Lecture-04 Design of RC Members for Shear and Torsion

Lecture-04 Design of RC Members for Shear and Torsion By: Prof. Dr. Qaisar Ali Civil Engineering Department UET Peshawar drqaisarali@uetpeshawar.edu.pk www.drqaisarali.com 1 Topics Addressed Design of RC Members for Shear Shear Stresses in Rectangular Beams Diagonal Tension in RC Beams Subjected to Flexure and Shear Loading Types of Cracks in RC Beams Shear Strength of Concrete Web Reinforcement Requirement ACI Code Provisions for Shear Design Effect of Axial Force on Shear Strength of Concrete 2 1

Topics Addressed Design of RC Members for Torsion Torsional Stresses in Solid Concrete Members Torsional Strength of Concrete Reinforcement Requirement ACI Requirements for Design of RC Members Subjected to Torsion Steps for Design of RC Member Subjected to Torsion Example 3 Design of RC Members for Shear 4 2

Shear Stresses in Rectangular Beams Shear stresses in homogeneous Elastic Rectangular Beams The shear stress (ν) at any point in the cross section is given by ν = VQ Ib Where; V = total shear at section I = moment of inertia of cross section about neutral axis b = width of beam at a given point Q = statical moment about neutral axis of that portion of cross section lying between a line through point in question parallel to neutral axis and nearest face (upper or lower) of beam 5 Shear Stresses in Rectangular Beams Shear stresses in homogeneous Elastic Rectangular Beams For the calculation of shear stress at level cd in the given figure, Q will be equal to A abcd y, where A abcd is area abcd and y is the moment arm. ν cd ν max 6 3

Shear Stresses in Rectangular Beams Shear stresses in homogeneous Elastic Rectangular Beams For shear at neutral axis ef: Q = A abef h/4 = bh 2 /8 a b I = bh 3 /12 Therefore, h e f h/4 ν = VQ/Ib = V(bh 2 /8)/{(bh 3 /12)b} ν max = (3/2)V/bh b 7 Shear Stresses in Rectangular Beams RC Beams in Non-Linear Inelastic Range When load on the beam is such that stresses are no longer proportional to strain, then equation v = VQ/Ib for shear stress calculation does not govern. The exact distribution of shear stresses over the depth of reinforced concrete member in such a case is not fully known. 8 4

Shear Stresses in Rectangular Beams RC Beams in Non-Linear Inelastic Range Tests have shown that the average shear stress prior to crack formation is: ν av = V/ bd 9 Diagonal Tension in RC Beams subjected to Flexure and Shear Loading 10 5

Diagonal Tension in RC Beams subjected to Flexure and Shear Loading Conclusions from Previous Discussion The tensile stresses are not confined to horizontal bending stresses that are caused by bending alone. Tensile stresses of various inclinations and magnitudes resulting from shear alone (at the neutral axis) or from the combined action of shear and bending, exist in all parts of a beam and can impair its integrity if not adequately provided for. It is for this reason that the inclined tensile stresses, known as diagonal tension stress must be carefully considered in reinforced concrete design. 11 Types of Cracks in Reinforced Concrete Beam These are of two types 1. Flexure Cracks 2. Diagonal Tension Cracks I. Web-shear cracks: These cracks are formed at locations where flexural stresses are negligibly small. II. Flexure shear cracks: These cracks are formed where shear force and bending moment have large values. 12 6

Shear Strength of Concrete Tensile Strength of Concrete The direct tensile strength of concrete ranges from 3 to 5 f c as given in table below: Table: Approximate range of tensile strengths of concrete Normal-Weight Concrete, psi Lightweight Concrete, psi Direct tensile strength f t ' 3 to 5 f c ' 2 to 3 f c ' Split-cylinder strength f ct 6 to 8 f c ' 4 to 6 f c ' Modulus of rupture f r 8 to 12 f c ' 6 to 8 f c ' 13 Shear Strength of Concrete Shear Strength of Concrete in Presence of Cracks A large number of tests on beams have shown that in regions where small moment and large shear exist (web shear crack location) the nominal or average shear strength is taken as 3.5 f c. However in the presence of large moments (for which adequate longitudinal reinforcement has been provided), the nominal shear strength corresponding to formation of diagonal tension cracks can be taken as: ν cr = 2 f c 14 7

