27 Lecture Note – 6 Moment-Curvature (M-φ) Relation - I M-φcharacteristics considering IS: 456: The actual moment-curvature relationship of R.C. prismatic section is obtained from stress-strain diagram of concrete and steel. Starting from the basic equations, expressions for the axial force and moment carrying capacity of the section are calculated in nondimensional form. The equation of the stress strain diagram of the parabolic part for concrete is 2 σ ε ⎛ε ⎞ = 2 − ⎜ ⎟ …………………………………(1) σ cu ε0 ⎝ ε0 ⎠ subject to the following limiting conditions: ( Figs. 1 and 2) 1. at ε = 0 , σ = 0, 2. at ε = ε0 , σ = σcu, dσ = 0. 3. at ε = ε0, dε The permissible compressive stress in concrete are considered as: σcu = 0.446fck Under any loading condition, the section undergoes strains and consequent stresses. A linear strain distribution over the depth of cross-section is assumed for stress distribution in concrete and steel(Figs.1 and 2), giving rise to two possible cases: 1. 0 ≤ ε4 ≤ ε0 and 2. ε0 ≤ ε 4 ≤ εu 2 ⎡ ε ⎛ ε ⎞ ⎤ ∴ σ = 0.446 f ck ⎢2 −⎜ ⎟ ⎥ For ε < 0.002 ⎢⎣ 0.002 ⎝ 0.002 ⎠ ⎥⎦ =0.446 f ck for 0.002 ≤ ε ≤ 0.0035 28 Fe250 Fig. 1 Stress strain diagram for steel (Fe 250) Fig. 2 Stress strain diagram for concrete by IS code 29 εy = 0.87 f y Fig 3 Stress strain diagram of steel (Fe415) For Fe415 For Fe500 ε σ (MPa) ε σ (MPa) 0.00000 0.00144 0.00163 0.00192 0.00241 0.00276 ≥ 0.0038 0.0 288.7 306.7 324.8 342.8 351.8 360.9 0.00000 0.00174 0.00195 0.00226 0.00277 0.00312 ≥ 0.00417 0.0 347.8 369.6 391.3 413.0 423.9 434.8 Es + 0.002 30 Case – I Case - II Xc Xt D Xc D Xt D Xc D D D (i) Cross section (i) Cross section ε3 ε 4 ε1 ε2 ε0 ε3 ε 4 ε1 ε 2 0 ≤ ε4 ≤ ε0 (ii) Strain diagram (ii) Strain diagram f3 f 2 ≤ f sy ε0 ≤ ε4 ≤ εu σ cu f4 f 2 ≤ f sy kD (iii) Stress diagram k 'D kD (iii) Stress diagram C1 C2 T (iv) Force diagram f3 C1 C3 T (iv) Force diagram Fig. 3 Stress strain distribution for case I and case II C2 f4 31 Lecture Note – 7 Moment-Curvature (M-φ) Relation - II Case – I ( 0 ≤ ε4 ≤ ε0 ) According to Fig. 3, k= ε4 ε2 + ε4 = x d Where k is the ratio of the neutral axis depth to the effective depth. ε1 and ε4 are the extreme fiber strains in cross section. For the given section, tensile force for the tensile reinforcement, T = Pt bDf2 = Pt bDEsε2 Where Es ε2 ≤ 0.87 f y Pt = Percentage of tensile steel b = Width of the section D = Overall depth of the section Es = Modulus of elasticity of steel f2 = Stress in tensile steel ε2 = strain in tensile steel f y = Characteristic strength of steel Compressive force due to parabolic part of the stress-strain diagram of concrete : ⎡ε ⎛ 250 ⎞ 2 ⎤ C1 = σ cu kbD × 10 3 ⎢ 4 − ⎜ ⎟ε 4 ⎥ ⎣2 ⎝ 3 ⎠ ⎦ Where σ cu = Ultimate 28 – days cube strength of concrete in compression. Compressive force due to presence of compression steel is, = Pc bD Es ε3 C2 = Pc bD f3 Where Es ε3 ≤ 0.87f y Pc = Percentage of compressive steel f3 = Stress in compressive steel ε3 = Strain in compressive steel Using non-dimensional force parameter one may get, 32 Pf N C1 + C 2 − T ⎡ε ' ⎛ 1 ⎞ ⎤ P f = = k ⎢ 4 − ⎜ ⎟ε ' 24 ⎥ + c 3 − t 2 ………….