ce2306 - design of rc elements

SUDHARSAN ENGINEERING COLLEGE

DEPARTMENT OF CIVIL ENGINEERING

SUBJ. CODDE AND NAME: CE 2306 – DESIGN OF REINFORCED CONCRETE

ELEMENTS

FACULTY NAME: S.ARUNKUMAR CLASS & SEC: III yr CIVIL

Academic year: 2013-’14 Semester: V

UNIT – 1

Part A

1. What are the assumptions made in the working stress method? (NOV-DEC 2012)

a) At any cross-section, plane sections before bending remain plain after bending.

b) All tensile stresses are taken up by reinforcement and none by concrete, except

as otherwise specifically permitted.

c) The stress-strain relationship of steel and concrete, under working loads, is a

straight line.

d) The modular ratio m has the value -280/3σbc.

2. Difference between Elastic method and limit state method. (NOV-DEC 2010)

Advantages of limit state method over the other methods

a. In the limit state method of analysis, the principles of both elastic as well as

plastic theories used and hence suitable for concrete structures.

b. The structure designed by limit state method is safe and serviceable under

design loads and at the same time it is ensured that the structure does not collapse

even under the worst possible loading conditions.

c. The process of stress redistribution, moment redistribution etc., are considered

in the analysis and more realistic factor of safety values are used in the design.

Hence the design by limit state method is found to be more economical.

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d. The overall sizes of flexural members (depth requirements) arrived by limit

state method are less and hence they provide better appearance to the structure

e. Because of the modified assumptions regarding the maximum compressive

strains in concrete and steel, the design of compressive reinforcement for double

reinforced beams and eccentrically loaded columns by limit state method gives

realistic valued which is not so in other methods.

3. Draw stress-strain curve for concrete in working stress design and mention the

salient points. (NOV-DEC 2010)

4. Define characteristic strength in limit state method. (NOV-DEC 2009) (APRIL

MAY 2012)

The term ‘characteristic strength’ means that value of the strength of the

material below which not more than 5 percent of the test results are expected to

fall.

5. What is meant by balanced section? (NOV-DEC 2009) (NOV-DEC 2012)

When the maximum stress in steel and concrete simultaneously reach their

allowable values, the section is said to be balanced section.in this section the

actual neutral axis depth is equal to the critical neutral axis.

6. Define : Limit state". (APRIL MAY 2012)

The acceptable limit for the safety and serviceability requirements before

failure occurs is called a ‘limit state’. The aim of design is td achieve acceptable

probabilities that the structure will not become unfit for the use for which it is

intended, that is, that it will not reach a limit state.

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7. What are the expressions recommended by the IS 456-2000 for Modulus of

Elasticity and Flexural Strength? (MAY JUNE 2009)

lexural strength f 0.7. fck /mm2

Where ‘fck’ is the characteristic cube compressive strength of concrete in N/mm2.

Where, E, is the short term static modulus of elasticity in N/mm2

8. Write the formula for the neutral axis depth factor 'K in working stress design.

(MAY JUNE 2009)

eutral axis depth factor ‘K’ σbc.m/(σbc.m + σst)

Where σbc permissible stress in concrete. σbc permissible stress in steel.

M = modular ratio.

Part B

1. Explain the limit state philosophy as detailed in the current IS code. (NOV-DEC

2012)

The Answer is in Page No.67 of IS 456:2000.

In the method of design based on limit state concept, the structure shall be

designed to withstand safely all loads liable to act on it throughout its life; it shall

also satisfy the serviceability requirements, such as limitations on deflection and

cracking. The acceptable limit for the safety and serviceability requirements

before failure occurs is called a ‘limit state’. The aim of design is td achieve

acceptable probabilities that the structure will not become unfit for the use for

which it is intended, that is, that it will not reach a limit state. 351.1 All relevant

limit states shall be considered in design to ensure an adequate degree of safety

and serviceability. In general, the structure shall be designed on the basis of the

most critical limit state and shall be checked for other limit states.

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35.1.2 For ensuring the above objective, the design should be based on

characteristic values for material strengths and applied loads, which take into

account the variations in the material strengths and in the loads to be supported.

The characteristic values should be based on statistical data if available; where

such data are not available they should be based on experience. The ‘design

values’ are derived from the characteristic values through the use of partial safety

factors, one for material strengths and the other for loads. In the absence of special

considerations these factors should have the values given in 36 according to the

material, the type of loading and the limit state being considered.

2. 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/mm

2.take

m=18. (NOV-DEC 2012)

Step 1: Design constants.

Modular ratio, m =18.

A Coefficient n σbc.m/(σbc.m + σst) 0.39

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

Moment of resistance Coefficient Q σbc/2. n. j 0.84

Step 2: Moment on the beam.

M = (w.l2)/8 = (6x6

2)/8 = 27kNm

M = Qbd2

d2

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

d = 245mm.

Step 3: Balanced Moment.

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Mbal = Qbd2 = 0.84x245x365

2 = 27.41kNm. > M. it can be designed as singly

reinforced section.

Step 4: Area of steel.

