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					Lecture 34 - Elastic Flexural
 Analysis for Serviceability
        November 20, 2001
           CVEN 444
            Lecture Goals

• Serviceability
• Crack width
• Moments of inertia
                  Introduction
Recall:

Ultimate Limit States         Lead to collapse
Serviceability Limit States   Disrupt use of Structures
                              but do not cause collapse
              Introduction
Types of Serviceability Limit States
      - Excessive crack width
      - Excessive deflection
      - Undesirable vibrations
      - Fatigue (ULS)
                Crack Width Control
Cracks are caused by tensile stresses due to loads moments,
shears, etc..
               Crack Width Control
Cracks are caused by tensile stresses due to loads moments,
shears, etc..
          Crack Width Control
Bar crack development.
         Crack Width Control
Temperature crack development
         Crack Width Control
Reasons for 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)
           Crack Width Control
Corrosion more apt to occur if (steel oxidizes       rust )

    • 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.
          Limits on Crack Width
ACI Code’s Basis
max.. crack width = 0.016 in.   for interior exposure
                    0.013 in.   for exterior exposure

 Cracking controlled in ACI code by regulating the
 distribution of reinforcement in beams/slabs.
           Limits on Crack Width
Gergely-Lutz Equation

                     = 0.076fs 3 dc A
 = Crack width in units of 0.001 in.
 = Distance from NA to bottom
    (tension) fiber, divided by
    distance to reinforcement.
    =(h-c)/(d-c)
fs = Service load stress in
     reinforcement in ksi
          Limits on Crack Width
Gergely-Lutz Equation
                     = 0.076fs 3 dc A
dc = Distance from extreme tension
     fiber to center of reinforcement
     located closest to it, (in.)
A = effective tension area of
    concrete surrounding tension
    bars (w/ same centroid)
    divided by # bars. (for 1 layer
    of bars A = (2dc b)/n
          Limits on Crack Width
   ACI Code Eqn 10-5 ( limits magnitude of z term )
                                 Note:  = 0.076 z
       z = f s 3 dc A                  ( =1.2 for beams)


Interior exposure: critical crack width = 0.016 in.
                   (  = 16 ) z = 175k/in
Exterior exposure: critical crack width = 0.013 in.
                   (  = 13 ) z = 145k/in
       Limits on Crack Width
Tolerable Crack Widths
                                         Tolerable
     Exposure Condition                 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)
          Limits on Crack Width
Thin one-way slabs: Use  =1.35

                       z = 155 k/in (Interior Exposure)
                        z = 130 k/in (Exterior Exposure)
fs = service load stress may be taken as
                          fy
    f s = 0.60 f y                 1.55 – average load factor
                        1.55   
                                     - strength reduction
                               
                        0.90      .     factor for flexure
            Example-Crack

Given: A beam with bw= 14 in. Gr 60 steel 4 #8
with 2 #6 in the second layer with a #4 stirrup.
Determine the crack width limit, z for exterior
and interior limits (145 k/in and 175 k/in.).
            Example-Crack
Compute the center of the steel for the given bars.


         As = 4 #8 bars  2 # 6 bars
            = 4  0.79 in 2 )  2  0.44 in 2 )
            = 4.04 in 2
             Example-Crack
The locations of the center of the bars are
                                  db
        y1 = cover  d stirrup 
                                   2
                                1.0 in.
           = 1.5 in.  0.5 in.
                                   2
           = 2.5 in.
                        db           d b2
       y2 = 2.5 in.   d b 
                         2            2
                       1.0 in.            0.75 in.
           = 2.5 in.           1.0 in.
                          2                  2
           = 4.375 in.
                       Example-Crack

Compute the center of the steel for the given bars.

y=
   y A    i       i

   A          i

       2.5 in.) 4  0.79 in 2 )   4.375 in.) 2  0.44 in 2 )
  =
                             4.04 in 2
  = 2.91 in.
            Example-Crack

Compute number of equivalent bars, n. Use the
largest bar.

          n=
                Ai   =
                        4.04 in 2
                                  = 5.11
                                2
               Abar     0.79 in
Compute the effective tension area