Shear Strength of Concrete Shear Strength of Concrete in Presence of Cracks The same has been adopted by ACI code (refer to ACI 22.5.2.1). This reduction of shear strength of concrete is due to the preexistence of flexural cracks. It is important to mention here that this value of shear strength of concrete exists at the ultimate i.e., just prior to the failure condition. 15 Web Reinforcement Requirement in Reinforced Concrete Beams Nominal Shear Capacity (V n ) of Reinforced Concrete Beam Now general expression for shear capacity of reinforced concrete beam is given as: V n = V c + V s Where, V c = Nominal shear capacity of concrete, V s = Nominal shear capacity of shear reinforcement. Note: in case of flexural capacity M n = M c + M s (M c = 0, at ultimate load) 16 8

Web Reinforcement Requirement in Reinforced Concrete Beams Nominal Shear Capacity (V n ) of Reinforced Concrete Beam Unlike moment, in expression for shear, the term V c 0 because test evidence have led to the conservative assumption that just prior to failure of a web-reinforced beam, the sum of the three internal shear components is equal to the cracking shear V cr. 17 Web Reinforcement Requirement in Reinforced Concrete Beams Nominal Shear Capacity (V n ) of Reinforced Concrete Beam This sum is generally (somewhat loosely) referred to as the contribution of the concrete to the total shear resistance, and is denoted by V c. Thus V c = V cr and V c = V cz + V d + V iy Where, V iy = Vertical component of sizable interlock forces. (V i ) have in fact been measured, amounting to about one-third of the total shear force. V cz = Internal vertical forces in the uncracked portion of the concrete. V d = Internal vertical forces across the longitudinal steel, acting as a dowel. 18 9

Web Reinforcement Requirement in Reinforced Concrete Beams Nominal Shear Capacity (V n ) of Reinforced Concrete Beam V c = V cz + V d + V iy = 2 (f c )b w d [ACI 22.5.5.1] V s = A v f y d/s V u = ΦV c + ΦV s V u = ΦV c + ΦA v f y d /s s = ΦA v f y d/ (V u ΦV c ) 19 ACI Code Provisions for Shear Design Strength Reduction Factor The strength reduction factor Φ is to be taken equal to 0.75 for shear. 20 10

ACI Code Provisions for Shear Design Spacing Requirements for Stirrups Note: (V u - Φ V c ) Φ 8 f c b w d ; otherwise increase the depth 21 ACI Code Provisions for Shear Design Location of Critical Section for Shear Design 22 11

ACI Code Provisions for Shear Design Location of Critical Section for Shear Design Typical support conditions Typical support conditions Concentrated load within distance d A beam loaded near its bottom edge, e.g. inverted T-beam End of a monolithic vertical element Fig a and b are the more frequent conditions while the fig c, d and e are the special conditions 23 Effect of Axial Force on Shear Strength of Concrete Members Axial Compression The ACI Code provides that for members carrying axial compression as well as bending and shear, the contribution of the concrete be taken as: V c = 2{1+ N u / (2000A g )} (f c )b w d N u Where, A g = Gross area of concrete N u = Axial load N u 24 12

Effect of Axial Force on Shear Strength of Concrete Members Axial Tension The ACI Code provides that, for members carrying significant axial tension as well as bending and shear, the contribution of the concrete be taken as: V c = 2{1+ N u / (500A g )} (f c )b w d N u but not less than zero, where N u is negative for tension. N u 25 Design of RC Members for Torsion 26 13

Introduction Torsional Stresses The shear stress induced due to applied torque on a member is called as torsional shear stress or torsional stress. Applied Torque, T 27 Circular Members Torsional Stresses Torsional stresses in solid circular members can be computed as: τ = Tρ/J Where, T = applied torque, ρ = radial distance, J = polar moment of inertia. 28 14

