singly r.c. beam

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GUIDED BY: Prof. Sunil Jaganiya

Prof. Pritesh Rathod

Sub : Elementary Structural Design

NAME ENROLL NO.Patel Jimi 131100106029Patel Milind 131100106035Patel Nirmal 131100106036Patel Viraj 131100106040Patel Yash 131100106042Shah Ashit 131100106051

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Singly R.C. beam

r.c.c as per is 456-2000

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Singly reinforced beam

In singly reinforced simply supported beams reinforcing steel bars are placed near the bottom of the beam where they are most effective in resisting the tensile stresses.

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Reinforcement in simply supported beam

COMPRESSION b

STEEL REINFORCEMENT D d

TENSION

SUPPORT SECTION A - A

CLEAR SPAN

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STRESS DIAGARM FOR SINGLY R.C.BEAM ( IS : 456 – 2000 , P.69 )

0.0035 0.446 fck

C = 0.36 fck * b* xu

T

Z = Lever ArmN A

X = X u max

0.87 fy Ast0.87fy

ES+ 0.002

Ast

b

d

SECTION STRAIN STRESS

0.42Xu

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Xu max = Maxi. Depth of Neutral axis b = width of beam d = effective depth Z = d-0.42xu = lever arm D = overall depth Xu = depth of N.A Ast = area of steel Resultant force of compression C = average stress X area

= 0.36 fck b xu

Resultant force of tension T = 0.87 fy Ast

Force of compression should be equal to force of tension,

0.36 fck b xu = 0.87 fy Ast

xu =

Where Ast = area of tension steel

0.87 fy Ast0.36 fck b

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Moment of resistance with respect to concrete= compressive force x lever arm

= 0.36 fck b x z

Moment of resistance with respect to steel = tensile force x lever arm

= 0.87 fy Ast x z

Maximum Depth Of Neutral Axis (IS 456 – 2000 , P.70)

fy ( N/mm2 ) Xu max

250 0.53 d

415 0.48 d

500 0.46 d

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If Xu = Xu max then it is balanced section If Xu < Xu max then it is under reinforced section If Xu > Xu max then it is over reinforced section

Xu max Xu < Xu max Xu > Xu max

Balanced section Under reinforced Over reinforced

b

D

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Moment of resistance Mu with respect to steel ( for under reinforced section)

Mu = T X Z ( IS 456-2000, P.96, Cl. G1.1 (b) )

where , Z = d - 0.42 xu = 0.87 fy Ast x (d - 0.42 xu ) T = 0.87 fy Ast

= 0.87 fy Ast d ( 1- fy Ast / fck b d )

Mu with respect to concrete ( for balanced section) Mu = C x Z ( IS 456-2000, P.96, Cl. G1.1 (c) ) = 0.36 fck b xu ( d – 0.42 xu ) where , C = 0.36 fck b xu

= 0.36 fck b xu d (1 – 0.42 xu / d)

= 0.36 fck b xu/d d2 (1 – 0.42 xu / d)

= 0.36 xu/d (1 – 0.42 xu / d) fck b d2

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Limiting value of tension steel and moment of resistance

Since the maximum depth of neutral axis is limited, the maximum value of moment of resistance is also limited.

Mu lim with respect to concrete = 0.36 fck b x Z

= 0.36 fck b Xu (d – 0.42 Xu)

Mu lim with respect to steel = 0.87 fck Ast (d – 0.42 Xu)

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Limiting moment of resistance

Grade of concrete

Grade of steelFe 250 steel Fe 450 steel Fe 500 steel

General 0.148 fck bd 0.138 fck bd 0.133 fck bd2 22

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Types of problem

a) Analysis of a section

b) Design of a section

13 Analysis of section

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For a limiting section 200 mm x 300 mm effective , determine the following , if it is reinforced with an effective cover of 50 mm

take M-20 conc. And Fe 250Sol :

b = 200 mm d = 300 mm

fck = 20 N / mm2

the given section is the given section is balanced section therefore , xu = xu max (IS 456 – 2000 , P 70)

b = 200

d= 300

Ast

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For Fe – 250 , xu max = 0.53 d = 0.53 x 300 = 159 mm xu = xu max = 159 mm 1. Maxi . Comp. stress in concrete = .446 x fck = .446 x 20 = 8.92 N / mm2

2 . Lever arm Z = d – 0.42 x

= 233.22 mm

3 . Total tension T = 0.87 fy Ast

Ast ;

xu = (IS 456 – 2000 , P.96)

Ast = 1052.6 mm2

0.87 fy Ast0.36 fck b

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T = 0.87 fy Ast

= 0.87 x 250 x 1052.6

= 228.96 kN

Total compression C = 0.36 fck b xu = 0.36 x 20 x 200 x 159 = 228.96 kN for limiting section C = T .

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Design of section

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Design a rectangular RC beam having width 250mm, simply supported with effective span of 4.5 m, it is loaded wit a udl of 20 kN/m including self weight . Use M20 concrete and Fe 415 steel. Check the beam for maximum and minimum steel and deflection Solution :

factored udl w = 1.5 x 20 b = 250 = 30 kN/m

Mu = wl2 / 2 = 75.94 kN/m Design as a balanced section , fy = 415 N/mmMu lim. = 0.138 fck bd2 75.94 x 106 = 0.138 x 20 x 250 x d2

AstD

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d= 331.75 mm provide d = 340 mm Therefore , effective size of beam is 250 mm x 340 mm to find Ast : pt = 50 fck / fy ( 1- (1- 4.6 Mu / fck bd2)1/2 ) pt = 0.896% Ast = pt bd /100 = 761.6 mm2 provide 4 – 16 dia. (Ast = 804 mm2) effective cover = 40 mm D = d + 40 = 380 mm

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Minimum steel in beam As / bd = 0.85 / fy As = 174.09 mm2 ( mini. Required) Ast > As therefore, ok Maximum steel in beam Ast = 0.4 bD = 0.4 x 250 x 380 = 3800 mm2 804 mm2 < 3800 mm2 therefore, ok

{ IS 456-2000 P.42 }

{ IS 456-2000 P.42 }

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Check for deflection actual l/d = 4500/340 = 13.23 Pt = 100 Ast / bd = 0.945 % Maxi l/d = 20 x M.F = 19.5 actual l/d < maxi. l/d therefore, ok o check for development length o Ld <= 1.3 M1 / V + L0

o d= 340 mm o 12 = 12 * 16 = 192 mmo taking larger of two values L0 = 340 mm o S F at support = wl/2 = (30 * 4.5) / 2 = 67.5 kN o Ast = 804 mm2

o M1 = 0.87 fy Ast d ( 1 – (fy Ast / fck bd))o M1 = 79.35 * 106 N.mm

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1.3 M1 / V + L0 = 1.3 *(79.35 * 106 / 67.5 * 103 ) + 340 = 1868 mm Ld = (0.87 fy) / 4 * bd = 1203.5 mm 1203.5 <= 1868 mm it is ok Check for shear v = Vu / bd = 67.5 * 103 / (250 * 340) = 0.794 N/mm2 For fck = 20 N/mm2 Ast = 804 mm2 pt =0.94% c =

0.62 N/mm2

for D = 380 K= 1 ’c = K * c = 0.62 v >’c it is not ok Therefor, increasing in Depth (d).

{ IS 456 - 2000 P.73 T.19 }

{ IS 456-2000 P.42 }

{ IS 456 - 2000 P.72 Cl 40.2.1.1 }

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