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Civil - Design of Reinforced Concrete Elements - Limit State Design For Flexure

1. Design a
one way slab with a clear span of 5m, simply supported on 230mm thick masonry
walls and subjected to a live load of 4kN/m^{2} and a surface

finish of
1kN/mm^{2}.Assume Fe 415 steel. Assume that the slab is subjected to
moderate exposure conditions.

__Step 1: Type of Slab.__

ly/lx = 5/1 = 5>2.it has to be
designed as one way slab.

__Step 2:Effective depth
calculation.__

d =
span/(basic value x modification factor) = 5000/(20x0.95) = 270mm D = 270 + 20
+ 10/2 = 295mm

__Step 3: Effective Span.__

Le =
clear span + effective depth = 5000 + 270 = 5.27m (or) Le =c/c distance b/w
supports = 5000 + 2(230/2) = 5.23m Adopt effective span = 5.23m least value.

__Step 4:
load calculation __Live load = 4kN/m^{2}

Dead load = 1x1x0.27x25 =
6.75kN/m^{2}

Floor Finish = 1kN/m^{2}

Total load = 11.75kN/m^{2}

Factored load = 11.75 x 1.5 =
17.625kN/m^{2}

__Step 5: Moment calculation.__

M = wl^{2}/8
= (17.625x5.23^{2})/8 = 60.26kNm __Step 6: Check for effective depth.__

M = Qbd^{2}

d^{2}
= M/Qb = 60.26/2.76x1 = 149.39mm say 150mm. For design consideration adopt d =

150mm.

__Step 7: Area of Steel.__

Mu = 0.87 fy Ast d (1- (fy ast)/(fck b d)) 60.26x10^{6}
= 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 300mm^{2}

Use 10mm dia bars

Spacing ,S = ast/Astx1000 = (78.53/300)1000 = 261mm say
260mmc/c Provide 10mm dia @260mm c/c.

2. Design a simply supported RC beam having an
effective span of 5m.the beam has

to carry
a load of 25 kN/m. sketch the reinforcement details.

__Step 1:
Effective length. __Effective span,le = 5m __Step 2: Size of the
beam.__

Effective
depth = le/10 = 5000/10 = 500mm Assume, b = 2/3d = 2/3x500 = 333.2mm say 340mm __Step
3: Load Calculation__

Live load = 25kN/m

Dead load
= 1x.340x.500x25 = 4.25kN/m Total load = 29.25kN/m

Factored
load = 29.25x1.5 = 43.85kN/m __Step 4: Moment Calculation.__

M = wl^{2}/8
= (43.85x5^{2})/8 = 137.08kNm __Step 5: Check for effective depth.__

M = Qbd^{2}

d^{2}
= M/Qb = 137.08/2.76x.340 = 382.2mm say 380mm. d = 380mm > 500mm

Hence it is safe.

__Step 5: Check for effective
depth.__

M_{bal} = Qbd^{2}
= 2.97x340x500^{2} = 252.06kNm > M

Hence it
can be designed as singly reinforced beam section. __Step 6: Area of Steel__

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

137.08x10^{6}
= 087x415xAstx500(1-(415 Ast)/(20x340x500)) Ast = 846.15mm^{2}

Use 20mm dia bars

No of bars = Ast/ast =
846.15/314.15 = 2.45 say 3nos Provide 3#20mm dia as tension reinforcement.

3. Design a
RC beam 350X700mm effective section, subjected to a bending moment of
300kNm.Adopt M20concrete and Fe415 steel.

__Step 1:
Size of the beam. __b = 350mm & D = 700mm d = 700-25-20/2 =665mm

__Step 2:
Moment Calculation. __M = 300kNm

__Step 3:
Check for effective depth. __

M_{bal}
= Qbd^{2} = 2.97x350x665^{2} = 459kNm > M

Hence it
can be designed as singly reinforced beam section. __Step 7: Area of Steel __

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

459x10^{6}
= 087x415xAstx665(1-(415 Ast)/(20x350x665)) Ast = 369.38 mm^{2}

Use 20mm
dia bars

No of
bars = Ast/ast = 369.39/314.15 = 1.45 say 2nos Provide 2#20mm dia as tension
reinforcement.

