Design&Drng Compiled G.S.suresh

7/28/2019 Design&Drng Compiled G.S.suresh

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e-Notes by Dr. G.S.Suresh, Professor, Civil Engineering Department,NIE, Mysore

1. LAYOUT DRAWING(For classes held on 26 and 27 th Feb 07)

1.1 Introduction to Structural drawing:

Drawing is the language of engineers, which conveys the idea of the engineer about

the shape, structural arrangement to builder. Structural drawings guide builders in theconstruction of apartment blocks, industrial buildings, highways, irrigation structures,

bridges and other important structures.

The guidelines given in IS 962 (Code of practice for architectural and buildings)

and SP 34 (Handbook on Concrete Reinforcement and Detailing) may be adoptedwhile preparing structural drawings for Reinforced Concrete Structures and its

elements.

Structural drawings are prepared in different sizes and depend on the number ofdetailed drawings to be presented. In large projects, the structural drawings are ofsame size. The preferred size of drawing sheets are given in table 1.1

Table 1.1 Drawing Sheet Size

Sl. No. Designation Trimmed Size

in mm

Un Trimmed Size

(Min) in mm

1 A0 841 x 1189 880 x 1230

2 A1 594 x 841 625 x 880

3 A2 420 x 594 450 x 625

4 A3 297 x 420 330 x 450

5 A4 210 x 297 240 x 330

6 A5 148 x 210 165 x 240

Margins and the divisions of drawings sheets into zones are given in Fig 1.1 a to f.

The title block is an important feature in a drawing and should be placed at the

bottom right-hand corner of the sheet, where it is readily seen when the prints are

folded in the prescribed manner. The size of the title block recommended by SP 34 is185 x 65 mm.

Layout of drawings is not standardized in detailing of reinforced concrete structures.However in practice, the key plan is placed in the upper left hand corner of the sheet,

with the elevations and details below and on to the right side of the plan. Schedulesand bending details are place in the upper right corner of the drawing. Fig. 1.2 gives a

broad outline of layout recommended. In large projects, the bending schedule can be

given separately and omitted in the structural drawing. Scale for drawings is selectedbased on the convenience to include all the details within workable size. Some

commonly used scales are :

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Plan:- 1:100, 1:50

Elevation:- 1:5, 1:30

Sections:- 1:50, 1:30, 1:25, 1:20,1:15,1:10

2

Fig.1.1a A0 Size Sheet


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3

Fig.1.1b A1 Size Sheet

Fig.1.1c A2 Size Sheet


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Fig.1.1d A3 Size Sheet Fig.1.1e A4 Size Sheet

Fig.1.1f A5 Size Sheet and Division of sheets


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Notes containing specifications of the concrete and steel to be used, size of chamfers and

fillets, concrete cover, live load, SBC of soil, lap lengths for different diameter of bars

etc,. Symbols and abbreviations to be adopted in the drawings are given below:

Symbols Relating to Cross-Sectional Shape and Size of Reinforcement

a) plain round bar or diameter of plain round bar;b) plain, square bar or side of plain square bar; andc) # deformed bar (including square twisted bar) or nominal size (equivalent diameter or

side) of the deformed bar .Symbols Relating to Shape of the Bar along its LengthsAlt Alternate bar

Bt Bent bar

B Bottom bar

min Minimummax Maximum

St Straight barStp Stirrup

Sp Spiral

Ct Column tieT Top bar

Symbols Relating to Position and DirectionEW Each way

@ Spacing centre-to-centreLimit of area covered by bars

Direction in which bars extend

Symbols Relating to Various Structural MembersBm orB BeamsCol Column(s)

Fg Footing(s)

GR GirdersJT Joints(s)

LL Lintel(s)

LB Lintel beam(s)Sb or S Slab(s)

WL Longitudinal wallWx Cross wallC Centre line

Graphical symbols given in Fig. 1.3 are recommended by SP34. Additional drawing

conventions for use on drawings for reinforcement as suggested in ISO 3766-1977 is

reproduced in Fig 1.4

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Fig.1.2 Typical Layout of a Drawing


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7

Fig. 1.3 Graphical symbols


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8

Fig. 1.4 Drawing conventions


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9

Fig. 1.4 Drawing conventions (Contd.)


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1.2 General Layout of Building:

After the preparation of architectural plan of the buildings, the structural planning of

the building frame is done. This involves determination of i) Positioning andorientation of columns, ii) Positioning of beams, iii) Spanning of slabs, iv) Layout of

stairs, v) Selecting proper type of footing.

Different structural members of a structure shall be marked using symbols,

abbreviations and notations. A key framing plan shall be prepared to a convenientscale and the two axes marked one side with alphabets A, B, C etc, and the other with

numbers. If the structural arrangement in all the floors is same then only one key plan

is prepared titling it as typical plan. If the arrangement varies for different floors a

separate key framing plan with grid arrangement and areas may be used for each ofthe floor. The floors shall be specified in accordance with the requirements of IS

2332-1973 (Specifications for nomenclature of floors and storeys). According to

this code BT symbol is used for Basement, MZ for Mezzanine, Floor 1, Floor 2 etcfor first, second etc floors respectively.

