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

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

Structural Analysis
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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(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
Transcribed Image Text:(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
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Answer 

We know that Nominal Moment capacity of singly reinforced beam is given by Mn = As*Fy*z

As = area of steel

Fy = yield strength of grade 60 steel

d = effective depth

z = arm

(1) we are given

As = 4 - #8

As = 3.14 in2 

Fy = 60000 psi       (grade 60 steel)

z = d - a2

d = 24 - 2.5

d= 21.5 in

depth of compression stress block is given by a = As*Fy0.85*f'c*b

f'c = 4000 psi

b = 12 in

a = 3.14*600000.85*4000*12

a = 4.62 in

substitute the values to get 'Mn'

Mn = As*Fy*(d-a2)

Mn = 3.14*60000*(21.5-4.622)

Mn = 3615396 lb-in

Mn = 301.283 K-ft

" Nominal Moment capacity  = 301.283 K-ft "

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