Page 1 :
SHEAR, BOND AND, DEVELOPMENT LENGTH (L.S.M.), , , , SULIEDTIES, , © Nominal Shear Stress in R.C. Section, Design Shear Strength of Concrete, Maximum Shear Stress,, Design of Shear Reinforcement. Minimum Shear Reinforcement, Forms of Shear Reinforcement. Bond, and Types of Bond. Bond Stress, Check for Bond Stress. Development Length in Tension and, Compression, Anchorage Value for Hooks 90° Bend and 45° Bend., , * Standard Lapping of Bars, Check for Development Length. Simple Numerical Problems on Deciding, whether Shear Reinforcement is required or not, Check for Adequacy of the Section in Shear. Design, of Shear Reinforcement; Minimum Shear Reinforcement in Beams; Determination of Development, Length required for Tension Reinforcement of Cantilever Beam and Slab, Check for Development, Length., , ¢ limit State of Collapse in Shear-Design Shear Strength of Concrete - Design Strengths, Vertical/Inclined Stirrups and Bent-up Bars in Shear Principle of Shear Design - Critical Sections of, Shear - Nominal Shear Stress Design of Vertical Stirrups, Inclined Stirrups and Bent-up Bars in, Rectangular Beams using Limit State Method - Simple Problems., , About This Chapter, , At the end of this chapter students will be able to:, , ¢ Explain the pattern of shear failure for given member in structure., , © Locate the zones of minimum shear reinforcement with sketch for the given loading conditions., Design shear reinforcement for the given conditions., , Judge the beam with given reinforcement on the basis of shear strength., , Compute the bond length for the beam with given reinforcement., , Identify the zone of minimum shear reinforcement in the given element., , , , , , , , , , , , , 5. (RODUCT!, Shear in a loaded beam is caused by the variation of bending moment al, The variation in bending stress at two cross-sections in, the same fibre gives rise to horizontal shear stress which, fl 7, in turn, produces vertical shear stress. (tet cn ae, A combination of shear stress and bending stress cs, produces principal tension in the section,, At supports principal tension is at an angle of 45° and it is, called ‘Diagonal Tension’. VA, , As concrete is weak in tension, large diagonal tensile stresses can produce cracking and even failure of the Fig. 5.1: Diagonal Tension in Beam, , concrete members., Hence the beams should always be checked for safety against 'shear failure’ in physical terms, , against diagonal tension., If the shear stress Is large, steel in the form of vertical stirrups or bent-up bars should be, , Provided to take up the tensile stresses., , oe ae, ong span., Beam, , , , (6-1)

Page 2 :
5.1.1 Nominal Shear Stress (w)], , QUESTION, , , , , , , , , , , , 1. What is nominal shear stress?, , by the following, * The nominal shear stress ty in beams of uniform depth shall be ee ", , equation:, Vu, w= ba, where, Vy = Shear force due to design loads h of the web, ‘Dy’ in, , b, , Breadth of the member (shall be taken as the breadt, case. of flanged beams)., , d = Effective depth., 5.1.2 Design Shear Strength of Concrete (‘)], , (A) Without shear reinforcement: fl, ¢ The design shear strength of concrete in beams without shear re!, , , , nforcement is given jp, , , , , , , , , , , , , , , , , , , , , , , , , , , Table 5.1. ., Table 5.1: Design Shear Strength of Concrete t- (N/mm ), soc = Concrete Grade, M15 M20 M 25 M 30, , (1) (2) (3) (4) (5), , < 0.15 0.28 0.28 0.29 0.29 0.29 ., 0:25 0.35 0.36 0.36 0.37 0.37 0.38, 0.50 0.46 0.48 0.49 0.50 0.50 0.51, 0.75 0.54 0.56 0.57 0.59 0.59 0.60, 1.00 0.60 0.62 0.64 0.66 0.67 0.68, 1.25 0.64 0.67 0.70 O:71 ot] 0235 mp = 0274, 1.50 0.68 0.72 0.74 0.76 0;78)\2 11 <0.79, 1.75 0.71 0.75 0.78 0.80 0.82 0.84 |, 2.00 0.71 0.79 0.82 | 0.84 0.86 | 0.88, Ze: 0.71 0.81 0.85 0.88 0.90 | 0.92, 2.50 0.71 0.82 0.88 0.91 0.93 | 0.95, 2.75 0.71 0.82 0.90 0.94 0.96 | 0.98, 3.00 0.71 0.82 0.92 0.96 0.99 1.01, , , , Note: The term A, is the area of longitudinal tension reinforcement which continues atleast one effective depth beyond, the section being considered except at supports where the full area of tension reinforcement may be used provided the bars, , are anchored properly according to codal requirements. }, The value of t, for intermediate % of steel can be obtained by interpolation formula as given, , , , , , , , , , , , , , , , , , , , below:, $2 — 94, CeO Te 1S { !, Tey — Te *o, o, = = t, or Te = Te, + PoP, =P, (P,- Py) } 7, L——___+, pty Dt,, Fig. 5.2, (B) For solid slabs, the design shear strength for concrete shall be +,. k where k has the values, given below:, Overall depth of 300 or 275 250 225 200 175 qe, slab, mm more or, k 1.00 1.05 | 110 | 1.15 [20] ox les,, = oa

