BOND • Adhesion between concrete and steel • Necessary to develop composite action between steel and concrete • Proper bond between steel and concrete needed to ensure No slip • Assumption “plane section remains plane” is valid only when there is no slip between steel and concrete or proper bond • Assumption “Strain in steel equal to strain in surrounding concrete” is also valid only when there is proper bond • It is the bond which results in variation in stress in steel along the span (if no bond, stress in steel would be uniform) • To develop full tension in steel, bars must be properly anchored on both side (otherwise slip will take place before developing full stress)
BOND STRENGTH OF CONCRETE •Bond strength of concrete depends on many factors such as: •The individual contributions of these factors are difficult to separate or quantify.
Design bond Stress, bd (For ribbed bars) = 1.6× Design bond stress for plain bars Grade of Concrete M15 M20 M25 M30 M35 M40 Average bond stress, bd (N/mm2) 1.60 1.92 2.24 2.40 2.72 3.04 For plain bars in tension Design Bond stress for bars in compression = 1.25× Design bond stress in tension Grade of Concrete M15 M20 M25 M30 M35 M40 Average bond stress, bd (N/mm2) 2.0 2.4 2.8 3.0 3.4 3.8 For plain bars in tension
•The quality of the concrete can be considered as major factors. •Code gives average bond stress as bd as Grade of Concrete
• As the BM varies along the span, bending stress also varies and as a result Tension in steel (i.e. bond stress) varies along the span • Shear stress developed at the interface of bar and concrete is known as bond stress • Bond stress is calculated as Force per unit of nominal surface area of Rebar • Average stress acting within anchorage length is called average anchorage bond stress • Bond is developed between steel and concrete due to: – Chemical adhesion between steel and concrete – Friction between steel and concrete – Mechanical interlock due to ribs in deformed bars and hooks in plain (mild steel) bars
M20 M25 M30 M35 M40
Average bond stress, bd (N/mm2) 1.2
1.4
1.5
1.7
1.9
Note: Above values of bd are for plain bars in tension Above values of bd will be increased by 60% for High deformed bars For bars in compression, above values further will be increased by 25% Thus bond strength of high deformed bars in compression = 1.25 1.60 bd
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Types of Bond • Anchorage bond or development bond arises in anchorage zone at the ends of beam which has pull out action
Anchorage length in direct Tension
Anchorage length in Tension
• Flexural bond Arises due to variation in BM which results in variation in axial tension along the length
Bar in tension Bar in compression
Anchorage length in Flexure
Anchorage length in Compression
Anchorage Bond • Bond between steel & concrete in tension or compression zone • Bond stress varies (parabolically) along the length
Development Length for Anchorage Bond: Development Length in Tension/Compression: Length required to develop bond strength equal to strength of bar Strength of bar (in tension or compression) = fs Ast = fs . /42 Bond strength developed in length Ld (referred as development length) = surface area Av. bond stress = ( ) Ld . bd Equating both, we get: fs . /42 = ( ) Ld . bd Ld = fs. / (4. bd) Where, fs = stress in the steel
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If area of steel provided is equal to as determined from design, then stress in steel, fs = 0.87 fy However, if area of steel provided is more than the required, then the stress in steel will be calculated as
Where, Ld is given as
Ld
0.87 f y .
4 u av For deformed bars in tension
0.87 f y .