Columns and foundations shall be specified by grid arrangements giving reference tothe floor. For example FG Col C2 with reference to Fig 1.5 indicates Footing for

column C2, Col 2C2 indicates column C2 at floor 2.

Beams, slabs and lintels and tie beams shall be consecutively numbered from left-hand top corner as shown in Fig. 1.5. Lay out building are generally drawn i)

Drawing showing position of columns along with excavation plan for footing and ii)

Key plan showing arrangements of beams and plans and called as form workdrawing. Fig 1.6 and 1.7 shows a typical lay out drawing developed for a building.

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Fig. 1.5 Typical Arrangement for the Key Framing Plan and marking Different

Structural Members


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Problem1:

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Fig. 1.6 Typical Drawing for a Multi-Storied Building showing column positionand excavation plan.

Fig. 1.7 Typical Drawing for a Multi-Storied Building showing Slab and Beam

alignment


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Prepare a general layout showing the positions and sizes of columns and footing to a

suitable scale for an industrial building:

A clear dimension of factory floor is 11.75 m x 19.75 mSpacing of columns 4m c/c

Size of columns 250 mm x 450 mm

Span of steel truss is 12.25 m c/cAt the ends to support the gable wall additional two RCC columns of size250 mm x 450 mm are to be provided at 4m c/c measured from end columns

All the walls all-round are 250 mm thick

Height of columns = 3mSize of footing 1.4m x 1.8m

Thickness of footing 300 mm uniform

Depth of foundation 1.2m below ground level

Also show the line of steel truss on the drawing

Solution:

External dimensions of the building:

Along X-direction = 11.75 + 2 x 0.25 = 12.25 m

Along Z-direction = 19.75 + 2 x 0.25 = 20.25 m

Centre line dimensions of the building:

Along X-direction = 12.25 0.45 = 11.80 m

Along Z-direction = 20.25 -0.25 = 20.00 m

Procedure for drawing the lay out plan

Note: Use Millimeter units for linear dimensions

1. Draw the centerline of the building having 11,800mm along X-direction and20,000 mm along Z-directions

2. Mark these centre lines as grid lines A and D for lines parallel to Z-directions

& grid lines 1 and 6 for lines parallel to X-axis

3. Measure 4000mm from grid lines A and D to get grid line B and C

4. Measure 4000mm c/c along Z-axis starting from grid line 1 to get grid lines 2,

3, 4 and 5

5. Draw rectangular filled box of size 250mm x 450mm at the intersection of

gird lines along A and D to indicate the position of column along these grid

lines as shown in Fig. 1.86. Draw rectangular filled box of size 250mm x 450mm at the intersection of

grid lines 1,6 with B and C respectively as shown in Fig. 1.8

7. Draw rectangles each of size 1400 mm x 1800mm symmetrically with respect

to centre of column to indicate excavation marking for all columns. Here theshorter side of this rectangle box is parallel to the shorter side of column as

shown in Fig. 1.8

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8. Sectional elevation and plan of the footing of column is drawn just to the right

of the key plan drawn as shown in Fig. 1.8

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Fig. 1.8a Layout Drawing


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16

Fig. 1.9 Layout Drawing


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2. DETAILING OF BEAMS(For class held on 5th March 07)

2.1 Introduction:

Beams are structural elements carrying transverse external loads that cause bending

moment, shear forces and in some cases torsion across their length. Concrete is strongin compression and very weak in tension. Steel reinforcement is used to take uptensile stresses in reinforced concrete beams. Mild steel bars of round section were

used in RCC work. But with the introduction of deformed and twisted bars, the use of

mild steel bars had declined. Deformed or High yield strength deformed bars (HYSD)have ribs on the surface and this increases the bond strength at least by 40%

compared to that of mild steel bar. Fig. 2.1 shows mild steel and deformed steel bars.

To facilitate construction process, good detailing of reinforcements with proper

drawings are essential at the site of construction. These drawing generally alsoinclude a bar bending schedule. The bar bending schedule describes the length and

number, position and the shape of the bar. The detailing is normally associated with

i) Size and number (or spacing) of bars, ii) Lap and curtailment (or bending) of bars,iii) Development length of bars, iv) Clear cover to the reinforcement and v) spacer

and chair bars.

Anchorage in steel bars is normally provided in the form of bends and hooks. Twisted

steel bars or deformed steel bars are not provided with hooks. The anchorage value ofbend of bar is taken as 4 times the diameter of bar for every 450 bend subjected to

maximum of 16 times the diameter of bar. Fig.2.2 shows the standard hooks and

bends. Bars are lapped over each other for increasing the length of bars. Minimum lap

length should be equal to development length. Development length for bars indifferent concrete mix is given tables 4.2 to 4.4 of SP34.

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Fig. 2.1


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II) A rectangular beam of cross section 300 x 450 mm is supported on 4 columns

which are equally spaced at 3m c/c. The columns are of 300 mm x 300 mm in section.The reinforcement consists of 4 bars of a6 mm diameter (+ve reinforcement) at mid

span and 4 bars of 16 mm diameter at all supports (-ve reinforcement). Anchor barsconsists of a 2-16 mm diameter. Stirrups are of 8 mm diameter 2 legged vertical at 200

c/c throughout. Grade of concrete is M20 and type of steel is Fe 415. Draw longitudinal

section and important cross sections.