Rectangular Members Torsional Stresses Torsional stress variation in rectangular members is relatively complicated. 29 Rectangular Members Torsional Stresses The largest stress occurs at the middle of the wide face a. The stress at the corners is zero. Stress distribution at any other location is less than that at the middle and greater than zero. τ max = T/ (αb 2 a) Table: Variation of α with ratio a/b. a/b 1 1.5 2 3 α 0.2 0.23 0.24 0.267 30 15

Rectangular Members Torsional Stresses Torsional stress close to the faces of the rectangular member is much greater than that of interior section. 31 Rectangular Members Torsional Stresses From the previous discussion it is concluded that torsional stresses are concentrated in a thin outer skin of the solid cross section. This leads to the concept of thin walled tube analogy. 32 16

Torsional Stresses Torsional Stresses From Thin Walled Tube Analogy According to ACI code, torsional stresses in a solid rectangular section are computed using Thin Walled Tube Analogy. 33 Torsional Stresses Torsional Stresses From Thin Walled Tube Analogy In the thin-walled tube analogy the resistance is assumed to be provided by the outer skin of the cross section roughly centered on the closed stirrups. According to thin walled tube analogy, shear stress and thus shear flow remains constant within the thin walls of the tube i.e., τ (shear stress)= shear force/ shear area = constant q (shear flow)= shear force per unit length = constant 34 17

Torsional Stresses Torsional Stresses From Thin Walled Tube Analogy τ 1 = V 1 /x o t (shear stress = shear force/ shear area) According to thin-walled tube theory, V 1 /x o t = V 2 /y o t = V 3 /x o t = V 4 /y o t = τ = constant Also as, q 1 = V 1 /x o {shear force/ unit length (shear flow)} Therefore, τ = q 1 /t OR q 1 = τt 35 Torsional Stresses Torsional Stresses From Thin Walled Tube Analogy As the terms V 1 through V 4 are induced shear and cannot be easily determined, therefore, they can be expressed in terms of torque. Taking moments about centerline of thin walled tube, T = V 4 x o /2 + V 2 x o /2 + V 1 y o /2 + V 3 y o /2 T = (V 4 + V 2 ) (x o /2) + (V 1 + V 3 ) (y o /2) As, V 4 = V 2 and V 3 = V 1, therefore, T = (2V 2 ) (x o /2) + (2V 1 ) (y o /2) As, q 1 = V 1 /x o and q 2 = V 2 /y o. Therefore, T = 2qy o x o /2 + 2qx o y o /2 = 2qA o As, q = τt, therefore, τ = T/ (2A o t) 36 18

Torsional Strength of Concrete Torsional Strength of Concrete as per ACI Code We know that in case of shear, shear strength of concrete is given as υ c = 2 (f c ), and As average shear stress (υ av ) is equal to V/bd, therefore V/bd = 2 (f c ) V = V c = 2 (f c )bd In case of torsion induced shear stresses (torsional stresses), ACI 22.7.5 says that cracking is assumed when tensile stresses reach 4 (f c ). τ c = 4 (f c )... (a) From the previous discussion on torsional stresses in thin walled tube: τ = T/ (2A o t). (b) Equating (a) and (b) gives: T c = 4 (f c ) (2A o t) 37 Torsional Strength of Concrete Torsional Strength of Concrete as per ACI Code According to ACI R22.7.5, A o = (2/3)A cp, t = (3/4)A cp / p cp t Where, A cp = xy, p cp = 2(x + y) {full section of the member} Substituting values of A o and t in eqn. for T c gives: T c = 4 (f c ) A cp2 / p cp x This equation represents the capacity of rectangular concrete member in torsion. y 38 19