4. Design a
one way slab for a clear span 4m simply supported on 230mm thick wall.
Subjected to a live load of 4kN/m^{2} and floor finish of 1kN/m^{2}.use
M20 concrete and F415 steel.

__Step 1: Type of Slab.__

ly/lx =
4/1 = 4>2.it has to be designed as one way slab. __Step 2:Effective depth
calculation.__

d =
span/(basic value x modification factor) = 4000/(20x0.95) = 270mm D = 270 + 20
+ 10/2 = 295mm

__Step 3: Effective Span.__

Le = clear span + effective depth
= 4000 + 270 = 4.27m (or)

Le =c/c
distance b/w supports = 4000 + 2(230/2) = 4.23m Adopt effective span = 4.23m
least value.

__Step 4:
load calculation __Live load = 4kN/m^{2}

Dead load = 1x1x0.27x25 =
6.75kN/m^{2}

Floor Finish = 1kN/m^{2}

Total load = 11.75kN/m^{2}

Factored load = 11.75 x
1.5 = 17.625kN/m^{2}

__Step 5: Moment
calculation.__

M = wl^{2}/8
= (17.625x4.23^{2})/8 = 60.26kNm __Step 6: Check for effective depth.__

M = Qbd^{2}

d^{2}
= M/Qb = 60.26/2.76x1 = 149.39mm say 150mm. For design consideration adopt d =
150mm.

__Step 7: Area of Steel.__

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

60.26x10^{6}
= 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 300mm^{2}

Use 10mm dia bars

Spacing
,S = ast/Astx1000 = (78.53/300)1000 = 261mm say 260mmc/c Provide 10mm dia
@260mm c/c.

5. Deign a
rectangular beam of cross section 230 x 600 mm and of effective span 6m.imposed
load on the beam is 40 kN/m. Use M20 concrete and Fe415 steel.

__Step 1:
Size of the beam. __

b = 230mm & D = 600mm d =
600-25-20/2 =565mm __Step 4: load calculation __Live load = 40kN/m^{2}

Dead load
= 1x.23x.565x25 = 3.245kN/m^{2}

Total
load = 43.24kN/m^{2}

Factored
load = 43.24 x 1.5 = 64.86kN/m^{2}

__Step 2:
Moment Calculation.__

M = wl^{2}/8
= (64.86x6^{2})/8 = 291.9kNm

__Step 3:
Check for effective depth.__

M_{bal}
= Qbd^{2} = 2.97x230x565^{2} = 218kNm < M

Hence it
can be designed as Doubly reinforced beam section.

__Step 7:
Area of Steel__

Ast = Ast_{1}
+ Ast_{2}

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

218x10^{6} = 087x415xAstx565 (1-(415 Ast)/
(20x230x565)) Ast = 1365mm^{2}

Use 20mm
dia bars, ast = ?/4 (20^{2}) = 314.15mm^{2}

No. of
bars = Ast/ast = 1365/314.15 = 4.47 say 5nos.

Ast_{2}
= (M-M_{bal})/(0.87fy(d-d^{1})) = (291x10^{6}-218x10^{6})/(361x(565-20))
=371.03mm^{2}

Use 20mm
dia bars, ast = ?/4 (20^{2}) = 314.15mm^{2}

No. of
bars = Ast/ast = 371.03/314.15 = 1.8 say 2nos.

__Step 5:
Area of Compression steel:__

Asc = (M-M_{bal}) / (fsc.(d-d^{1})) = (291x10^{6}-218x10^{6})
/ (351.8x(470-30))= 1580.65 mm^{2}

Use 20mm
dia bars, ast = ?/4 (20^{2}) = 314.15mm^{2}

No. of
bars = Ast/ast = 1580.65/314.15 = 5.5 say 6nos.

Provide 6#20mm dia bars as compression
reinforcement.

A hall has clear dimensions 3 m x 9m with wall thickness 230
mm the live load on the slab is 3kN/m^{2} and a finishing load of 1kN/m^{2}
may be assumed. Using M20 concrete and Fe415 steel, design the slab.