Solution:

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3.CANTILEVER BEAMSI) Draw to scale of 1:20 the Longitudinal section and two cross-section of a

cantilever beam projecting 3.2 from a support using following data

Clear span =3.2mOverall depth at free end = 150 mmOverall depth at fixed end = 450 mm

Width of cantilever beam = 300 mm

Main steel = 4-28 mm dia with two bars curtailed at 1.5m from

supportAnchor bars = 2 Nos. 16 mm dia

Nominal stirrups = 6mm dia at 40 mm c/c

Bearing at fixed end = 300 mmUse M20 concrete and Fe 415 steel

Solution:

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II) A cantilever beam with 3.2m length is resting over a masonry wall and supporting

a slab over it. Draw to a suitable scale Longitudinal section, two cross-sections

and sectional plan with the following data:Size of beam = 300 mm x 350 mm at free end and 300 mm x 450 mm at fixed end

and in the wall up to a length of 4.8m

Main steel: 4 nos. of 25 mm dia bars, two bars curtailed at 1.2m from free end

Hanger bars: 2 nos. 16mm.Stirrups: 6mm dia 2 legged stirrups @ 200 mm c/c the support length and @100

mm c/c from fixed end up to length of 1m @ 150mm c/c up to curtailed bars and

remaining @ 200 c/c.Use M20 concrete and Fe 415 steel

Solution:

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3. DETAILING OF SLABS(For class held on 6th March 07)

3.1 Introduction:

A structural member used for covering spaces in the form of roof or floor is called a

slab. Slab may be supported on walls or beams or columns. Slab supported directly by

columns are called flat slab. Detailing of flat slab is out of scope in this syllabus.

Slab supported on two sides and bending takes place predominantly in one directiononly is called One Way Slab. On the other hand, when slab is supported on all four

sides and bending take place in two directions are said to be Two Way Slab. The

slabs having ratio of longer length to its shorter length (Ly/Lx) greater than 2 is called

one way slab otherwise as two way slab. In one way slab main reinforcement isparallel to shorter direction and the reinforcement parallel to longer direction is called

distribution steel. In two way slab main reinforcement is provided along both

direction.

Slabs could be simply supported, continuous or cantilever. In two way slab the

corners may be held down by restraints or may be allowed to lift up. Additionaltorsion reinforcement is required at corners when it is restrained against uplifting as

shown in Fig.3.1

Thickness of the slab is decided based on span to depth ratio specified in IS456-2000.

Min reinforcement is 0.12% for HYSD bars and 0.15 % for mild steel bars. The

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Fig. 3.1


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diameter of bar generally used in slabs are: 6 mm, 8 mm, 10 mm, 12 mm and 16 mm.

The maximum diameter of bar used in slab should not exceed 1/8 of the total

thickness of slab. Maximum spacing of main bar is restricted to 3 times effectivedepth or 300 mm which ever is less. For distribution bars the maximum spacing is

specified as 5 times the effective depth or 450 mm which ever is less.

Minimum clear cover to reinforcements in slab depends on the durability criteria andthis is specified in IS 456-200. Generally 15 mm to 20 mm cover is provided for themain reinforcements. Alternate main bars can be cranked near support or could be

bent at 1800 at the edge and then extended at the top inside the slab as shown in

Fig.3.1. Curtailment and cranking of bars and is shown in Fig. 3.2

Torsion reinforcement shall be provided at any corner where the slab is simplysupported on both edges meeting at that corner and is prevented from lifting unless

the consequences of cracking are negligible. It shall consist of top and bottom

reinforcement, each with layer of bars placed parallel to the sides of the slab andextending from the edges a minimum distance of one fifth of the shorter span. The

area of reinforcement per unit width in each of these four layers shall be three-

quarters of the area required for the maximum mid-span moment per unit width in theslab. Torsion reinforcement equal to half that described above shall be provided at a

corner contained by edges over only one of which the slab is continuous. Torsion

reinforcement to be provided is shown in Fig. 3.3

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Fig. 3.2


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Problems:

1. Prepare a detailed structural drawing of one way continuous slab for a hall of

clear dimensions 7m wide and 11.77 m long, use following dataCentre to centre distance of supporting beams = 3.0 m

Span of the beams = 7.23m

Beams are supported on walls of 0.23 m thicknessC/s of beam = 230 x 450 mmGrade of concrete : M20

Type of steel : Fe415

Clear cover : 20 mmSlab thickness: 150 mm

Beam depth is inclusive of slab depth, The hall is having walls on all 4 sides

Main positive reinforcement @ end span = 8mm diameter @100 c/c

Main reinforcement in other interior panels = 8 mm diameter @ 200 c/c

Negative reinforcement @ all supports = 8mm diameter @ 200 c/cDistribution steel= 8mm diameter @ 200 c/c

Solution:

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Live load wL = 1.50 kN/m2 (Given in the data)

Maximum service load moment at interior support =9

Lw

10

Lw 2L2

d + = 8.5 kN-m

Mu=1.5 x 8.5 = 12.75 kN-m/mMulim=Qlimbd

2= 2.76 x 1000 x 1002 / 1 x 106 = 27.6 kN-m > 12.75 kN-m (Qlim=2.76)

