reinforced concrete serviceability

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Elastic Flexural Analysis for ServiceabilityLecture GoalsServiceabilityCrack widthMoments of inertia

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  • Elastic Flexural Analysis for Serviceability

    Lecture Goals

    Serviceability

    Crack width

    Moments of inertia 1

  • Introduction

    Ultimate Limit States Lead to collapse

    Serviceability Limit States Disrupt use of Structures

    but do not cause collapse

    Recall:

    Types of Serviceability Limit States

    - Excessive crack width

    - Excessive deflection

    - Undesirable vibrations

    - Fatigue (ULS) 2

  • Crack Width Control

    Cracks are caused by tensile stresses due to loads

    moments, shears, etc..

    3

  • Crack Width Control

    Cracks are caused by tensile stresses due to loads

    moments, shears, etc..

    4

  • Crack Width Control Bar crack development.

    5

  • Crack Width Control

    Temperature crack development

    6

  • Crack Width Control

    Appearance (smooth surface > 0.01 to 0.013 public concern)

    Leakage (Liquid-retaining structures)

    Corrosion (cracks can speed up occurrence of corrosion)

    Reasons for crack width control?

    7

  • Crack Width Control

    Chlorides (other corrosive substances) present

    Relative Humidity > 60 %

    High Ambient Temperatures (accelerates chemical reactions)

    Wetting and drying cycles

    Stray electrical currents occur in the bars.

    Corrosion more apt to occur if (steel oxidizes rust )

    8

  • Limits on Crack Width

    0.4mm (0.016 in.) for interior exposure

    0.33mm (0.013 in.) for exterior exposure

    max.. crack width =

    ACI Codes Basis

    Cracking controlled in ACI code by regulating the

    distribution of reinforcement in beams/slabs.

    9

  • Limits on Crack Width

    Tolerable Crack Widths

    Tolerable

    Crack Width

    Dry air or protective membrane - 0.016 in.

    Humidity, moist air, soil - 0.012 in.

    Deicing chemicals - 0.007 in.

    Seawater and seawater spray - 0.006 in.

    wetting and drying

    Water-retaining structures - 0.004 in.

    (excluding nonpressure pipes)

    Exposure Condition

    10

  • 11

  • Deflection Control

    Visual Appearance

    ( 25 ft. span 1.2 in. )

    Damage to Non-structural Elements

    - cracking of partitions

    - malfunction of doors /windows

    (1.)

    (2.)

    Reasons to Limit Deflection

    1 are generally visible

    250l

    12

  • Deflection Control

    Disruption of function

    - sensitive machinery, equipment

    - ponding of rain water on roofs

    Damage to Structural Elements

    - large s than serviceability problem

    - (contact w/ other members modify

    load paths)

    (3.)

    (4.)

    13

  • Allowable Deflections

    14

  • Allowable Deflections

    15

  • Deflection Response of RC Beams (Flexure)

    A- Ends of Beam Crack

    B - Cracking at midspan

    C - Instantaneous

    deflection under service

    load

    C - long time deflection

    under service load

    D and E - yielding of

    reinforcement @ ends &

    midspan

    12

    2wlM

    12

    2wlM

    2

    16

    wlM

    16

  • Moment of Inertia

    For normal weight

    concrete

    psi 33 c1.5

    cc fE

    For wc = 90 to 155 lb/ft3

    psi 57000 cc fE

    (ACI 8.5.1)

    17

  • EIM

    EI

    M

    slope

    Moment Vs curvature plot

    18

  • Moment Vs Slope Plot

    The cracked beam starts to lose strength as the amount of

    cracking increases

    19

  • Variation of EI along the length of the beam

    20

  • Moment of Inertia for Deflection Calculation

    For (intermediate values of EI) gtecr III

    Brandon

    derived

    Cracking Moment =

    Moment of inertia of transformed cross-section

    Modulus of rupture =

    r g

    t

    f I

    y

    c7.5 f

    Mcr =

    Igt =

    fr =

    3 3

    cr cre g cr

    a a

    1M M

    I I IM M

    21

  • Moment of Inertia for Deflection Calculation

    Distance from centroid to extreme tension fiber

    maximum moment in member at loading stage

    for which Ie is being computed or at any

    previous loading stage

    Moment of inertia of concrete section neglect

    reinforcement

    yt =

    Ma =

    Ig =

    3 3

    cr cre g cr

    a a

    1M M

    I I IM M

    3

    cre cr g cr

    a

    MI I I I

    M

    22

  • Deflection Response of RC Beams (Flexure)