(2) N0 σ cu bD σ cu ⎣ 2 ⎝ 12 ⎠ ⎦ σ cu Es ε3 ≤ 0.87 f y Es ε2 ≤ 0.87 f y Axial thrust Ultimate axial thrust ( = σ cu bD) υ= Where f3 f2 N N0 = = = = Similarly, nondimensional moment parameter, M M µ= = M 0 σ cu bD 2 = k2 2 ⎡⎛ 2 ⎞ ⎛1⎞ 2 ⎧1 1 ⎫ ⎤ (1 − X t − X c ) Pc f 3 ...(3) ⎢⎜ 3 ⎟ ε '4 − ⎜ 8 ⎟ ε '4 + {2(1 − X t )ε '1 − ε '4 } ⎨ 2 − 12 ε '4 ⎬ ⎥ + σ cu ⎝ ⎠ ⎩ ⎭⎦ ⎣⎝ ⎠ Where ε ' 4 = ε4 x 10 3 ε '1 = ε1 x 10 3 M = Bending moment M0 = Ultimate bending moment (σ cu bD2 ) Case II ( ε0 ≤ ε 4 ≤ εu ) According to Fig. 3, k ' = ε0 ε2 + ε4 = ε0 k ε4 For the given section compressive force due to parabolic part of the stress-strain diagram 2 k’ σ cubD of concrete is: C1 = 3 Compressive force due to compression steel, if present, is: C2 = Pc bD f3 Compressive force due to straight part of the stress-strain diagram: C3 = σ cu(k-k’)bD Tensile force due to tensile steel: T = Pt bDf2 From the above values of C1, C2 , C3 and T, one may get P f Pf 1 N = k − k '+ c 3 − t 2 …………….……………(4) ν = σ cu σ cu 3 N0 and, µ = M ⎛ 5 ⎞ '2 ⎛ 2 ⎞ ⎛1⎞ = ⎜ ⎟ k + ⎜ ⎟ k ' ( (1 − X t ) − k ) + ⎜ ⎟ ( K − K ')( 2 × (1 − X t ) + K '− K ) + M 0 ⎝ 12 ⎠ ⎝ 3⎠ ⎝ 2⎠ (1 − X t − X c ) pC f 3 σ cu ………………….……(5) To satisfy the equations (2), (3), (4) and (5) following conditions are to be fulfilled. 1. Es ε3 ≤ 0.87 f y 33 2. Es ε2 ≤ 0.87 f y Putting different values of ν in the equations (2) and (4) and assigning some specific values of ε4 (ε4 ≤ 0.0035), corresponding values are calculated. The strain of concrete being known or assumed, the rotation capacity of the structure is calculated. Corresponding to this rotation capacity, moment carrying capacity of the structure is calculated from the equations (3) and (5). For specific values of compression and tensile steel, the value of ∆ε (ε1+ε4) and µ ( = M / M0 ) are calculated. Governing equations for M − Φ characteristics considering ACI Code: Reference: P. Desayi and S. Krishnan, “Equation for the stress – strain curve of concrete” Journal of American Concrete Institute, Vol. 61, pp. 345-350, 1964. The equation of the stress-strain curve given by ACI Code is = Eε ⎛ε ⎞ 1+ ⎜ ⎟ ⎝ ε0 ⎠ Where, E = 2σ u ε0 2 = ⎛ε ⎞ 2σ u ⎜⎜ ⎟⎟ ⎝ ε 0 ⎠ .......................................... (6) ⎛ε ⎞ 1 + ⎜⎜ ⎟⎟ ⎝ ε0 ⎠ 2 = initial tangent modulus, assumed to be twice the secant modulus at Stress(N/mm2) maximum stress σ u , ε0 = strain at maximum stress. Strain( ×10−3 ) Fig 4 Stress strain diagram for