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 barsA. As'= 7 of 20mm and As=14 of 20mmB. As' = 9 of 20mm and As=18 of 20mmC. As'= 5 of 20mm and As=10 of 20mmD. As' = 3 of 20mm and As=6 of 20mm Please provide neat and clear solutions Note: kindly check below answer M₁ = 1-2 MPL + 1.6MLLM₁ = 1-2 (200) + 1.6 (300)Arsuming $=0.9Mn =SmaxMu$Smaxcomputing beam's0.85 féfy0.018 =assuming maximum possible tensile steel andandnominal momentstrength³,= 800 KN-M0.85 f.fyEcuEcu + EcAs,3max(ACI Load Combination)= 720 KN m1.-F(0.85)(27.5)414= 0·018R = 6.26 MPa(0.85)(27.5)414H= Smax bd = (0·018 )(350) (500)2Rn0.8511no compression steel(0.85) (-=-2 Rn(0.85)(27.5)0.0030.003 +0.00 52= 3150mm² A'sRna =As2=Es=Mui&bd²As, fy0.85 b( =Mn 2f'sNumber ofMn₂Mni³₁(d-d)=bars==Mai$Mn-Mn₁6.26 =159.40.85(0.003)(3150) (414)(0.85)(27.5)(350)TT492.9750.9Mui(0-9)(350)(500)²252.25x106(280) (500-100)2252-23x20²A's ts (2827.43) (280)fy.414= 187.53 mm== 800 547.75 = 252.25 KN.m187.53-100187-53= 159.4 mm.547 75 KN-m~ 9= 2252.23 mm²2Mu,2= 1912-27 mm²- 492 975 KN.m(0.003) = 0.0014 <barst₁ = (0-0014) (200000) = 280 траEs=414200000(As = 2827-43 mm²)= 0.002.07As = As, + As₂ = 3150 + 2827-43As = 5062.27 mm²(18 bars)

Answered: Design the reinforcement of a…

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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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

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

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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