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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/mm^{2} and 140 N/mm^{2}.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.l^{2})/8 = (6x6^{2})/8 = 27kNm

M = Qbd^{2}

d^{2} = M/Qb = (27x10^{6})/ (0.84x2/3xd)

d = 245mm.

__Step 3: Balanced Moment.__

M_{bal} = Qbd^{2} = 0.84x245x365^{2} = 27.41kNm. > M. it can be designed as
singly

reinforced section.

__Step 4: Area of steel.__

Ast = M_{bal} / (?st.j.d) 616.72mm^{2}

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

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/mm^{2} and 140 N/mm^{2}.

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 = Qbd^{2}

D = 500mm, b = 240mm

d = 500-30mm = 470mm

.__Step 3: Balanced Moment.__

M_{bal} = Qbd^{2} = 0.84x240x470^{2} =
44.53kNm. < M. it can be designed as doubly reinforced section.

__Step 4: Area of Tension steel.__

Ast = Ast_{1} + Ast_{2}

Ast_{1 }= M_{bal} / (?st.j.d) (44.53x10^{6})/(140x0.87x470)
= 777.87mm^{2}

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

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

Ast_{2 }= (M-M_{bal}) / (?st.(d-d^{1})) = (80x10^{6}-44.53x10^{6})/(140x(470-30)) = 575.8mm^{2}

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

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

__Step 5: Area of Compression steel:__

Asc = (M-M_{bal}) / (?sc.(d-d^{1})) =
(80x10^{6}-44.53x10^{6})/(51.8x(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.

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/mm^{2} and 140 N/mm^{2}.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 = Qbd^{2}

d^{2} = M/Qb =
(40x10^{6})/ (0.84x1/2xd)

d = 456.2 say 460 mm.

b = ½
d = 0.5x460 = 230mm

__Step 3: Balanced Moment.__

M_{bal} = Qbd^{2} =
0.84x230x460^{2} = 40.88kNm. > M. it can be
designed as singly reinforced section.

__Step 4: Area of steel.__

Ast = M_{bal} / (?st.j.d)
(40.88x10^{6})/(140x0.87x460) =
729.64mm^{2}

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

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/mm^{2} and
5N/mm^{2}.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 xa^{2}/2 =
m.Ast.(d- xa)

160. xa^{2}/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.l^{2})/8

W = (8 x 15.74)/5^{2} = 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/m^{2}.the slab
carries 100mm thick lime concrete (density 19kN/m^{3}).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/m^{2}

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

Dead load
= 1x1x0.1x19 = 1.9kN/m^{2}

Floor Finish = 1kN/m^{2}

Total load = 12.15kN/m^{2}

Factored load = 12.15 x
1.5 = 18.225kN/m^{2}

__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 = 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.

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 xa^{2}/2 = m.Ast.(d- xa)

300. xa^{2}/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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