CHAPTER 2. Design Formulae for Bending

Transcription

1 CHAPTER 2 Design Formulae for Bending Learning Objectives Appreciate the stress-strain properties of concrete and steel for R.C. design Appreciate the derivation of the design formulae for bending Apply the formulae to determine the steel required for bending CONTENTS 2.1 Material Stress-strain Relations Concrete Reinforcement Example Design Ultimate Capacity for Axial Compression 2.2 Design Formulae for Bending Limit to Neutral Axis Examples Effective Depth Simplified Stress Block Design Formulae for Singly Reinforced Section Limits of the Lever Arm The Balanced Section Examples Singly Reinforced Section Design Formulae for Doubly Reinforced Section Examples Doubly Reinforced Section 2.3 Flanged Section Effective Flange Width Examples Flanged Section 2.4 Limits to Bar Spacing and Steel Ratio Bar Spacing Maximum and Minimum Percentage of Steel Chapter 2 1

2 2.1 Material Stress-strain Relations The stress-strain curve of a material describes the deformation of the material in response to the load acted upon it. In generalized terms, deformation is presented in terms of change in length per unit length, i.e. strain, while load is in terms of force per unit area, i.e. stress. Strain, Ɛ = Deformation Length (Dimensionless) Stress, f = Force Area (in N/mm 2 or MPa) The stress-strain curves of concrete and steel provides the fundamental knowledge to understand the behaviour of the R.C. composite under loads and for deriving the necessary formulae for R. C. design Concrete Concrete is comparatively very weak in tension. Its tensile strength is about 1/10 of the compressive strength. It is usually ignored in the design. Therefore, the stress-strain curve of concrete is usually referring to concrete under compression. Typical stress-strain curves of concrete are shown in Figure 2.1 below. The shape of stress-strain curves varies with the strength of the concrete. The elastic modulus, i.e. the slope of the initial part of the curves, is higher for higher strength concrete. In addition, the higher is the strength of the concrete, the more sudden the failure of the concrete, i.e. more brittle. Chapter 2 2

3 Stress (MPa) Strain Figure 2.1 Typical Stress-strain Curves of Concrete In order to facilitate the derivation of design formulae, an idealized stress-strain curve in parabolic-rectangular shape is adopted in the design code as shown in Figure 2.2 below. Figure 2.2 Idealized Stress-strain Curve of Concrete for Design (Figure 3.8 of HKCP-2013) Chapter 2 3

4 Take note of the following points: (a) The design ultimate strength of concrete is: 0.67 f cu / m = 0.67 f cu / f cu where 0.67 is to account for the differences between the testing condition and the actual effect on concrete in the structure. 1.5 is the partial factor of safety for the strength of concrete under bending or axial load. (b) The ultimate compressive strain of concrete is: 1 Ɛ cu = The concrete crushes when it deforms to this value and the failure is brittle and sudden. This value defines the ultimate limit state, ULS, of R. C structure Reinforcement Steel is much stronger and more ductile than concrete as illustrated in the typical stress-strain curves of steel in Figure 2.3. The initial part of the curve is linear and the slope, i.e. the elastic module, is constant disregard of the strength. The following value of elastic modulus is adopted in R.C. design. E s = 200 kn/m 2 or N/m 2 1 This value is for concrete not higher than Grade C60. Concrete becomes more brittle when its strength is higher, and therefore the value of Ɛ cu is lower. Details refer to the design code. Chapter 2 4

5 Stress (MPa) % Proof Stress Yield Stress A typical stress-strain curve of concrete Strain Figure 2.3 Typical Stress-strain Curves of Steel Most of the grades of steel exhibit a definite yield point at which strain increases suddenly without increase in stress. The stress at this point, i.e. yield stress, f y, is adopted for design. For steel without yield, 0.2% proof stress is adopted. Beyond this point, the strength of steel continues to increase but with substantial increase in strain. Steel deforms substantially before rupture, and the ultimate strength increases by more than 8% above the yield, and the ultimate strain is more than This strength hardening and ductility properties render steel a good structural material. The tensile and compressive strength properties of steel are the same. The design code provides an idealized stress-strain curve for design as given in Figure 2.4. Chapter 2 5

6 Ɛ y Figure 2.4 Idealized Stress-strain Curve of Reinforcement for Design (Figure 3.9 of HKCP-2013) Take note of the following points: (a) The design yield strength is: f y / m = f y / 1.15 = 0.87 f y (b) Within the elastic range, i.e. before yield: Stress = E s Ɛ s = Ɛ s (c) The yield strain, i.e. beyond which the stress of steel is 0.87f y, is: Ɛ y = 0.87f y / For grade 500 steel, Ɛ y = 0.87 x 500 / = Chapter 2 6

