CEE 254-1 RC Introduction

CEE 254 ‐ Spring 2015

Nagi Abo‐Shadi, PhD, SE, PEng 1

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Advanced Reinforced Concrete Behavior and DesignCEE 254 Spring 2015

University of California, IrvineThe Henry Samueli School of Engineering Department of Civil & Environmental Engineering

Nagi Abo-Shadi, PhD, PE, SE, PEng

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Course OutlineI. Review of Reinforced Concrete Design

II. Ductility of unconfined and confined members with axial loads.A. Unconfined Members with Axial LoadsB. Confined Members with Axial Loads

III. Shear strength and deformation1. Shear strength of reinforced concrete members2. Shear friction, and corbel design3. Interaction of shear‐flexure‐axial loads4. Effect of cyclic loading

IV. Torsional strength and deformation1. Torsional strength of reinforced concrete members2. Interaction of torsion‐shear‐flexural loads3. Effect of cyclic loading



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Course Outline, Cont’dV. Reinforced Concrete Shear Walls

1. Behavior of tall and squat shear walls2. Behavior of flanged shear walls3. Boundary element requirements4. Interaction of shear walls with:

1. Each other2. Concrete moment frames

5. Design of shear wall for seismic forces6. Seismic detailing of shear walls

VI. Reinforced Concrete Ductile Moment Frames1. Behavior of concrete moment frames2. Plastic hinge formation3. Plastic hinge degradation4. Moment redistribution5. Design of concrete moment frames for seismic forces6. Seismic detailing of moment frames

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Course Outline, Cont’d

VII. Slab Design1. Two‐way slabs2. Flat slabs3. Slab detailing

VIII. Reinforced Concrete DetailingFor: slabs, beams, columns, moment frames, beam‐column joints and shear walls



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Course ProjectThe course project will be assigned after the midterm exam.The project will involve the use of computer software. Teamwork is essential. Your performance will affect the grade ofyour team. Be an asset not a liability to your team. Apresentation will be required at submittal of the project. Neatsketches and typed reports shall be submitted.

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TextbooksReinforced Concrete: Mechanics and Design (6th Edition),James K. Wight, James G. MacGregor, Pearson Education,ISBN-10: 0132176521 | ISBN-13: 978-0132176521

“Building Code Requirements For Structural Concrete andCommentary (ACI 318-11),” American Concrete Institute,Farmington Hills, MI, 2011.

ReferencesReinforced Concrete Structures, R. Park, and T. Paulay, Wiley,1975, ISBN-10: 0471659177 | ISBN-13: 978-0471659174.

Technical papers, ACI Structural Journal, American ConcreteInstitute



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Assessment and Grading Policy

Homework & Quizzes 15%

Course Project 15%

One Mid‐term Exam 30% February 12, 2015 9:30am to 10:50am

Final Examination 40% March 19m 2015 8:00am to 10:00am

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Concrete PropertiesThe concrete stress‐strain relationship is shown

The unit weight of thenormal‐weight concreteis 150 pcf .

The unit weight of thelight‐weight concrete is110 pcf.

The Building Code Requirements for Structural Concrete, ACI 318 assumes a maximum ultimate concrete strain of 0.003 to be used for design consideration. Typical concrete strength ranges from 2,000 psi to 6,000 psi.



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Reinforcing Steel PropertiesThe steel yield strength, the steel modulus of elasticity, the ultimatesteel strength and the maximum steel strain are the steelmechanical properties that are needed for the design purpose. Thesteel stress‐strain relationship is shown.

Reinforcing steel typically meet the ASTM A615 and A706 requirements. The typical yield strength is 40 ksi, 60 ksi and 75 ksi.

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ASTM Standard Reinforcing Bars

Bar size no. Nominal diameter, in Nominal area, in2

3 0.375 0.114 0.500 0.205 0.625 0.316 0.750 0.447 0.875 0.608 1.000 0.799 1.128 1.0010 1.270 1.2711 1.410 1.56



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Strength Design

Required StrengthThe ACI‐318 states that the Required Strength, U, shall be at leastequal to the effects of factored loads in the following Equations 1.1to 1.4.

