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

Design Problems: Civil - Methods Of Design Of Concrete Structures

   Posted On :  16.07.2016 10:02 pm

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/m2 and a surface

 

finish of 1kN/mm2.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/m2

Dead load = 1x1x0.27x25 = 6.75kN/m2

 

Floor Finish = 1kN/m2

 

Total load = 11.75kN/m2

 

Factored load = 11.75 x 1.5 = 17.625kN/m2

 

Step 5: Moment calculation.

 

M = wl2/8 = (17.625x5.232)/8 = 60.26kNm Step 6: Check for effective depth.

M = Qbd2

 

d2 = 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)) 60.26x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 300mm2

 

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 = wl2/8 = (43.85x52)/8 = 137.08kNm Step 5: Check for effective depth.

M = Qbd2

 

d2 = M/Qb = 137.08/2.76x.340 = 382.2mm say 380mm. d = 380mm > 500mm

 

Hence it is safe.

 

Step 5: Check for effective depth.

 

Mbal = Qbd2 = 2.97x340x5002 = 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))

 

137.08x106 = 087x415xAstx500(1-(415 Ast)/(20x340x500)) Ast = 846.15mm2

 

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.

 

Mbal = Qbd2 = 2.97x350x6652 = 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))

 

459x106 = 087x415xAstx665(1-(415 Ast)/(20x350x665)) Ast = 369.38 mm2

 

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/m2 and floor finish of 1kN/m2.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/m2

Dead load = 1x1x0.27x25 = 6.75kN/m2

 

Floor Finish = 1kN/m2

 

Total load = 11.75kN/m2

 

Factored load = 11.75 x 1.5 = 17.625kN/m2

 

Step 5: Moment calculation.

 

M = wl2/8 = (17.625x4.232)/8 = 60.26kNm Step 6: Check for effective depth.

M = Qbd2

 

d2 = 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))

 

60.26x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 300mm2

 

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/m2

Dead load = 1x.23x.565x25 = 3.245kN/m2

 

Total load = 43.24kN/m2

 

Factored load = 43.24 x 1.5 = 64.86kN/m2

 

Step 2: Moment Calculation.

 

M = wl2/8 = (64.86x62)/8 = 291.9kNm

 

Step 3: Check for effective depth.

 

Mbal = Qbd2 = 2.97x230x5652 = 218kNm < M

 

Hence it can be designed as Doubly reinforced beam section.

 

Step 7: Area of Steel

 

Ast = Ast1 + Ast2

 

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

 

218x106 = 087x415xAstx565 (1-(415 Ast)/ (20x230x565)) Ast = 1365mm2

 

Use 20mm dia bars, ast = ?/4 (202) = 314.15mm2

 

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

 

Ast2 = (M-Mbal)/(0.87fy(d-d1)) = (291x106-218x106)/(361x(565-20)) =371.03mm2

 

Use 20mm dia bars, ast = ?/4 (202) = 314.15mm2

 

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

 

Step 5: Area of Compression steel:

 

Asc = (M-Mbal) / (fsc.(d-d1)) = (291x106-218x106) / (351.8x(470-30))= 1580.65 mm2

 

Use 20mm dia bars, ast = ?/4 (202) = 314.15mm2

 

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/m2 and a finishing load of 1kN/m2 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/m2

Dead load = 1x1x0.27x25 = 6.75kN/m2

 

Floor Finish = 1kN/m2

 

Total load = 11.75kN/m2

 

Factored load = 11.75 x 1.5 = 17.625kN/m2

 

Step 5: Moment calculation.

 

M = wl2/8 = (17.625x3.232)/8 = 60.26kNm

 

Step 6: Check for effective depth. M = Qbd2

 

d2 = 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))

 

60.26x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 300mm2

 

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/m2

Floor finish = 1kN/m2

 

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/m2

 

Dead load = 1x1x0.27x25 = 6.75kN/m2

Floor Finish = 1kN/m2

 

Total load = 11.75kN/m2

 

Factored load = 11.75 x 1.5 = 17.625kN/m2

 

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 = Qbd2

 

d2 = 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))

 

4.425x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 180mm2

 

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))

 

9.49x106 = 087x415xAstx150(1-(415 Ast)/(20x1000x150)) Ast = 200mm2

 

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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