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
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
24 comments Page 1 of 3.

PIYUSH SINGH said:   1 decade ago
w         |         w           
---------------------------|---------------------------
A (L/3) B (L/6) | (L/6) C (L/3) D


Reaction Ra = W.

Reaction Rb = W.

Bending Moment at Center of the Beam = [W * (L/3 + L/6) ]-[W* (L/6) ].

= WL/3.

Now Moment due to prestressing force = P*h (h= maximum dip).

Dip required => P*h = WL/3.

Therefore h = WL/3P.
(17)

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.
(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.

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)).

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.
(1)

Rajesh said:   8 years ago
Thank u@piyush.
Done very cleverly....

Lokesh said:   9 years ago
Good and thanks @Piyush Singh.

SANAT KUMAR ROUT said:   6 years ago
Thanks for explaining @Piyush.

Umesh said:   7 years ago
Very nice explanation Thanks.

Sushmitha said:   1 decade ago
Please give the explanation.


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