275.1100x1000

10x75.12

bd

M2

6

2

u ==

From table 2 of SP16 pt=0.384; Ast=(0.384 x 1000 x 100)/100= 384 mm2

Spacing of 10 mm dia bars = (78.54 x 1000)/384= 204.5 mm c/c

Provide #10 @ 200 c/c

Area of distribution steel Adist=0.12 x 1000 x 120 / 100 = 144 mm2

Spacing of 8 mm dia bars = (50.26 x 1000)/144= 349 mm c/c

Provide #8 @ 340 c/c. Main and dist. reinforcement in the slab is shown in Fig.6.3

Step2: Preliminary design of beams and columns

Beam:

Effective span = 8mEffective depth based on deflection criteria = 8000/12 = 666.67mm

Assume over all depth as 700 mm with effective depth = 650mm, breadth b = 400mmand column section equal to 400 mm x 600 mm.

Step3: Analysis

Load on framei) Load from slab = (3.63+1.5) x 4 =20.52 kN/m

ii) Self weight of rib of beam = 0.4x0.58x24 = 5.56 kN/m

Total 27.00 kN/m

The portal frame subjected to the udl considered for analysis is shown in Fig. 6.4

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Design moments:

Service load end moments: MB=102 kN-m, MA=51 kN-mDesign end moments MuB=1.5 x 102 = 153 kN-m, MuA=1.5 x 51=76.5 kN-m

Service load mid span moment in beam= 27x82/8 102 =114 kN-m

Design mid span moment Mu+

=1.5 x 114 = 171 kN-mMaximum Working shear force (at B or C) in beam = 0.5 x 27 x 8 = 108kN

Design shear force Vu = 1.5 x 108 = 162 kN

Step4:Design of beams:

The beam of an intermediate portal frame is designed. The mid span section of this beam

is designed as a T-beam and the beam section at the ends are designed as rectangular

section.Design of T-section for Mid Span :

Design moment Mu=171 kN-m

Flange width bf= fwo

D6b

6

L++ , Here Lo=0.7 x L = 0.7 x 8 =5.6m

bf= 5.6/6+0.4+6x0.12=2m

bf/bw=5 and Df /d =0.2 Referring to table 58 of SP16, the moment resistance factor isgiven by KT=0.459,

Mulim=KT bwd2 fck = 0.459 x 400 x 600

2 x 20/1x106 = 1321.92 kN-m > Mu Safe

The reinforcement is computed using table 2 of SP16

Mu/bd2 = 171 x 106/(400x6002)1.2 for this pt=0.359

Ast=0.359 x 400x600/100 = 861.6 mm2

No of 20 mm dia bar = 861.6/(x202/4) =2.74Hence 3 Nos. of #20 at bottom in the mid span

Design of Rectangular-section for End Span :

Design moment MuB=153 kN-mMuB/bd

2= 153x106/400x60021.1 From table 2 of SP16 pt=0.327Ast=0.327 x 400 x 600 / 100 = 784.8

No of 20 mm dia bar = 784.8/(x202/4) =2.5Hence 3 Nos. of #20 at the top near the ends for a distance of o.25 L = 2m from face

of the column as shown in Fig 6.6

Check for Shear

Nominal shear stress = 675.0600400

10x162

bd

V 3uv =

==

pt=100x 942/(400x600)=0.390.4

Permissible stress for pt=0.4 from table 19 c=0.432 < v Hence shear reinforcement isrequired to be designedStrength of concrete Vuc=0.432 x 400 x 600/1000 = 103 kNShear to be carried by steel Vus=162-103 = 59 kN

Spacing 2 legged 8 mm dia stirrup sv=

3671059

60050241587.0

V

dAf87.03

us

svy=

=

Two legged #8 stirrups are provided at 300 mm c/c (equal to maximum spacing)

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Step5:Design of columns:Cross-section of column = 400 mm x 600 mm

Ultimate axial load Pu=1.5 x 108 = 162 kN (Axial load = shear force in beam)

Ultimate moment Mu= 1.5 x 102 = 153 kN-m ( Maximum)

Assuming effective cover d = 50 mm; d/D 0.1

053.060040020

10153

bDf

M2

6

2

ck

u =

=

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033.060040020

10162

bDf

P 3

ck

u =

=

Referring to chart 32 of SP16, p/fck=0.03; p=20 x 0.03 = 0.6

Minimum steel in column should be 0.8 %, Hence min steel percentage shall be adopted

Ast=0.8x400x600/100 = 1920 mm2

No. of bars required = 1920/314 = 6.1

Provide 8 bars of #20

8mm diameter tie shall have pitch least of the following

i) Least lateral dimension = 400 mmii) 16 times diameter of main bar = 320 mm

iii) 48 times diameter of tie bar = 384

iv) 300mm

Provide 8 mm tie @ 300 mm c/c

Step6:Design of Footing:

