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ChpChp--6:Lecture Goals6:Lecture Goals
• Serviceability
• Deflection calculation
• Deflection example • 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
:Time dependent Factor
:Amplification factor
ζ
Solution:Solution:
See Example 2.2
Ma: Max.
Service
Load Moment
Reinforced Concrete Sections Reinforced Concrete Sections -- ExampleExample
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 f = 4 ksi and f = 60 ksi.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 Reinforced Concrete Sections -- ExampleExample
The components of the beam
( )( )
2 2
s
2 2
2 0.6 in 1.2 in
4 0.6 in 2.4 in
A
A
′ = =
= =( )2 2
s
c c
4 0.6 in 2.4 in
1 k57000 57000 4000
1000 lb
3605 ksi
A
E f
= =
= =
=
Reinforced Concrete Sections Reinforced Concrete Sections -- Example Example
The compute the n value and the centroid, I uncracked
s
c
29000 ksi8.04
3605 ksi
En
E= = =
n area (in2) n*area (in2) yi (in) yi *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
c 3605 ksiE
Reinforced Concrete Sections Reinforced Concrete Sections -- ExampleExample
The compute the centroid and I uncracked
3i i i
2
3840.38 in12.26 in.
313.34 in
y n Ay
n A= = =∑
∑ 2
i i
2 4 4
i i i i
4
12.26 in.313.34 in
I 13824 in 2266.7 in
16090.7 in
yn A
I d n A
= = =
= + = +
=
∑∑ ∑
Reinforced Concrete Sections Reinforced Concrete Sections -- Example Example
The compute the centroid and I for a cracked doubly
reinforced beam.
( ) ( )0
212212 ssss2 =+′−
−+′−
+b
dnAAny
b
nAAny
bb
( )( ) ( )( )
( )( ) ( )( )( )
2 2
2
2 2
s
2
2 7.04 1.2 in 2 8.04 2.4 in
12 in.
2 7.04 1.2 in 2 8.04 2.4 in 21.5 in.0
12 in.
4.624 72.664 0
y y
y y
++
+− =
+ − =
Reinforced Concrete Sections Reinforced Concrete Sections -- ExampleExample
The compute the centroid for a cracked doubly reinforced
beam.
2 4.624 72.664 0y y+ − =
( ) ( )24.624 4.624 4 72.664
2
6.52 in.
y− + +
=
=
Reinforced Concrete Sections Reinforced Concrete Sections -- Example Example
The compute the moment of inertia for a cracked doubly
reinforced beam.
( ) ( ) ( )2
s
2
s
3
cr 1
3
1ydnAdyAnybI −+′−′−+=
3
( )( )
( )( )( )
( )( )( )
3
cr
22
22
4
112 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.
5575.22 in
I =
+ −
+ −
=
Reinforced Concrete Sections Reinforced Concrete Sections -- Example Example
The critical ratio of moment of inertia
4
cr
4
5575.22 in0.346
16090.7 in
I
I= =
4
g
cr g
16090.7 in
0.35
I
I I≈
Reinforced Concrete Sections Reinforced Concrete Sections -- ExampleExampleFind the components of the beam
( )( ) ( )
( )
c c
s
0.85 0.85 4 ksi 12 in. 0.85 34.68
2.5 in. 0.00750.003 0.003
C f ba c c
c
c cε
= = =
− ′ = = −
( )
( ) ( )
s
s s s
2
s s s c
0.003 0.003
0.0075 217.529000 0.003 87
217.50.85 1.2 in 87
261100.32
c c
f Ec c
C A f fc
c
ε
ε
′ = = −
′= = − = −
′= − = −
= −
Reinforced Concrete Sections Reinforced Concrete Sections -- ExampleExample
Find the components of the beam
( )( )2
c s
2.4 in 60 ksi 144 k
261
T
T C C
= =
= +
( ) ( ) ( )( )
2
2
261144 k 34.68 100.32 34.68 43.68 261 0
43.68 43.68 4 261 34.68
2 34.68
3.44 in.
c c cc
c
= + − ⇒ − − =
+ +=
=
The neutral axis
Reinforced Concrete Sections Reinforced Concrete Sections -- ExampleExample
The strain of the steel
( )
( )
s
3.44 in. 2.5 in.0.003 0.0008 0.00207
3.44 in.
21.5 in. 3.44 in.0.003 0.0158 0.00207
ε
ε
− ′ = =
− = =
�
�
Note: At service loads, beams are assumed to act
elastically.
( )s 0.003 0.0158 0.002073.44 in.