U = 1.4 (D+F) Equation 1.1U = 1.2 (D+F+T) + 1.6(L+H) + 0.5( Lr or S or R) Equation 1.2U = 1.2 D +1.6 (Lr or S or R) + (1.0L or 0.8W) Equation 1.3U = 1.2 D +1.6L + 0.5(Lr or S or R) Equation 1.4

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All structural members and sections must be proportioned tomeet the above criterion under the most critical loadcombination for all possible actions (flexural, axial load, shear,etc.):

Pn ≥ Pu Equation 1.5Mn ≥ Mu Equation 1.6Vn ≥ Vu Equation 1.7Tn ≥ Tu Equation 1.8

Where: = strength reduction factorPn = nominal strength



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Strength Reduction Factor, The strength reduction factor is based on the type of strength andthe expected more of failure. The following table shows differentvalues for the strength reduction factor.

Type of Strength Tension‐controlled sections 0.9Compression‐controlled sections

Members with spiral reinforcement conforming to 10.9.3Other reinforced members

0.700.65

Shear and torsion 0.75Bearing on concrete (except for post‐tensioned anchorage zones) 0.65Post‐tensioned anchorage zones 0.85Strut‐and‐tie models (Appendix A) 0.75

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The ACI 318‐05 requires the strength reduction factor for tension‐controlled sections to behigher that for compression‐controlled sections. The strain in the tensile steel differentiatesbetween the tension and compression controlled sections. The figure above includes achart that relates the strength reduction factor to the tensile steel strain



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Analysis of Concrete SectionsTo simplify the analysis of concrete sections, the following assumptions need to be considered: 1) Plan sections remain plan after bending

2) The compressive stress distribution is assumed to be uniform (Block)3) Failure occurs at an ultimate concrete strain of 0.0034) The block stress is equal to: 0.85 f’c5) The depth of the block, a, is equal to 1 c

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1 = a factor that relates the depth of the block to the depth of the neutral axis

1 = 0.85 when f’c ≤ 4,000 psi1 = 0.65 when f’c ≥ 8,000 psi

1 = 0.85 – (f’c – 4ksi) 0.05 when 8,000 psi ≥ f’c ≥ 4,000 psi



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Minimum Reinforcement of Flexural MembersReinforced concrete beams

As provided shall not be less than that given by

But not less than:

Reinforced concrete slabs and foundationIn reinforced concrete one-way slabs, 2-was slabs and foundation; the minimum flexural reinforcement is given by the following equation:

As = 0.0018 Ac

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Flexural StrengthThe flexural strength of a concrete section is equal to:

Mn = As fy (d - a / 2)

Where:T = As fyC = 0.85 fc’ b aT = C

As fy = 0.85 f’c b a



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ExampleThe concrete section shown has a concrete strength, f’c =4,000 psi and steel yield strength, fy = 60,000 psi. Forsimplicity, neglect hanger bars (compression reinforcement).

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1- Define rectangular concrete stress distribution.d = dt = 16 - 2.5 = 13.5 in.As = 3 x 0.79 = 2.37 in2, assuming εs > εy,a = As fy / (0.85 f’c b) = 142.2 / (0.85 x 4 x 10) = 4.18 in

2- Determine net tensile strain εs and c = a/β1 = 4.18/ 0.85 = 4.92 in.εs = (dt - c) x 0.003/c = (13.50 - 4.92) x 0.003/4.92 = 0.00523 > 0.005



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Therefore, section is tension-controlled, = 0.9This also confirms that εs > εy at nominal strength

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3- Determine the nominal moment strength, Mn, and the design moment strength, Mn.

Mn = As fy (d – a/2) = 142.2 (13.50 – 2.09) = 1,622.3 kip-in = 135.2 kip-ft

Mn = 0.9 (135.2) = 121.7 kip-ft



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Engineering Software

- PCA Column- CSI Column- ETABS or SAP2000 or RAM Structural- Response 2000 for Moment-Curvature Analysis

CEE 254-1 RC Introduction

Documents

Transcript of CEE 254-1 RC Introduction