Chp-6:Lecture Goals
• Serviceability
• Deflection calculation
• Deflection example
• Structural Design Profession is concerned with:
Limit States Philosophy:
Strength Limit State
(safety-fracture, fatigue, overturning buckling
etc.-)
Serviceability
Performance of structures under normal service
load and are concerned the uses and/or
occupancy of structures
;
λ :Amplification factor
ζ :Time dependent Factor
Solution:
See Example 2.2
Ma: Max.
Service
Load Moment
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
3840.38
13824
313.344
2266.74
Reinforced Concrete Sections - Example
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
 16090.7 in 4
4
i
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
2
y 
y
0
b
b
y2 

2  7.04  1.2 in 2   2  8.04   2.4 in 2 
2  7.04  1.2 in 2 
s
y
12 in.
 2  8.04   2.4 in 2   21.5 in.
12 in.
y 2  4.624 y  72.664  0
0
Reinforced Concrete Sections - Example
The compute the centroid for a cracked doubly reinforced
beam.
y 2  4.624 y  72.664  0
y
4.624 
 6.52 in.
 4.624 
2
2
 4  72.664 
Reinforced Concrete Sections - Example
The compute the moment of inertia for a cracked doubly
reinforced beam.
1 3
2
2


I cr  by  n  1As  y  d   nAs d  y 
3
1
3
I cr  12 in. 6.52 in.
3
  7.04  1.2 in
  6.52 in.  2.5 in.
  8.04   2.4 in   21.5 in.  6.52 in.
2
2
 5575.22 in 4
2
2
Reinforced Concrete Sections - Example
The critical ratio of moment of inertia
4
I cr 5575.22 in

 0.346
4
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


0.0075 
217.5

fs  Es s  29000  0.003 
  87 
c 
c

217.5 

Cs  As  fs  0.85 f c   1.2 in   87 

c


261
 100.32 
c
2
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
c
43.68 
 3.44 in.
 43.68  4  261 34.68 
2  34.68 
2
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
M cr   
M cr   
f r I 0.4743 ksi 16090.7 in


y
 24 in.  12.26 in.
fr I


y
4
  650.2 k-in.
0.4743 ksi 16090.7 in 4 
12.26 in.
 622.6 k-in.
Calculate the Deflections
(1) Instantaneous (immediate) deflections
(2) Sustained load deflection
Instantaneous Deflections
due to dead loads( unfactored) , live, etc.
Calculate the Deflections
Instantaneous Deflections
Equations for calculating Dinst for common cases
Calculate the Deflections
Instantaneous Deflections
Equations for calculating Dinst for common cases
Calculate the Deflections
Instantaneous Deflections
Equations for calculating Dinst for common cases
Calculate the Deflections
Instantaneous Deflections
Equations for calculating Dinst for common cases
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
reinforcement (reduces
creep strain in concrete)
Sustained Load Deflections
Sustain load deflection = l Di
l
 

