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Design and drawing of RC Structures CV 61 Dr. G. S. Suresh Civil Engineering Design and drawing of RC Structures CV 61 Dr. G. S. Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: [email protected] com 1

DETAILING OF BEAMS 2 DETAILING OF BEAMS 2

Learning out Come • • • Introduction Simply supported rectangular beams Continuous rectangular beams Learning out Come • • • Introduction Simply supported rectangular beams Continuous rectangular beams Cantilever rectangular beams Flanged beams 3

Introduction • • • Carries Transverse External Loads That Cause Bending Moment, Shear Forces 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% 4 Dr. G. S. Suresh

Introduction 5 Dr. G. S. Suresh Introduction 5 Dr. G. S. Suresh

Introduction • Drawinsg generally include a bar bending schedule • The bar bending schedule Introduction • Drawinsg generally include a bar bending schedule • The bar bending schedule describes the length and number, position and the shape of the bar Sl. No. Type of bar and mark Shape Weight per unit length Length in m in Kg No. 6

Introduction • Anchorage in steel bars is normally provided in the form of bends Introduction • 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. 7

Standard hooks 8 Standard hooks 8

 • • Introduction The beams are classified as: According to shape: Rectangular, T, • • Introduction 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 9

 • • Introduction Minimum cover in beams must be 25 mm or shall • • Introduction 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 16 A of IS 456 -2000 should be used to satisfy the durability criteria. 10

Introduction Generally a beam consists of following steel reinforcements: • Longitudinal reinforcement at tension Introduction 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) 11

Specification for the reinforcement in beams is given in clause 8. 1 to 8. Specification for the reinforcement in beams is given in clause 8. 1 to 8. 6 of SP 34 12

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Introduction • While drawing the details of a beam following convention representation of bars Introduction • 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. 17

Different forms of stirrups used in beams 18 Different forms of stirrups used in beams 18

Typical drawing of a simply supported beam 19 Typical drawing of a simply supported beam 19

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PROBLEM No. 1 Draw the Longitudinal section, cross section and prepare bar bending schedule 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. 5 m Width of beam = 220 mm Overall depth of beam = 300 mm 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, M 20 grade concrete 21

PROBLEM No. 1 22 PROBLEM No. 1 22

PROBLEM No. 1 contd. Bar Bending Schedule: Bottom straight bar (12 dia)= Total length PROBLEM No. 1 contd. Bar Bending Schedule: Bottom straight bar (12 dia)= Total length of beam +2 x 16 -2 x 3 -2 x end cover = (3500+2 x 200)+26 x 12 -2 x 25 =4162 4200 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 =4372 4400 mm Length of hanger bar (10 dia)= Length of straight bar =4162 4200 mm Stirrups: Number of stirrups = Length of bar (end to end)/c/c distance of stirrup= [(3500+2 x 200)-2 x 25]/200 = 17 Length of stirrup = 2 ( A+B)+24 of stirrup = 2 x(250+170)+24 x 6 = 984 mm 1000 mm 23

PROBLEM No. 1 contd. 24 PROBLEM No. 1 contd. 24

PROBLEM No. 2 Draw the Longitudinal section, cross section and prepare bar bending schedule 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. 5 m Width of beam = 250 mm Overall depth of beam = 300 mm Main reinforcement = 5 Nos -18 mm diameter bars with 2 bars bent up at 900 mm 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, M 20 grade concrete 25

PROBLEM No. 2 26 PROBLEM No. 2 26

PROBLEM No. 2 contd. Bar Bending Schedule: Bottom straight bar (18 dia)= Total length 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+2 x 200)-2 x 25]/200 = 24. 25 Length of stirrup = 2 ( A+B)+24 of stirrup = 2 x(250+200)+24 x 6 = 1044 mm 1100 mm 27

PROBLEM No. 2 contd. 28 PROBLEM No. 2 contd. 28

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PROBLEM No. 3 Draw the Longitudinal section and two cross sections one near the 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 = 3 m each Width of beam = 200 mm Overall depth of beam = 300 mm 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 M 20 grade concrete 30

PROBLEM No. 3 31 PROBLEM No. 3 31

PROBLEM No. 4 A rectangular beam of cross section 300 x 450 mm is PROBLEM No. 4 A rectangular beam of cross section 300 x 450 mm is supported on 4 columns which are equally spaced at 3 m c/c. The columns are of 300 mm x 300 mm in section. The reinforcement consists of 4 bars of a 6 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 M 20 and type of steel is Fe 415. Draw longitudinal section and important cross sections. 32

PROBLEM No. 4 33 PROBLEM No. 4 33

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PROBLEM No. 5 Draw to scale of 1: 20 the Longitudinal section and two PROBLEM No. 5 Draw to scale of 1: 20 the Longitudinal section and two crosssection of a cantilever beam projecting 3. 2 from a support using following data Clear span =3. 2 m 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. 5 m from support Anchor bars = 2 Nos. 16 mm dia Nominal stirrups = 6 mm dia at 40 mm c/c Bearing at fixed end = 300 mm Use M 20 concrete and Fe 415 steel 35

PROBLEM No. 5 36 PROBLEM No. 5 36

PROBLEM No. 6 A cantilever beam with 3. 2 m length is resting over PROBLEM No. 6 A cantilever beam with 3. 2 m 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. 8 m Main steel: 4 nos. of 25 mm dia bars, two bars curtailed at 1. 2 m from free end Hanger bars: 2 nos. 16 mm. Stirrups: 6 mm dia 2 legged stirrups @ 200 mm c/c the support length and @100 mm c/c from fixed end up to length of 1 m @ 150 mm c/c up to curtailed bars and remaining @ 200 c/c. Use M 20 concrete and Fe 415 steel 37

PROBLEM No. 2 38 PROBLEM No. 2 38

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PROBLEM No. 7 A beam has following data Clear span = 4 m Support PROBLEM No. 7 A beam has following data Clear span = 4 m Support width = 300 mm Size of web = 350 x 400 Size of flange = 1200 x 120 mm 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: 2 L-8 tor @ 200 c/c Use M 20 concrete and Fe 415 steel Draw longitudinal and cross section if the beam is 1. T-beam 2. Inverted T-beam 3. L-Beam 40

PROBLEM No. 5 41 PROBLEM No. 5 41

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PROBLEM No. 5 43 PROBLEM No. 5 43

PROBLEM No. 5 44 PROBLEM No. 5 44

Do it Yourself 1. Draw the longitudinal section and typical cross sections ( at 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 5 m supported on walls of 0. 3 m, use a suitable scale Reinforcements: Main: 4 No. 16 mm dia with 2 No. cranked at 1 m 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 2 m 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. 45

Do it Yourself 2. Prepare the bar bending schedule and estimate quantity of steel 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 2 m from free end Hanger bars: 1 -12 mm dia Two legged stirrups 8 mm dia @ 140 mm c/c for full length. 46

Dr. G. S. Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Dr. G. S. Suresh Civil Engineering Department The National Institute of Engineering Mysore-570 008 Mob: 9342188467 Email: [email protected] com 47