Load:Axial Working load on column = 108 kN

Self weight of footing @10% = 11 kN

Total load= 119120 kNWorking load moment at base = 51 kN-m

Approximate area footing required = Load on column/SBC= 108/150 =0.72 m2

However the area provided shall be more than required to take care of effect of

moment. The footing size shall be assumed to be 2mx3m (Area=6 m2)

45

2m

1.2m

3m

0.6m

0.4m

X

X

600

Tie #8 @300 c/c8-#20

400


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Step1:Design of slab

Assume over all depth of slab as 120mm and effective depth as 100mmSelf weight of slab = 0.12 x 24 = 2.88 kN/m2

Weight of roof finish = 0.50 kN/m2 (assumed)

Ceiling finish = 0.25 kN/m2

(assumed)Total dead load wd = 3.63 kN/m2

Live load wL = 1.50 kN/m2 (Given in the data)

Maximum service load moment at interior support =9

Lw

10

Lw 2L2

d + = 8.5 kN-m

Mu=1.5 x 8.5 = 12.75 kN-m/m

Mulim=Qlimbd2= 2.76 x 1000 x 1002 / 1 x 106 = 27.6 kN-m > 12.75 kN-m (Qlim=2.76)

275.1100x1000

10x75.12

bd

M2

6

2

u ==

From table 2 of SP16 pt=0.384; Ast=(0.384 x 1000 x 100)/100= 384 mm2

Spacing of 10 mm dia bars = (78.54 x 1000)/384= 204.5 mm c/c

Provide #10 @ 200 c/c

Area of distribution steel Adist=0.12 x 1000 x 120 / 100 = 144 mm2

Spacing of 8 mm dia bars = (50.26 x 1000)/144= 349 mm c/c

Provide #8 @ 340 c/c. Main and dist. reinforcement in the slab is shown in Fig.6.9

Step2: Preliminary design of beams and columns

Beam:

Effective span = 10mEffective depth based on deflection criteria = 10000/13 = 769.23mm

Assume over all depth as 750 mm with effective depth = 700mm, breadth b = 450mm

and column section equal to 450 mm x 600 mm.

Step3: Analysis

Load on frame

i) Load from slab = (3.63+1.5) x 4 =20.52 kN/mii) Self weight of rib of beam = 0.45x0.63x24 = 6.80 kN/m

Total 28.00 kN/mHeight of beam above hinge = 4+0.1-075/2 =3.72 m

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Moment Distribution Table

Joint A B C D

Members AB BA BC CB CD DC

DF - 0.5 0.5 0.5 0.5 -

FEM 0 0 -233 233 0 0Balance - 116.5 116.5 -116.5 -116.5 -

Carry

over

- - -58.25 58.25 - -

Balance - 29.13 29.13 -29.13 -29.13 -

Carryover

- - -14.57 14.57 - -

Balance - 7.29 7.29 -7.29 -7.29 -

Carry

over

- -3.65 3.65 - -

Balance - 1.83 1.83 -1.83 -1.83 -

Carryover

- - -0.92 0.92 - -

Balance - 0.46 0.46 -0.46 -0.46 -

Total - 155.21156

-155.21-156

155.21156

-155.21-156 -

Bending Moment diagram

Fig. 6.11

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Design moments:

Service load end moments: MB=156 kN-m,

Design end moments MuB=1.5 x 156 = 234 kN-m,Service load mid span moment in beam= 28x102/8 102 =194 kN-m

Design mid span moment Mu+=1.5 x 194 = 291 kN-m

Maximum Working shear force (at B or C) in beam = 0.5 x 28 x 10 = 140kNDesign shear force Vu = 1.5 x 140 = 210 kN

Step4:Design of beams:

The beam of an intermediate portal frame is designed. The mid span section of this beam

is designed as a T-beam and the beam section at the ends are designed as rectangular

section.

Design of T-section for Mid Span :Design moment Mu=291 kN-m

Flange width bf= fwo

D6b6

L++ , Here Lo=0.7 x L = 0.7 x 10 =7m

bf= 7/6+0.45+6x0.12=2.33mbf/bw=5.2 and Df /d =0.17 Referring to table 58 of SP16, the moment resistance factor

is given by KT=0.43,Mulim=KT bwd

2 fck = 0.43 x 450 x 7002 x 20/1x106 = 1896.3 kN-m > Mu Safe

The reinforcement is computed using table 2 of SP16

Mu/bd2 = 291 x 106/(450x7002)1.3 for this pt=0.392

Ast=0.392 x 450x700/100 = 1234.8 mm2

No of 20 mm dia bar = 1234.8/(x202/4) =3.93Hence 4 Nos. of #20 at bottom in the mid span

Design of Rectangular-section for End Span :

Design moment MuB=234 kN-m

MuB/bd2= 234x106/450x70021.1 From table 2 of SP16 pt=0.327Ast=0.327 x 450 x 700 / 100 = 1030

No of 20 mm dia bar = 1030/(x202/4) =3.2Hence 4 Nos. of #20 at the top near the ends for a distance of o.25 L = 2.5m fromface of the column as shown in Fig 6.12

