Lecture 06 Design of Reinforced Concrete Beam for Shear By: Civil Engineering Department UET Peshawar drqaisarali@uetpeshawar.edu.pk Topics Addressed Shear Stresses in Rectangular Beams Diagonal Tension in RC Beams subjected to Flexure and Shear Loading Types of Cracks in Reinforced Concrete Beam Shear Strength of Concrete Web Reinforcement Requirement in Reinforced Concrete Beams ACI Code Provisions for Shear Design Example 2 1
Shear Stresses in Rectangular Beam Shear stresses in homogeneous Elastic Rectangular Beam The shear stress (ν) at any point in the cross section is given by ν ν max 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 3 Shear Stresses in Rectangular Beam Shear stresses in homogeneous Elastic Rectangular Beam 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 4 2
Shear Stresses in Rectangular Beam Shear stresses in homogeneous Elastic Rectangular Beam For shear at neutral axis ef: a b Q=A abef h/4 = bh 2 /8 I = bh 3 /12 h e f h/4 Therefore, ν = VQ/Ib = V(bh 2 /8)/{(bh 3 /12) b} ν max = (3/2)V/bh b 5 Shear Stresses in Rectangular Beam Shear stresses in RC Beam The shear stresses in reinforced concrete member cannot be computed from v = VQ/Ib, because concrete is not an elastic homogeneous material. 6 3
Shear Stresses in Rectangular Beam Shear Stresses in RC Beam Tests have shown that the average shear stress in a RC beam can be expressed by : ν av =V/bd ν av = V/bd The maximum value, which occurs at the neutral axis, will exceed this average by an unknown but moderate amount. ν max 7 Diagonal Tension in RC Beams subjected to Flexure and Shear Loading 8 4
Diagonal Tension in RC Beams subjected to Flexure and Shear Loading Types of Cracks in Reinforced Concrete Beam 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. 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 11 Web Reinforcement Requirement in Reinforced Concrete Beam Nominal Shear Capacity (V n ) of Reinforced Concrete Beam The general expression for shear capacity of reinforced concrete beam is given as: V n =V c +V s ( ACI 22.5.1.1) 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) 12 6
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 shear capacity of concret V c. 13 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 cz +V d +V iy Where, V iy = Vertical component of sizable interlock forces. 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. 14 7
Web Reinforcement Requirement in Reinforced Concrete Beams Nominal Shear Capacity (V n ) of Reinforced Concrete Beam V n =V c +V s According to the ACI 22.5.5.1, V c =V cz +V d +V iy = 2 (f c ) b w d 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 V n =V c +V s V s =na v f y The number of stirrups n spaced a distance s apart depend on the length p of the horizontal projection of the diagonal crack, as shown, hence, n = p/s. This length p is conservatively assumed to be equal to the effective depth d of the beam; thus n=d/s Therefore; V s =A v f y d / s (ACI 22.5.10.5.3) 16 8
Web Reinforcement Requirement in Reinforced Concrete Beams Design Shear Capacity (Φ V n ) of Reinforced Concrete Beam ΦV n =ΦV c +ΦV s To avoid shear failure ΦV n V u ΦV n =V u A v is the cross sectional area of web reinforcement within a distance s, for single loop stirrups (2 legged), A v =2A s A s = cross sectional area of the stirrup bar ΦV c +ΦV s =V u ΦV c +ΦA v f y d / s= V u a a s Φ Φ At section a-a, if #3 bar is used A s =0.11in 2, A v = 2 x 0.11= 0.22 in 2 Φ = 0.75 (ACI 21.2.2) 17 ACI Code Provisions for Shear Design Location of Critical Section for Critical Shear V u In most of the cases, for the design of shear, critical shear is taken at a distance d from the support instead of maximum shear at the face of the support. 18 9
