Civil Engineering - RCC Structures Design - Discussion
Discussion Forum : RCC Structures Design - Section 5 (Q.No. 20)
20.
A singly reinforced concrete beam of 25 cm width and 70 cm effective depth is provided with 18.75 cm2 steel. If the modular ratio (m) is 15, the depth of the neutral axis, is
Discussion:
6 comments Page 1 of 1.
Anbarasan muthaiah said:
1 decade ago
0.5*b n^2 = m Ast (d-n).
0.5 25*n^2 = 15 *18.75 (70-n).
12.5*n^2 = 19425-281.25n.
By solving n = 30cm.
0.5 25*n^2 = 15 *18.75 (70-n).
12.5*n^2 = 19425-281.25n.
By solving n = 30cm.
(2)
Pawan said:
6 years ago
Thanks @Anbarasan Muthaiah.
Heiro said:
6 years ago
I don't get the answer. Please explain me.
Hussein said:
6 years ago
Take the moment about N.A axis equal to zero.
0.5 * b n^2 = m Ast (d-n).
0.5 25*n^2 = 15 * 18.75 (70-n).
12.5*n^2 = 19425-281.25n.
n = 30 cm.
0.5 * b n^2 = m Ast (d-n).
0.5 25*n^2 = 15 * 18.75 (70-n).
12.5*n^2 = 19425-281.25n.
n = 30 cm.
Gsasikumar said:
6 years ago
Thanks for solving this sum.
Vasikh Ali said:
4 years ago
Moment of compression area wrt neutral axis = moment of equivalent tensile area wrt NA.
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