Check for Shear

Nominal shear stress = 67.0700450

10x210

bd

V 3uv =

==

pt=100x 1256/(450x700)=0.390.4Permissible stress for pt=0.4 from table 19 c=0.432 < v Hence shear reinforcement is

required to be designedStrength of concrete Vuc=0.432 x 450 x 700/1000 = 136 kN

Shear to be carried by steel Vus=210-136 = 74 kNSpacing 2 legged 8 mm dia stirrup sv=

53.3411074

70050241587.0

V

dAf87.03

us

svy=

=

Two legged #8 stirrups are provided at 300 mm c/c (equal to maximum spacing)

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Cross-Sections of Beam

Cross-Section of Column

LONGITUDINAL ELEVATION

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Reference Books

N.Krishna Raju Advanced Reinforced concrete Design

Jaikrishna and O.P.Jain Plain and reinforced concrete Vol2

B.C.Punmia Reinforced Concrete Structures Vol2

Problems for Practice

1. A portal frame ABCD has fixed supports at A and D. The columns AB and

CD are 5m in height while the beam BC is 10 m in length. The frames arespaced at 3.5m intervals. The live load on the roof slab which is 100 mm thick

may be taken as 1.5 kN/m2. Design the beam, column and footing and sketch

the details of reinforcements. Adopt M-20 concrete, Fe-415 steel andSBC=200 kN/m2

2. The roof of an assembly hall 30m long and 12 m wide between centres of

columns, consists of a continuous reinforced concrete slab over rectangular

portal frames spaced 3m apart. The columns are provided with independent

footings and hinged at the bottom. The ceiling height is 3.5m above the hingelevel. Adopting M-20 concrete and Fe-415 for steel, design the continuous

roof slab and the portal frame and foundation footing for the columns assumesafe bearing capacity of the soil as 150 kN/m2. Sketch the details of

reinforcements in the portal frame.

**************

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jd

MA

st

st

= ; Let pt be the percentage of steel expressed as

st

cbc

st

sttbal

k50

bd

1

jd

M100

bd

A100p

=

==

Design constants for balanced section is given in table 6.1

Table 6.1 Design constants

Concrete

Grade

Steel

Gradecbc st k j Qbal ptbal

M20Fe250 7 140 0.4 0.87 1.21 1.00

Fe415 7 230 0.29 0.9 0.91 0.44

M25Fe250 8.5 140 0.4 0.87 1.48 0.68

Fe415 8.5 230 0.29 0.9 1.1 0.533

6.3 Liquid Retaining Members subjected to axial tension only:

When the member of a liquid retaining structure is subjected to axial tension only, themember is assumed to have sufficient reinforcement to resist all the tensile force and

the concrete is assumed to be uncracked.

For analysis purpose 1m length of wall and thickness t is considered. The tension inthe member is resisted only by steel and hence

st

st

TA

= and T 1000 t ct+(m-1)Astst or

st

ct

ct

)1m(11000

Tt

Minimum thickness of the member required is tabulate in table 6.2Table 6.2 Minimum thickness of members under direct tension (Uncracked condition)

Grade of

concrete

Thickness of members in mm for force T in N

Mild steel HYSDM20 T/1377 T/1331

M25 T/1465 T/1423

M30 T/1682 T/1636

6.4 Liquid Retaining Members subjected to Bending Moment only:

For the members subjected to BM only with the tension face in contact with water or

for the members of thickness less than 225 mm, the compressive stress and tensile

stresses should not exceed the value given in IS 3370. For the member of thicknessmore than 225 mm and for the face away from the liquid, this condition need not be

satisfied and higher stress in steel may be allowed. The bending analysis is done for

cracked and uncracked condition.Cracked condition: The procedure of designing is same as in working stress methodexcept that the stresses in steel are reduced. The design coefficients for these reduced

stresses in steel is given in Table 6.3

Table 6.3 Design constants for members in bending (Cracked condition)

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Concrete

Grade

Steel

Gradecbc st k j Qbal ptbal

For members less than 225 mm thickness and tension on liquid face

M20Fe250 7 115 0.445 0.851 1.33 1.36

Fe415 7 150 0.384 0.872 1.17 0.98

For members more than 225 mm thickness and tension away from liquid face

M20Fe250 7 125 0.427 0.858 1.28 1.2

Fe415 7 190 0.329 0.89 1.03 0.61

Uncracked condition: In this case, the whole section is assumed to resist the moment.

Hence the maximum tensile stress in concrete should not be more than permissible value.The section is designed as a homogenous section.

Taking moments of transformed areas about NA

b kD kD/2 = b (D-kD) (D-kD)/2 + (m-1) Ast (d-kD)Substituting Ast = pt bD /100 and simplifying

)1m(p2200

)1m(D

dp2100

kt

t

+

+=

Moment of inertia Ixx=bD3/12 + bD (kD-D/2)2 + (m-1) Ast (d-kD)

2

substituting Ast = pt bD /100 and simplifying

Ixx=(1/3 k(1-k)+(d/D-k)2 (m-1) pt/100)bD3

The moment of resistance may be expressed using Bernoulis equation

)k1(DkDDI

M cbccb

xx

=

= and

)k1(D

IM xxcbt

=

6.5 Liquid Retaining Members subjected to Combined axial tension and

Bending Moment :

cbt

cbc

kD

d

b

SectionStress

Diagram

Fig. 6.1 Singly Reinforced Section

D

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66

Elevated Circular

Rectangular Spherical

Intz Conical Bottom


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6.6.1 Circular Tanks resting on ground :

