017)|
|Raj Chavda said: (Nov 15, 2017)|
Now moment at centre=[W*2L/3-W*L/3] = WL/3.
Now moment due to prestressing = P*h.
Now moment P*h should be > = for dip required.
P*h> = WL/3.
h> = WL/3P.
Generally h>or = WL/3P.
|Saravanaraja said: (Dec 18, 2017)|
|Bending Moment at Center of the Beam = [W * (L/3 + L/6) ]-[W* (L/6) ]. How this come? Please tell me, I am not understanding.
Explain me briefly @Piyush Singh.
|Saikiran said: (Jan 14, 2018)|
|L/3+L/6 is L/2.
Multiply reaction with L/2 and subtract moment due to load ( load(W)*distance from load to center(L/6)).
|Lakhan Bhavnani said: (Jan 18, 2018)|
|Anil Kum said: (Jan 25, 2018)|
|Good, Thanks @Piyush.|
|Bidyut said: (Mar 1, 2018)|
|Thanks for the explanation.|
|Ayesha said: (Jun 26, 2018)|
|Caluculate : W(L/3+L/6)-W(L/6) = WL/3.
M = WL/3,
Equate both you got h value @Saravanaraja.
|Umesh said: (Dec 24, 2018)|
|Very nice explanation Thanks.|
|Srihari said: (Apr 26, 2019)|
|Sanat Kumar Rout said: (Sep 11, 2019)|
|Thanks for explaining @Piyush.|
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