      2 yb 2  2.91 in.)14 in.)
   A=     =                      = 15.93 in 2
       n           5.11
              Example-Crack

The effective service load stress is

   fs = 0.60 f y = 0.6  60 ksi ) = 36 ksi

Compute the effective tension area

   z = f s 3 d c A =  36 ksi ) 3  2.5 in.) 15.93 in 2 )
     = 122.9 k/in.
            Example-Crack

The limits magnitude of z term.
122.9 k/in. < 145 k/in. - Interior exposure
122.9 k/in. < 175 k/in. - Exterior exposure
Crack width is
                    w = 0.076  z
                      = 0.076 1.2 )122.9 )
                      = 11.2
or  = 0.0112 in.
             Deflection Control
Reasons to Limit Deflection
  (1.)   Visual Appearance
                     1
                      * l are generally visible
                    250
         ( 25 ft. span  1.2 in. )
  (2.)   Damage to Non-structural Elements
             - cracking of partitions
             - malfunction of doors /windows
          Deflection Control

(3.)   Disruption of function
             - sensitive machinery, equipment
             - ponding of rain water on roofs
(4.)   Damage to Structural Elements
           - large ’s than serviceability problem
           - (contact w/ other members        modify
              load paths)
        Allowable Deflections
ACI Table 9.5(a) = min. thickness unless ’s are
computed
ACI Table 9.5(b) = max. permissible computed
deflection
           Allowable Deflections
Flat Roofs ( no damageable nonstructural elements
             supported)


                                 l
                LLinst ) 
                               180
          Allowable Deflections
Floors ( no damageable nonstructural elements
         supported )

                              l
            LLinst ) 
                           180
            Allowable Deflections
Roof or Floor elements (supported nonstructural elements
                        likely damaged by large ’s)


                            l
                 
                         480
           Allowable Deflections
Roof or Floor elements ( supported nonstructural elements
                         not likely to be damaged by large
                         ’s )


                              l
                    
                           240
         Allowable Deflections

         Deflection occurring after attachment of
          nonstructural elements


allow    Need to consider the specific structures
          function and characteristics.
Moment of Inertia for Deflection Calculation

       For   I cr  I e  I gt        (intermediate values of EI)

 Brandon             M 
                                 a
                                                         
                                                           a


                Ie =  cr           * I gt  1 -  M cr   * I
 derived                                                cr
                      Ma                      Ma  
                                                            
                                        fr Ig
 Mcr = Cracking Moment =
                                yt
 Igt   = Moment of inertia of transformed cross-section
 fr    = Modulus of rupture = 7.5 f c
Moment of Inertia for Deflection Calculation
              M 
                       a
                                               
                                                 a


        I e =  cr        * I gt  1 -  M cr   * I
                                              cr
               Ma                   Ma  
                                                  
 yt = Distance from centroid to extreme tension fiber
Ma = maximum moment in member at loading stage for
     which Ie (  ) is being computed or at any previous
     loading stage
 Ig = Moment of inertia of concrete section neglect
      reinforcement
Moment of Inertia for Deflection Calculation

           M 
                    3
                                  
                                             3
                                                 
            cr  * I  1 -  M cr 
      Ie =                                      *I
                    g                         cr
            Ma          Ma 
                                                
                                                 
          or
                                         3
                                M 
      I e = I cr  I g - I cr )
                                 cr 
                                     
                                 Ma 
Moment Vs curvature plot




               M                M
          =         slope =       = EI
               EI               
          “Moment Vs Slope” Plot
The cracked beam starts to lose strength as the amount
of cracking increases
Moment of Inertia
          For wc = 90 to 155 lb/ft3

            Ec = c 33 f c
                  1.5
                                 psi)
         For normal weight concrete

             Ec = 57000 f c      psi)
           (ACI 8.5.1)
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 & Note: Stiffness (slope) decreases
midspan                      as cracking progresses
Deflection Response of RC Beams (Flexure)




                     wl 2                wl 2 
                                         M =      
         2
  M = wl         M =      
                                              