Torsional Strength of Concrete Definition of various terms A o = gross area enclosed by shear flow path. A oh = area enclosed by centerline of the outermost closed transverse torsional reinforcement; According to thin-walled analogy, as the resistance is assumed to be provided by the outer cross section roughly centered on the closed stirrups, A o should be equal to A oh ; however to account for cracking ACI R22.7.6.1 requires A o to be calculated as A o = 0.85A oh. Therefore A oh = x o y o & A o = 0.85A oh. 39 Torsional Strength of Concrete Definition of various terms A cp = area enclosed by outside perimeter of concrete cross section p cp = outside perimeter of the concrete cross section p h = perimeter of centerline of outermost closed transverse torsional reinforcement Cross hatched area =A oh b eff 40 20

Reinforcement Requirement Review of Reinforcement Requirement for Flexure To understand the reinforcement requirement for torsion, recall the concept of flexural design of RC beam. Elastic Range Flexural Capacity For an uncracked concrete beam, the flexural stresses are given by: f = My/I According to ACI code, at f = f r =7.5 f c, concrete is at the verge of failure in tension. Therefore capacity of section at that stage is: M = M cr = f r y/i 41 Reinforcement Requirement Review of Reinforcement Requirement for Flexure Ultimate Flexural Capacity For a cracked RC beam at ultimate stage, the flexural capacity is given as: M n = M c + M s Where, M c = Nominal flexural capacity of concrete in tension, M s = Nominal flexural capacity of tension steel. 42 21

Reinforcement Requirement Review of Reinforcement Requirement for Flexure Ultimate Flexural Capacity As concrete is weak in tension (Refer ACI 9.5.2.1 for concrete tensile strength), M c 0, therefore, M n M s = A s f y (d a/2) So the tension reinforcement along with the concrete in compression acts as a couple to resists the flexural demand on the member. 43 Reinforcement Requirement Review of Reinforcement Requirement for Shear Similarly, recall the concept of shear design of RC beam. Elastic Range Shear Capacity For an uncracked concrete beam, the shear stresses are given by: υ = VQ/Ib According to ACI code, υ = υ cr = 2 f c is the nominal shear strength corresponding to formation of diagonal tension cracks. Therefore shear capacity of section at that stage is: V cr = υ cr Ib/Q = 2 f c Ib/Q 44 22

Reinforcement Requirement Review of Reinforcement Requirement for Shear Ultimate Shear Capacity For a cracked RC beam at ultimate stage, the shear capacity is given as: V n = V c + V s Where, V c = Nominal shear capacity of concrete, V s = Nominal shear capacity of shear reinforcement. 45 Reinforcement Requirement Review of Reinforcement Requirement for Shear Ultimate Shear Capacity Unlike flexure, the term V c 0 because from test evidence: V c = V cz + V d + V iy = 2 (f c )b w d [ACI 22.5.5.1] Therefore shear steel (stirrups) along with the contribution of concrete (V c ) acts together to resists the shear demand due to applied load on the member. 46 23

Reinforcement Requirement Reinforcement requirement for Torsion In the case of Torsion ΦT n = ΦT c + ΦT s ACI Code requires that ΦT c shall be taken equal to zero. Therefore, ΦT n = ΦT s We will see in the next slides that how ΦT s is calculated. 47 Reinforcement Requirement Reinforcement requirement for Torsion Reinforcement requirement for reinforced concrete members subjected to torsion is determined using thin walled tube and space truss analogies. As discussed earlier, from thin walled tube analogy, internal effects in the form of induced shear forces (V 1 to V 4 ) will generate due to applied torque (T) as shown. 48 24

Reinforcement Requirement Space Truss Analogy Such internally induced shear forces will crack the member as shown. Due to cracks the member splits up into diagonal compressive portions or struts. 49 Reinforcement Requirement Space Truss Analogy If the compressive force in the strut is C, then it can be resolved into two components: C H = Horizontal component C V = Vertical component C H C C V Longitudinal reinforcement shall be provided to resist C H and vertical stirrups shall be provided to resist C V. This leads to space truss analogy. 50 25