Ast = Mbal / (σst.j.d) 616.72mm2

Use 20mm dia bars ast π/4 (202) = 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.

3. 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. (NOV-DEC 2010)

Assume the permissible stressed in the concrete and steel are not to exceed

5N/mm2 and 140 N/mm

2.







M = 80kNm

M = Qbd2

D = 500mm, b = 240mm

d = 500-30mm = 470mm

.Step 3: Balanced Moment.

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Mbal = Qbd2 = 0.84x240x470

2 = 44.53kNm. < M. it can be designed as doubly

reinforced section.

Step 4: Area of Tension steel.

Ast = Ast1 + Ast2

Ast1 = Mbal / (σst.j.d) (44.53x106)/(140x0.87x470) = 777.87mm

2


2


Ast2 = (M-Mbal) / (σst.(d-d1)) = (80x10

6-44.53x10

6)/(140x(470-30)) = 575.8mm

2


2


Step 5: Area of Compression steel:

Asc = (M-Mbal) / (σsc.(d-d1)) = (80x10

6-44.53x10

6)/(51.8x(470-30))=1580.65 mm

2


2


Provide 6#20mm dia bars as compression reinforcement.

4. Design a beam subjected to a bending moment of 40kNm by working stress

design. Adopt width of beam equal to half the effective depth. (NOV-DEC 2010)

Assume the permissible stressed in the concrete and steel are not to exceed

5N/mm2 and 140 N/mm

2.take m=18.



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M = 40kNm

M = Qbd2

d2

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

d = 456.2 say 460 mm.

b = ½ d = 0.5x460 = 230mm

Step 3: Balanced Moment.

Mbal = Qbd2 = 0.84x230x460

2 = 40.88kNm. > M. it can be designed as singly

reinforced section.

Step 4: Area of steel.

Ast = Mbal / (σst.j.d) (40.88x106)/(140x0.87x460) = 729.64mm

2


2


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

5. 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. (NOV-DEC 2009)

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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 σbc.d/(σst/.m + σcbc) 117.39mm < Xa 159.42mm

it is Over reinforced Section.

Step 3: Moment of Resistance

M (b. xa/2 .σcbc )(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

6. 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

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Adopt effective span = 3.23m least value.

For longer span:


Le =c/c distance b/w supports = 6000 + 2(230/2) = 6.23m


Step 4: load calculation

Live load = 2.5kN/m2

Dead load = 1x1x0.27x25 = 6.75kN/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 αx . w . lx 0.103x18.225x3.23 = 9.49kNm

My αy . 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:


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

Ast = 200mm2

Use 10mm dia bars



7. Differentiate between working stress method and limit state method. (APRIL

MAY 2012)

In the limit state method of analysis, the principles of both elastic as well as

plastic theories used and hence suitable for concrete structures.

The structure designed by limit state method is safe and serviceable under

design loads and at the same time it is ensured that the structure does not

collapse even under the worst possible loading conditions.

The process of stress redistribution, moment redistribution etc., are considered

in the analysis and more realistic factor of safety values are used in the design.

Hence the design by limit state method is found to be more economical.

The overall sizes of flexural members (depth requirements) arrived by limit

state method are less and hence they provide better appearance to the structure

Because of the modified assumptions regarding the maximum compressive

strains in concrete and steel, the design of compressive reinforcement for double

reinforced beams and eccentrically loaded columns by limit state method gives

realistic valued which is not so in other methods.

8. Explain the following terms :

a. characteristic strength and characteristic loads.


b. partial safety factors.


c. Balanced section and under reinforced section. (APRIL MAY 2012)

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When the maximum stress in steel and concrete simultaneously reach their

allowable values, the section is said to be balanced section.in this section the

actual neutral axis depth is equal to the critical neutral axis.

When the percentage of steel in the section is less than that required for a

balanced section. In this section the actual neutral axis depth is equal to the critical

neutral axis.

9. Derive the expressions for the depth of Neutral axis and Moment of resistance of a

Rectangular Singly reinforced balanced beam section under flexure and obtain the

design constants K, j and Q for M 20 grade concrete and Fe 415 grade steel. Use

working stress method. (MAY JUNE 2009)

10. 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. (MAY JUNE 2009)

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 σbc.d/(σst/.m + σcbc) = 194.66mm > Xa = 117.81mm

it is Under reinforced Section.

Step 3: Moment of Resistance

For steel:

M = (Ast.σst )(d- xa/3) = (1963.5x230)(550-117.81/3) = 230.64kNm

For concrete:

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

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SUDHARSAN ENGINEERING COLLEGE

DEPARTMENT OF CIVIL ENGINEERING

SUBJ. CODDE AND NAME: CE 2306 – DESIGN OF REINFORCED CONCRETE

ELEMENTS

FACULTY NAME: S.ARUNKUMAR CLASS & SEC: III yr CIVIL

Academic year: 2013-’14 Semester: V

UNIT – 2

Part A

1. Explain the check for deflection control in the design of slabs? (NOV-DEC 2012)

The deflection of a structure or part thereof shall not adversely affect the

appearance or efficiency of the structure or finishes or partitions. The deflection

shall generally be limited to the following:

a) The final deflection due to all loads including the effects of temperature, creep

and shrinkage and measured from the as-cast level of the , supports of floors,

roofs and all other horizontal members, should not normally exceed span/250.

b) The deflection including the effects of temperature, creep and shrinkage

occurring after erection of partitions and the application of finishes should not

normally exceed span/350 or 20 mm whichever is less.

