Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 4 (Q.No. 41)
41.
For M 150 grade concrete (1:2:4) the moment of resistance factor is
Discussion:
23 comments Page 1 of 3.
Vinod Yadav said:
4 years ago
Resistance factor = 0.7 * √fck.
=0.7 * √150.
= 8.57.
=0.7 * √150.
= 8.57.
Vinod Yadav said:
4 years ago
Resistance factor = 0.7 * √fck.
=0.7 * √150.
= 8.57.
=0.7 * √150.
= 8.57.
Vinod kumar yadav said:
4 years ago
Resistance factor = 0.7 * √(fck).
0.7 * √{150) = 8.5.
0.7 * √{150) = 8.5.
(1)
Vinod kumar yadav said:
4 years ago
Resistance factor = 0.7 * √(fck).
0.7 * √{150) = 8.5.
0.7 * √{150) = 8.5.
Jay said:
4 years ago
0.7 X √(150) = 8.57.
(1)
Raihan said:
5 years ago
0.87 is the correct answer.
(2)
Sachin Singh said:
5 years ago
WSM mtd Mr = Qbd2.
M150 Xc= 0.39d, z=0.87d.
MOR = bx sigma z/2 = b* 0.39d* 50* .87d/2.
MOR= 8.5bd2.
σ = 50 kg/cm2.
M150 Xc= 0.39d, z=0.87d.
MOR = bx sigma z/2 = b* 0.39d* 50* .87d/2.
MOR= 8.5bd2.
σ = 50 kg/cm2.
Sushil Jaiswal said:
5 years ago
No, it is not right because MOR factor can get only SP-16 graph which is given in IS456:1998 and value is in table 70 (8.5 N/sqm).
Jayaraj said:
6 years ago
Mur = Q.bd^2
= 0.148 fck bd^2. For Fe250
Q = 0.148x15 = 2.22.
Why this is not correct?
= 0.148 fck bd^2. For Fe250
Q = 0.148x15 = 2.22.
Why this is not correct?
Vishal said:
6 years ago
0.87 is the correct answer.
(1)
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