Reinforcement Requirement Space Truss Analogy In space truss analogy, the concrete compression diagonals (struts), vertical stirrups in tension (ties), and longitudinal reinforcement (tension chords) act together as shown in figure. The analogy derives that torsional stress will be resisted by the vertical stirrups as well as by the longitudinal steel. Vertical Reinforcement acts as ties C H C C V Concrete Compression strut Long. Reinforcement Acts as tension chord 51 Reinforcement Requirement Vertical Stirrup Reinforcement Refer to figure (a), from space truss analogy, C v = V 4 Refer figure (b), V 4 = na t f yv = (y o cotθ/s) A t f yv [where n = no of stirrups] As V 4 = V 2, therefore V 2 = (y o cotθ /s) A t f yv And V 1 = V 3 = (x o cotθ /s) A t f yv (b) C H C V 4 C V N4 (a) Face 4 of the member. 52 26

Reinforcement Requirement Vertical Stirrup Reinforcement If V 1 to V 4 are known, A t can be determined from previous eqns. However as discussed earlier, it is convenient to express V 1 to V 4 in terms of T by taking moments about centerline of thin walled tube, T n = V 4 x o /2 + V 2 x o /2 + V 1 y o /2 + V 3 y o /2 With V 4 = V 2 and V 1 = V 3, we have, T n = 2V 2 x o /2 + 2V 1 y o /2 OR T n = V 2 x o + V 1 y o Putting values of V 1 and V 2 derived earlier, T n = {(y o cotθ/s)a t f yv }x o + {(x o cotθ/s)a t f yv }y o T n = 2(A t /s)f yv x o y o cotθ With x o y o = A o, we have, T n = 2(A t /s)f yv A o cotθ.. (i) 53 Reinforcement Requirement Vertical Stirrup Reinforcement For no failure, torsional capacity of the member shall be greater than or equal to torsional demand i.e., ΦT n T u [ with Φ = 0.75 ] For ΦT n = T u, equation (i) becomes, T u = Φ2(A t /s)f yv A o cotθ Where, A t = Steel area in one leg of stirrup, therefore from above equation: A t (one legged) = T u s/ (Φ2f yv A o cotθ) For two legs, we have, A t (2 legged) = 2T u s/ (Φ2f yv A o cotθ) A t (2 legged) = T u s/ (Φf yv A o cotθ) 54 27

Reinforcement Requirement Vertical Stirrup Reinforcement If θ = 45 o (for non-prestressed members, ACI 22.7.6.1), then: A t (2 legged) = T u s/ (Φf yv A o ) [where A o = 0.85A oh (ACI R22.7.6.1)] This expression can be used to find shear reinforcement requirement due to torsion. The total shear reinforcement requirement is therefore the sum of the shear reinforcement requirements due to direct shear and torsion both i.e., A total = A t (2 legged) + A v (2 legged) Where, A v (2 legged) = (V u ΦV c )s/ (Φf yv d) 55 Reinforcement Requirement Longitudinal Steel Reinforcement Refer figure (b) and (c), for face 4 we have, ΔN 4 = V 4 cot θ = V 4 [ for θ = 45 o ] Similarly, ΔN 1 = V 1 ; ΔN 2 = V 2 ; ΔN 3 = V 3 ; Total longitudinal reinforcement for torsion is: N = N 1 + N 2 + N 3 + N 4 (c) N = V 1 + V 2 + V 3 + V 4 As V 1 = V 3 and V 2 = V 4, therefore, N = 2V 1 + 2V 4 C H C V 4 C 4 V N (a) Face 4 of the member. 56 28

Reinforcement Requirement Longitudinal Steel Reinforcement N = 2V 1 + 2V 4 With, V 4 = (y o /s)a t f yv, and, V 1 = (x o /s)a t f yv N = 2 (x o /s)a t f yv + 2 (y o /s)a t f yv (c) N = 2 (x o + y o )A t f yv /s As ΔN = A l f yl,and therefore, A l f yl = 2 (x o + y o )A t f yv /s As (2x o + 2y o ) = p h, therefore, A l = (A t /s) p h (f yv /f yl ) C H C V 4 N C 4 V (a) Face 4 of the member. 57 Reinforcement Requirement Longitudinal Steel Reinforcement Now as derived earlier, A t (one legged) = T u s/ (Φ2f yv A o ) Therefore expression for A l becomes, A l = {(T u s/ (Φ2f yv A o ) /s} p h (f yv /f yl ) A l = T u p h / (Φ2f yl A o ) This expression can be used to find the longitudinal reinforcement requirement due to torsion. 58 29