ACI Code Provisions for Shear Design Location of Critical Section for Critical Shear V u 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 19 ACI Code Provisions for Shear Design A typical half shear force diagram for a simply supported beam, subjected to uniformly distributed load is shown below. The critical shear V u,φv c and ΦV c /2areplottedonthediagram.BasedonthevalueofV u, following three conditions are possible. ΦV c V u ΦV c <V u ΦV c /2 < V u 20 10
ACI Code Provisions for Shear Design 1. When ΦV c /2 > V u, no web reinforcement is required. Where V u is the critical shear taken at a distance d from the face of the support. 2. When ΦV c /2 < V u but ΦV c V u, theoretically no web reinforcement is required. However ACI 22.5 recommends that minimum web reinforcement in the form of Maximum spacing s max shall be minimum of: A v f y /(50b w ), d/2 24 inches A v f y / {0.75 f b w } 21 ACI Code Provisions for Shear Design 3. When ΦV c < V u, web reinforcement is required: Required Spacing, s d = Φ A v f y d / (V u -ΦV c ) If s d > s max ; use s max Maximum Spacing s max is minimum of: A v f y /(50b w ), d/2 24 inches A v f y / {0.75 f b w } 22 11
ACI Code Provisions for Shear Design 4. Check for Depth of Beam: if ΦV s Φ8 f b w d (ACI 22.5.1.2), depth of beam, ok! If ΦV s > Φ8 f b w d, increase depth of beam. 5. Check for Spacing: if ΦV s > Φ4 f b w d; Divide spacing of s max equation by 2 ; otherwise use s max 23 ACI Code Provisions for Shear Design Placement of Shear Reinforcement When ΦV c < V u, use Design Spacing, s d When ΦV c > V u, use Maximum Spacing, s max V u is the shear force at distance d from the face of the support. ΦV c and ΦV c /2 are plotted on shear force diagram. 24 12
Design the beam shown below as per ACI 318-14. W D = 0.75 kip/ft (Excluding self-weight of beam) WL = 0.75 kip/ft 20-0 Take f c = 3 ksi & f y = 40 ksi 25 Step No. 01: Sizes. For 20 length, h min = l /16 = 20x12/16 = 15 For grade 40, we have = h min =15 x (0.4 + 40,000/100,000) = 12 This is the minimum requirement of the code for depth of beam. However we select 18 deep beam. Generally the minimum beam width is 12, therefore, width of the beam is taken as 12 The final selection of beam size depends on several factors specifically the availability of formwork. 26 13
Step No. 01: Sizes. Depth of beam, h = 18 h = d + ȳ; ȳ is usually taken from 2.5 to 3.0 inches For ȳ = 2.5 in; d = 18 2.5 = 15.5 Width of beam cross section (b w ) = 12 In RCD, Width of beam is usually denoted d by b w instead of b. 18 y 12 27 Step No. 02: Loads. Self weight of beam = γ c b w h = 0.15 (12 18/144) = 0.225 kips/ft W u = 1.2W D + 1.6W L = 1.2 (0.225 + 0.75) + 1.6 0.75 = 2.37 kips/ft 28 14
Step No. 03: Analysis. Flexural Analysis: M u = W u l 2 /8 = 2.37 (20) 2 12/8 = 1422 in-kips Analysis for Shear in beam: V = 23.7 kips, To find V u at a distance d from face of Support, d = 15.5 = 1.29 using similarity of triangles: V u / (10-1.29) = 23.7 / 10 V u = 23.7 (10 1.29) / 10 = 20.64 k 2.37 kip/ft 20.64 kips 23.7 1422 in-kips SFD BMD 29 Step No. 04: Design. Design for flexure: ΦM n M u (ΦM n is M design or M capacity ) For ΦM n = M u ΦA s f y (d a/2) = M u A s = M u / {Φf y (d a/2)} Calculate A s by trial and success method. 30 15