Due to hydrostatic pressure, the tank has tendency to increase in diameter. This increasein diameter all along the height of the tank depends on the nature of joint at the junction

of slab and wall as shown in Fig 6.5

Tank with flexible base

Tank with rigid base

Fig. 6.5

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The thickness is assumed as t = 30H+50=149 160 mmStep 2: Design of Vertical wall

Max hoop tension at bottom kN5.2142

133.310

2

DHT =

==

Area of steel

23

ststst

mm1430150

105.214TTA =

=

=

=

Minimum steel to be provided

Ast min=0.24%of area of concrete = 0.24x 1000x160/100 = 384 mm2

The steel required is more than the minimum required

Let the diameter of the bar to be used be 16 mm, area of each bar =201 mm2

Spacing of 16 mm diameter bar=1430x 1000/201= 140.6 mm c/c

Provide #16 @ 140 c/c as hoop tension steel

Step 3: Check for tensile stress

Area of steel provided Ast provided=201x1000/140 = 1436.16 mm2

Modular ratio m= 33.1373

280

3

280

cbc

=

=

Stress in concrete2

3

st

c N/mm2.11436)133.13(1601000

105.214

A)1m(t1000

T=

+

=

+=

Permissible stress cat=0.27fck= 1.2 N/mm2

Actual stress is equal to permissible stress, hence safe.

Step 4: Curtailment of hoop steel:

Quantity of steel required at 1m, 2m, and at top are tabulated. In this table the maximumspacing is taken an 3 x 160 = 480 mm

Height from top Hoop tension

T =HD/2 (kN)Ast= T/st Spacing of #16

mm c/c

2.3 m 149.5 996 2001.3 m 84.5 563.33 350

Top 0 Min steel (384 mm2) 400

Step 5: Vertical reinforcement:

For temperature and shrinkage distribution steel in the form of vertical reinforcement is

provided @ 0.24 % ie., Ast=384 mm2.

Spacing of 10 mm diameter bar = 78.54x1000/384=204 mm c/c 200 mm c/c

Step 6: Tank floor:

As the slab rests on firm ground, minimum steel @ 0.3 % is provided. Thickness of slab

is assumed as 150 mm. 8 mm diameter bars at 200 c/c is provided in both directions atbottom and top of the slab.

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6.6.1.2 Design of Circular Tanks resting on ground with rigid base:

Due to fixity at base of wall, the upper part of the wall will have hoop tension and lower

part bend like cantilever. For shallow tanks with large diameter, hoop stresses are verysmall and the wall act more like cantilever. For deep tanks of small diameter the

cantilever action due to fixity at the base is small and the hoop action is predominant.

The exact analysis of the tank to determine the portion of wall in which hoop tension ispredominant and the other portion in which cantilever action is predominant, is difficult.

Simplified methods of analysis are

i) Reissners methodii) Carpenters simplified method

iii) Approximate method

iv) IS code method

Use of IS code method for analysis and design of circular water tank with rigid base isstudied in this course.

IS code method

Tables 9,10 and 11 of IS 3370 part IV gives coefficients for computing hoop tension,

moment and shear for various values of H

2

/DtHoop tension, moment and shear is computed as

T= coefficient ( wHD/2)M= coefficient (wH3)V= coefficient (wH2)Thickness of wall required is computed from BM consideration ie.,

Qb

Md =

where,

Q= cbcjk

stcbc

cbc

m

m

k +

=

j=1-(k/3)

b = 1000mmProviding suitable cover, the over all thickness is then computed as t = d+cover.

Area of reinforcement in the form of vertical bars on water face is computed as

jd

MA

st

st

= . Area of hoop steel in the form of rings is computed asst

1st

TA

=

Distribution steel and vertical steel for outer face of wall is computed from minimum

steel consideration.

Tensile stress computed from the following equation should be less than the permissible

stress for safe design

st

cA)1m(t1000

T

+= and the permissible stress is 0.27 fck

Base slab thickness generally varies from 150mm to 250 mm and minimum steel is

distributed to top and bottom of slab.

Design Problem No.1:

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Design Problem No.2:

Design a circular water tank to hold 5,50,000 liters of water. Assume rigid joints between

the wall and base slab. Adopt M20 concrete and Fe 415 steel. Sketch details ofreinforcements.

Solution:Step 1: Dimension of tank

Volume of tank V=550 m3

Assume H= 4.5A=550/4.5 = 122.22 m2

D= (4 x 122.22/) = 12.47 12.5 mStep 2: Analysis for hoop tension and bending moment

One meter width of the wall is considered and the thickness of the wall is estimated ast=30H+50 = 185 mm. The thickness of wall is assumed as 200 mm.