7 2.1.3 Example Design Ultimate Capacity for Axial Compression The design ultimate capacity of a concrete section subject to axial compression is given by Eqn 6.5 of HKCP : N uz = Compression Resistance of Concrete + Compression Resistance of Steel = 0.45 f cu A nc f y A sc where A nc = Net cross-sectional area of concrete A sc = Area of steel in compression Question Determine the design ultimate capacity for axial compression of the following concrete section: Concrete : C40 Dimensions : 400mm x 400mm Rebars : 4T25 vertical bars fully restrained by links Solution A sc = 4 x 491 = mm 2 A nc = 400 x = mm 2 N uz = 0.45 f cu A nc f y A sc = (0.45 x 40 x x 500 x 1964) x 10-3 = = kn 2 The application of this equation is subject to the following conditions: (i) the column is subject to axial load only, without eccentricity and moment, (ii) the rebars restrained from buckling, and (iii) the column is not slender. Design of column will be discussed in another chapter. Chapter 2 7

8 2.2 Design Formulae for Bending When a beam is under downward bending as shown in Figure 2.5 below, the upper part of the beam is in compression and the lower part is in tension. If the plane section remains plane after deformation as shown in figure (a), the strain distribution will be linear as shown in figure (b), with zero strain at the neutral axis and increasing linearly outward towards the top and bottom fibres of the section. 3 a d b c Elevation of a Beam under Load Ɛ cc a d M M d x Neutral Axis Rebar b c Ɛ st (a) Deformation of a-b-c-d (b) Strain (c) Stress (Elastic) (d) Stress (Plastic) Figure 2.5 Distribution of Stresses and Strains across a Beam Section When the load is small and the material is still linear elastic, the stress will be in linear proportion to the strain. The distribution of the compressive stress above the neutral axis is then in triangular shape as shown in figure (c). The concrete below the neutral axis is assumed unable to take up any tensile stress, and rebars are provided to take up the tension. 3 The "plane section remains plane" assumption is usually valid in beam design, except under some circumstances, for examples, deep beam with span-to-depth ratio is smaller than 4, at section under very high shear force, etc. It is out of the scope of this chapter. Chapter 2 8

9 If the load is further increased until the section become plastic, the compressive stress block of concrete will become parabolic in shape and the tension steel become yielded as shown in figure (d) Limits to the Neutral Axis Considering the compatibility of strain in figure (b) above, and assuming there no slip at the interface of concrete and steel bar, the relationship of the maximum strain of concrete compression, Ɛ cc, and the strain of steel in tension, Ɛ st, is given by: Ɛ st = Ɛ cc (d x) x where d = The effective depth of the section. It is the depth measured from the top of the section (for sagging moment) to the centroid of the tension reinforcement. x = The depth of the neutral axis, above which (for sagging moment) the section is in compression while below which the section cracks under tension. In order to ensure ductility, it is desirable to have the tension reinforcement yielded before the concrete crushes. That is Ɛ st when Ɛ cc = Therefore, (d x) x Re-arranging, it becomes x d In other words, the section should be designed such that depth of neutral axis should not exceed the limit to ensure ductility. Chapter 2 9

10 HKCP-2013 limits the depth of neutral axis to: 4 x 0.5 d [2.1] If moment redistribution is more than 10%, i.e. β b < 0.9, the depth of neutral axis is limited to: x (β b - 0.4) d [2.1a] For example, if β b = 0.8, the limiting neutral axis is 0.4d Examples Effective Depth Question A Determine the effective depth of the following section: Overall beam depth, h = 500 mm Concrete cover = 40 mm Size of link: 10 h d Bottom bars: 3T32 in one layer 3T32 Solution Effective Depth, d = /2 = 434 mm d Bar size / 2 Link size Cover Question B Determine the effective depth of the following section: Overall beam depth, h = 650 mm Concrete cover = 45 mm h d Size of link: 12 Bottom bars: 2T40 + 3T32 in two layers 2T32 2T40 + T32 4 This limit is for concrete not higher than Grade C45. For higher grade of concrete the limit is more stringent. Details refer to the design code. Chapter 2 10

11 Solution The clear spacing between two layers of bars should not be less than (vide Cl.8.2 of HKCP-2013): (a) maximum bar size (b) aggregate size + 5 mm (c) 20 mm In this case, the maximum bar size controls, i.e. 40mm. Distance to the bottom T40 = = 573 mm Distance to the bottom T32 = = 577 mm Distance to the 2 nd layer T32 = = 497 mm 1257 x 2 x x x 2 x 497 Effective Depth, d = 1257 x x 3 = 549 mm Alternatively, the effective depth is simply taken to the "center", instead of the centroid, of the two layers of rebars as follows. Effective depth, d = = 533 mm (the deviation is about 2.9% only) Unless rigorous checking is required, this method is in general acceptable for manual calculation in design office. In fact, during the initial design stage, the amount steel required is unknown. Assumption has to be made on the bar size, based on which to estimate the effective depth for calculating the steel required and then the number and size of bars. Once the bar size is known, the initial assumption on effective depth has to verified. If the initial assumption is on conservative side and does not deviate too much from actual value, the result will then be treated as acceptable and the calculation would not be re-done Simplified Stress Block? Q.1 Q.4 After the steel has yielded, the beam continues to deform until the top concrete crushes at the ultimate strain, Ɛ cu, and the distribution of compressive stress in the compression zone, i.e. above the neutral axis, will then be in the shape of rectangular-parabolic as shown in (b) of Figure 2.6 below. In order to make it more manageable in deriving the design formula for bending, a simplified rectangular stress block as shown figure (c) of Figure 2.6 is adopted (Figure 6.1 of HKCP-2013). Chapter 2 11

