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Civil - Design of Reinforced Concrete Elements - Methods Of Design Of Concrete Structures

Design Problems: Civil - Methods Of Design Of Concrete Structures

   Posted On :  16.07.2016 08:00 pm

Civil - Design of Reinforced Concrete Elements - Methods Of Design Of Concrete Structures


 

Design Problems:

 

1. Design a R.C beam to carry a load of 6 kN/m inclusive of its own weight on an effect span of 6m keep the breath to be 2/3 rd of the effective depth .the permissible stressed in the concrete and steel are not to exceed 5N/mm2 and 140 N/mm2.take

 

m=18. Step 1: Design constants.

    Modular ratio, m =18.

 

       A Coefficient n  ?bc.m/(?bc.m + ?st)   0.39

 

    Lever arm Coefficient, j=1-(n/3) = 0.87

 

    Moment of resistance Coefficient Q   ?bc/2. n. j   0.84

 

Step 2: Moment on the beam.

 

M = (w.l2)/8 = (6x62)/8 = 27kNm

 

M = Qbd2

 

d2 = M/Qb = (27x106)/ (0.84x2/3xd)

 

d = 245mm.

 

Step 3: Balanced Moment.

 

Mbal    = Qbd2  = 0.84x245x3652  = 27.41kNm. > M. it can be designed as singly

reinforced section.

 

Step 4: Area of steel.

 

Ast = Mbal / (?st.j.d)   616.72mm2

 

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

 

No. of bars = Ast/ast = 616.72/314.15 = 1.96 say 2nos.

 

Provide 2#20mm dia bars at the tension side.

 

2.      Design a doubly reinforced beam of section 240X500mm to carry a bending moment of 80kNm.Assume clear cover at top a bottom as 30mm and take m=18.adopt working stress method. Assume the permissible stressed in the concrete and steel are not to exceed 5N/mm2 and 140 N/mm2.

 

Step 1: Design constants.

    Modular ratio, m =18.

 

       A Coefficient n  ?bc.m/(?bc.m + ?st)   0.39

 

    Lever arm Coefficient, j=1-(n/3) = 0.87

 

    Moment of resistance Coefficient Q   ?bc/2. n. j   0.84

 

Step 2: Moment on the beam.

 

M = 80kNm

 

M = Qbd2

 

D = 500mm, b = 240mm

 

d = 500-30mm = 470mm

 

.Step 3: Balanced Moment.

Mbal    = Qbd2  = 0.84x240x4702 = 44.53kNm. < M. it can be designed as doubly reinforced section.

 

Step 4: Area of Tension steel.

 

Ast = Ast1 + Ast2

Ast1 = Mbal / (?st.j.d) (44.53x106)/(140x0.87x470) = 777.87mm2

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

No. of bars = Ast/ast = 777.87/314.15 = 2.47 say 3nos.

Ast2 = (M-Mbal) / (?st.(d-d1)) = (80x106-44.53x106)/(140x(470-30)) = 575.8mm2

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

 

No. of bars = Ast/ast = 575.8/314.15 = 1.8 say 2nos.

 

Step 5: Area of Compression steel:

 

Asc = (M-Mbal) / (?sc.(d-d1)) = (80x106-44.53x106)/(51.8x(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.

 

3.      Design a beam subjected to a bending moment of 40kNm by working stress design. Adopt width of beam equal to half the effective depth.

 

 

Assume the permissible stressed in the concrete and steel are not to exceed 5N/mm2 and 140 N/mm2.take m=18.

 

Step 1: Design constants.

 

    Modular ratio, m =18.

 

          A Coefficient n  ?bc.m/(?bc.m + ?st)   0.39

Lever arm Coefficient, j=1-(n/3) = 0.87

Moment of resistance Coefficient Q ?bc/2. n. j   0.84

Step 2: Moment on the beam.

 

M = 40kNm

 

M = Qbd2

 

d2 = M/Qb = (40x106)/ (0.84x1/2xd)

 

d = 456.2 say 460 mm.