concrete (ACI Code) The values of stresses and strains are taken as per ACI code. 34 Strain distribution over the depth of section is considered as linear. According to Fig. 4, ε4 k= ε2 + ε4 Where k = Ratio of the natural axis depth to the effective depth. (ε2 and ε4 are the extreme fiber strains in cross section.) For the given section, tensile force for the tensile reinforcement, T = Pt bdf2 = Pt bdEsε2 Where Es ε2 ≤ 0.87 f y Pt = Percentage of tensile steel b = Width of the section D = Overall depth of the section Es = Modulus of elasticity of steel f2 = Stress in tensile steel ε2 = Strain in tensile steel f y = Characteristic strength of steel Compressive force due to parabolic part of the stress-strain diagram of concrete : 2 σ u kbd ⎡ ⎛ ε ⎞ ⎤ C1 = ln ⎢1 + ⎜⎜ ⎟⎟ ⎥ ε0 ⎢⎣ ⎝ ε 0 ⎠ ⎥⎦ Where σu = Ultimate 28 – days cube strength of concrete in compression. Compressive force due to the presence of compression steel, is C2 = Pc bd f3 = Pcbd Esε3 Where Es ε3 ≤ 0.87f y Pc = Percentage of compressive steel f3 = Stress in compressive steel ε3 = Strain in compressive steel Using non-dimensional force parameter, 2 ε 0 ⎡ ⎛ ε 4 ⎞ ⎤ Pc f 3 Pt f 2 N C1 + C 2 − T − = = k ln ⎢1 + ⎜⎜ ⎟⎟ ⎥ + ………….(7) ν= ε4 ⎢ ⎝ ε0 ⎠ ⎥ σ u σu σ u bd No ⎣ ⎦ Where f3 = Es ε3 ≤ 0.87 f y f2 = Es ε2 ≤ 0.87 f y N = Axial thrust N0 = Ultimate axial thrust ( = σ cu bD) Similarly, nondimensional moment parameter, 35 µ = = M M = M 0 σ cu bD 2 (1 − X t ) k ln ⎡⎢1 + ⎛ ε 4 ⎞ ε4 ⎣⎢ ⎤ 2 k 2ε 2 ⎡ ⎤ (1 − X c − X t ) Pc Es 2 −1 ⎛ ε 4 ⎞ 0 ⎢ε 4 − ε 0 tan ⎜ ⎜ ⎟ ⎥− ⎟⎥ + 2 ε4 ⎣ σu ⎝ ε 0 ⎠ ⎦⎥ ⎝ ε 0 ⎠⎦ ……………………………(8) 2 Where, M = Bending moment M0 = Ultimate bending moment (σ cu bD2 ) To satisfy the equations (7) and (8) following conditions are to be fulfilled. 1. Es ε3 ≤ 0.87 f y 2. Es ε2 ≤ 0.87 f y 36 Lecture Note – 8 Behavior of RC Member: Flexure Equivalent Compression Block Total compression, C= Area of (ABFE-EFD) = 0.446 f ck xu b − 0.446 f ck Thus, C = 0.364 f ck xu b ≈ 0.36 f ck xu b Taking moment about EF 0.36 f ck xu b( x − k2 xu ) = −0.446 f ck xu b. xu (0.57 xu ) 2 − 0.45 f ck b 2 12 ⇒ k 2 = 0.416 ≈ 0.42 To find depth of NA T=C ⇒ f st Ast = 0.36 f ck xb ⇒ xu = f st Ast 0.36 f ck b For balanced section, xu 0.87 f st Ast ----------------------(i) = d 0.36 f ck bd 0.57 xu b 3 37 Limiting value of Xu /d fy ε su = + 0.002 ; 1.15 Es Again, Thus, Where Ε s = 2 × 10 5 Mpa xu ε cu 0.0035 = = d ε cu + ε su 0.0035 + ε su xu can be obtained as given below: d Type of steel ε su Fe250 Fe415 Fe500 0.0031 0.0038 0.0042 xu d 0.53 0.48 0.46 The compressive force due to concrete will be: C = 0.36 f ck xu d = 0.36 f ck Where, F = 0.36 xu d Type of steel Fe250 F 0.192 Fe415 0.175 Fe500 0.167 Moment capacity M u = 0.87 f y Ast ( L. A.) = 0.87 f y Ast ( d − 0.416 xu ) = 