ε = =
�
3.44 in.
y 6.52 in.
c =
=
Reinforced Concrete Sections Reinforced Concrete Sections -- ExampleExample
Using a linearly varying ε and σ = Eε along the NA
is the centroid of the area for an elastic center
My
Iσ = −
The maximum tension stress in tension is
r c7.5 7.5 4000
474.3 psi 0.4743 ksi
f f= =
= ⇒
I
Reinforced Concrete Sections Reinforced Concrete Sections -- ExampleExample
The uncracked moments for the beam
( )40.4743 ksi 16090.7 in
My IM
I y
f I
σσ = ⇒ =
( )( )
( )
( )( )
r
cr
4
r
cr
0.4743 ksi 16090.7 in650.2 k-in.
24 in. 12.26 in.
0.4743 ksi 16090.7 in622.6 k-in.
12.26 in.
f IM
y
f IM
y
+
−
= = =−
= = =
Calculate the DeflectionsCalculate the Deflections
(1) Instantaneous (immediate) deflections
(2) Sustained load deflection
Instantaneous Deflections
due to dead loads( unfactored) , live, etc.
Calculate the DeflectionsCalculate the DeflectionsInstantaneous Deflections
Equations for calculating ∆inst for common cases
Calculate the DeflectionsCalculate the DeflectionsInstantaneous Deflections
Equations for calculating ∆inst for common cases
Calculate the DeflectionsCalculate the DeflectionsInstantaneous Deflections
Equations for calculating ∆inst for common cases
Calculate the DeflectionsCalculate the DeflectionsInstantaneous Deflections
Equations for calculating ∆inst for common cases
Sustained Load DeflectionsSustained Load Deflections
Creep causes an increase
in concrete strainCurvature
increases⇒
Compression steel Increase in compressive Compression steel
present
Increase in compressive
strains cause increase in
stress in compression
reinforcement (reduces
creep strain in concrete)Helps limit this
effect.
Sustained Load DeflectionsSustained Load Deflections
Sustain load deflection = λ ∆i
Instantaneous deflection
ρ
ξλ
′+=
501 ACI 9.5.2.5ρ ′+ 501 ACI 9.5.2.5
bd
As′
=′ρ at midspan for simple and continuous beams
at support for cantilever beams
Sustained Load DeflectionsSustained Load Deflections
ξ = time dependent factor for sustained load
5 years or more
12 months
6 months1.4 ⇒1.2 ⇒
2.0 ⇒
6 months
3 months
1.2 ⇒1.0 ⇒
Also see Figure 9.5.2.5 from ACI code
Sustained Load DeflectionsSustained Load Deflections
For dead and live loads
( ) ( )
( ) ( )
total DL inst LL inst
DL L.T. LL L.T.
∆ = ∆ + ∆
+ ∆ + ∆DL and LL may have different ξ factors for LT ( long
term ) ∆ calculations
( )instDLtotal
components N/S
of attachmentafter total
∆−∆=∆
Sustained Load DeflectionsSustained Load Deflections
The appropriate value of Ic must be used to calculate
∆ at each load stage.
( )instDL∆ Some percentage of DL (if given)( )
( ) ( )instDLinstLL
instDL
∆+∆
∆ Some percentage of DL (if given)
Full DL and LL
Serviceability Load Deflections Serviceability Load Deflections --
ExampleExample
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 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 Serviceability Load Deflections -- ExampleExample
fc = 5 ksi
fy = 60 ksi
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
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 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 Serviceability Load Deflections -- ExampleExample
Deflection before glass partition is installed (85 % of DL)
( )e
35ft 12 in/ft105 in
lb ≤ = =
( )( ) ( )
e
w
105 in4 4
8t 2 8 4.5 in 2 12 in = 84 in
s = 10 ft = 120 in.
b
b
≤ = =
≤ + = +
≤
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Compute the centroid and gross moment of inertia, Ig.
b h Area yi Ai * yi
Flange 84 4.5 378 17.75 6709.5
Web 15.5 12 186 7.75 1441.5
564 8151
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
The moment of inertia
3
2
8151 in14.45 in 14.5 in
564 in
i i
i
y Ay
A= = = ≅∑
∑3 21
I bh Ad = +∑
( )( ) ( )( )
( )( ) ( )( )
3 2
g
3 22
3 22
4 4
1
12
184 in 4.5 in 378 in 17.75 in - 14.45 in
12
112 in 15.5 in 186 in 7.75 in - 14.45 in
12
16,950 in 16900 in
I bh Ad = +
= +
+ +
= ≅
∑
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
The moment capacity
( )( )
r c
c c
4
7.5 7.5 5000 530 psi
E 57000 57000 5000 /1000 lbs=4030 ksi
530 psi 16900 in
f f
f
f I
= = =
= =
( )( )
( )( )( )( )
( )( )
( ) ( )( ) ( )
4
r g
cr -
t -
4
r g
cr +
t +
530 psi 16900 inM 1610 k-in
y 5.55 in 1000 lbs/kip
530 psi 16900 inM 618 k-in
y 14.5 in 1000 lbs/kip
f I
f I
= = =
= = =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
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 Serviceability Load Deflections -- ExampleExample
The moments at the two locations
( )( )
( )
2
A D n
2
0.85 /11
0.85*1.00k/ft* 33.67 ft /11
M w l= −
= −
( )( )
( )
2
B D n
2
87.6 k-ft 1050 k-in
0.85 /16
0.85*1.00k/ft* 33.67 ft /16
60.2 k-ft 723 k-in
M w l
= ⇒
=
= −
= ⇒
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Moment at C will be set equal to Ma for simplicity, as
given in the problem statement.