12 Ɛ cu 0.45f cu 0.45f cu x s s/2 d Neutral Axis z Ɛ st Section (a) Strain at Ultimate Limit State (b) Parabolic Stress Block (c) Simplified Rectangular Stress Block Figure 2.6 Stress and Strain Distribution at Ultimate Limit State In the simplified stress block, a uniform compressive stress of 0.67 f cu / m = 0.45f cu is adopted over a depth of: 5 s = 0.9 x and, the lever arm, z, between the centroid of the compression force in the concrete and the tension force of rebars is: z = d s/2 Rearranging, s = 2(d z) Design Formulae for Singly-Reinforced Section The objective of the design formulae is to determine the steel area, A s, with the following information given: The design ultimate moment: M 5 It is for concrete not higher than Grade C45. The value of "s" is smaller for higher grade of concrete. Details refer to the design code. Chapter 2 12

13 Grade of concrete : Grade of steel : Breadth of section : Effective depth : f cu f y b d b 0.45f cu d x Neutral Axis s = 0.9 x F cc z s/2 M F st A s Section Simplified Rectangular Stress Block Figure 2.7 Simplified Stress Block for R.C. Design Compression in concrete F cc = 0.45 f cu (b s) = 0.9 f cu b (d - z) Tension in the rebar F st = 0.87 f y A s Take moment about the rebar, and by equilibrium of moment: M = F cc z = 0.9 f cu b (d - z) z Divide both sides by (bd 2 f cu ), and let K = M/(bd 2 f cu ) [2.2] the equation becomes: Chapter 2 13

14 (z/d) 2 (z/d) + K/0.9 = 0 Solve for z/d, the level arm factor: z/d = [0.5 + (0.25 K/0.9) 0.5 ] and then, the level arm: z = [0.5 + (0.25 K/0.9) 0.5 ] d [2.3] This is Eqn 6.10 of HKCP Take moment about the centroid of the compression force, and by equilibrium of moment: M = F st z = 0.87 f y A s z Rearranging, A s = M / (0.87 f y z) [2.4] This is Eqn 6.12 of HKCP Limits of the Lever Arm, z The limit to the depth of neutral axis, i.e. x 0.5 d, imposes a lower limit to the lever arm 6 : z d 0.9(0.5d) / 2 z d [2.5] If moment redistribution is more than 10%, i.e. β b < 0.9 z ( β b ) d [2.5a] 6 This limit is for concrete not higher than Grade 45. This limit is more stringent for higher grade of concrete. Refer to the design code for details. Chapter 2 14

15 In addition, the design code also provides an upper limit to the lever arm (Cl (c) of HKCP2013): z 0.95 d When putting z = 0.95d into the equation of lever arm [2.3], we can find that the corresponding value of K is In other words, If K z = 0.95 d [2.6] The Balanced Section, K' If the amount of reinforcement is provided such that the depth of the neutral axis is just at the limit of 0.5d, the section will then fail by crushing of the concrete immediately after the steel has yielded. This beam section is called a balanced section. The corresponding level arm is 0.775d. Putting this value into the equation of equilibrium of moment about the tension force, the moment of resistance of the balanced section is: M bal = F cc z = 0.9 f cu b (d 0.775d) 0.775d Therefore, M bal = f cu bd 2 and, K' = M bal / (bd 2 f cu ) = If the design moment of a section is larger than M bal, it will fail by crushing of concrete before yielding of steel no matter how much tension steel is provided, that is undesirable; unless compression steel is provided as discussed in below. In other words, if a section is over-reinforced, the neutral axis will exceed the upper limit leading to failure without ductility, though the moment capacity is increased, as illustrated in following figure. Chapter 2 15

16 Moment A s > A s,bal (w/o compression steel) M bal A s A s,bal A s < A s,bal 0 Deflection HKCP-2013 specifies the K-value for the balanced section as: 7 K' = (for β b 0.9) [2.7] If moment redistribution is more than 10%, i.e. β b < 0.9, the value of K' is reduced to: K' = 0.402(β b 0.4) (β b 0.4) 2 [2.7a] For example, if β b = 0.8, K' = Examples Singly Reinforced Section In summary, the procedures to determine the area of tension steel are: 1. Calculate the K value K = M / (bd 2 f cu ) 2. Check balanced section K < K' 3. Calculate the lever arm z = [0.5 + (0.25 K/0.9) 0.5 ] d 7 It is for concrete not higher than Grade C45. The value of K' is smaller for higher grade of concrete. Details refer to the design code. Chapter 2 16