__Step 1: Type of Slab.__

ly/lx =
9/3 = 3>2.it has to be designed as one way slab. __Step 2:Effective depth
calculation.__

d =
span/(basic value x modification factor) = 3000/(20x0.95) = 270mm D = 270 + 20
+ 10/2 = 295mm

__Step 3: Effective Span.__

Le =
clear span + effective depth = 3000 + 270 = 3.27m (or) Le =c/c distance b/w
supports = 3000 + 2(230/2) = 3.23m Adopt effective span = 3.23m least value.

__Step 4:
load calculation __Live load = 4kN/m^{2}

Dead load = 1x1x0.27x25 =
6.75kN/m^{2}

Floor Finish = 1kN/m^{2}

Total load = 11.75kN/m^{2}

Factored load = 11.75 x 1.5 =
17.625kN/m^{2}

__Step 5: Moment calculation.__

M = wl^{2}/8 =
(17.625x3.23^{2})/8 = 60.26kNm

__Step 6:
Check for effective depth. __M = Qbd^{2}

d^{2}
= M/Qb = 60.26/2.76x1 = 149.39mm say 150mm. For design consideration adopt d =
150mm.

__Step 7: Area of Steel.__

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

60.26x10^{6}
= 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 300mm^{2}

Use 10mm dia bars

Spacing ,S = ast/Astx1000
= (78.53/300)1000 = 261mm say 260mmc/c

Provide 10mm dia @260mm
c/c.

7. Design a
two way slab panel for the following data. Size = 7mx5m

Width of
Supports = 230 mm Edge condition = interior Live load = 4kN/m^{2}

Floor
finish = 1kN/m^{2}

Consider
M 20 grade concrete and Fe 415 grade steel. (MAY JUNE 2009) __Step 1: Type of
Slab. __

ly/lx =
7/5 = 1.4>2.it has to be designed as two way slab. __Step 2:Effective depth
calculation. __

For
Economic consideration adopt shorter span to design the slab.

d =
span/(basic value x modification factor) = 5000/(20x0.95) = 270mm D = 270 + 20
+ 10/2 = 295mm

__Step 3:
Effective Span. For shorter span: __

Le =
clear span + effective depth = 5000 + 270 = 5.27m (or) Le =c/c distance b/w
supports = 5000 + 2(230/2) = 5.23m

Adopt
effective span = 5.23m least value. __For__ __longer span:__

Le =
clear span + effective depth = 7000 + 270 = 7.27m (or) Le =c/c distance b/w
supports = 7000 + 2(230/2) = 7.23m Adopt effective span = 7.23m least value.

__Step 4: load calculation__

Live load = 4kN/m^{2}

Dead load = 1x1x0.27x25 =
6.75kN/m^{2}

Floor
Finish = 1kN/m^{2}

Total
load = 11.75kN/m^{2}

Factored
load = 11.75 x 1.5 = 17.625kN/m^{2}

__Step 5:
Moment calculation.__

Mx = ?x . w .
lx = 0.103x17.625x5.23 = 9.49kNm My =

?y . w . lx = 0.048 x17.625x5.23 = 4.425kNm
__Step
6: Check for effective depth.__

M = Qbd^{2}

d^{2} = M/Qb = 9.49/2.76x1 = 149.39mm say 150mm. For
design consideration adopt d = 150mm.

__Step 7: Area of Steel. For longer span:__

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

4.425x10^{6} = 087x415xAstx150(1-(415
Ast)/(20x1000x150)) Ast = 180mm^{2}

Use 10mm
dia bars

Spacing ,S = ast/Astx1000 = (78.53/300)1000 = 261mm say
260mmc/c Provide 10mm dia @260mm c/c.

__For
shorter span:__

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

9.49x10^{6} = 087x415xAstx150(1-(415 Ast)/(20x1000x150))
Ast = 200mm^{2}

Use 10mm
dia bars

Spacing ,S = ast/Astx1000 =
(78.53/300)1000 = 281mm say 300mmc/c Provide 10mm dia @300mm c/c.

Tags : Civil - Design of Reinforced Concrete Elements - Limit State Design For Flexure

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