Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 1 (Q.No. 8)
8.
A prestressed rectangular beam which carries two concentrated loads W at L/3 from either end, is provided with a bent tendon with tension P such that central one-third portion of the tendon remains parallel to the longitudinal axis, the maximum dip h is
Discussion:
24 comments Page 2 of 3.
Subhra said:
8 years ago
Thanks @Piyush.
Raj chavda said:
8 years ago
Ra-A........L/3...........!(w)..........2L/3...........B-Rb
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.
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.
(1)
SARAVANARAJA said:
8 years ago
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.
Explain me briefly @Piyush Singh.
SaiKiran said:
8 years ago
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)).
Multiply reaction with L/2 and subtract moment due to load ( load(W)*distance from load to center(L/6)).
Lakhan bhavnani said:
8 years ago
Thanks @Piyush.
Anil kum said:
8 years ago
Good, Thanks @Piyush.
Bidyut said:
8 years ago
Thanks for the explanation.
(1)
Ayesha said:
7 years ago
Caluculate : W(L/3+L/6)-W(L/6) = WL/3.
M = WL/3,
M= h*P.
Equate both you got h value @Saravanaraja.
M = WL/3,
M= h*P.
Equate both you got h value @Saravanaraja.
(1)
Umesh said:
7 years ago
Very nice explanation Thanks.
Srihari said:
6 years ago
Dip means?
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