Reinforced Concrete Serviceability 2x16b

Elastic Flexural Analysis for Serviceability Lecture Goals Serviceability Crack width Moments of inertia 1

Introduction Recall: Ultimate Limit States

Lead to collapse

Serviceability Limit States

Disrupt use of Structures but do not cause collapse Types of Serviceability Limit States - Excessive crack width - Excessive deflection - Undesirable vibrations - Fatigue (ULS)

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Crack Width Control Cracks are caused by tensile stresses due to loads moments, shears, etc..

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Crack Width Control Cracks are caused by tensile stresses due to loads moments, shears, etc..

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Crack Width Control Bar crack development.

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Crack Width Control Temperature crack development

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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)

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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.

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Limits on Crack Width ACI Code’s Basis max.. crack width = 0.4mm (0.016 in.) for interior exposure 0.33mm (0.013 in.) for exterior exposure

Cracking controlled in ACI code by regulating the distribution of reinforcement in beams/slabs.

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Limits on Crack Width Tolerable Crack Widths Tolerable Crack Width

Exposure Condition Dry air or protective membrane Humidity, moist air, soil Deicing chemicals Seawater and seawater spray

-

0.016 in.

-

0.004 in.

0.012 in. 0.007 in. 0.006 in.

wetting and drying Water-retaining structures (excluding nonpressure pipes)

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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

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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 - ( w/ other modify load paths)

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Allowable Deflections

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Allowable Deflections

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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

2    M  wl    12  

 wl 2  M   16  

2    M  wl    12  

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Moment of Inertia For wc = 90 to 155 lb/ft3

Ec  c1.533 f c

psi

For normal weight concrete Ec  57000 f c

psi

(ACI 8.5.1) 17

Moment Vs curvature plot



M EI

 slope 

M



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 EI

“Moment Vs Slope” Plot The cracked beam starts to lose strength as the amount of cracking increases

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Variation of EI along the length of the beam

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Moment of Inertia for Deflection Calculation For

I cr  I e  I gt

Brandon derived

(intermediate values of EI)

3   M cr   M cr   Ie    I g  1     I cr   M a    Ma  3

Mcr = Cracking Moment =

fr Ig

Igt

yt = Moment of inertia of transformed cross-section

fr

= Modulus of rupture = 7.5 f c 21

Moment of Inertia for Deflection Calculation

3   M cr   M cr   Ie    I g  1     I cr   M a    Ma  3 3

 M cr  I e  I cr   I g  I cr     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

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Deflection Response of RC Beams (Flexure) For Continuous beams ACI 9.5.2.4 ACI Com. 435 Weight Average

 I eavg  0.50I emid   0.25I e1  I e2 

2 ends continous :   I e avg  0.70 I emid   0.15I e1  I e2  1 end continous :   I e avg  0.85I e mid   0.15I e1 

I e mid   I e@ midspan I e1  I e@ end 1 I e2  I e@ end 2

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Uncracked Transformed Section Part Concrete A’s As

(n) =Ej /Ei 1 n n

Area bwh A’s As

 yi  ni A y *  ni Ai

* i

n×Area bwh (n-1)A’s (n-1)As  n A

yi 0.5×h d’ d

yi(n)A 0.5bwh2 (n-1)A’sd’ (n-1)Asd  yi  n  Ai

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

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Cracked Transformed Section

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Cracked Transformed Section Finding the centroid of singly Reinforced Rectangular Section  y by    nAs d   yi Ai  y  2  by  nAs  Ai

Solve for the quadratic for

y

 y by  nAs y  by    nAs d   2 2

b 2   y  nAs y  nAs d  0    2 y2 

2nAs b

y

2nAs d

0

b

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Cracked Transformed Section Singly Reinforced Rectangular Section y  2

2nAs

y

2nAs d

b

0

b

I cr  by 3  nAs d  y  3 1

Note: n 

2

Es Ec 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

Es 29000 ksi n  8 Ec 3625 ksi

A #7 bar has an As = 0.6 in2 As  4  0.6 in 2   2.4 in 2

The uncracked centroid is bh 2   n  1 As d y 2 bh   n  1 As

12 in 18 in  

2

  8  1  2.4 in 2  15.5 in 

2 12 in 18 in   8  1  2.4 in 2 

2204.4 in 3   9.47 in 2 232.8 in 29

Example The uncracked moment of inertia 2

bh3  h 2  I    y  bh   d  y   n  1 As 12  2 

12 in 18 in   12

3

2

2  18 in    9.47 in  12 in 18 in   15.5 in  9.47 in  8  1  2.4 in 2   2 

 6491 in 4

The cracked centroid is defined by: As 2.4 in 2    0.0129 bd 12 in 15.5 in  2 y  n   2n   n       d   y  0.3627 15.5 in   5.62 in

 8 0.0129  

2

 2  8  0.0129   8  0.0129   0.3627 30

Example The cracked moment of inertia is by 3 2 I   d  y  nAs 3 1 3 2  12 in  5.62 in   15.5 in  5.62 in   8   2.4 in 2  3  2584.2 in 4