    For Continuous beams

    ACI 9.5.2.4

    ACI Com. 435

    Weight Average

    e21emideavge 25.050.0 IIII

    e21emideavge 15.070.0

    :continous ends 2

    IIII

    1emideavge 15.085.0

    :continous end 1

    III

    2 end @

    1 end @

    midspan @

    ee2

    ee1

    emide

    II

    II

    II

    23

  • Uncracked Transformed Section

    Part (n) =Ej /Ei Area nArea yi yi(n)A

    Concrete 1 bwh bwh 0.5h 0.5bwh2

    As n As (n-1)As d (n-1)Asd

    As n As (n-1)As d (n-1)Asd

    n A i iy n A

    *

    i i i

    *

    i i

    y n Ay

    n A

    Note: (n-1) is to remove area of

    concrete

    24

  • Cracked Transformed Section

    25

  • Cracked Transformed Section

    s

    s

    i

    ii 2

    nAyb

    dnAy

    yb

    A

    Ayy

    Finding the centroid of singly Reinforced Rectangular

    Section

    022

    0

    2

    2

    ss2

    ss

    2

    ss

    2

    b

    dnAy

    b

    nAy

    dnAynAyb

    dnAy

    ybynAyb

    Solve for the quadratic for y

    26

  • Cracked Transformed Section

    022 ss2 b

    dnAy

    b

    nAy

    Note:

    c

    s

    E

    En

    Singly Reinforced Rectangular Section

    2s3

    cr

    3

    1ydnAybI

    27

  • Example (Rectangular - Single)

    For the following example find centroid and moment

    of inertia for an uncracked and cracked section and

    compare the results.

    Es = 29000 ksi

    Ec = 3625 ksi

    d = 15.5 in b = 12 in. h = 18 in.

    Use 4 # 7 bars for the steel.

    28

  • Example

    A #7 bar has an As = 0.6 in2

    s

    c

    E 29000 ksi8

    E 3625 ksin

    2 2s 4 0.6 in 2.4 inA

    The uncracked centroid is

    2

    s

    s

    2

    2

    2

    3

    2

    12

    1

    12 in 18 in8 1 2.4 in 15.5 in

    2 12 in 18 in 8 1 2.4 in

    2204.4 in 9.47 in

    232.8 in

    bhn A d

    ybh n A

    29

  • Example The uncracked moment of inertia

    232

    s

    3 22 2

    4

    112 2

    12 in 18 in 18 in9.47 in 12 in 18 in 15.5 in 9.47 in 8 1 2.4 in

    12 2

    6491 in

    bh hI y bh d y n A

    The cracked centroid is defined by:

    2

    s 2.4 in 0.012912 in 15.5 in

    A

    bd

    222 8 0.0129 2 8 0.0129 8 0.0129 0.3627

    0.3627 15.5 in 5.62 in

    yn n n

    d

    y

    30

  • Example The cracked moment of inertia is

    32

    s

    3 2 2

    4

    3

    112 in 5.62 in 15.5 in 5.62 in 8 2.4 in

    3

    2584.2 in

    byI d y nA

    Notice that the centroid changes from 9.47 in. to 5.62

    in. and the moment of inertia decreases from 6491 in4

    to 2584 in4 . The cracked section loses more than half

    of its strength.

    31

  • Cracked Transformed Section

    's s s s2 2 1 2 2 1 2 0n A nA n A d nA d

    y yb b

    Note:

    c

    s

    E

    En

    Doubly Reinforced Rectangular Section

    2s2

    s

    3

    cr 1

    3

    1ydnAdyAnybI

    32

  • Uncracked Transformed Section

    steel

    2

    s

    2

    s

    concrete

    2

    3

    gt

    11

    212

    1

    dyAndyAn

    hybhbhI

    Note: 3g

    12

    1bhI

    Moment of inertia (uncracked doubly reinforced beam)

    33

  • Example (Rectangular - Double)

    34

  • 35

  • 36

  • 37

  • 38

  • 39

  • Cracked Transformed Section

    Finding the centroid of doubly reinforced T-Section

    0

    212

    2122

    w

    ss

    2

    we

    w

    sswe2

    b

    dnAAntbb

    y

    b

    nAAnbbty

    40

  • Cracked Transformed Section

    Finding the moment of inertia for a doubly reinforced

    T-Section

    233

    cr e e w

    beamflange

    2 2

    s s

    steel

    1 1

    12 2 3

    1

    tI b y b t y b y t

    n A y d nA d y

    41

  • Reinf. Concrete Sections - Example

    Given a doubly reinforced beam with h = 24 in, b = 12

    in., d = 2.5 in. and d = 21.5 in. with 2# 7 bars in

    compression steel and 4 # 7 bars in tension steel. The

    material properties are fc = 4 ksi and fy= 60 ksi.

    a) Determine Igt, Icr , Mcr(+), Mcr(-)

    b) Determine the NA of the beam at ultimate Load.

    42

  • Reinf. Concrete Sections - Example

    The components of the beam

    2 2

    s

    2 2

    s

    c c

    2 0.6 in 1.2 in

    4 0.6 in 2.4 in

    1 k57000 57000 4000

    1000 lb

    3605 ksi

    A

    A

    E f

    43

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