Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 2 (Q.No. 38)
38.
If a rectangular prestressed beam of an effective span of 5 meters and carrying a total load 3840 kg/m, is designed by the load balancing method, the central dip of the parabolic tendon should be
Discussion:
11 comments Page 1 of 2.
AJyles said:
6 years ago
For this assume force transmitted by the tendon as 120000kg and use load balancing concept as
P*e = External B.M.
or, 120000*e = (w*l^2)/8.
By Solve taking w =3840 kg/m to get the value of central dip (e).
P*e = External B.M.
or, 120000*e = (w*l^2)/8.
By Solve taking w =3840 kg/m to get the value of central dip (e).
Sunil said:
7 years ago
For UDL h=wl^2/8P and P should be 1200KN.
Kanhaiya pal said:
7 years ago
Due to UDL total load will divide by 8. And 3840/8=480.
480 will be a prestressing force.
480 will be a prestressing force.
Sonu said:
7 years ago
From where 480came? @Qasim.
Qasim said:
7 years ago
h = WL/4p,
h = 38.4 x 500/ ( 4 x 480).
h= 10 cm.
h = 38.4 x 500/ ( 4 x 480).
h= 10 cm.
(3)
Abhishek said:
8 years ago
If the load is total 3840 then how its unit is kg/m?
Manikandan sivakasi said:
8 years ago
P*e=M.
3840*e=(load*perpendicular distance),
3840*e=(3840*5),
e=19200/3840,
e=5.
So 5 is correct answer I think.
3840*e=(load*perpendicular distance),
3840*e=(3840*5),
e=19200/3840,
e=5.
So 5 is correct answer I think.
Hit said:
8 years ago
Which formula used for calculation?
Kruti said:
9 years ago
wl/4
3.84 * 5 / 4 = 4.8
So, ans is 5.
3.84 * 5 / 4 = 4.8
So, ans is 5.
Saksham said:
10 years ago
Yes no force is given.
I think according to answer they should give force as 480.
I think according to answer they should give force as 480.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers