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Unformatted text preview: Chapter Two
_____________________________________________________________________ Analysis of Sections of Reinforced Concrete Members 2.1 Design StressStrain Curves Shortterm stressstrain 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. 2.11 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. 2.12 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. 2.21 (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. 2.22 (a) Cracked beam; (b) loaddeflection 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 crosssection at failure Fig. 2.23 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 crossaction 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. 2.24 Characteristics of Hognestad et al.’s stress block (2) Whitney’s equivalent rectangular block Fcc = Accfcu Fst = Asfs Fig. 2.25 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. 2.26 (a) Section in bending with a rectangularparabolic 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. 2.26a),
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. 2.27 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. 2.28 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. 2.26(b) Fig. 2.28(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.
 Spring '06
 JSKUANG

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