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.
Pabitra Basnet said:
9 years ago
How this value occur?
Hussain said:
9 years ago
Please explain this.
Ubi said:
8 years ago
Please anyone explain this?
Bala said:
8 years ago
Please explain this problem.
Vikas Tomar said:
8 years ago
.87 is the correct answer.
Vinod said:
8 years ago
How .87 is correct? Can you explain?
Gaurav said:
8 years ago
If grade of steel not how xu max determined and then how mu limit will be determined?
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.
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.
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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers