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1 Design and drawing of RC Structures CV61 Dr. G.S.Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: [email protected]

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Design and drawing of RC Structures

CV61

Dr. G.S.Suresh

Civil Engineering Department

The National Institute of Engineering

Mysore-570 008

Mob: 9342188467Email: [email protected]

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DETAILING OF BEAMS

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Learning out Come

• Introduction

• Simply supported rectangular beams

• Continuous rectangular beams

• Cantilever rectangular beams

• Flanged beams

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Dr.G.S.Suresh4

Introduction

• Carries Transverse External Loads That Cause Bending Moment, Shear Forces And In Some Cases Torsion

• Concrete is strong in compression and very weak in tension.

• Steel reinforcement is used to take up tensile stresses in reinforced concrete beams.

• Mild steel bars or Deformed or High yield strength deformed bars (HYSD)

• HYSD bars have ribs on the surface and this increases the bond strength at least by 40%

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Dr.G.S.Suresh5

Introduction

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•Drawinsg generally include a bar bending schedule

•The bar bending schedule describes the length and number, position and the shape of the bar

Introduction

Sl.No.Type of bar and mark

Shape No. Length in m

Weight per unit lengthin Kg

Weight

in Kg

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

•The anchorage value of bend 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.

Introduction

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Standard hooks

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• The beams are classified as:

• According to shape: Rectangular, T, L, Circular etc

• According to supporting conditions: Simply supported, fixed, continuous and cantilever beams

• According to reinforcement: Singly reinforced and doubly reinforced

Introduction

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• Minimum cover in beams must be 25 mm or shall not be less than the larger diameter of bar for all steel reinforcement including links.

• Nominal cover specified in Table 16 and 16A of IS456-2000 should be used to satisfy the durability criteria.

Introduction

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Generally a beam consists of following steel reinforcements:•Longitudinal reinforcement at tension and compression face.•Shear reinforcements in the form of vertical stirrups and or bent up longitudinal bars are provided. •Side face reinforcement in the web of the beam is provided when the depth of the web in a beam exceeds 750 mm. (0.1% of the web area and shall be distributed equally on two faces at a spacing not exceeding 300 mm or web thickness whichever is less)

Introduction

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Specification for the reinforcement in beams is given in clause 8.1 to 8.6 of SP34

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•While drawing the details of a beam following convention representation of bars are used•Mild steel bars : ; HYSD bars: # or •Main bars are shown by thick single line. •Hanger bars are shown by medium thick lines.

Introduction

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Different forms of stirrups used in beams

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Typical drawing of a simply supported beam

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PROBLEM No. 1 Draw the Longitudinal section, cross section and prepare bar bending schedule of a rectangular simply supported RCC beam with the following data:

Clear span =3.5m

Width of beam = 220mm

Overall depth of beam = 300mm

Bearing width in support = 200 mm

Main reinforcement = 5 Nos -12 mm diameter bars with 2 bars bent up at L/7 from centre of support

Anchor/hanger bars= 2-10 mm diameter

Stirrups = 6 mm diameter @ 200 mm c/c.

Materials : Mild steel, M20 grade concrete

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PROBLEM No. 1

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PROBLEM No. 1 contd.

Bar Bending Schedule:Bottom straight bar (12 dia)= Total length of beam +2 x16 -2 x 3 -2 x end cover

= (3500+2 x 200)+26 x 12-2 x 25 =41624200 mm

Length of bent up bar (12 dia)= Length of straight bar +2 x (0.42 x depth of bend) =4162+2 x 0.42 x 250 =43724400 mm

Length of hanger bar (10 dia)= Length of straight bar =41624200 mm

Stirrups:

Number of stirrups = Length of bar (end to end)/c/c distance of stirrup= [(3500+2x200)-2x25]/200 = 17

Length of stirrup = 2 ( A+B)+24 of stirrup = 2x(250+170)+24 x 6 = 984 mm 1000 mm

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PROBLEM No. 1 contd.