      12             24                  12 

  The maximum moments for distributed load acting
  on an indeterminate beam are given.
Deflection Response of RC Beams (Flexure)
 For Continuous beams
 ACI 9.5.2.4       I e avg) = 0.50 I e m id)  0.25 I e1  I e2 )
 ACI Com. 435                       2 ends continous :
                       
 Weight Average I eavg) = 0.70I emid)  0.15I e1  I e2 )
                                  1 end continous :
                        
                        I e avg) = 0.85I emid)  0.15I e1 )
 I e m id) = I e@ midspan
 I e1 = I e@ end 1
 I e2 = I e@ end 2
   Uncracked Transformed Section
  Part       (n) =Ej /Ei           Area     n*Area      yi       yi*(n)A
Concrete          1                bw*h      bw*h     0.5*h    0.5*bw*h2
  A’s             n                 A’s    (n-1)A’s     d’    (n-1)*A’s*d’
   As             n                 As     (n-1)As      d      (n-1)*As*d
                                           n* A               yi * n * Ai


           y=
               y *n A
                     i         i
                                   *
                                   i
                                          Note: (n-1) is to remove area
                                                of concrete
               n A        i
                               *
                               i
     Cracked Transformed Section
Finding the centroid of singly Reinforced Rectangular
Section
                  y                             y
              by    nAs d     by  nAs y = by    nAs d
                                     2
                  
y=
    yi Ai =  2            
                                                  
                                                 2
     Ai         by  nAs        b 2
                                   y  nAs y - nAs d = 0
                                  
                                  2
Solve for the quadratic for y          2nAs    2nAs d
                                  y 
                                   2
                                            y-         =0
                                        b         b
      Cracked Transformed Section
Singly Reinforced Rectangular Section

       2nAs        2nAs d
 y 
  2
              y-            =0
        b        b
 I cr = by  nAs d - y )
       1 3               2

       3

                        Es
      Note:        n=
                        Ec
   Cracked Transformed Section
Doubly Reinforced Rectangular Section
        2n - 1)As  2nAs        2n - 1)As  2nAs d
  y 
   2
                             y-                          =0
                 b                         b

                        y - d )2  nAs d - y )2
  I cr = by  n - 1)As
        1    3

        3
  Note:               Es
                 n=
                      Ec
 Uncracked Transformed Section
Moment of inertia (uncracked doubly reinforced beam)
                                 2
     1            h
I gt = bh  bh y - 
            3
                   
      12  
           2    
                concrete


                            n - 1)As  y - d )  n - 1)As  y - d )
                                                2                     2
                                                        
                                                 steel
    Note:
                           1
                Ig =            bh   3

                           12
       Cracked Transformed Section
    Finding the centroid of doubly reinforced T-Section



      2t be - bw )  2n - 1)As  2nAs
y 
2
                                           y
                     bw

       -
         be - bw ) t 2  2n - 1)As  2nAs d   =0
                          bw
     Cracked Transformed Section
 Finding the moment of inertia for
 a doubly reinforced T-Section


                                 2
                      t 1
     1
I cr = be y  bet
            3      y -   bw  y - t )3
                       
             
      12   
                      2 3  
                flange               beam


    n - 1)As  y - d )  nAs d - y )
                            2          2
                          
                         steel
  Reinforced 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.
Determine Igt, Icr , Mcr(+), Mcr(-), and compare to the NA of
the beam.
    Reinforced Concrete Sections - Example

 The components of the beam
As = 2  0.6 in 2 ) = 1.2 in 2
As = 4  0.6 in 2 ) = 2.4 in 2


                             1k 
Ec = 57000 f c = 57000 4000          
                             1000 lb 
   = 3605 ksi
     Reinforced Concrete Sections - Example

The compute the n value and the centroid, I uncracked
                  Es 29000 ksi
               n=    =          = 8.04
                  Ec   3605 ksi

        n     area (in2) n*area (in2)   y i (in)   y i *n*area (in2)   I (in4)   d (in)   d2*n*area(in4)
A's    7.04      1.2        8.448        2.5            21.12            -       -9.756      804.10
As     7.04      2.4       16.896       21.5           363.26            -       9.244      1443.75
Ac      1       288          288          12          3456.00          13824     -0.256      18.89