17 4. Check z / d Calculate the steel area A s = M / (0.87 f y z) Question A Determine the rebars for the following beam section: Design ultimate moment, M = 350 kn-m β b = 1.0 Breadth, b = 350 mm Effective depth, d = 480 mm Concrete, f cu = 35 MPa Steel, f y = 500 MPa Solution K = M / (bd 2 f cu ) = 350 x 10 6 / (350 x x 35) = β b = 1.0 < (Singly reinforced) Lever arm, z = [0.5 + (0.25 K/0.9) 0.5 ] d = [0.5 + ( /0.9) 0.5 ] x 480 = x 480 = 401 mm Tension steel req'd, A s = M / (0.87 f y z) = 350 x 10 6 / (0.87 x 500 x 401) = mm 2 (Provide 2T32 + 1T25) A s,pro = 2 x = 2099 mm 2 Question B For the section in Example A, find the moment of resistance of the balanced section and the corresponding amount of steel. Solution M bal = K' f cu bd 2 Chapter 2 17

18 = x 35 x 350 x x 10-6 = 440 kn-m z = d = x 480 = 372 mm A s = M / (0.87 f y z) = 440 x 10 6 / (0.87 x 500 x 372) = 2719 mm 2 Question C Determine the ultimate moment of resistance of the following beam section: Breadth, b = 350 mm Effective depth, d = 480 mm Area of steel provided, As = 2412 mm 2 (i.e. 3T32) Concrete, f cu = 35 MPa Steel, f y = 500 MPa Solution (The design formulae in the design 0.45f cu code are given in the form to facilitate the determination of steel area from the design moment. This example, however, requires you to work back the ultimate moment capacity of the d x N.A. s = 0.9 x F cc z s/2 M section from the steel area provided.) F st Simplified Rectangular Stress Block For equilibrium of compression and tension forces: F cc = F st 0.45 f cu (b s) = 0.87 f y A s (assuming steel has yielded) s = 0.87 f y A s / (0.45 f cu b) = 0.87 x 500 x 2412 / (0.45 x 35 x 350) = mm Depth of neutral axis, x = s / 0.9 = / 0.9 Chapter 2 18

19 = mm < 0.5 x 480 = 240 mm (steel has yielded as assumed) Lever arm, z = d s/2 = / 2 = mm Take moment about the centroid of compression zone: Moment of resistance, M = F st z = 0.87 f y As z = 0.87 x 500 x 2412 x x 10-6 = 404 kn-m? Q.5 Q Design Formulae for Doubly Reinforced Section When the design moment becomes so large that K > K', compression reinforcement is required to provide additional compressive resistance in the compression zone of the section. If the area of compression steel, A' s, is located at d' from the top of the section as shown in Figure 2.8, the formulae for determining A s and A' s are derived as follows. The neutral axis cannot be further lowered. It remains at the limiting depth to retain ductility. Therefore, z = ( β b ) d F cc z = K' f cu bd 2 For β b < 0.9, z = d F cc z = f cu bd 2 Chapter 2 19

20 Ɛ cu 0.45f cu Ɛ sc d' F sc A s ' x s A s d Neutral Axis F cc z = d s/2 d d' M F st Section Ɛ st (a) Strain at Ultimate Limit State (b) Simplified Rectangular Stress Block Figure 2.8 Stress and Strain Distribution for Doubly Reinforced Section Compression in the compression reinforcement: F sc = f sc A' s where f sc = stress in the compression steel Take moment about the tension steel, and by equilibrium of moment: M = F cc z + F sc (d - d') = f cu bd 2 + f sc A' s (d - d') Rearranging, it becomes A' s = M f cu bd 2 f sc (d - d') Chapter 2 20

21 or A' s = (K K') f cu bd 2 [2.8a] f sc (d - d') By equilibrium of forces F st = F cc + F sc 0.87 f y A s = f cu bd 2 / z + f sc A s ' Rearranging, it becomes A s = f cu bd 2 f sc + A s ' 0.87 f y z 0.87 f y or A s = K' f cu bd 2 f sc + A s ' 0.87 f y z 0.87 f y [2.9a] The value of f sc can be determined from the strain distribution in Figure 2.8(a) above, that is Ɛ sc / (x - d') = Ɛ cu / x Rearranging, d' / x = 1 - Ɛ sc / Ɛ cu If Ɛ sc > at Ɛ cu = , the compression steel has yielded at the ultimate limit state, i.e. f sc = 0.87 f y : In other words, if d' / x < / = 0.38 [2.10] A' s = (K K') f cu bd 2 [2.8] 0.87 f y (d - d') and, A s = K' f cu bd 2 + A' s [2.9] 0.87 f y z These are Eqns 6.14 and 6.15 in HKCP On the other hand, Chapter 2 21

22 if d' / x > 0.38 the compression bars are so close to the neutral axis that they has not yielded and the stress in the compression bars has to be calculated by: f sc = E s Ɛ sc Examples Doubly Reinforced Section In summary, the procedures to determine the steel areas for doubly reinforced section are (for β b > 0.9): 1. Provide comp'n steel, if K > Calculate the lever arm z = 0.775d 3. Calculate the neutral axis x = 0.5d 4. Check d' / x 0.38 to ensure rebar has yielded 5. Calculate compression steel A' s = (K K') f cu bd f y (d - d') 6. Calculate tension steel A s = K' f cu bd + A' s 0.87 f y z Question A Determine the steel required for the following beam section: Design ultimate moment, M = 500 kn-m β b = 1.0 Breadth, b = 350 mm Effective depth of tension steel, d = 480 mm Effective depth to comp'n steel, d' = 70 mm Concrete, f cu = 35 MPa Steel, f y = 500 MPa Chapter 2 22

