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
0.87
8.50
7.50
5.80
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
23 comments Page 2 of 3.

Tirupathi said:   7 years ago
M/I=f/y=E/R.
Use M/I=f/y.

M=f/y*I.
f=15.
I=a4/12 (bd3/12) here b&d both are same b&d=a.
Y bcz cube is square.
y=a/2.

As per standard dimension of the concrete cube for testing, compression is 150mm=a.
M=15*150*150*150*150*2/12*150=8.4375=8.5

Consider fck is 15 full strength.

Pawan said:   6 years ago
Thanks @Tirupathi.

Sandip das said:   6 years ago
As per IS code 456- 2000 R is the moment of the resistance factor, does not depend on dimensions, it's called design constant, the value of R for M15;
1. mild steel =0.867.
2. Fe 415=0.65.
3. Fe 500=0.58.

Vishal said:   6 years ago
0.87 is the correct answer.
(1)

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?

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).

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.

Raihan said:   5 years ago
0.87 is the correct answer.
(2)

Jay said:   4 years ago
0.7 X √(150) = 8.57.
(1)

Vinod kumar yadav said:   4 years ago
Resistance factor = 0.7 * √(fck).
0.7 * √{150) = 8.5.


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