Page 3 :
on stabt reinforcement (ec, max)?, 5 ae reinforcement, shall the nominal shear stress in, , um Shear Stress, tc, max N/mm?, Th M25 M 30 M35 M 40, © Nominal sh, 3.1 3.5 3.7 4.0, far stress shall not exceed half the appropriate values given, , , , , , , , for solid slabs:, in Table 5.2., , , , , , , , , , , , , , , , , , , , at, i.e. wt Pa, 5.1.3 Maxi, : ima, m Spacing of Shear Reinforcement, 4, State maximum spacing of stirrups in even, , The maximum spacing of shi, , ., exceed 0.75 d for vertica| Stirru E ,, depth of the section under consideraticn a inclined stirrups at 45°, where d is the effective, , : No case shall the spacing exceed 300 mm., -1.4 Minimum Shear Reinforcement, , QUESTION, , inforcement giving its expression as per IS., mum shear reinforcement in the form of stirrups for the, , e, ‘ar reinforcement Measured along the axis of the member shall not, , , , 4 State the condition for minimum shear rei, « 1S, code specifies the provision of mini, , following reasons :, , o TO prevent sudden failure,, , ¢ To prevent premature failure if the bond between main steel and concrete is lost., © Toact as a tie for holding the beam reinforcement., e To confine the concrete., * Minimum shear reinforcement in the form of stirrups shall be provided such that:, Asv . _0.4, bSy = 0.87 fy, where, Acy = Total cross-sectional area of stirrup legs effective in shear., Sy = Stirrups spacing along the length of the member., b = Breadth of the beam or breadth of the web of flanged beam and, characteristic strength of the stirrup reinforcement in N/mm2., Which shall not be taken greater than 415 N/mmz2, (that is, even for Fe 500,, fy in above relation shall be taken as 415 N/mm2),, , However, in members of minor structural importance, such as lintels or where the maximum, Shear stress calculated is less than half the permissible value, this provision need not be, compiled with., , ut, , fy, , , , , , , , , , , , , , , , , , , , 1. State various forms of shear reinforcement giving expression for strength as per IS., * When wy exceeds + given in Table shear reinforcement shall be provided in any of the following, C, forms;, , (8) Verti Bent-up bars alongwith stirrups, and (c) Inclined stirrups, the ain > rarely used in practice because of high fabrication labour cost,, , _ The common types of vertical stirrups are shown in Fig. 5.3.

Page 5 :
(Sibi, , 1.8 Zo), nes of Mi, in, , qreré are three following zones g imum Shear Reinforcement, pee 4, Design shear reinforcement z, , , Minimum shear reinforcemen|, , "Nominal ee reinforcement zone, Consider an R.C.C. beam subjec:, rhe aesign shear can be rieadoe design u.d./, ‘wa! over entire span., 4, Concrete alone. However /, , zone III. , Nominal shear reinforcement is required to be provided in, 2, Concrete and minimum shear reinf, ‘0, , 3. ics tat a shear ralntGteantenht'n zone 1. ne, The for! pacing fo :, discussed in Article Eta ee shear reinforcement and design shear reinforcement are, , , , , , 7, , Ail ear reinforcement along the span of the beam. ., , t zone, , , , , 1, ' Centre of beam, 1, , , , , , , , , , , , , , , , , , | bs i, Design shear Minimi ' . i, jum shear Nominal shear 1, reinforcement zone reinforcement zone reinforcement zone j, U2, Z 0.87 fy Asv.d Se 0.87 fy Asv i, Vus < 0.46, Fig. 5.4, , , , , , 5.1.9 Procedure for Design of Beams for Shear, , , , , , QUESTION, 4. Write entire procedure step-by-step for providing shear reinforcement in an R.C.C. bi, , * Following steps are adopted for design of beams of shear., wh, 1, Calculate shear force VA >, , eam according to LSM., , , , Design or factored shear force, Vu = Vx 9 (for D. L. and L. L. yp = 1.5), , 2. Calculate nominal shear stress, , Vu, ty = bd, It should be less than tc, max:, A,, 3. Find percentage of steel, % Pr = bd * 100, , From % p, calculate shear strength of concrete tc from Table 5.1 by interpolation., , 4. If ty > t¢ shear reinforcement Is required., , (However if ty < tc then provide nominal shear reinforcement)., , 5. Calculate shear force for which shear reinforcement is required,, Vus = Vur te. bd, , 8. If bent-up bars are used, calculate shear carried by bent-up bar, , Vusb = 0.87 fy Asb + sina, it, , and should be < 0.5 Vus .