23 Solution K = M / (bd 2 f cu ) = 500 x 10 6 / (350 x x 35) = β b =1.0 > (Compression steel is required) Lever arm, z = d = x 480 = 372 mm Depth to neutral axis, x = 0.5 d = 0.5 x 480 = 240 mm Check d' / x = 70 / 240 = 0.29 < 0.38 (f sc = 0.87f y ) (K K') f cu bd 2 Compression steel req'd, A' s = 0.87 f y (d - d') ( ) x 35 x 350 x = 0.87 x 500 x (480 70) = 332 mm 2 (Provide 2T16 Top Bars) A' s,pro = 2 x 201 = 402 mm 2 Tension steel req'd, A s = K' f cu bd 2 + A s ' 0.87 f y z = x 35 x 350 x x 500 x 372 = = 3053 mm 2 (Provide 4 T32 Bottom Bars) A s, pro = 4 x 804 = 3216 mm 2 Chapter 2 23

24 Question B Determine the strain and stress of the compression reinforcement for a doubly reinforced concrete section with the following information: d = 350 mm d' = 70 mmm d' Ɛ sc Ɛ cu Solution For doubly reinforced section, x = 0.5 d = 0.5 x 350 d x Neutral Axis = 175 mm d' / x = 70 / 175 = 0.40 > 0.38 (Compression steel has not yielded) Refer to the strain diagram, the relationship of the strain of concrete and the strain of steel: Ɛ sc / (x - d') = Ɛ cu / x Ɛ sc = x (1 d'/x) = x (1-0.40) = Stress of comp'n steel, f sc = E s Ɛ s = x = 420 N/mm 2? Q.9 Q Flanged Section Reinforced concrete beams are usually constructed monolithically with the floor slab, and therefore they will act in integral to resist sagging moment as shown in the following figure. The slab acts as the top flange of the beam to share the flexural compressive stress. As the slab is much wider than the breadth of the beam, the compressive zone can be achieved by a much Chapter 2 24

25 shallower neutral axis, which, in most circumstances, falls within the flange without trespassing into the web of the beam. On the other hand, the flange does not assist in resisting the hogging moment at the supports, where the compression zone is at the bottom of the section. Figure 2.9 Slab Acting as the Flange of a Beam (Isometric View) (For clarity of illustration, the main beams supporting beam A-B are not shown) Chapter 2 25

26 b eff b eff Compression h f d h Tension b w b w T Section L Section Figure 2.10 Flanged Sections Effective Flange Width The flexural compressive stress in the flange is assumed to be uniformly distributed over an effective width, b eff, which depends on the dimensions of the beam & slabs and the length of the sagging moment, pi l, which may be obtained from the following figures. Notes: (a) The length of the cantilever, l3, should be less than half the adjacent span. (b) The ratio of the adjacent spans should lie between 2/3 and 1.5. Figure 2.11 Definition of pi l for Calculation of Effective Flange Width (Figure 5.1 of HKCP-2013) For simply-supported beam, the whole span is under sagging moment, and therefore, pi l = effective span of the simply-supported beam (L). Chapter 2 26

27 c/c distance of adjacent slabs Figure 2.12 Effective Flange Width (Figure 5.2 of HKCP-2013) The design of flanged section for bending can be simply treated as the design of rectangular section by putting: b = b eff provided that the neutral axis is within the flange, i.e. x h f or (d z) / 0.45 h f [2.11] where h f is the thickness of the flange, i.e. the slab Examples Flanged Section Question A Determine the effective flange width for an interior span of a continuous beam with approximately equal spans with the following information: Breadth, b w = 350 mm Chapter 2 27

28 Effective span, L = Clear spacing btw adjacent beams = mm 2 500mm (same on both sides) Solution Internal span of ctu beam pi l = 0.7 x 6700 = mm b 1 = b 2 = / 2 = 1250 mm b eff,1 = b eff,2 = Min (0.2x x4690 or 0.2x4690 or 1250) = Min (719 or 938 or 1250) = 719 mm b eff = 2 x = 1788 mm Note: As 719 > 0.1 x 4690, the shear stress between the web and flange has to be checked, i.e. Note 1 of Figure 2.12, which is outside the scope of this chapter and is ignored for the purpose of this course. Question B Determine the effective flange width for the following simply-supported beam: Breadth, b w = 300 mm Effective span, L = mm Clear spacing btw adjacent beams = 2 000mm (same on both sides) Solution Simply supported beam pi l = 9000 b 1 = b 2 = 2000 / 2 = 1000 mm b eff,1 = b eff,2 = Min (0.2x x9000 or 0.2x9000 or 1000) = Min (1100 or 1800 or 1000) = 1000 mm b eff = 2 x = 2300 mm Chapter 2 28

