Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 3 (Q.No. 39)
39.
An R.C.C beam of 25 cm width has a clear span of 5 metres and carries a U.D.L. of 2000 kg/m inclusive of its self weight. If the lever arm of the section is 45 cm., the beam is
Discussion:
7 comments Page 1 of 1.
MJM Gcek said:
4 years ago
Yes agreed with All, safe with stirrups is right.
Hardik. said:
5 years ago
@ Gubendhiran.
Its wl/2..
=(2000 * 5)/2 =5000 kg..
Its wl/2..
=(2000 * 5)/2 =5000 kg..
(2)
Xiya said:
7 years ago
According to me, the Answer should be B.
Safe with stirrups.
Safe with stirrups.
(2)
Gubendhiran said:
8 years ago
Shear stress=SF/(leverage X d).
SF=wl/4 = (2000*5)/4 =5000kg.
Shear stress=5000/(45*25)=4.44kg/cm^2 which is less than 5kg/cm^2 hence SAFE in shear stress.
SF=wl/4 = (2000*5)/4 =5000kg.
Shear stress=5000/(45*25)=4.44kg/cm^2 which is less than 5kg/cm^2 hence SAFE in shear stress.
(6)
Shailendra Kumar said:
9 years ago
SF = 2000 ÷ (500 * 100) = 0.04 kg/sq cm.
Safe shearing stress = L/A
Load = 20 * 500 = 10000 kg.
Area = 25 * 45 = 1125 sq cm.
Therefore L/A = 8.88 kg/sqcm.
Hence Safe shearing stress > shear force.
Result : Beam is safe in shear.
Safe shearing stress = L/A
Load = 20 * 500 = 10000 kg.
Area = 25 * 45 = 1125 sq cm.
Therefore L/A = 8.88 kg/sqcm.
Hence Safe shearing stress > shear force.
Result : Beam is safe in shear.
(3)
Ankit said:
9 years ago
Someone please solve it?
Engbaseem said:
9 years ago
How to check that?
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