1  50  
Instantaneous deflection
ACI 9.5.2.5
As
at midspan for simple and continuous beams
bd
at support for cantilever beams
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
Also see Figure 9.5.2.5 from ACI code
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
Sustained Load Deflections
The appropriate value of Ic must be used to calculate
D at each load stage.
D DLinst
Some percentage of DL (if given)
D LLinst  D DLinst
Full DL and LL
Serviceability Load Deflections Example
Show in the attached figure is a typical interior span of a floor
beam spanning between the girders at locations A and C. Partition
walls, which may be damaged by large deflections, are to be
erected at this level. The interior beam shown in the attached
figure will support one of these partition walls. The weight of the
wall is included in the uniform dead load provided in the figure.
Assume that 15 % of the distributed dead load is due to a
superimposed dead load, which is applied to the beam after the
partition wall is in place. Also assume that 40 % of the live load
will be sustained for at least 6 months.
Serviceability Load Deflections - Example
fc = 5 ksi
fy = 60 ksi
Serviceability Load Deflections - Example
Part I
Determine whether the floor beam meets the ACI Code maximum
permissible deflection criteria. (Note: it will be assumed that it is
acceptable to consider the effective moments of inertia at location
A and B when computing the average effective moment of inertia
for the span in this example.)
Part II
Check the ACI Code crack width provisions at midspan of the
beam.
Serviceability Load Deflections - Example
Deflection before glass partition is installed (85 % of DL)
l 35ft 12 in/ft 
be  
 105 in
4
4
  8t  2   bw  8  4.5 in  2  12 in = 84 in
 s = 10 ft = 120 in.
Serviceability Load Deflections - Example
Compute the centroid and gross moment of inertia, Ig.
b
Flange
Web
h
84
15.5
4.5
12
Area
378
186
564
yi
17.75
7.75
Ai * yi
6709.5
1441.5
8151
Serviceability Load Deflections - Example
The moment of inertia
 yi Ai
8151 in 3
y

 14.45 in  14.5 in
2
 Ai 564 in
1 3

I g    bh  Ad 2 
 12

1
3
2
2
  84 in  4.5 in    378 in  17.75 in - 14.45 in 
12
1
3
2
2
 12 in 15.5 in   186 in   7.75 in - 14.45 in 
12
 16,950 in 4  16900 in 4
Serviceability Load Deflections - Example
The moment capacity
f r  7.5 f c  7.5 5000  530 psi
E c  57000 f c  57000 5000 /1000 lbs=4030 ksi
 530 psi  16900 in 4 
M cr - 

 1610 k-in
y t  -  5.55 in 1000 lbs/kip 
 530 psi  16900 in 4 
fr Ig
M cr +  

 618 k-in
y t  +  14.5 in 1000 lbs/kip 
fr Ig
Serviceability Load Deflections - Example
Determine bending moments due to initial load (0.85
DL) The ACI moment coefficients will be used to
calculate the bending moments Since the loading is not
patterned in this case, This is slightly conservative
Serviceability Load Deflections - Example
The moments at the two locations
M A    0.85wD  ln  /11
2
 0.85*1.00k/ft*  33.67 ft  /11
2
 87.6 k-ft  1050 k-in
M B   0.85wD  ln  /16
2
 0.85*1.00k/ft*  33.67 ft  /16
2
 60.2 k-ft  723 k-in
Serviceability Load Deflections - Example
Moment at C will be set equal to Ma for simplicity, as
given in the problem statement.
M A  1050 k-in <M cr  -
 1610 k-in  Use Ig @ supports
M B  723 k-in>M cr  + 
 618 k-in  Use Icr @ midspan
Serviceability Load Deflections - Example
Assume Rectangular Section Behavior and calculate the
areas of steel and ratio of Modulus of Elasticity
Es 29000 ksi
n

 7.2
Ec
4030 ksi
As  3#5  3  0.31 in 2   0.93 in 2
As  4 # 7  4  0.6 in 2   2.40 in 2
Serviceability Load Deflections - Example
Calculate the center of the T-beam
 2  n  1 As  2nAs 
 2  n  1 Asd   2nAs d 
y 
 y 
0
b
b




 2  6.2   0.93 in 2   2  7.2   2.4 in 2  
y
y2  
84 in.


 2  6.2   0.93 in 2   2.5 in.  2  7.2   2.4 in 2  17.5 in. 
0

84 in.


y 2  0.549 y  7.54  0
2
0.549  30.47
y
 2.49 in.
2
Serviceability Load Deflections - Example
The centroid is located at the A’s < 4.5 in. = tf Use
rectangular section behavior
I cr  + 
1 3
2
2
 by   n  1 As  y  d    nAs  d  y 
3
1
3
  84 in  2.49 in 
3
  6.2   0.93 in
2
  7.2   2.4 in
 17.5 in  2.49 in 
 4330 in 4
2
  2.5 in  2.49 in 
2
2
Serviceability Load Deflections - Example
The moment of inertia at midspan
3