29 Question C Determine the steel required for the following beam section: Design ultimate moment, M = 500 kn-m (sagging) Breadth, b w = 350 mm Slab thickness, h f = 150 mm Effective flange width, b eff = 1780 mm Effective depth of tension steel, d = 480 mm Concrete, f cu = 35 MPa Steel, f y = 500 MPa Solution b = b eff = 1780mm K = M / (bd 2 f cu ) = 500 x 10 6 / (1780 x x 35) = < (No compression steel required) K < Lever arm, z = 0.95d = 0.95 x 480 = 456 mm Check x = ( ) / 0.45 = 53 < 150 mm (N.A. is within the flange) Tension steel req'd, A s = M / (0.87 f y z) = 500 x 10 6 / (0.87 x 500 x 456) = mm 2 (Provide 2T40 + 1T20) A s,pro = 2 x = 2828 mm 2 This example is similar to Question A in except that it is a flanged section. As the value of b in this case is 5 times that of the rectangular section in the previous example, the value of K is therefore reduced also by almost 5 times and becomes much smaller than K' and, as a result, the upper bound value of z is adopted. When the total steel area required (2521 mm 2 ) is compared with that required for rectangular section ( = 3365 mm 2 ), there is a saving of 25%.? Q.14 Q.17 Chapter 2 29

30 2.4 Limits to Bar Spacing and Steel Ratio There are lower and upper limits to the amount of steel and the spacing between bars in reinforced concrete. The lower bound is to prevent unsightly cracking due to shrinkage, temperature effect, restrained action and brittle failure. On the other hand, the upper bound is to prevent congestion of reinforcement bars that would affect the proper compaction of concrete Bar Spacing Adequate clear spacing should be provided between bars such that concrete can be placed and compacted satisfactorily around the bars. The clear distance (horizontal and vertical) between individual or horizontal layers of parallel bars should not be less than (Cl.8.2 of HKCP-2013): i. maximum bar size ii. aggregate size + 5 mm iii. 20 mm On the other hands, reinforcement bars cannot be placed too far apart; they have to be placed close enough to distribute the cracks on the surface of the concrete element. The maximum spacing of the bars is determined by the service stress in the rebars, their distance from the concrete surface and the thickness of the concrete element. Detailed requirements for beams and slabs can be found in Cl and Cl of HKCP They will not be covered in this chapter. For simplicity, the following rules of thumb can be adopted for preliminary design: For beam, the maximum bar spacing requirement can in general be complied with by providing one bar for every 100 to 150 mm width of the beam. Chapter 2 30

31 Example For a beam of 400 mm wide, provide 3 to 4 bars at the outer layer depending on the size of the bars. For slab, under most circumstances, limit the spacing of main bars to not more than 2h or 250 mm whichever is lesser Maximum and Minimum Percentage of Steel The maximum and minimum limits for Grade 500 steel commonly used in R.C. design are summarized in the table below: 8 Elements Minimum (%) Maximum (%) Beam Flexural tension steel Rectangular section 0.13 Flanged section (b w /b < 0.4) Flexural compression steel Rectangular section 0.20 Column Wall Vertical bars Horizontal bars 0.25 Table 2.1 Minimum and Maximum Percentage of Reinforcement (Grade 500) In calculating the steel ratio for the above table, the gross area of the concrete, A c, is adopted. 8 There are more specific requirements on the steel ratios, like compression steel in the flange of flanged beam, steel area at the lapping of rebars, cantilever slab, ductility requirements for members resisting lateral load, etc. Refer to the design code for details. Chapter 2 31

32 For rectangular section, A c = bh For flanged section, A c = b w h Example For a beam of 600 (h) x 300 (b), the minimum flexural tension steel is 0.13 x 600 x 300 / 100 = 234 mm 2. Overall depth, h, instead of effective depth, d, is used to check steel ratio.? Q.18 Q.19 Chapter 2 32

33 Key Concepts/Terms Design Ultimate Strength of Concrete 0.45 f cu Ultimate Strain of Concrete Ɛ cu = Design Yield Strength of Steel Effective Depth 087f y d Depth of Neutral Axis and Its Limit x < 0.5d K and Balanced Section K < Lever Arm and Its Limits 0.775d < z < 0.95d Effective Flange Width b eff Maximum and Minimum Steel Ratios Chapter 2 33

34 Self-Assessment Questions Q.1 Find the effective depth, d, of the following beam section. Overall beam depth, h = 600 mm Referred size of link: 10 Concrete cover = 40 mm Preferred size of main bars: 40 A. 510 mm B. 530 mm C. 265 mm D. 550 mm Q.2 Find the effective depth, d, of the following beam section. Overall beam depth, h = 625 mm Referred size of link: 12 Concrete cover = 40 mm Main bars: 2 layers of 32 A. 557 mm B. 541 mm C. 525 mm D mm Q.3 Find the effective depth, d, of the following slab section. Overall slab thickness, h = 175 mm Concrete cover = 25 mm Preferred size of main bars: 12 A. 159 mm B. 138 mm C. 144 mm D. 150 mm Chapter 2 34