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PROBLEM No. 2 Draw the Longitudinal section, cross section and prepare bar bending schedule of a rectangular simply supported RCC beam with the following data:Clear span =4.5m

Width of beam = 250mm

Overall depth of beam = 300mm

Main reinforcement = 5 Nos -18 mm diameter bars with 2 bars bent up at 900mm from inside of each end support

Anchor/hanger bars= 2-12 mm diameter

Stirrups = 6 mm diameter @ 200 mm c/c.

Concrete cover = 25 mm

Materials : HYSD bars, M20 grade concrete

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PROBLEM No. 2

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PROBLEM No. 2 contd.

Bar Bending Schedule:Bottom straight bar (18 dia)= Total length of beam -2 x end cover

= (4500+2 x 200) -2 x 25 =4850 mm

Length of bent up bar (18 dia) = Length of straight bar +2 x (0.42 x depth of bend) =4850+2 x 0.42 x 250 =5050 mm

Length of hanger bar (12 dia)= Length of straight bar =4850 mm

Stirrups:

Number of stirrups = Length of bar (end to end)/c/c distance of stirrup = [(4500+2x200)-2x25]/200 = 24.25 25

Length of stirrup = 2 ( A+B)+24 of stirrup

= 2x(250+200)+24 x 6 = 1044 mm 1100 mm

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PROBLEM No. 2 contd.

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PROBLEM No. 3 Draw the Longitudinal section and two cross sections one near the support and other near the mid span of a RCC continuous beam with the following data:

Clear span of beams = 3m each

Width of beam = 200mm

Overall depth of beam = 300mm

Width in intermediate supports = 200 mm

Main reinforcement = 4 Nos -12 mm diameter bars with 2 bars bent up

Anchor/hanger bars= 2-10 mm diameter

Stirrups = 6 mm diameter @ 300 mm c/c.

Materials : HYSD bars and M20 grade concrete

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PROBLEM No. 3

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PROBLEM No. 4 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 bars consists 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.

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PROBLEM No. 4

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PROBLEM No. 5 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 =3.2m

Overall depth at free end = 150 mm

Overall 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 support

Anchor bars = 2 Nos. 16 mm dia

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

Bearing at fixed end = 300 mm

Use M20 concrete and Fe 415 steel

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PROBLEM No. 5

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PROBLEM No. 6A 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

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PROBLEM No. 2

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PROBLEM No. 7 A beam has following data

Clear span = 4m

Support width = 300mm

Size of web = 350 x 400

Size of flange = 1200 x 120mm

Main reinforcement in two layers : 3-20 tor + 3-16 tor and to be curtailed at a distance 400 mm from inner face of support

Hanger bars: 3- 20 tor

Stirrups: 2L-8 tor @ 200 c/c

Use M20 concrete and Fe 415 steel

Draw longitudinal and cross section if the beam is

1. T-beam

2. Inverted T-beam

3. L-Beam

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PROBLEM No. 5

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PROBLEM No. 5

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PROBLEM No. 5

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Do it Yourself

1. Draw the longitudinal section and typical cross sections ( at centre and support), and show the reinforcement details in a simply supported rectangular beam of size 300 mm x 500 mm, clear span 5m supported on walls of 0.3m, use a suitable scale

Reinforcements:

Main: 4 No. 16mm dia with 2 No. cranked at 1m from centre of support. Stirrup holders 2 Nos. of 12 mm dia

Stirrups: 2 legged 8 mm dia stirrups at 250 mm c/c in the central 2m span and 2 legged 8 mm dia stirrups at 150 mm c/c in the remaining portion. Assume concrete M 20 grade and steel Fe 415, and suitable cover. Prepare the bar bending schedule and calculate quantity of steel and concrete required.

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Do it Yourself

2. Prepare the bar bending schedule and estimate quantity of steel and concrete after drawing the longitudinal and cross section. Other details are

Span of beam = 4.2 m

Cross section at support end 300 x 600 mm and cross section at free end 300 x 150 mm

Reinforcements:

Main tension steel: 4-20 mm dia, 2 bars are curtailed at a distance of 2m from free end

Hanger bars: 1-12 mm dia

Two legged stirrups 8mm dia @ 140 mm c/c for full length.

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Dr. G.S.Suresh

Civil Engineering Department

The National Institute of Engineering

Mysore-570 008

Mob: 9342188467 Email: [email protected]