0.87 415 47 4 bd 4 1.92 Grade of Concrete M15 M20 M25 M30 M35 M40 bd (average) (N/mm2) 1.60 1.92 2.24 2.40 2.72 3.04 34 45 39 36 32 29 (Ld/) for Fe-250 56 47 40 38 33 30 (Ld/) for Fe-415 68 57 49 45 40 36 (Ld/) for Fe-500
Ld for M20 concrete and Fe415 deformed bars in tension: Ld
fs = 0.87fy . (Ast required) / (Ast provided)
Flexural bond •Consider a beam segment of length x. •Moments acting on this segments are M and M+ M and •Tensile Forces in steel bar are T and T+ T •Moment at left and right faces may be calculated as •On left face M = T.z ; and On right face M+ M = (T+ T).z •On subtracting, above two equation, we get •(M+ M) - M = (T+ T)z - Tz or M = T. z If O is the perimeter of the bar and bd is the bond stress then For the length x, for adequate bond between concrete and steel, T = O. x. bd Substituting value of T, M = T. Z = O. x. bd . z Or bd = (M/ x) . 1/(O .z) = V/(O .z)
Thus, we have bd
Check for Development Length Development length is checked at following locations • Sections of maximum BM • at supports • at Points of cut-off • at Points of inflection (where BM changes sign)
V O. z
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Check of Ld at supports Let the beam is to be checked for development length at a section located at distance Ld from the support At this section, if shear force is V and lever arm is z, Then the bond stress developed between bars (s) and concrete will be u = V/(O .z) For the development length, Ld , total force developed will be T = u. O. Ld If Tension force in bars (s) at this section is T and the moment carrying capacity beam at this section is M1 , which may be calculated as M1 = T.z = (u O Ld).z = V/(O .z) O Ld).z = V. Ld or Ld = M1 / V
Check for Ld at points of Max BM • Specially needed for small members such as footings • Need to check Ld on both side of Max BM (i.e. face of column) • If available length is less than required, • Decrease the diameter of bar more surface area reduction in required Ld • If required, additional development length is provided by anchoring the bars (bends, hooks)
Thus, if at any point, if Moment carrying capacity of the section with available bars (stressed to full strength i.e. 0.87 fy,) is M1 and Maximum shear force at section is V, then available development length, Ld = (M1/V)
Note: If the bars are confined by compression reaction, Ld may be increased by 30%. In this case Ld = 1.3 (M1 / V) Further, if the bars are continued beyond the (inner edge of the) support to a length (called anchorage length), La
Determination of La
Available development length = Ld + La = 1.3 (M1 / V) + La Thus at any section, the available development length [= La + 1.3 (M1 / V)] must not be less than the required development length, Ld (which is needed to develop the full strength of the bars). M1 If the Bars are confined by Tension reaction La Ld at s fd V 1.3 M1 If the Bars are confined by compression reaction La Ld at f V s
d
Bars are bent in such a way that k = 4 for Mild steel; k = 2 for High strength steel If bars are bent as per above mentioned guidelines Max. allowable La in Tension : 900 Bend 8, 1800 Hook 16 Max. allowable La in compression: Projected length of bend=(k + )
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Curtailment of Bars As we move from Mid-span to support, BM decreases, Steel may be reduced for economical design For SS beams, as per code, at least one third of +ve moment reinforcement must be continued to support on the same face For Continuous beams, as per code, at least one fourth of +ve moment reinforcement must be continued to support on the same face Remaining steel is cut-off at intermediate locations called curtailment Need to check the development length for the remaining +ve Moment reinforcement as well as there is a Need to check the development length for the curtailed bars
Determination of Theoretical Cut-off Point Let there are n bars at any section and few bars are to curtailed such that remaining bars are m. The theoretical cut-off point may be determined such that the moment carrying capacity of the section with remaining bars (i.e. m bars), M1 is equal to the moment coming at that section due to applied loads, M Let x is the distance of the point of curtailment measured from the nearest support, then
M x Vu x
wu x2 M1 2
M1 Vu x
wu x2 2
Where, Vu is shear force at section due to applied loads. Solve above equation for x.
However, the bars are not curtailed at theoretical cut-off point Code recommends, every bar that is to be curtailed, should continue at least for ‘d’ or 12 whichever greater beyond theoretical cut-off point Check for Development Length for bars in Tension at Point of Cut-off
Total available development length at the section = M1/ V + Lo Where, Lo = sum of anchorage beyond the center of the support and the equivalent value of any hook or mechanical anchorage at simple support (which is minimum of ‘d’ or 12) M1 = Moment carrying capacity of section after curtailment V = Shear force at that section At least one-third the positive moment reinforcement in simple members and one fourth the positive moment reinforcement in continuous members shall extend along the same face of the member into the support, to a length equal to Ld /3.
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Example on Check for Development Length and curtailment of Bars A reinforced concrete beam 720 mm deep (effective) and 350 mm wide is simply supported over the span of 6 m (effective). The beam is subjected to a uniformly distributed load including its self weight of 60 kN/m. Design the beam for flexure. Apply a check for development length near support if only two third of the bars are continued to the support. Factored load = 1.5×60 = 90 kN/m Max BM at mid span = wL2/8 = 90×62/8 = 405 kNm Maximum Shear Force, V= 90×6/2 = 270 kN Max moment carrying capacity of the section as singly reinforced = 0.138fck bd2 = 0.138×20×35×7002 = 473 kNm Section may be designed as singly reinforced. Steel Required:
415 A st 405 10 6 0 .87 415 A st 720 1 20 350 720 8 .23 10 5 Ast Ast2 Ast 1558 A st 1835 mm 2
No. of 28 mm bars required 2.97 3 bars Let one of these bars are curtailed between support and mid-span If we assume that Neutral axis does not changes significantly then Moment of resistance of remaining two bars = 405×(2/3) = 270 kNm However, actual moment of resistance of two bars is found as 287 kNm The theoretical cut-off point may be determined as wx2 M x Vx 287 270x 90 x 2 / 2 2 45x 2 270x 287 0 x 1.38 m from sup port Actual cut-off point: at ‘12’ or at ‘d’ which ever more. Actual cut-off point = 1.38 – (720 mm or 12×28= 336 mm which ever more) = 0.66 m from support. Thus one of the three bars may be curtailed at 0.66 m from the center of the support and remaining may be continued into the support.