35 Q.4 Determine the allowable depth of the neutral axis, x, of the following rectangular beam section. Overall beam depth, h = 600 mm Referred size of link: 10 Concrete cover = 35 mm Preferred size of main bars: 32 β b = 1.0 A. 322 mm B. 300 mm C. 539 mm D mm Q.5 Determine the K value for the following rectangular beam section. Design ultimate moment, M = 350 kn-m β b = 1.0 Breadth, b = 300 mm Effective depth, d = 454 mm Overall depth, h = 520mm Concrete, f cu = 40 MPa A B C D Chapter 2 35

36 Q.6 Determine the amount of tension steel, A s, required for the following rectangular beam section. Design ultimate moment, M = 422 kn-m β b = 1.0 Breadth, b = 325 mm Effective depth, d = 534 mm Concrete, f cu = 40 MPa Steel, f y = 500 MPa A mm 2 B mm 2 C mm 2 D mm 2 Q.7 Determine the amount of tension steel, A s, required for the following rectangular beam section. Design ultimate moment, M = 153 kn-m β b = 1.0 Breadth, b = 350 mm Effective depth, d = 534 mm Concrete, f cu = 30 MPa Steel, f y = 500 MPa A. 701 mm 2 B mm 2 C mm 2 D. 693 mm 2 Q.8 Determine the amount of tension steel, A s, required for the following rectangular beam section. Design ultimate moment, M = 26 kn-m β b = 0.8 Breadth, b = 1000 mm Effective depth, d = 169 mm Concrete, f cu = 30 MPa Steel, f y = 500 MPa A. 372 mm 2 B. 366 mm 2 C. 327 mm 2 D. 701 mm 2 Chapter 2 36

37 Q.9 Determine the lever arm, z, for the following rectangular beam section. K = β b = 1.0 Breadth, b = 350 mm Effective depth, d = 634 mm Concrete, f cu = 30 MPa Steel, f y = 500 MPa A. 462 mm B. 426 mm C. 317 mm D. 491 mm Q.10 Determine the compression steel, A ' s, required for the following rectangular beam section. K = β b = 1.0 Breadth, b = 350 mm Overall depth, h = 700 mm Effective depth, d = 634 mm d' = 70 mm Concrete, f cu = 35 MPa Steel, f y = 500 MPa A mm 2 B. 642 mm 2 C mm 2 D. 491 mm 2 Chapter 2 37

38 Q.11 Determine the steel required for the following rectangular beam section. Design ultimate moment, M = 683 kn-m β b = 1.0 Breadth, b = 350 mm Overall depth, h = 575 mm Effective depth, d = 505 mm d' = 70 mm Concrete, f cu = 40 MPa Steel, f y = 500 MPa A. A' s = 563 mm 2 and A s = 3525 mm 2 B. A' s = 563 mm 2 and A s = 4483 mm 2 C. A' s = 666 mm 2 and A s = 3272 mm 2 D. A' s = 666 mm 2 and A s = 3938 mm 2 Q.12 Determine the steel required for the following rectangular beam section. Design ultimate moment, M = 766 kn-m β b = 1.0 Breadth, b = 400 mm Overall depth, h = 545 mm Effective depth, d = 475 mm d' = 70 mm Concrete, f cu = 40 MPa Steel, f y = 500 MPa A. A' s = 1151 mm 2 and A s = 4668 mm 2 B. A' s = 1151 mm 2 and A s = 3516 mm 2 C. A' s = 666 mm 2 and A s = 3272 mm 2 D. A' s = 666 mm 2 and A s = 3938 mm 2 Q.13 Determine the lever arm, z, for the following rectangular beam section. K = β b = 0.8 Breadth, b = 350 mm Effective depth, d = 634 mm Concrete, f cu = 30 MPa Steel, f y = 500 MPa A. 462 mm B. 520 mm C. 317 mm D. 491 mm Chapter 2 38

39 Q.14 Determine the effective flange width, b eff, for the end span of a continuous beam with approximately equal spans with the following information: Breadth, b w = Effective span, L = Clear distance between adjacent beams = 300 mm mm 2 600mm (same on both sides) A mm B mm C mm D mm Q.15 Determine the effective flange width, b eff, for the following simply-supported beam: Breadth, b w = Effective span, L = c/c distance between adjacent beams = 250 mm (same for adjacent beams) mm 3000mm (same on both sides) A mm B mm C mm D mm Chapter 2 39

40 Q.16 Determine the lever arm, z, for the following flanged beam section. Design ultimate moment, M = 666 kn-m (sagging) β b = 1.0 Breadth, b w = 400 mm Effective flange width, b eff = 2150 mm Effective depth, d = 485 mm h f = 200 mm Concrete, f cu = 40 MPa Steel, f y = 500 MPa A. 457 mm B. 355 mm C. 376 mm D. 461 mm Q.17 Determine the steel area required for the following flanged beam section. Design ultimate moment, M = 668 kn-m (sagging) β b = 1.0 Breadth, b w = 350 mm Effective flange width, b eff = 2050 mm Effective depth, d = 486 mm h f = 190 mm Concrete, f cu = 40 MPa Steel, f y = 500 MPa A. A' s = 0 mm 2 and A s = 3326 mm 2 B. A' s = 0 mm 2 and A s = 4077 mm 2 C. A' s = 823 mm 2 and A s = 4900 mm 2 D. A' s = 823 mm 2 and A s = 3326 mm 2 Q.18 In the design of a 600mm (h) x 400mm (b) reinforced concrete beam, if the amount of steel required for resisting the design moment by the rectangular section is found to be 210 mm 2, which of the following reinforcement is most appropriate? A. 2T12 B. 2T16 C. 4T12 D. 1T20 Chapter 2 40