 M cr 
 M cr  

 I g   1  
  I cr
Ma 
 Ma 



3
I e midspan 
3
 618 k-in 
4

 16900 in 
 723 k-in 
  618 k-in 3 
4
 1  
4330 in 



  723 k-in  


4
 12200 in
Serviceability Load Deflections - Example
Calculate average effective moment of inertia, Ie(avg) for
interior span (for 0.85 DL) For beam with two ends
continuous and use Ig for the two ends.
I e avg   0.7 I e mid   0.15  I e1  I e2 
 0.7 12200 in 4 
0.15 16900 in 4  16900 in 4 
 13600 in
4
Serviceability Load Deflections - Example
Calculate instantaneous deflection due to 0.85 DL:
Use the deflection equation for a fixed-fixed beam but use
the span length from the centerline support to centerline
support to reasonably approximate the actual deflection.
D DLinst  
l
4
384 EI
0.85 1.00 k/ft  35 ft  12 in/ft 
4

 0.105 inches
384  4030 ksi  13600 in 4 
3
Serviceability Load Deflections - Example
Calculate additional short-term Deflections (full DL & LL)
 1.62 k/ft  33.67 ft 
  D   L  l

MA 
11
11
 167 k-ft  2000 k-in
2
n
MB
D  L  l


2
n
1.62 k/ft  33.67 ft 


16
 115 k-ft  1380 k-in
16
2
2
Serviceability Load Deflections - Example
Calculate additional short-term Deflections (full DL & LL)
Let Mc = Ma = - 2000 k-in for simplicity see problem
statement
M c  M A  2000 k-in  M cr -  1610 k-in
 cracking at supports
M B  1380 k-in  M cr  +   618 k-in
 cracking at midspan
Serviceability Load Deflections - Example
Assume beam is fully cracked under full DL + LL,
therefore I= Icr (do not calculate Ie for now).
n  7.20
Icr for supports
As  2  4   0.20 in 2   3  0.6 in 2 
 3.40 in 2
d  20 in - 2.5 in = 17.5 in
As  3  0.6 in 2   1.80 in 2
d   2.5 in
Serviceability Load Deflections - Example
Class formula using doubly reinforced rectangular section
behavior.
 2  n  1 As  2nAs 
 2  n  1 Asd   2nAs d 
2
y 
 y 
0
b
b




 2  6.2  1.80 in 2   2  7.2   3.4 in 2  
y
y2  
12 in


 2  6.2  1.80 in 2   2.5 in   2  7.2   3.4 in 2  17.5 in  
0

12 in


Serviceability Load Deflections - Example
Class formula using doubly reinforced rectangular section
behavior.
 2  n  1 As  2nAs 
 2  n  1 Asd   2nAs d 
y 
 y 
0
b
b




2
y  5.94 y  76.05  0
2
5.94  340
y
 6.25 in.
2
Serviceability Load Deflections - Example
Calculate moment of inertia.
I cr  + 
1 3
2
2
 by   n  1 As  y  d    nAs  d  y 
3
1
3
 12 in  6.25 in 
3
  6.2  1.80 in
2
  2.5 in  6.25 in 
2
  7.2   3.4 in
 17.5 in  6.25 in 
2
 4230 in 4
2
Serviceability Load Deflections - Example
Weighted Icr
I cr  0.7 I cr  mid   0.15  I cr1  I cr2 
 0.7  4330 in
 4300 in
4
4
  0.15  4230 in
4
 4230 in
4

Serviceability Load Deflections - Example
Instantaneous Dead and Live Load Deflection.
D DLinst  
D LLinst 
 Dl
4
384 EI
1.00 k/ft  35 ft  12 in/ft 