81.82.05.12

5.4

Dt

H 22=

=

Referring to table 9 of IS3370 (part IV), the maximum coefficient for hooptension = 0.575

Tmax=0.575 x 10 x 4.5 x 6.25 =161.72 kN

Referring to table 10 of IS3370 (part IV), the maximum coefficient for

bending moment = -0.0146 (produces tension on water side)Mmax= 0.0146 x 10 x 4.5

3=13.3 kN-m

Step 3: Design of section:

For M20 concrete cbc=7, For Fe415 steel st=150 MPa and m=13.33 for M20 concreteand Fe415 steel

The design constants are:

39.0m

mkstcbc

cbc =+

=

j=1-(k/3)=0.87

Q= cbcjk = 1.19

Effective depth is calculated as mm7.1051000x19.1

10x3.13

Qb

Md

6

===

Let over all thickness be 200 mm with effective cover 33 mm dprovided=167 mm

26

st

stmm27.610

167x87.0x150

10x3.13

jd

MA ==

=

Spacing of 16 mm diameter bar = c/mmc36.32927.610

1000x201 = (Max spacing

3d=501mm)

Provide #16@300 c/c as vertical reinforcement on water face

Hoop steel:2

3

st

1st mm13.1078150

10x72.161TA ==

=

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Tension in short wall is computed as Ts=pL/2

Tension in long wall TL=pB/2

Horizontal steel is provided for net bending moment and direct tensile force

Ast=Ast1+Ast2; jd

'MA

st

1st

= ; M=Maximum horizontal bending moment T x; x= d-D/2

Ast2=T/st

Design problem No.1

Design a rectangular water tank 5m x 4m with depth of storage 3m, resting on ground

and whose walls are rigidly joined at vertical and horizontal edges. Assume M20

concrete and Fe415 grade steel. Sketch the details of reinforcement in the tankSolution:

Step1: Analysis for moment and tensile force

i)Long wall:

L/a=1.671.75; at y=0, x/a=1, Mx=-0.074; at y=b/2, x/a=1/4, My=-0.052Max vertical moment = Mxwa3 = -19.98Max horizontal moment = Mywa3 = -14.04; Tlong=wab/2=60 kNii) Short wall:

B/a=1.331.5; at y=0, x/a=1, Mx=-0.06; at y=b/2, x/a=1/4, My=-0.044

Max vertical moment = Mxwa3 = -16.2Max horizontal moment = Mywa3 = -11.88; Tshort=waL/2=75 kN

E

B

A

F

D

C

Free

a=H=3m

b=4m

L=5m

Fixed

79

xD/2

d T


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CA

11.88

14.4

B

Step2: Design constants

cbc=7 MPa, st=150 MPa, m=13.33

38.0m

mk

stcbc

cbc =+

=

j=1-(k/3)=0.87

Q= cbcjk = 1.15Step3: Design for vertical moment

For vertical moment, the maximum bending moment from long and short wall(Mmax)x=-19.98 kN-m

mm8.1311000x15.1

10x98.19

Qb

Md

6

===

Assuming effective cover as 33mm, the thickness of wall is

t=131.88+33=164.8 mm170 mmdprovided=170-33=137mm

26

st

stmm54.1117

137x87.0x15010x98.19

jdMA ==

=

Spacing of 12 mm diameter bar = c/mmc2.10154.1117

1000x113= (Max spacing 3d=411mm)

Provide #12 @ 100 mm c/cDistribution steel

Minimum area of steel is 0.24% of concrete area

Ast=(0.24/100) x1000 x 170 = 408 mm2

Spacing of 8 mm diameter bar = c/mmc19.123408

1000x24.50=

Provide #8 @ 120 c/c as distribution steel.Provide #8 @ 120 c/c as vertical and horizontal distribution on the outer face.

Step4: Design for Horizontal moment

Horizontal moments at the corner in long and short wall produce unbalanced moment at

the joint. This unbalanced moment has to be distributed to get balanced moment usingmoment distribution method.

56.020/9

4/1DF

44.020/9

5/1DF

20

9K;

5

1K;

5

1K

AB

AC

ACAC

==

==

===

Moment distribution Table

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Joint A

Member AC AB

DF 0.44 0.56

FEM -14 11.88

Distribution 0.9328 1.1872

Final Moment -13.0672 13.0672The tension in the wall is computed by considering the section at height H1 from the base.

Where, H1 is greater of i) H/4, ii) 1m, ie., i) 3/4=0.75, ii) 1m; H1= 1mDepth of water h=H-H1=3-1-2m; p=wh=10 x 2= 20 kN/m2

Tension in short wall Ts=pL/2=50 kNTension in long wall TL=pB/2= 40 kN

Net bending moment M=M-Tx, where, x= d-D/2=137-(170/2)=52mm

M=13.0672-50 x 0.052=10.4672 kN-m

26

1st mm46.585137x87.0x150

10x4672.10A ==

23

2st mm33.333150

10x50A ==

Ast=Ast1+Ast2=918.79 mm2

Spacing of 12 mm diameter bar = c/mmc12374.918

1000x113= (Max spacing 3d=411mm)

Provide #12@120 mm c/c at corners

Step5: Base Slab:

The slab is resting on firm ground. Hence nominal thickness and reinforcement is

provided. The thickness of slab is assumed to be 200 mm and 0.24% reinforcement is

provided in the form of #8 @ 200 c/c. at top and bottom

A haunch of 150 x 150 x 150 mm size is provided at all corners

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Documents

Design&Drng Compiled G.S.suresh