41 Q.19 In the design of a 400mm x 400mm reinforced concrete column, if the amount of steel required for resisting the design axial force is found to be 1250 mm 2, which of the following reinforcement is most appropriate? A. 4T20 B. 3T25 C. 4T25 D. 12T32 Q.20 Which of the following is the most appropriate rebars for the flanged beam under sagging moment as described below? Breadth, b w = 600 mm Effective flange width, b eff = 2050 mm Effective depth, d = 350 mm h f = 150 mm Overall depth, h = 400 mm The amount of bottom steel required to resist the sagging moment = 380 mm 2 A. 2T20 B. 2T16 C. 4T12 D. 5T12 Chapter 2 41

42 Answers: Q1 B d = /2 = 530 mm Q2 C d = /2 = 525 mm Q3 C d = /2 = 144 mm Q4 D max x = 0.5d = 0.5 x ( /2) = mm Q5 A K = 350 x 10 6 / (300 x x 40) = Q6 D K = 0.114; z = x 534 = 454.7mm; A s = 422 x 10 6 / (0.87 x 500 x 454.7) = 2134 mm 2 Q7 A K = ; z = x 534 = 501.7mm; A s = 153 x 10 6 / (0.87 x 500 x 501.7) = 701 mm 2 Q8 A K = ; z = 0.95 x 169 = 160.6mm, A s = 26 x 10 6 / (0.87 x 500 x 160.6) = 372 mm 2 Q9 D K = > 0.156, therefore, z = x 634 = 492 mm Q10 B A s ' = ( )x35x350x634 2 / (0.87x500x(634-70)) = 643 mm 2 Q11 D K = ; A s ' = ( )x40x350x505 2 / (0.87x500x(505-70)) = 666 mm 2 ; A s = 0.156x40x350x505 2 / (0.87x500x0.775x505) = 3938 mm 2 Q12 A K = ; A s ' = ( )x40x400x475 2 / (0.87x500x(475-70)) = 1151 mm 2 ; A s = 0.156x40x400x475 2 / (0.87x500x0.775x475) = 4668 mm 2 Q13 B β b = 0.8 < 0.9, max x = (β b 0.4)d = 0.4 x 634 = 253.6; z = x = 520mm Q14 C b i = 2600/2 = 1300mm; l pi = 0.85x7700 = 6545mm; b eff = 2 x min(0.2x x6545, 0.2x6545, 1150) = 2129 mm Q15 C b i = ( )/2 = 1375mm; l pi = 8700mm; b eff = 2 x min(0.2x x8700, 0.2x8700, 1150) = 2540 mm Q16 D K = < ; z = 0.95 x 485 = 461 mm Q17 A K = < ; z = 0.95 x 486 = mm; A s = 668 x 10 6 / (0.87 x 500 x 461.7) = 3326 mm 2 Q18 C Min A s = 0.13x600x400/100 = 312 > 210 mm 2 ; preferably provide 4 bars over 400mm; therefore use 4T12, A s = 452 mm 2 Q19 C (A) 100A s /bh = < 0.8 unacceptable; (B) At least 4 bars; (D) 100A s /bh = 6.03 > 3 unacceptable Q20 D Min A s = 0.18x600x400/100 = 432 > 380; although the steel area 4T12 is ok, it is preferable to provide at least 5 6 bars over the width of 600mm. Chapter 2 42

43 Tutorial Questions [Present your answers with detailed working steps in a neat, tidy and logical manner.] AQ1 Given the following information of a rectangular beam section: Design ultimate moment, M = 425 kn-m β b = 1.0 Breadth, b = 600 mm Concrete cover = 35mm Overall depth, h = 400 mm Preferred Link size: 10 Concrete, f cu = 40 MPa Preferred Bar size: 32 Steel, f y = 500 MPa Determine the following: (a) The effective depth (b) The K value (c) The lever arm, z (d) The steel area required (e) Number and size of reinforcement bars, and steel area provided (f) Check if the steel provided comply with the max. & min. limits AQ2 Given the following information of a flanged beam section of a simply supported beam: Design ultimate moment, M = 425 kn-m β b = 1.0 Breadth, b w = 600 mm Concrete cover = 35mm Overall depth, h = 400 mm Preferred Link size: 10 Slab thickness, hf = 150 mm Preferred Bar size: 32 Effective span, L = mm Steel, f y = 500 MPa c/c distance btw adjacent beams = 2 500mm Concrete, f cu = 40 MPa Concrete cover = 35mm Determine the following: (a) The effective depth (b) The effective flange width (c) The K value (d) The lever arm, z Chapter 2 43

44 (e) (f) (g) The steel area required Number and size of reinforcement bars, and steel area provided Check if the steel provided comply with the max. & min. limits AQ3 Identify the assumptions that have been made in deriving the formulae for bending. Chapter 2 44