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 2 of 3.
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.
1. mild steel =0.867.
2. Fe 415=0.65.
3. Fe 500=0.58.
Pawan said:
6 years ago
Thanks @Tirupathi.
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.
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:
7 years ago
The moment of resistance Mr=Qbd^2.
Where is, Q is moment resistance factor.
Q=1/2(X1 * sigma cb * Z1 )----> eqn (1).
x1 is natural axix factor.
Z1 is lever arm factor.
Put the direct value of x1= 0.391 and z1=0.871 and,
for m150, σ cb is 5 N/mm2.
For M150 and fe250.
From eqsn 1,
Q =(1*5*0.391*0.870)/2.
Q= 0.8504 answer.
Where is, Q is moment resistance factor.
Q=1/2(X1 * sigma cb * Z1 )----> eqn (1).
x1 is natural axix factor.
Z1 is lever arm factor.
Put the direct value of x1= 0.391 and z1=0.871 and,
for m150, σ cb is 5 N/mm2.
For M150 and fe250.
From eqsn 1,
Q =(1*5*0.391*0.870)/2.
Q= 0.8504 answer.
Manoj Dethe said:
8 years ago
M=Es/Ec.
Ec = 5700 Sq.root FCK.
M = 210000/22066.
M = 9.51 greater value is given 8.5.
Ec = 5700 Sq.root FCK.
M = 210000/22066.
M = 9.51 greater value is given 8.5.
Naveen Kallan said:
8 years ago
Mu = 0.138 fck b d^2 for Fe415.
0.138 X 15 = 2.07.
0.138 X 15 = 2.07.
Gaurav said:
8 years ago
If grade of steel not how xu max determined and then how mu limit will be determined?
Vinod said:
8 years ago
How .87 is correct? Can you explain?
Vikas Tomar said:
8 years ago
.87 is the correct answer.
Bala said:
8 years ago
Please explain this problem.
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