0.87 f y Ast d (1 − 0.416 = 0.87 f y Ast d [1 − 0.416 = 0.87 f y Ast d [1 − 1.005 xu ) d 0.87 f y Ast 0.36 f ck bd f y Ast f ck bd ] ] [From eqn (i)] xu bd = Ff ck bd d 38 = 0.87 f y pbd 2 [1 − ∴ pf y f ck ] ⎡ pf y ⎤ Mu f p = 0.87 ⎢1 − ⎥ y bd 2 f ck ⎦ ⎣ Mu in terms of concrete M u = 0.36 f ck xu b(d − 0.416 xu ) = 0.36 f ck xu d x ⎛ ⎜1 − 0.416 u d ⎝ ⎞ 2 ⎟bd ⎠ Thus, M u = Kf ck bd 2 where, K = 0.36 Type of steel Fe250 Fe415 Fe500 xu d xu ⎞ ⎛ ⎜ 1 − 0.416 d ⎟ ⎝ ⎠ K 0.15 0.14 0.13 Minimum depth for given M u M u = Kf ck bd 2 ⇒d = Mu Kf ck b For Fe250, K = 0.15, d = 6.66M u f ck b For Fe415, K = 0.14, d = 7.1M u f ck b Expression for Steel Area for Balanced Singly Reinforced Section From the equilibrium condition: T = C ∴ Ast 0.36 f ck xu f = ⋅ = k / ck bd 0.87 f y d fy ⇒ 0.87 f y Ast = 0.36 f ck bx u 39 Now, assuming p = Ast f p × 100 ; ∴ = k / ck bd 100 fy fy x 0.36 ×100 xu ⋅ = 41.3 u 0.87 d d Thus, p f ck = 100k / = Thus the percentage of balanced reinforcement pB will be as shown in the table below: Steel x/d Fe250 Fe415 Fe500 0.53 0.48 0.46 ⎛f ⎞ p⎜ y ⎟ f ck ⎠ ⎝ 21.97 19.82 18.87 pB (for M20) 1.758 0.955 0.755 Where p B = Balanced percentage of steel Expression for x /d for a given b, d & Mu M u = 0.36 f ck x⎛ x⎞ 2 ⎜ 1 − 0.42 ⎟ bd d⎝ d⎠ 2 ⎛ x⎞ ⎛ x ⎞ 6.68M u ⇒ ⎜ ⎟ − 2.4⎜ ⎟ + =0 2 ⎝d⎠ ⎝ d ⎠ f ck bd ⎡ 6.68M u ⎤ x 2 ⇒ = 1.2 − ⎢(1.2) − ⎥ d f ck bd 2 ⎦ ⎣ Thus, 1 2 x can be found out from M u , f ck , b & d d Doubly Reinforced Beam This type of beam is necessary if 1. Depth is restricted for architectural point of view 2. Moment is high 3. Moving load 4. Ductlity is required 5. Seismic resistant 6. At support of continuous beam. The strain in steel can be calculated from the relation of: ε s = 0.0035(1 − d / x ) 40 Two cases may arise regarding the stress in reinforcement 1. Strain of compression steel is reached at yield strain. 2. Strain of concrete is reached to yield strain of 0.0035 & strain of compression steel is below of yield strain. Analysis & Design of Doubly Reinforced Beam 1. Strain compatibility method using basic equation. 2. Steel Beam Theory. 3. Use of design aids (SP-16) Strain Compatibility Method Step1: Choose the value of x. Assume the concrete fails at a compressive strain of 0.0035. Then draw the strain distribution of the section. Step2: Calculate strain & corresponding stress in the tensile steel. T = f st Ast Step3: Calculate strain & corresponding stress in the compressive steel. C s = f sc Asc Step4: Calculate the compressive force in the concrete block C c = 0.36 f ck xb Step5: Calculate total compressive force as C = C s + C c Step6: Check if T = C , then assumed neutral axis depth x is OK. Otherwise choose another suitable x so that T = C Step7: M u = C c (d − k 2 x ) + C s (d − d / ) 41 Lecture Note – 9 Behavior of RC Member: Flexure Steel Beam Theory Mu1 =Maximum moment of the concrete beam can carry. Mu2 =Moment capacity of the steel Beam The total moment will be: M u = M u1 + M u 2 where M u 2 = f sc Asc ((d − d / ) Singly reinforced beam reaches ultimate strain in t Steel Beam Theory Steps for analysis Step1: Calculate M u for concrete failure as a singly reinforced beam. Thus, M u1 = 0.149 f ck bd 2 (Fe250) = 0.138 f ck bd 2 (Fe415) = 0.133 f ck bd 2 (Fe500) Step2: Determine the balanced “p” ⎛ 21.97 f ck pt = ⎜ ⎜ fy ⎝ ⎞ ⎟ for Fe250 ⎟ ⎠ 42 ⎛ 19.82 f ck =⎜ ⎜ fy ⎝ ⎞ ⎟ for Fe415 ⎟ ⎠ ⎛ 18.87 f ck ⎞ ⎟ for Fe500 =⎜ ⎟ ⎜ f y ⎠ ⎝ p bd Ast1 = t 100 Step3: Determine f sc , Then C s = f sc Asc Step4: Find additional moment, M u 2 = f sc Asc ( d − d / ) Step5: Find total moment from compression failure M u = M u1 + M u2 = Muc Step6: Find Ast 2 by tension failure; Ast 2 = Ast − Ast1 ( ) Step7: Find the total moment by tension failure, M u = M u1 + Ast 2 (0.87 f y ) d − d / = M ut Step8: M u = lesser value of M uc & M ut For given value of Mu, b, d and grade of concrete and steel, find out Asc & Ast Step-1. Find M ul = kf ck bd 2 for concrete failure. Step-2. Calculate Ast1 considering balanced reinforced section. ( Step-3. Find M u 2 = M u − M u1 = 0.87 f y Ast 2 d − d / Step-4. Calculate Ast 2 = M u2 0.87 f y d − d / ( ) ) ∴ Ast = Ast1 + Ast 2 Step-5. Find f sc which depends on d ' . Find the corresponding stresses from table F of d SP-16. ⎛ d/ ⎞ Otherwise ε sc = 0.0035 ⎜1 − ⎟ . The value of fsc then can be ⎝ xu ⎠ found from the stress-strain diagram of steel. 43 Step-6. Find Asc = Ast 2 (0.87 f y ) f sc Thus Ast & Asc can be found out Design Aids (SP-16) • Tables are easier than chart • Charts 19 &20 give Ast2 value for (d-d/) and Fe250 only. • For other grade of steel, it is to be modified by table G on page 13 of SP16 • Table (45-56) give values for direct design of doubly reinforced beams. A p c = sc × 100 bd A pt = st × 100 where , Ast = Ast1 + Ast 2 bd p t = p t1 + p t 2 p c & p t are obtained from the following expression. We know, M u = M u (lim) + Or, pt 2 bd (0.87 f y )(d − d / ) 100 / M u M u (lim) pt 2 = + (0.87 f y )(1 − d ) 2 2 d 100 bd bd Now, p t = p t1(lim) + p t 2 and ⎛ 0.87 f y p c = pt 2 ⎜⎜ ⎝ f sc ⎞ ⎟⎟ ⎠ Maximum & minimum Tension steel in Beam: Maximum steel M u = 0.87 f y fy p ⎞ 2 p ⎛ ⎜1 − ⎟ bd f ck 100 ⎠ 100 ⎝ Here M u increases with p in a parabolic relation. For M20 & Fe415, M u = 3.6 p − 0.75 p 2 44 ⎛x⎞ From T = C we can have: 0.36 f ck b ⎜ ⎟ d = f st Ast ⎝d ⎠ ⇒ f st x p = d 0.36 f ck 100 x = 0.58 p d Ductility can be measured by curvature. For Fe415 & M20: ⇒ εc x ϕ dθ εs Thus, 1 Strain in compression fibre 0.0035 = = R Depth of NA x 1 dθ ε c 0.0035 0.006 =φ = = = = R dx x 0.58 pd pd Thus, Ductility ∝ 1 p Hence, though M u increases with the increase of p but ductility (i.e., curvature) decreases. Hence, IS code (Clause 26.5.1) has put the upper limit of tension & compression steel as 4%. Minimum Steel Minimum steel is required to take care ductility & shrinkage of concrete. As per clause 26.5.1 45 As 0.85 = bd fy = 0.00205 = 0.2% for Fe415 = 0.0035 = 0.35% for Fe250 Necessity of minimum steel for shear Clause:26.5.1.6 Minimum steel is necessary to 1. Prevent brittle shear failure, which can occur without shear steel. 2. Guard against any sudden failure of a beam if concrete cover bursts and the bond to the tension is lost. 3. Prevent failure that can be caused by tension due to shrinkage and thermal stresses and internal cracking in the beam. 4. Hold the reinforcements in place while pouring the concrete. 5. Act as the necessary ties for the compression steel and make them effective. Minimum spacing Asv 0.4 = bSv 0.87 f y ⎛ AS ⎞ Sv = 2.175 f y ⎜ v ⎟ ⎝ b ⎠ ⎛ AS = 902 ⎜ v ⎝ b ⎞ ⎟ for Fe415 ⎠ ⎛ AS = 544 ⎜ v ⎝ b ⎞ ⎟ for Fe250 ⎠ Maximum spacing Sv ⊄ 0.75d for vertical stirrups ⊄ d for inclined ( 450 ) ⊄ 300 in any case Asv= total area of stirrups Sv=Spacing b=breadth of web at level of tension 46 Lecture Note – 10 Behavior of RC Beam: Shear Types of shear Failure Shear strength of RC Beam (Without web Reinforcement) Total resistance = vcz + vay + vd 47 vcz → Shear in compression zone vay → Aggregate interlock forces vd → Dowel action from longitudinal bars 1. 2. 3. 4. 5. Tensile strength of concrete Æ Affect inclined cracking load Longitudinal reinforcement (p) ÆRestrain cracks Increase in the depth of beamÆ Reduced shear stress at inclined cracking Axial tension ÆDecrease inclined cracking load Axial compressionÆ Increase inclined cracking load Function & strength of web reinforcement Function of web Reinforcement • Web reinforcement is provided to ensure that the full flexural capacity is developed. • Acts as ‘clamps’ to keep shear cracks from widening. • Shear resisted by v s apart from vcz , v ay & v d . • v s increases as cracks widen until yielding of stirrups & then stirrups provide constant resistant. Flexural Cracking: Shear is resisted by Vcz , Vay , Vd & V s Designing to Resist Shear Shear Strength (ACI 318 Sec 11.1) φVn ≥ Vu [i.e., Capacity ≥ demand] 48 Vu Æfactored shear force at section Vn ÆNormal shear strength φ Æ0.85(Shear)-strength reduction factor V n = Vc + V s Vc ÆNormal shear resistance provided by concrete Vs Æ Normal shear provided by the shear reinforcement Shear Strength Provided by Concrete Bending only Vc = 2 f c bw d Eqn [11.3] Simple formula: ≤ 3.5 f c bw d More detailed: Vc ⎛ ⎛ V d ⎞⎞ = ⎜ 1.9 f c + 2500 pw ⎜ u ⎟ ⎟ bw d ⎜ ⎟ ⎝ Mu ⎠⎠ ⎝ ≤ 3.5 f c bw d Eqn [11.6] Bending and Axial Compression Simple formula ⎛ N u ⎞⎟ Vc = 2 ⎜1 + f b d Eqn [11.4] ⎜ 2000 A ⎟ c w g ⎠ ⎝ ≤ 3.5 f c bw d 1 + Nu 500 Ag N u is positive for compression and Nu Eqn [11.8] Ag are in psi. More detailed ⎛ 4h − d ⎞ M m = M u − Nu ⎜ ⎟ Use M m in Eqn [11.6] with no limits ⎝ 8 ⎠ Vc ≤ 3.5 f c bw d 1 + Nu Eqn [11.8] 500 Ag N u is positive for compression and N u Ag are in psi. ⎛V d ⎞ Note ⎜⎜ u ⎟⎟ ≤ 1 ⎝ Mu ⎠ 49 Bending and Axial Tension Simple formula Vc = 0 Design shear reinforcement for all shear ⎛ Nu ⎞ f b d = 2 ⎜1 + ⎜ 500Ag ⎟⎟ c w ⎝ ⎠ N u is negative for tension and N u are in psi. Ag Vc Lightweight Concrete: Shear Strength Provided by Shear Reinforcement Minimum Shear Reinforcement: (11.5.5) 1 Required when Vu ≥ φ Vc 2 Except: (a)Slabs & Footings (b)Concrete Joist Construction (defined 8.11) ⎧ 10" ⎪ (c )Beams with h ≤ larger of ⎨ 2.5 t f ⎪1/2 b w ⎩ 50 Typical Shear Reinforcement Stirrup - perpendicular to axis of Members (minimum labor - more material) Av f y d (sin α + cos α ) Vs = ACI eqn 11-15 s A f d α = 90o ⇒ Vs = v y s Bent Bars (more labor - minimum material) see required in 11.5.6 Vs = Av f y d (sin α + cos α ) s α = 45o ⇒ Vs = 1.41Av f y d s ACI 11-5.6 51 Design Procedure for Shear 1. Calculate Vu 2. Calculate φVc Eqn 11-3 or 11-5 (no axial force) 3. Check Vu ≥ 4. If ⎧ If yes, add web reinforcement (go to 4) 1 φVc → ⎨ 2 ⎩ If no, done 1 φVc ≤ Vu ≤ φVc → Provide minimum shear reinforcement 2 Av (min ) = 50 bw s fy A f ⎞ ⎛ ⎜⎜ or smax = v ys for min Av ⎟⎟ 50bw ⎠ ⎝ Also done d smax ≤ ≤ 24" (11.5.4) 2 5. If Vu ≥ φVc , calculate Vs (required) Vu ≤ φVn = φVc + φVs ⇒ φVs = Vu − φVc ⇒ Vs = Vu φ − Vc Check, Vs ≤ 8 f c′ bw d (otherwise illegal) (11.5.4) 6. Solve for required stirrup spacing (strength) Assume # 3, #4, or #5 stirrups Av f ys d from 11-15 s≤ Vs 7. Check minimum steel requirement (eqn 11-13) smax = Av f ys 50bw 8. Check maximum spacing requirement (ACI 11.5.4) d If Vs ≤ 4 f c′ bw d → smax ≤ ≤ 24" 2 d If Vs ≥ 4 f c′ bw d → smax ≤ ≤ 12 " Note: Vs ≥ 8 f c′ bw d 4 9. Use smallest spacing from steps 6,7,8 Note: A practical limit to minimum stirrup spacing is 4 inches. (Illegal) 52 Location of Maximum Shear for Beam Design Non-pre-stressed members: Sections located less than a distance d from face of support may be designed for same shear, Vu, as the computed at a distance d. Compression fan carries load directly into support. Compression fan carries load into support Location of Maximum Shear for Beam Design When: 1. The support reaction introduces compression into the end regions of the member. 2. No concentrated load occurs within d from face of support. Location of Maximum Shear for Beam Design
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