2. When do you do for doubly reinforced beams? (NOV-DEC 2012) (NOV-DEC

2010) (APRIL MAY 2012)

The section reinforced in both tension and compression zone is known as

doubly reinforced section. The doubly reinforced beams are adopt when the

balanced moment is smaller than the Actual moment.

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3. What type of slabs are usually used in practice, under reinforced or over

reinforced? (NOV-DEC 2009)

The depth of slab chosen from deflection requirements will be usually

greater than the depth required for balanced design. Hence the area of steel

required will be less than the balanced amount. So, the slab is designed as under

reinforced section.

4. Why is necessary to provide transverse reinforcement in one way slab? (APRIL

MAY 2012)

Since the one way slab bends in one direction and also in shorter direction,

so it is necessary to provide transvers reinforcement in one way slabs. These slabs

adopted when availability of two supports in one direction.

5. Distinguish between under reinforced and over reinforced sections. (MAY JUNE

2009)

A beam reaches its permissible stress in steel under the working moment

before concrete reaches its stress is called as Under reinforced section.

A beam reaches its permissible stress in concrete under the working

moment before steel reaches its stress is called as Over reinforced section.

6. Sketch the edge and middle strips of a two way slab. (MAY JUNE 2009)

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Part B

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. (NOV-DEC 2012)


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



D = 270 + 20 + 10/2 = 295mm






Live load = 4kN/m2






M = wl2/8 = (17.625x5.23

2)/8 = 60.26kNm


M = Qbd2

d2 = M/Qb = 60.26/2.76x1 = 149.39mm say 150mm.




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

Ast = 300mm2

Use 10mm dia bars



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. (NOV-DEC 2010)

(NOV-DEC 2012)

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 = (43.85x5

2)/8 = 137.08kNm


M = Qbd2

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

d = 380mm > 500mm

Hence it is safe.


Mbal = Qbd2 = 2.97x340x500

2 = 252.06kNm > M

Hence it can be designed as singly reinforced beam section.

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Step 7: Area of Steel


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. (NOV-DEC 2009)


b = 350mm & D = 700mm

d = 700-25-20/2 =665mm


M = 300kNm


Mbal = Qbd2 = 2.97x350x665

2 = 459kNm > M

Hence it can be designed as singly reinforced beam section.



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/m

2.use M20 concrete

and F415 steel. (NOV-DEC 2009)

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D = 270 + 20 + 10/2 = 295mm






Live load = 4kN/m2






M = wl2/8 = (17.625x4.23

2)/8 = 60.26kNm


M = Qbd2

d2 = M/Qb = 60.26/2.76x1 = 149.39mm say 150mm.




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

Ast = 300mm2

Use 10mm dia bars



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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.

(APRIL MAY 2012)


b = 230mm & D = 600mm

d = 600-25-20/2 =565mm


Live load = 40kN/m2

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




M = wl2/8 = (64.86x6

2)/8 = 291.9kNm


Mbal = Qbd2 = 2.97x230x565

2 = 218kNm < M

Hence it can be designed as Doubly reinforced beam section.


Ast = Ast1 + Ast2


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

Ast = 1365mm2

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

2

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

Ast2 = (M-Mbal)/(0.87fy(d-d1)) = (291x10

6-218x10

6)/(361x(565-20)) =371.03mm

2


2


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Step 5: Area of Compression steel:

Asc = (M-Mbal) / (fsc.(d-d1)) = (291x10

6-218x10

6) / (351.8x(470-30))= 1580.65

mm2


2


Provide 6#20mm dia bars as compression reinforcement.

6. 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/m

2 may be assumed. Using M20

concrete and Fe415 steel, design the slab. (APRIL MAY 2012)





D = 270 + 20 + 10/2 = 295mm






Live load = 4kN/m2






M = wl2/8 = (17.625x3.23

2)/8 = 60.26kNm


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M = Qbd2

d2 = M/Qb = 60.26/2.76x1 = 149.39mm say 150mm.




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

Ast = 300mm2

Use 10mm dia bars



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)


ly/lx = 7/5 = 1.4>2.it has to be designed as two way slab.


For Economic consideration adopt shorter span to design the slab.


D = 270 + 20 + 10/2 = 295mm


For shorter span:



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For longer span:





Live load = 4kN/m2






Mx = αx . w . lx = 0.103x17.625x5.23 = 9.49kNm

My = αy . w . lx = 0.048 x17.625x5.23 = 4.425kNm


M = Qbd2

d2 = M/Qb = 9.49/2.76x1 = 149.39mm say 150mm.



For longer span:


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

Ast = 180mm2

Use 10mm dia bars



For shorter span:


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

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Ast = 200mm2

Use 10mm dia bars



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ce2306 - design of rc elements

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