Lecture_5

Lecture_5 - Chapter Two

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Unformatted text preview: Chapter Two _____________________________________________________________________ Analysis of Sections of Reinforced Concrete Members 2.1 Design Stress-Strain Curves Short-term stress-strain curves in BS 8110: Part 1: 1997 are presented in Figs 2.1 and 2.2. These curves are in an idealised form and are used in the analysis of R.C. member sections. 2.1.1 Concrete 0.67 f cu γm Strain Fig. 2.1-1 BS8110 design stress/strain curves for concrete in compression • Maximum stress: 0.67 f cu γm (when γm = 1.5, 0.67⋅fcu / 1.5 = 0.45fcu ) The factor 0.67 allows for the difference between the bending strength and the cube crashing strength of concrete. • Concrete is assumed to fail at an ultimate strain, εcu , of 0.0035. 33 2.1.2 Reinforcing steel fy / γ m Es = 200 kN/mm2 fy / γ m Fig. 2.1-2 BS8110 design stress/strain curves for steel reinforcement • Maximum stress: fy /γm (when γm = 1.05, fy /1.05 = 0.95fy ) (Hong Kong code: γm = 1.15; fy /1.15 = 0.87fy ) Design yield strain: ε y = f y /γ m Es (For fy = 460 N/mm2, εy = 0.00219) 2.2 General Behaviour of R.C. Beams in Bending 2.2.1 Types of cross sections The three common types of R.C. beam sections: (1) Rectangular section with tension steel only; (2) Rectangular section with both tension and compression steel; (3) Flanged sections of T or L shapes. Fig. 2.2-1 (a) Rectangular beam and slab, tension steel only; (b) rectangular beam, tension and compression steel; (c) flanged beams. 34 2.2.2 Structural behaviour in bending (a) Yield load Crack load (b) (c) Fig. 2.2-2 (a) Cracked beam; (b) load-deflection curve; (c) cracked section and compressive stress distributions of concrete • Concrete − resist compression Reinforcing steel − resists tension due to bending • Path of failure: Concrete cracking Steel yielding Concrete crashing 35 2.2.3 A general theory for ultimate flexural strength Current design methods for R.C. beams in British and American codes are based on the general theory described below. • Basic assumptions (a) Plane sections remain plane. (b) The ultimate limit state of collapse is reached when concrete strain reaches εcu = 0.0035. (ACI code: εcu = 0.003) (c) Tensile strength of concrete is ignored. • Strain and stress distributions of cross-section at failure Fig. 2.2-3 Strain and stress distributions at failure Notation: ′ As − tension reinforcement; As − compression reinforcement; εcu − ultimate strain of concrete (0.0035); εs − strain in steel; d − effective depth (h − overall depth); x − neutral axis depth. 36 • Notes o Form Assumption (a), the strain distribution in a beam cross-action is linear. o Form Assumption (b), the maximum concrete compressive strain has a specified value εcu at the instant of collapse. The corresponding values of εs and ε’s are εs = d−x ε cu x (2.1) ′ εs = x − d′ ε cu x (2.2) o From Assumption (c), the forces on the beam section are Concrete tension: Fct = 0 Concrete compression: Fcc = k1fcubx Reinforcement tension: Fst = Asfs Reinforcement compression: Fsc = As′ f s′ For the condition of equilibrium, ′ k1 f cu bx + As f s′ = As f s (2.3) Taking moments about the level of tension steel, ′ M u = (k1 f cu bx)(d − k 2 x) + As f s′(d − d ′) (2.4) Or taking moments about the centroid of the concrete stress block, ′ M u = As f s (d − k 2 x) + As f ′(k 2 x − d ′) (2.5) where Mu is the ultimate flexural strength or maximum moment of resistance. 37 2.2.4 Stress blocks (1) Hognestad et al.’s block Fig. 2.2-4 Characteristics of Hognestad et al.’s stress block (2) Whitney’s equivalent rectangular block Fcc = Accfcu Fst = Asfs Fig. 2.2-5 Whitney’s equivalent rectangular stress block 38 o Both BS 8110 and the ACI Building Code make use of the concept of an equivalent rectangular stress block, which was pioneered by Whitney. o The actual stress block may be replaced by a fictitious rectangular block of intensity of 0.85 f c′ with a depth xw: Area of 0.85 f c′x w = that of the actual block Centroids of two blocks are very nearly at the same level o For equilibrium: Fcc = Fst or 0.85 f c′bxw = As f s (2.6) When fs = fy xw = As f y 0.85 f c′b = fy 0.85 f c′ ρd (2.7) where ρ is reinforcement ratio, defined by ρ= As bd (2.8) The ultimate moment of resistance is M u = As f y (d − xw ) 2 (2.9) This is a theoretical value of the maximum moment of resistance – No any safety factor has been included. 39 (3) BS 8110 stress block o The rectangular−parabolic stress block (a) d (b) Fig. 2.2-6 (a) Section in bending with a rectangular-parabolic stress block; (b) BS 8110 design stress block for ultimate limit state o Determine the mean concrete stress, k1 From the strain diagram x ε cu = w With εcu = 0.0035 and εo = 2.4 × 10 w= xε 0 ε cu = (2.10) ε0 −4 f cu 1.5 x f cu 17.86 (2.11) 40 0.45fcu For the stress block (Fig. 2.2-6a), k1 = Area ( pqrs ) − Area (rst ) f cu x k1 = 0.45 f cu x − 0.45 f cu w / 3 f cu x or (2.12) Substituting for w from Eq. (2.11) into Eq. (2.12) gives k1 = 0.45 [1 − f cu / 53.5] (2.13) o Determine the depth of the centroid k2x Taking moments about the neutral axis: (area of stress block)⋅( x − k 2 x ) x 2 = Area (pqrs)⋅ − Area (rst)⋅ ∴ x − k2 x = w 4 0.45( x 2 / 2 − w 2 / 12) 0.45 x 2 = k1 x k1 x f cu ⎤ ⎡ ⎢0.5 − 3828 ⎥ ⎣ ⎦ Hence, k2 = 1 − f cu ⎤ 0.45 ⎡ ⎢0.5 − 3828 ⎥ k1 ⎣ ⎦ (2.14) 41 Fig. 2.2-7 Characteristic values of k1 and k2 for different concrete grades o BS 8110 simplified rectangular block (Clause 3.4.4.4) (a) Strain (b) Stress Fig. 2.2-8 Simplified stress block for concrete at ultimate limit state The simplified rectangular block is a special case of BS 8110 stress block when k1 = 0.405 and k2 = 0.45. Fig. 2.2-6(b) Fig. 2.2-8(b) k1fcubx = (0.45fcu)b·0.9x = 0.405fcubx k2 x = 0 .9 x = 0.45 x 2 i.e. k1 = 0.405 and k2 = 0.45. 42 ...
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This note was uploaded on 12/24/2011 for the course CIVL 232 taught by Professor Jskuang during the Spring '06 term at HKUST.

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