                             313.344                  3840.38          13824                2266.74
  Reinforced Concrete Sections - Example

The compute the centroid and I uncracked

    y=
        yi ni Ai
              =
                3840.38 in 3
                             = 12.26 in.
        ni Ai 313.34 in  2


    I =  I i   di2 ni Ai =13824 in 4  2266.7 in 4
     = 16090.7 in 4
  Reinforced Concrete Sections - Example
The compute the centroid and I for a cracked doubly
reinforced beam.
         2n - 1)As  2nAs    2n - 1)As  2nAs d
     y 
      2
                            y-                      =0
                 b                      b
            2  7.04 ) 1.2 in 2 )  2 8.04 )  2.4 in 2 )
    y2                                                          y
                                     12 in.
        2  7.04 ) 1.2 in 2 )        2  8.04 )  2.4 in 2 )  21.5 in.)
    -                            s
                                                                             =0
                           12 in.
    y 2  4.624 y - 72.664 = 0
  Reinforced Concrete Sections - Example
The compute the centroid for a cracked doubly reinforced
beam.

                y 2  4.624 y - 72.664 = 0


               -4.624     4.624 )        4  72.664 )
                                      2

          y=
                               2
            = 6.52 in.
  Reinforced Concrete Sections - Example
The compute the moment of inertia for a cracked doubly
reinforced beam.
        I cr = by  n - 1)As  y - d )  nAs d - y )
               1 3                       2              2

               3
             1
     I cr = 12 in.) 6.52 in.)
                                 3

             3
          7.04 ) 1.2 in  )  6.52 in. - 2.5 in.)
                             2                    2



          8.04 )  2.4 in )  21.5 in. - 6.52 in.)
                             2                         2



       = 5575.22 in 4
  Reinforced Concrete Sections - Example
The critical ratio of moment of inertia

                               4
           I cr 5575.22 in
               =           4
                             = 0.346
           I g 16090.7 in


           I cr  0.35I g
  Reinforced Concrete Sections - Example
Find the components of the beam
   Cc = 0.85 f cba = 0.85  4 ksi )12 in.) 0.85 ) c = 34.68c


           c - 2.5 in.                      0.0075
    s =                0.003) = 0.003 -
                c                              c

               = 29000  0.003 -
                                     0.0075          217.5
    fs = Es s                               = 87 -
                                       c               c
                                       2        217.5 
   Cs = As  fs - 0.85 f c ) = 1.2 in )  87 -       
                                                  c 
                     261
        = 100.32 -
                      c
  Reinforced Concrete Sections - Example
Find the components of the beam
        T =  2.4 in 2 )  60 ksi ) = 144 k


         T = Cc  Cs
                              261
    144 k = 34.68c  100.32 -      34.68c 2 - 43.68c - 261 = 0
                               c
The neutral axis
               43.68     43.68)  4  261) 34.68 )
                                  2

          c=
                             2  34.68 )
            = 3.44 in.
  Reinforced Concrete Sections - Example
The strain of the steel

         3.44 in. - 2.5 in. 
    =
  s                         0.003) = 0.0008 0.00207
            3.44 in.       
        21.5 in. - 3.44 in. 
  s =                         0.003) = 0.0158 0.00207
             3.44 in.        

Note: At service loads, beams are assumed to act
      elastically.
                          c = 3.44 in.
                          y = 6.52 in.
  Reinforced Concrete Sections - Example
Using a linearly varying  and s = E along the NA is the
centroid of the area for an elastic center
                          My
                     s =-
                           I
The maximum tension stress in tension is
                f r = 7.5 f c = 7.5 4000
                   = 474.3 psi  0.4743 ksi
   Reinforced Concrete Sections - Example
The uncracked moments for the beam
   My      sI
s=    M =
    I       y
               fr I     0.4743 ksi 16090.7 in 4 )
M cr   )   =      =                                = 650.2 k-in.
                y           24 in. - 12.26 in.)
               fr I     0.4743 ksi 16090.7 in 4 )
M cr  - )   =      =                                = 622.6 k-in.
                y               12.26 in.
        Homework-12/2/02

Problem 8.7