 

b =   d = 0.5x460 = 230mm

 

Step 3: Balanced Moment.

 

Mbal = Qbd2 = 0.84x230x4602 = 40.88kNm. > M. it can be designed as singly reinforced section.

 

Step 4: Area of steel.

 

Ast = Mbal / (?st.j.d)   (40.88x106)/(140x0.87x460) = 729.64mm2

 

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

 

No. of bars = Ast/ast = 729.64/314.15 = 2.96 say 3nos.

 

Provide 3#20mm dia bars at the tension side.

 

4.Determine the moment of resistance of a singly reinforced beam 160X300mm effective section, if the stress in steel and concrete are not to exceed 140N/mm2 and 5N/mm2.effectve span of the beam is 5m and the beam carries 4 nos of 16mm dia bars. Take m=18.find also the minimum load the bam can carry. Use WSD method.

Step 1: Actual NA.

b xa2/2 = m.Ast.(d- xa)

160. xa2/2 = 18 X 804.24(300 xa)

Xa = 159.42mm

Step 2: Critical NA.

xc ?bc.d/(?st/.m + ?cbc)   117.39mm <  Xa   159.42mm

it is Over reinforced Section.

Step 3: Moment of Resistance

M (b. xa/2 .?cbc )(d- xa/3) = (160x159.42/2x5)(300-159.42/3) = 15.74kNm

Step 4: Safe load.

 

M = (w.l2)/8

 

W = (8 x 15.74)/52 = 5.03 kN/m

 

5. Design an interior panel of RC slab 3mX6m size, supported by wall of 300mm thick. Live load on the slab is 2.5kN/m2.the slab carries 100mm thick lime concrete (density 19kN/m3).Use M15 concrete and Fe 415 steel. (NOV-DEC 2009)

 

 

 

Step 1: Type of Slab.

 

ly/lx = 6/3 = 2 = 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) = 3000/(20x0.95) = 270mm D = 270 + 20 + 10/2 = 295mm

 

Step 3: Effective Span. For shorter 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. For longer span:

 

Le = clear span + effective depth = 6000 + 270 = 6.27m (or) Le =c/c distance b/w supports = 6000 + 2(230/2) = 6.23m Adopt effective span = 6.23m least value.

 

Step 4: load calculation Live load = 2.5kN/m2

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

 

Dead load = 1x1x0.1x19 = 1.9kN/m2

 

Floor Finish = 1kN/m2

 

Total load = 12.15kN/m2

 

Factored load = 12.15 x 1.5 = 18.225kN/m2

 

Step 5: Moment calculation.

 

Mx    ?x . w . lx   0.103x18.225x3.23 = 9.49kNm

 

My    ?y . w . lx   0.048 x18.225x3.23 = 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.

 

 

6. A reinforced concrete rectangular section 300 mm wide and 600 mm overall depth is reinforced with 4 bars of 25 mm diameter at an effective cover of 50 mm on the tension side. The beam is designed with M 20 grade concrete and Fe 415 grade steel. Determine the allowable bending moment and the stresses developed in steel and concrete under this moment. Use working stress method.

 

 

Step 1: Actual NA.

 

b xa2/2 = m.Ast.(d- xa)

 

300. xa2/2 = 18 X 1963.50(550 xa) Xa = 117.81mm

 

Step 2: Critical NA.

 

xc ?bc.d/(?st/.m + ?cbc) = 194.66mm > Xa = 117.81mm it is Under reinforced Section.

 

Step 3: Moment of Resistance For steel:

 

M = (Ast.?st )(d- xa/3) = (1963.5x230)(550-117.81/3) = 230.64kNm For concrete:

M        (b. xa/2 .?cbc )(d- xa/3) = (300x117.81/2x7)(550-117.81/3) = 63.17kNm



Tags : Civil - Design of Reinforced Concrete Elements - Methods Of Design Of Concrete Structures
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