Please solve the following! (1)(3)21(4)4.#82000 12.5"1224"4BAL6"4#910"3"14"4#712"4#92415"1/24I 3""/3%$ 2.5"3"4%Ifé= 4000 psi, NWCGrade 60 steel, 4 #8.Mn = ?fé = 3 ksiGrade 60 steel. 49Mn=?37NWCfé = 3 ksi, NWCGrade 60 steel4#7fé= 3 ksi, NwCGrade 60. steel.4#9

Answered: féz 4000 psi, NWC Grade 60 steel, Mn…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Please solve the following!

(1)
(3)
21
(4)
4.#8
2000 12.5"
12
24"
4
BAL
6"
4#9
10"
3"
14"
4#7
12"
4#9
24
15"
1/
24
I 3"
"/
3%
$ 2.5"
3"
4%
I
fé= 4000 psi, NWC
Grade 60 steel, 4 #8.
Mn = ?
fé = 3 ksi
Grade 60 steel. 49
Mn=?
37
NWC
fé = 3 ksi, NWC
Grade 60 steel
4#7
fé= 3 ksi, NwC
Grade 60. steel.
4#9

Expert Solution

Step 1

Answer

We know that Nominal Moment capacity of singly reinforced beam is given by M_n = A_s*F_y*z

A_s = area of steel

F_y = yield strength of grade 60 steel

d = effective depth

z = arm

(1) we are given

A_s = 4 - #8

A_s= 3.14 in²

F_y = 60000 psi (grade 60 steel)

z = d - $\frac{a}{2}$

d = 24 - 2.5

d= 21.5 in

depth of compression stress block is given by a = $\frac{A_{s} * F_{y}}{0.85 * f' c * b}$

f'c = 4000 psi

b = 12 in

a = $\frac{3.14 * 60000}{0.85 * 4000 * 12}$

a = 4.62 in

substitute the values to get 'M_n'

M_n = A_s*F_y*(d- $\frac{a}{2}$ )

M_n = 3.14*60000*(21.5- $\frac{4.62}{2}$ )

M_n = 3615396 lb-in

M_n = 301.283 K-ft

" Nominal Moment capacity = 301.283 K-ft "

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féz 4000 psi, NWC Grade 60 steel,  Mn = ?

féz 4000 psi, NWC Grade 60 steel, Mn = ?