ACI Requirements Concrete Part in Torsional Capacity of RC Member According to ACI R22.7.6, in the calculation of T n, all the torque is assumed to be resisted by stirrups and longitudinal steel with T c = 0. At the same time, the shear resisted by concrete V c is assumed to be unchanged by the presence of torsion. 59 ACI Requirements Size of Cross-Section: ACI 22.7.7.1 The cross-section dimensions for solid section shall be such that: [{V u / (b w d)} 2 + {T u p h / (1.7A oh2 )} 2 ] Φ {2 (f c ) + 8 (f c )} Shear stress Torsional shear stress Shear capacity ACI restriction ACI R22.7.7.1 The size of a cross section is limited for two reasons, first to reduce unsightly cracking and second to prevent crushing of the surface concrete due to inclined compressive stresses due to shear and torsion. If the above condition does not meet, increase cross-section. For Hollow section: V u / (b w d) + T u p h / (1.7A oh2 ) Φ {2 (f c ) + 8 (f c )} Shear stress Torsional shear stress Shear capacity ACI restriction 60 30

ACI Requirements Reinforcement Requirement for Torsion: ACI 22.7.3 Statically Indeterminate Case For statically indeterminate case, no torsional reinforcement is required if ΦT c T u 61 ACI Requirements Reinforcement Requirement for Torsion: ACI 22.7.3 Statically Determinate Case: For statically determinate case, no torsional reinforcement is required if ΦT c /4 T u T c = 4 (f c )A cp2 /p cp. 62 31

ACI Requirements Maximum Spacing of the Stirrups (Torsional Stirrups) (ACI 9.7.6.3.3) Least of s max = p h /8, 12 inches or A total f y /(50b w ) [A total = A v + A t (two legged) ] Minimum Longitudinal Reinforcement: ACI 9.6.4.3 A lmin =minimum of 5 (f c )A cp /f yl (A t /s)p h f yv /f yl and 5 (f c )A cp /f yl (25b w / f yv )p h f yv /f yl where A t is single legged reinforcement. If A t has been calculated for 2-legged, then use A t /2 in the above equation, and, A t /s 25b w /f yv, with f yv in psi. Spacing of longitudinal bars along depth of beam shall be 12 in and should be distributed around perimeter. Dia. of longitudinal bars 3/8, Dia. of longitudinal bar > 0.042s {s = stirrups spacing} 63 Steps for Design of Reinforced Concrete Members Subjected to Torsion Determine ΦT cr, if ΦT cr /4 < T u for statically determinate case or ΦT cr < T u for indeterminate case, torsional reinforcement is required. Check dimensions according to ACI: 22.7.7.1. Calculate A t (2 legged) = T u s/ (Φf yv A o ) {where A o = 0.85A oh ) and A v (2 legged) = (V u ΦV c )s/ (Φf y d) A total = A t + A v. Then determine spacing. Check spacing with maximum spacing requirements of ACI. A l = T u p h / (Φ2A o f yl ) Check maximum spacing, minimum reinforcement and diameter of longitudinal bars. 64 32

Example Design of Cantilever Beam: A cantilever beam supports its own dead load plus a concentrated load. The beam is 54 inches long, and the concentrated load acts at a point 6 inches from the end of the beam and 6 inches away from the centroidal axis of the member. The un-factored concentrated load consists of a 20-kip dead and 20 kip live load. The beam also supports an un-factored axial compressive dead load of 40 kip. Use normal weight concrete with f c = 3000 psi and both f y and f yt = 60000 psi. 65 Example Design of Cantilever Beam 20 kips DL & 20 kips LL 66 33