384  4030 ksi   4300 in 4 
 0.390 inches
0.62 k/ft
 D DLinst  *
1.00 k/ft
 0.242 inches
4
3
Serviceability Load Deflections - Example
Long term Deflection at the midspan
3  0.31 in 
As  t 
  t  

 0.00221
2bw d 2 12 in 17.5 in 
2
Dead Load (Duration > 5 years)

2.0
l

 1.80
1  50   1  50  0.00221
D DL L.T.  lDL D DLinst   1.8  0.390 in   0.702 in
Serviceability Load Deflections - Example
Long term Deflection use the midspan information
3  0.31 in 2 
As  t 
  t  

 0.00221
2bw d 2 12 in 17.5 in 
Live Load (40 % sustained 6 months)

1.2
l

 1.08
1  50   1  50  0.00221
D LL L.T.  lLL D LLinst   1.08  0.242 in  0.40 
 0.105 in
Serviceability Load Deflections - Example
Total Deflection after Installation of Glass Partition Wall.
D total  D DL inst   D LLinst   D DL L.T.  D LL L.T.
D after attachment
D permissible
 0.390 in  0.242 in  0.702 in  0.105 in
= 1.44 in
 D total  D DLinst 
 1.44 in  0.105 in  1.33 in
l

480
35 ft 12 in/ft 

 0.875 in < 1.33 in  NO GOOD!
480
Serviceability Load Deflections - Example
Check whether modifying Icr to Ie will give an acceptable
deflection:
3

 M cr 
 M cr  

 I g  1  
  I cr
Ma 
 Ma 



3
I e midspan 
3
 618 k-in 
4

 16900 in 
 1380 k-in 
  618 k-in 3 
4
 1  
4330
in

  1380 k-in   


4
 5460 in
Serviceability Load Deflections - Example
Check whether modifying Icr to Ie will give an acceptable
deflection:
3
 1610 k-in 
4
I esupport   
16900
in

 
 2000 k-in 
  1610 k-in 3 
4
 1  
4230
in

  2000 k-in   


 10800 in 4
I e avg   0.7 I e mid   0.15  I e1  I e2 
 0.7  5460 in
 7060 in 4
4
  0.15 10800 in
4
 10800 in
4

Serviceability Load Deflections - Example
Floor Beam meets the ACI Code Maximum permissible
Deflection Criteria. Adjust deflections:
D DLinst 
D LLinst 
D DL L.T.
D LL L.T.
 4300 in 4 
 0.390 in 
 0.238 in
4 
 7060 in 
 4300 in 4 
 0.242 in 
 0.147 in
4 
 7060 in 
 4300 in 4 
 0.702 in 
 0.428 in
4 
 7060 in 
 4300 in 4 
 0.105 in 
 0.064 in
4 
 7060 in 
Serviceability Load Deflections - Example
Adjust deflections:
D total  D DLinst   D LLinst   D DL L.T.  D LL L.T.
 0.238 in.  0.147 in.  0.428 in.  0.064 in.
 0.877 in.
D after attachment  D total  D DLinst 
 0.877 in.  0.105 in.
 0.772 in.  D permissible  0.875 in.
 OKAY!
Serviceability Load Deflections - Example
Part II: Check crack width @ midspan
z  fs 3 dc A
d c  ds  2.5 in
A e  2dsb  2  2.5 in 12 in   60 in 2
Ae
60 in 2
A

 15 in 2
# bars
4
Serviceability Load Deflections - Example
Assume
fs  0.6 f y  0.6  60 ksi   36 ksi
 ACI 10.6.4
z   36 ksi  3  2.5 in  15 in 2 
 120 k/in < 175 k/in  OK 
For interior exposure, the crack width @ midspan
is acceptable.
http://civilengineeringreview.com/book/theorystructures/beam-deflection-method-superposition
http://911review.org/WTC/con
crete-core.html
http://iisee.kenken.go.jp/quakes/kocaeli/index.htm