Check for Development Length at support for remaining 2 bars: Moment of resistance of two remaining bars = 287 kNm Available development length at the support 1 .3 M 1 L 0 if any
Check for development Length at point of inflexion
Required development length 0.87 f y 0.87 415 φ 28 1316 mm = 4τ 4 1.6 1.2
• Point of Inflexion occurs near the continuous support
V
bd
For safe design M 287 1 .3 1 L 0 L d or 1 .3 L0 1 .316 V 270 L0 = 0.0165 m = 16.5 mm Let a bend is provided. L0 for the bend = 8 = 8×28 = 224 mm > 16.5 mm OK
(bent-up bars at near top face of beam)
• At the point of Inflexion: BM changes the sign • Need to check the Development length for tension reinforcement (in tension zone beyond pt of inflection) • Check that available development length M1/ V + actual La (but not > ‘d’ or 12 whichever greater) is equal to or greater that the required development length • Note: here steel is not confined by compression reaction, therefore, 1.3 factor not to be used for (M1/V).
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Conditions for Terminating Tension Reinforcement Bars should extend at least ‘d’ or 12 (which ever greater) beyond Theoretical Cut-off point (TCP) Actual shear capacity available at TCP not less than required shear strength i.e. (Vuc + Vus) > or = 1.5 Vu OR For bars with diameter ≤ 36mm, if the moment carrying capacity of section is double than moment at section and (Vuc + Vus) 1.33 Vu If above conditions are not satisfied Need to provided additional shear reinforcement as follows Determine area of additional shear reinforcement as Asv = 0.4 bs/fy (b= breadth of beam; s= spacing of shear rein.) The Resultant spacing not greater than d / (8 b) where, b = area of bars cut-off / total area of bars at that section
Bearing Stresses at Bends
Reinforcement Splicing As per IS 456 2000 clause 26.2.5 • Reinforcement is needed to be joined to make it longer • Reinforcement is joined by: – overlapping sufficient length bars Lap Splicing – welding to develop its full design bond stress Weld splicing – Socket type joints Mechanical splicing • Bar joints must be staggered. • The joint of reinforcing bars should be away from the sections of maximum stress • Splices in flexural members should not be at sections where the bending moment > 50% of the moment of resistance of section • Not more than half the bars shall be spliced at a section.
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Lap Splices (As As per IS 456 2000 clause 26.2.5.1) • Force is transferred by bond between bar and concrete • They should be used for bar diameters up to 36 mm. • They should be considered as staggered if the C/C distance of the splices > 1.3 lap length (as mentioned Below) • The lap length including anchorage value of hooks for bars In flexural tension Ld or 30 , whichever is greater In direct tension 2Ld or 30 , whichever is greater In compression Ld (in compression) or 24 whichever is greater • • The lap length shall be calculated on the basis of diameter of the smaller bar when bars of two different diameters are to be spliced. • Lap splices of bundled bars shall be made by splicing one bar at a time and all such individual splices within a bundle shall be staggered. • Due to splicing high shear stress in concrete cracking in concrete • In direct tension enclose splice by 6mm spirals @ 100mm (max)
Welded Splicing & Mechanical Connections (As As per IS 456 2000 clause 26.2.5.1) • Becomes necessary for jointing the bars of diameter > 36 mm • The strength of welded splices and mechanical connections shall be taken as 100% of the design strength of joined bars for compression splices. • For tension splices, such strength of welded bars shall be taken as 80% of the design strength of welded bars. • However, it can go even up to 100% if – welding is strictly supervised and – if at any cross-section of the member not more than 20 per cent of the tensile reinforcement is welded. • For mechanical connection of tension splice, 100% of design strength of mechanical connection shall be taken.
Side face reinforcement As per Clause 26.5.1.3 of IS 456:2000 If depth of the web in a beam > 750 mm, side face reinforcement shall be provided along the two faces Total area of such reinforcement 0.1% of the web area This 0.1% steel shall be distributed equally on two faces Spacing of side reinforcement ≤ 300 mm or web thickness (whichever less)
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