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 Doubly Reinforced Rectangular Section '   2 n  1 A  2 nA 2 n  1 A d  2nAs d     s s s 2 y  y 0 b b 1 2 2 I cr  by 3  n  1As  y  d   nAs d  y  3

Note: n

Es Ec 32

Uncracked Transformed Section Moment of inertia (uncracked doubly reinforced beam) 2

  h I gt  bh  bh y     12 2   1

3

concrete 2 2    n  1As  y  d   n  1As  y  d   steel

Note: I g 

1

bh 3

12 33

Example (Rectangular - Double)

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Cracked Transformed Section Finding the centroid of doubly reinforced T-Section y  2

2t be  bw   2n  1As  2nAs

y

bw

 be  bw  t 2  2n  1As  2nAs d 

0

bw

40

Cracked Transformed Section Finding the moment of inertia for a doubly reinforced T-Section 2

1 t 1 3  3 I cr  be y  bet  y    bw  y  t  12 2 3  beam

flange

  n  1 As  y  d    nAs  d  y  2

2

steel

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 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

43

Reinf. Concrete Sections - Example Compute the n value and the centroid, I uncracked Es 29000 ksi n   8.04 Ec 3605 ksi area (in2) n×area (in2)

n

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

3840.38

13824

313.344

2266.74

The compute the centroid and I uncracked y n A 3840.38 in  y   12.26 in.  n A 313.34 in I   I   d n A 13824 in  2266.7 in 3

i i

i

2

i

i

i

2 i i

4

i

4

 16090.7 in 4

44

Reinf .Concrete Sections - Example Compute the centroid and I for a cracked doubly reinforced beam. '   2 n  1 A  2 nA 2 n  1 A d  2nAs d     s s s 2 y  y 0 b b

y2  

2  7.04  1.2 in 2   2  8.04   2.4 in 2 

y

12 in. 2  7.04  1.2 in 2  (2.5)  2  8.04   2.4 in 2   21.5 in.

12 in. y 2  4.624 y  72.664  0

y

4.624 

 4.624  2

2

 4  72.664 

 6.52 in.

0

45

Reinf. Concrete Sections - Example Compute the moment of inertia for a cracked doubly reinforced beam. 1 2 2 I cr  by 3  n  1As  y  d   nAs d  y  3 1 3 2 2 I cr  12 in. 6.52 in.   7.04  1.2 in   6.52 in.  2.5 in. 3

  8.04   2.4 in

2

  21.5 in.  6.52 in.

2

 5575.22 in 4

The critical ratio of moment of inertia I cr 5575.22 in 4   0.346 4 I g 16090.7 in



I cr  0.35I g 46

Reinf. Concrete Sections - Example The maximum tension stress in tension is f r  7.5 fc  7.5 4000  474.3 psi  0.4743 ksi

The uncracked moments for the beam My I  M  I y M cr    M cr   

4 f r I 0.4743 ksi 16090.7 in     650.2 k-in. y  24 in.  12.26 in.

4 f r I 0.4743 ksi 16090.7 in     622.6 k-in. y 12.26 in.

47

Reinf. Concrete Sections - Example b) Find the components of the beam (at ultimate) Cc  0.85 f cba  0.85  4 ksi 12 in. 0.85  c  34.68c 0.0075  c  2.5 in.  0.003  0.003     c c   0.0075  217.5   fs  Es s  29000  0.003    87  c  c 

 s  

217.5 261   Cs  As  fs  0.85 f c   1.2 in 2   87   0.85(4)   100.32  c c  

Find the components of the beam T   2.4 in 2   60 ksi   144 k 261 T  Cc  Cs  144 k  34.68c  100.32   34.68c 2  43.68c 48261  0 c

Reinf. Concrete Sections - Example The neutral axis

43.68 

c

 43.68  4  261 34.68 2

2  34.68 

 3.44 in.

The strain of the steel  3.44 in.  2.5 in.    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.  

 s  

Note: At service loads, beams are assumed to act elastically. c  3.44 in. y  6.52 in. 49

Calculate the Deflections (1) Instantaneous (immediate) deflections (2) Sustained load deflection Instantaneous Deflections due to dead loads (unfactored) , live, etc.

50

Instantaneous (Immediate) deflections

51

Instantaneous (Immediate) deflections

52

53

Deflection Coefficients

54

Sustained Load Deflections Creep causes an increase in concrete strain Compression steel present Helps limit this effect.



Curvature increases

Increase in compressive strains cause increase in stress in compression reinf. (reduces creep strain in concrete)

55

Sustained Load Deflections Sustained load deflection = D Di

D 

 1  50  

Instantaneous deflection ACI 9.5.2.5

 

As

at mid-span for simple and continuous beams

bd

at for cantilever beams

56

Sustained Load Deflections  = time dependent factor for sustained load

5 years or more 12 months 6 months 3 months

 2.0  1.4  1.2  1.0

57

58

Sustained Load Deflections For dead and live loads

D total  D DLinst   D LLinst   D DL L.T.  D LL L.T. DL and LL may have different  factors for LT ( long term) D calculations

D

 after attachment of   total  N/S components 

 D total  D DL inst 59

Example 9-5:

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Example 9-6:

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