Solution: Example Step No. 01: Sizes. According to ACI 9.5.2.1, table 9.5 (a): Minimum thickness of beam (cantilever) = h min = l/8 As l = 54, therefore h min = (54/8) = 6.75 (Minimum requirement by ACI 9.5.2.1). Though any depth of beam greater than 6.75 can be taken as per ACI minimum requirement, we will use a depth equal to 24. Let the width of beam section (b w ) be equal to 14. Therefore, b w = 14 ; h = 24 ; d = h 2.5 = 21.5 67 Solution: Example Step No. 02: Loads. Factored self weight of beam per running foot = 1.2(14 24/144) 0.15 = 0.42 kip/ft Factored concentrated load = 1.2DL + 1.6LL = 1.2 20 + 1.6 20 = 56 kips Factored axial load (N u ) = 1.2 40 = 48 kips 68 34

Solution: Example Step No. 03: Analysis. 69 Solution: Example Step No. 03: Analysis. Analysis for Shear: V max = 57.9 kips V u = 57.1 kips Analysis for Flexure: M u = - 226 kip-ft Analysis for Torsion: T u = 28.0 kip-ft 70 35

Solution: Example Step No. 04: Design. Check if P u 0.1 f c A g P u = 48 kip 0.1f c A g = 0.1 3 14 24 = 100.8 kip P u < 0.1 f c A g Therefore, axial force effect can be neglected in flexural design. Incase P u > 0.1 f c A g, member shall be designed for bending and axial load both. 71 Solution: Example Step No. 04: Design. Design for Flexure: M u = 228 ft-kip = 2736 in-kip By trial and success method, A s = 2.62 in 2 (6 #6 bars) A smax = ρ max b w d ρ max = 0.85 0.85 (3/60) {0.003/ (0.003 + 0.005)} = 0.01355 A smax = 0.01355 14 21.5 = 4.07 in 2 A smin = ρ min b w d = (greater of 3 f c /f y OR 200/f y )b w d = 0.0033 14 21.5 = 0.993 in 2 A smin <A s <A smax, O.K. 72 36

Solution: Example Step No. 04: Design. Design for Torsion and Shear both: The design parameters required for torsion design are as follows, b w = 14 ; d = 21.5 ; h = 24 ; h f = 6 A cp, p cp, A oh, p h, A o : A cp = b w h = 14 24 = 336 in 2 p cp = 2h +2b w = 2 24 + 2 14 = 76 in With 1 ¾ in cover to the centre of the stirrup bars from all faces x o = 14 2 1 ¾ = 10.5 in y o = 24 2 1 ¾ = 20.5 in 73 Solution: Example Step No. 04: Design. Design for Torsion and Shear both: The design parameters required for torsion design are as follows, Thus, A oh = x o y o = 10.5 20.5 = 215.25 in 2 A o = 0.85A oh = 182.96 in 2 p h = 2 (x oh + y oh ) = 2 (10.5 + 20.5) = 62 inches Finally, A cp = 336 in 2 ; p cp = 76 inches; A oh = 215.25 in 2 ; p h = 62 inches; A o = 182.96 in 2 74 37

Solution: Example Step No. 04: Design. Design for Torsion and Shear both: The design parameters required for torsion design are as follows, (i) Check for size of beam: ACI 11.5.3, [{V u / (b w d)} 2 + {T u p h / (1.7A oh2 )} 2 ] Φ {2 (f c ) + 8 (f c )} Where, V u and T u at critical section. Therefore, [{57.1/(21.5 12)} 2 +{28 12 62/(1.7 215.25 2 )} 2 ] 0.75 10 (3000)/1000 0.345 ksi 0.411 ksi Therefore size of the beam is O.K. 75 Solution: Example Step No. 04: Design. Design for Torsion and Shear both: (ii) Check if ΦT c /4 T u (For statically determinate case): ΦT c = 0.75 4 (f c )A cp2 /p cp = 0.75 4 { (3000)/ (12 1000)} 336 2 / 76 = 20.34 ft-kip No reinforcement is required if ΦT c /4 T u ΦT c /4 = 20.34/4 = 5.085 ft-kip As ΦT c /4 < T u, therefore torsional reinforcement is required. 76 38

