Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 1 (Q.No. 28)
28.
The reinforced concrete beam which has width 25 cm, lever arm 40 cm, shear force 6t/cm2, safe shear stress 5 kg/cm2 and B.M. 24 mt,
Discussion:
9 comments Page 1 of 1.
Henok Manaye said:
3 years ago
The Shear force must be given in t only or kg only.
Fazeel Nonari said:
5 years ago
No, Totally wrong, how to share force can be in kg/cm2 or t/cm2?
(1)
ARABINDA MONDAL said:
6 years ago
The maximum shear stress:
q=shear force/(width*lever arm).
=(6*1000)/(25*40),
=6kg/cm2 >5kg/cm2(safe shear stress).
So, it is unsafe.
q=shear force/(width*lever arm).
=(6*1000)/(25*40),
=6kg/cm2 >5kg/cm2(safe shear stress).
So, it is unsafe.
(4)
Rajesh said:
8 years ago
Very clear it should not be lever arm. It will be depth and given shear stress also shear force.
Suresh said:
9 years ago
Force is given in load per area.
(1)
VIJAY said:
10 years ago
It is unsafe in shear because 6 is greater than 5.
Varun said:
1 decade ago
But @Abhishek lever arm is not considered to be effective depth, Lever arm is the distance between the compressive force and the centroid of Ast, so how can we come to a conclusion to this.
Abhishek said:
1 decade ago
Shear force = V/bd.
= 6000/(24)(40).
= 6>5.
So its unsafe.
= 6000/(24)(40).
= 6>5.
So its unsafe.
(1)
Anu said:
1 decade ago
As the shear force=6t/sq.cm is greater than safe shear stress=5kg/sq.cm, hence unsafe in shear.
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