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Lecture_12 - 2.6 Axial Load Plus Uniaxial Bending (Sections...

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Unformatted text preview: 2.6 Axial Load Plus Uniaxial Bending (Sections of equivalent eccentrically-loaded short columns) 2.6.1 Axial load plus bending at ultimate limit state The combination of an axial force N and bending moment M is equivalent to the axial load N applied at eccentricity e. e= M N Fig. 2.6-1 Section subjected to bending and axial load Uniaxial bending Biaxial bending Fig. 2.6-2 Uniaxial bending and biaxial bending 95 Fig. 2.6-3 represents the cross-section of a member with typical strain and stress distributions for varying positions of the neutral axis at the ultimate limit state. ε ’s Fs ε’s Fig. 2.6-3 Bending plus axial load with varying positions of neutral axis where Fcc = compressive force developed in the concrete and acting through the centroid of stress block Fsc = compressive force in the reinforcement area A’s and acting through its centroid Fs = tensile or compressive force in the reinforcement area As and acting through its centroid 96 2.6.2 Design formulae (1) Equilibrium of forces on the section The applied axial compressive force N = Fcc + Fsc + Fs or N = 0.45 f cu bs + f sc As′ + f s As (2.64) where fsc = compressive stress in reinforcement A’s fs = tensile or compressive stress in reinforcement As . fs is negative when in tension. (2) Moment of resistance of forces developed within the cross-section By taking moment about the mid-depth of the section, hs h h M = Fcc ( − ) + Fsc ( − d ′) − Fs (d − ) 22 2 2 or hs h h M = 0.45 f cu bs ( − ) + f sc As′ ( − d ′) − f s As ( d − ) 22 2 2 (2.65) (3) For a symmetrical arrangement of reinforcement in the cross-section (A's = As = Asc /2, where Asc represents the total amount of vertical reinforcement), Eqs (2.64) and (2.65) become A′ A N 0.405 f cu x = + f sc s + f s s bh h bh bh (2.66) A′ 1 d ′ Ad1 M 0.202 f cu x x (1 − 0.9 ) + f sc s ( − ) − f s s ( − ) (2.67) = 2 bh h h bh 2 h bh h 2 97 2.6.3 Design charts The direct solutions of Eqs (2.66) and (2.67) for the design of column reinforcement would be very tedious. A set of design charts has then been constructed and presented in BS 8110: Part 3. (1) BS 8110 column design charts (based on BS 8110 stress block) Example 2.6-1: A short braced column section 300×300 mm, d/h = 0.85. Design axial load = 1480 kN, design moment = 54 kN-m, fcu = 30 N/mm2, fy = 460 N/mm2. Determine the total amount of steel reinforcement Asc. N 1480 × 10 3 M 54 × 10 6 = = 16.4 , = = 2.0 . bh 300 2 bh 2 300 3 Using the chart, 100Asc /bh = 2.1. Asc = 2.1×3002/100 = 1890 mm2. Provide 4T25 (Asc = 1963 mm2) 98 (2) I.Struct.E. Design charts (From “Manual for the design of reinforced concrete building structures”, 2nd edition, July 2002, The Institution of Structural Engineers, UK) The design charts are prepared based on the simplified stress block. The advantages: o A relative small number of charts will cover the design needs. o One chart can be used for different values of fcu and fy by introducing a reinforcement index ρ fy f cu . 2.6.4 Types of failure For the sections subjected to axial load plus uniaxial bending, the relative magnitude of M and N governs the modes of failure of the section, i.e. failing in tension mode or in compression mode. 99 ε s = ε cu Steel strains: x − d′ d−x ′ and ε s = ε cu . x x Tension failure, but ε’s < fy /Es (compression steel has not yielded.) Tension failure εy ε's < fy /Es Compression failure εcu = 0.0035 xbal Fig. 2.6-4 Strain distributions with different modes of failure (1) Balanced failure (εc = εcu and εs = εy ): x = xbal Failure occurs with yielding of tensile steel and crushing of concrete. The depth of neutral axis: xbal = d 1+ εy ε cu ⎧ xbal ⎪ ⎪ ⎨ ⎪ xbal ⎪ ⎩ High-yield steel = 0.615d for f y = 460 N/mm 2 , where ε y = 0.00219 = 0.746d for f y = 250 N/mm 2 , where ε y = 0.00119 From Eqs (2.64) and (2.65) N bal = 0.45 f cu b ( 0.9 xbal ) + f sc As′ − 095 f y As (2.68) h⎞ ⎛h ⎞ ⎛ M bal = 0.45 fcu b ( 0.9 xbal ) + f sc As′ ⎜ − d ′ ⎟ + 0.95 f y As ⎜ d − ⎟ (2.69) 2⎠ ⎝2 ⎠ ⎝ where f sc ≤ 0.95 f y 100 (2) Tension failure (εc = εcu and εs > εy ): x < xbal It is associated with large eccentricity and small depths of neutral axis: Failure begins with yielding of the tensile steel, followed by crushing of the concrete. (3) Compression failure (εc = εcu and εs < εy ): x > xbal It is associated with small eccentricity and large depths of neutral axis: Failure begins with crushing of the concrete. Special case in the compression failure: When ε s′ = ε y , x= d′ = 2.67 d ′ (for high-yield steel) 1 − ε y / ε cu (2.70) 2.6.5 M-N interaction diagram A solution of Eqs (2.64) and (2.65) can be generally represented by the curve shown in Fig. 2.6-5: M-N interaction diagram Mbal Fig. 2.6-5 Bending moment-axial force interaction diagram • A pint on the curve represents a state of incipient failure; inside the curve: a safe combination of M and N; outside the curve: an unacceptable combination of M and N. 101 • Balanced failure occurs when N = Nbal (x = xbal) Tension failure occurs when N < Nbal (x < xbal) Compression failure occurs when N > Nbal (x > xbal) • Strain states for different types of failure a εcu εcu εcu or εs < εy c εcu εy d εcu Mbal Ne e εs > εy εs ≥ εy Point a: pure compression failure; Point b: balance failure Point c: compression failure; Point d: tension failure Point e: Pure tension failure (reference point) Fig. 2.6-6 Strain states for different types of failure *Construction of column interaction diagrams (optional): See the details in Reinforced and Prestressed Concrete by Kong and Evans, Chapter 7. 102 ...
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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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