Step No. 04: Design. Design for flexure: First Trial: Assume a = 4 A s = 1422 / [0.9 40 {15.5 (4/2)}] = 2.92 in 2 a = A s f y / (0.85f c b w ) = 2.92 40/ (0.85 3 12) = 3.82 inches 31 Step No. 04: Design. Design for flexure: Second Trial: Third Trial: A s = 1422 / [0.9 40 {15.5 (3.82/2)}] = 2.90 in 2 a = 2.90 40/ (0.85 3 12) = 3.79 inches A s = 1422 / [0.9 40 {15.5 (3.79/2)}] = 2.90 in 2 a = 4.49 40/ (0.85 3 12) = 3.79 inches Close enough to the previous value of a so that A s = 2.90 in 2 O.K 32 16
Step No. 04: Design. Design for flexure: Check for maximum and minimum reinforcement allowed by ACI: A smin = 3 ( f /f y ) b w d (200/f y ) b w d 3 ( f /f y ) b w d= 3 ( 3000 /40000) b w d = 0.004 x 12 x15.5 = 0.744 in 2 (200/f y ) b w d = (200/40000) x12 15.5 = 0.93 in 2 A smin = 0.93 in 2 33 Step No. 04: Design. Design for flexure: A smax =0.27(f c /f y )b w d = 0.27 x (3/40) x 12 15.5 = 3.76 in 2 A smin (0.93) < A s (2.90) < A smax (3.76) O.K 34 17
Step No. 04: Design. Design for flexure: Bar Placement: 5,#7 bars will provide 3.0 in 2 of steel area which is slightly greater than required. Other options can be explored. For example, 7,#6 bars (3.08 in 2 ), 4,#8 bars (3.16 in 2 ), or combination of two different size bars. 35 Step No. 04: Design. Design for Shear: V u = 20.64 kips ΦV c = (Capacity of concrete in shear) = Φ2 f b w d = 0.75 2 3000 12 15.5/1000 = 15.28 kips As ΦV c < V u, Shear reinforcement is required. 36 18
Step No. 04: Design. Design for Shear: Assuming #3, 2 legged (0.22 in 2 ), vertical stirrups. Spacing required (S d ) = ΦA v f y d/ (V u ΦV c ) = 0.75 0.22 40 15.5/ (20.64 15.28) 19.1 37 Step No. 04: Design. Design for Shear: Maximum spacing and minimum reinforcement requirement as permitted by ACI is minimum of: s max =A v f y /(50b w ) =0.22 40000/(50 12) = 14.66 ( f y is in Psi) s max = d/2 = 15.5/2 = 7.75 s max = 24 A v f y / 0.75 (f c )b w = 0.22 40000/ {(0.75 (3000) 12} =17.85 Therefore s max =7.75 38 19
Step No. 04: Design. Design for Shear: Other checks: Check for depth of beam: ΦV s Φ8 f b w d Φ8 f b w d = 0.75 8 3000 12 15.5/1000 = 61.12 k ΦV s =V u ΦV c = 20.64 15.28 = 5.36 k < 61.12 k, O.K. Therefore depth is O.K. If not, increase depth of beam. 39 Step No. 04: Design. Design for Shear: Other checks: Check if ΦV s Φ4 f b w d 5.36 kips < 30.56 kips O.K. ΦV s Φ4 f b w d, the maximum spacing (s max ) is O.K. Otherwise reduce spacing by one half. 40 20
Step 05: Drafting (Shear Reinforcement) V u = 20.64 kips ΦV c = 15.28 kips ΦV c /2 = 7.64 kips 23.7 kips 20.64kips 2.37 kip/ft 15.28 kips 7.64 kips s d s max s d = 19.12 smax = 7.75 x 1 = (7.64)(10)/(23.7) 3.22 ft x 2 = (15.28)(10)/(23.7) 6.44 ft #3, 2 legged stirrup @ 7.75 c/c Theoretically no stirrups needed #3, 2 legged stirrups @ 12 c/c x 1 x 2 Note: As S d S max, we will provide S max from the support up to 6.78 ft. Beyond this point, theoretically no reinforcement is required, however, we will provide #3 2- legged stirrups @ 12 in c/c. 41 Step 05: Drafting (Flexural Reinforcement) 42 21
Practice Examples Design 12 x 18, 20 feet long beam for shear using the following data: S.No. Concrete Compressive Strength f c (ksi) Rebar Tensile Strength f y (ksi) Shear force V u (kips) 1. 3 60 35 2. 4 60 45 3. 4 40 30 43 Practice Examples Design 12 x 24, 20 feet long beam for shear using the following data: S.No. Concrete Compressive Strength f c (ksi) Rebar Tensile Strength f y (ksi) Shear force V u (kips) 1. 3 60 35 2. 4 60 45 3. 4 40 30 44 22
References Design of Concrete Structures 14 th /15 th edition by Nilson, Darwin and Dolan. Building Code Requirements for Structural Concrete (ACI 318-14) 45 23