Solution: Example Step No. 04: Design. Design for Torsion and Shear both: (a) Torsional Reinforcement (at critical section, two legged): A t (2 legged) = T u s/(φf yv A o ) Therefore, A t (2 legged) = (28 12) s/ (0.75 60 182.96) = 0.041s 77 Solution: Example Step No. 04: Design. Design for Torsion and Shear both: (b) Shear Reinforcement: As an axial load of 48 kips is acting on the beam, therefore ΦV c shall be calculated using V c = 2{1+ N u / (2000A g )} (f c )b w d, where A g is the gross area and N u is the axial force. ΦV c = Φ2{1+ N u / (2000A g )} (f c )b w d = 0.75 2 {1+48/(2000 336)} (3000) 14 21.5/1000 = 24.73 kips 78 39

Solution: Example Step No. 04: Design. Design for Torsion and Shear both: (b) Shear Reinforcement: A v (2 legged) = (V u ΦV c )s/ (Φf yv d) A v (2 legged) = {57.1 24.73}s/ (0.75 60 21.5) A v (2 legged) = 0.03346s 79 Solution: Example Step No. 04: Design. Design for Torsion and Shear both: A total = A t (2 legged) + A v (2 legged) = 0.041s + 0.03346s A total = 0.07446s Assuming 3/8 Φ, 2 legged with total bar area A b = 0.22 in 2 s design = 0.22/0.07446 = 2.95 inches Trying 1/2 Φ, 2 legged with total bar area A b = 0.40 in 2 s d = 0.40/0.07446 = 5.37 inches We will use 1/2 Φ bars at 5 c/c 80 40

Solution: Example Step No. 04: Design. Design for Torsion and Shear both: (c) Maximum Spacing Requirement (ACI 11.5.5) For torsion s = p h /8 or 12 = 62/8 = 7.75 or 12 For shear s = d/2 or 24 = 21.5/2 = 10.75 or 24 Therefore maximum spacing allowed = 7.75 First stirrup is to be placed at a distance (s d /2 = 5/2 2.5 ) from the face of the support. Therefore 2.5 inches from face of the support provide s = 5 inches throughout. 81 Solution: Example Step No. 04: Design. Design for Torsion and Shear both: Longitudinal reinforcement. A l = T u p h / (Φ2A o f yl ) = 28 12 62/ (0.75 2 182.96 60) = 1.265 in 2 A lmin = 5{ (f c )/f yl }A cp (A t / 2s)p h f yv / f yl = 5{ (3000) 336/60000} (0.041s/2s) 62 60/60 = 0.262 in 2 82 41

Solution: Example Step No. 04: Design. Design for Torsion and Shear both: Longitudinal reinforcement. According to the ACI Code, the spacing must not exceed 12 in., and the bars may not be less than No.3 (No. 10) in size nor have a diameter less than 0.042s. d min = 0.042s = 0.042 5 = 0.21 inch 83 Solution: Example Step No. 04: Design. Design for Torsion and Shear both: Longitudinal reinforcement. Reinforcement will be placed at the top, mid depth, and bottom of the member, each level to provide not less than 1.265/3 = 0.422 in 2. Two No. 6 bars will be used at mid depth, and reinforcement to be placed for flexure will be increased by 0.422 in 2 at the top and bottom of member. Therefore final flexural reinforcement is: A s, main = 2.62 + 0.422 = 3.042 in 2 (7 #6 bars) Whereas 2 #6 bars will be provided for bottom and mid depth reinforcement each. 84 42

Solution: Example Step No. 05: Drafting. (5+2) #6 bars #4, 2 legged @ 5 c/c 54 (5+2) #6 bars 2 #6 bars 24 2 #6 bars #4, 2 legged @ 5 c/c 14 85 References Design of Concrete Structures (14 th / 15 Ed.) by Nilson, Darwin and Dolan. Reinforced Concrete Mechanics and Design [6 th Ed.] by James MacGregor. ACI 318-14. 86 43

The End 87 44