Design the reinforcement of a rectangular beam to resist a dead load moment of 200kN.m (including its own weight) and a live load moment of 300kN.m. The beam is limited in size to 350mm by 600mm overall depth. Steel covering from centroid of bars to outermost fiber is 100mm for both tension and compression reinforcement. Use fc'=27.5Mpa, fy = 414MPa, 20mm bars A. As'= 7 of 20mm and As=14 of 20mm B. As' = 9 of 20mm and As=18 of 20mm C. As'= 5 of 20mm and As=10 of 20mm D. As' = 3 of 20mm and As=6 of 20mm
Design the reinforcement of a rectangular beam to resist a dead load moment of 200kN.m (including its own weight) and a live load moment of 300kN.m. The beam is limited in size to 350mm by 600mm overall depth. Steel covering from centroid of bars to outermost fiber is 100mm for both tension and compression reinforcement. Use fc'=27.5Mpa, fy = 414MPa, 20mm bars A. As'= 7 of 20mm and As=14 of 20mm B. As' = 9 of 20mm and As=18 of 20mm C. As'= 5 of 20mm and As=10 of 20mm D. As' = 3 of 20mm and As=6 of 20mm
Engineering Fundamentals: An Introduction to Engineering (MindTap Course List)
5th Edition
ISBN:9781305084766
Author:Saeed Moaveni
Publisher:Saeed Moaveni
Chapter12: Electric Current And Related Variables In Engineering
Section12.2: Electrical Circuits And Components
Problem 6BYG
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Design the reinforcement of a rectangular beam to resist a dead load moment of 200kN.m (including its own weight) and a live load moment of 300kN.m. The beam is limited in size to 350mm by 600mm overall depth. Steel covering from centroid of bars to outermost fiber is 100mm for both tension and compression reinforcement. Use fc'=27.5Mpa, fy = 414MPa,
20mm bars
A. As'= 7 of 20mm and As=14 of 20mm
B. As' = 9 of 20mm and As=18 of 20mm
C. As'= 5 of 20mm and As=10 of 20mm
D. As' = 3 of 20mm and As=6 of 20mm
Please provide neat and clear solutions
Note: kindly check below answer

Transcribed Image Text:M₁ = 1-2 MPL + 1.6MLL
M₁ = 1-2 (200) + 1.6 (300)
Arsuming $=0.9
Mn =
Smax
Mu
$
Smax
computing beam's
0.85 fé
fy
0.018 =
assuming maximum possible tensile steel and
and
nominal moment
strength
³,
= 800 KN-M
0.85 f.
fy
Ecu
Ecu + Ec
As,
3max
(ACI Load Combination)
= 720 KN m
1.-F
(0.85)(27.5)
414
= 0·018
R = 6.26 MPa
(0.85)(27.5)
414
H
= Smax bd = (0·018 )(350) (500)
2Rn
0.8511
no compression steel
(0.85) (-=-
2 Rn
(0.85)(27.5)
0.003
0.003 +0.00 5
2
= 3150mm²

Transcribed Image Text:A's
Rn
a =
As2=
Es
=
Mui
&bd²
As, fy
0.85 b
( =
Mn 2
f's
Number of
Mn₂
Mni
³₁
(d-d)
=
bars
=
=
Mai
$
Mn-Mn₁
6.26 =
159.4
0.85
(0.003)
(3150) (414)
(0.85)(27.5)(350)
TT
492.975
0.9
Mui
(0-9)(350)(500)²
252.25x106
(280) (500-100)
2252-23
x20²
A's ts (2827.43) (280)
fy.
414
= 187.53 mm
=
= 800 547.75 = 252.25 KN.m
187.53-100
187-53
= 159.4 mm.
547 75 KN-m
~ 9
= 2252.23 mm²
2
Mu,
2
= 1912-27 mm²
- 492 975 KN.m
(0.003) = 0.0014 <
bars
t₁ = (0-0014) (200000) = 280 тра
Es
=
414
200000
(As = 2827-43 mm²)
= 0.002.07
As = As, + As₂ = 3150 + 2827-43
As = 5062.27 mm²
(18 bars)
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Step 1: Introducing Given Data
VIEWStep 2: Calculate nominal moment
VIEWStep 3: Calculate As1
VIEWStep 4: Calculate Mu1
VIEWStep 5: Calculate depth of stress block (a)
VIEWStep 6: Calculate stress in steel (fs)
VIEWStep 7: Calculate provided area of compression steel (As')
VIEWStep 8: Calculate area of tension steel
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