( )A cr -1050 k-in <M
1610 k-in Use I @ supports
M =
= →
( )
g
B cr +
cr
1610 k-in Use I @ supports
723 k-in>M
618 k-in Use I @ midspan
M
= →
=
= →
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Assume Rectangular Section Behavior and calculate the
areas of steel and ratio of Modulus of Elasticity
( )
s
c
29000 ksi7.2
4030 ksi
En
E= = =
( )( )
c
2 2
s
2 2
s
3#5 3 0.31 in 0.93 in
4#7 4 0.6 in 2.40 in
A
A
′ = = =
= = =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExampleCalculate the center of the T-beam
( ) ( )
( )( ) ( )( )
s s s s2
2 2
2
2 1 2 2 1 20
2 6.2 0.93 in 2 7.2 2.4 in
84 in.
n A nA n A d nA dy y
b b
y y
′ ′ ′− + − + + − =
+ +
( )( )( ) ( ) ( )( )2 2
2
84 in.
2 6.2 0.93 in 2.5 in. 2 7.2 2.4 in 17.5 in.0
84 in.
0.549 7.54 0
0.549 30.472.49 in.
2
y y
y
+ − =
+ − =
− ±= =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
The centroid is located at the A’s < 4.5 in. = tf Use
rectangular section behavior
( ) ( ) ( ) ( )
( )( )
2 23
s scr +
11
3
1
I by n A y d nA d y′ ′= + − − + −
( )( )
( )( )( )
( )( )( )
3
22
22
4
184 in 2.49 in
3
6.2 0.93 in 2.5 in 2.49 in
7.2 2.4 in 17.5 in 2.49 in
4330 in
=
+ −
+ −
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
The moment of inertia at midspan
( )
( )
3 3
cr crg cre midspan
a a
3
1
618 k-in
M MI I I
M M
= + −
( )
( )
3
4
3
4
4
618 k-in16900 in
723 k-in
618 k-in1 4330 in
723 k-in
12200 in
=
+ −
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
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.
( ) ( ) ( )e1 e2e avg e mid0.7 0.15I I I I= + +( ) ( )
( )( )
e avg e mid
4
4 4
4
0.7 12200 in
0.15 16900 in 16900 in
13600 in
=
+ +
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
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.
( )( )( ) ( )
( )( )
4 34
DL inst 4
0.85 1.00 k/ft 35 ft 12 in/ft
384 384 4030 ksi 13600 in
0.105 inches
l
EI
ω∆ = =
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Calculate additional short-term Deflections (full DL & LL)
( ) ( )( )22
D L n
A
1.62 k/ft 33.67 ft
11 11
167 k-ft 2000 k-in
lM
ω ω− + −= =
= − = −
( ) ( )( )22
D L n
B
1.62 k/ft 33.67 ft
16 16
115 k-ft 1380 k-in
lM
ω ω+= =
= = −
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Calculate additional short-term Deflections (full DL & LL)
Let Mc = Ma = - 2000 k-in for simplicity see problem
statement
2000 k-in 1610 k-inM M M= = − > = −( )
( )
c A cr -
B cr +
2000 k-in 1610 k-in
cracking at supports
1380 k-in 618 k-in
cracking at midspan
M M M
M M
= = − > = −
⇒
= > =
⇒
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Assume beam is fully cracked under full DL + LL,
therefore I= Icr (do not calculate Ie for now).
Icr for supports ( )( ) ( )2 2
s
7.20
2 4 0.20 in 3 0.6 in
n
A
=
= +( )( ) ( )
( )
s
2
2 2
s
2 4 0.20 in 3 0.6 in
3.40 in
20 in - 2.5 in = 17.5 in
3 0.6 in 1.80 in
2.5 in
A
d
A
d
= +
=
=
′ = =
′ =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Class formula using doubly reinforced rectangular section
behavior.
( ) ( )
( )( ) ( )( )
s s s s2
2 2
2 1 2 2 1 20
2 6.2 1.80 in 2 7.2 3.4 in
n A nA n A d nA dy y
b b
′ ′ ′− + − + + − =
+( )( ) ( )( )
( )( )( ) ( )( )( )
2 2
2
2 2
2 6.2 1.80 in 2 7.2 3.4 in
12 in
2 6.2 1.80 in 2.5 in 2 7.2 3.4 in 17.5 in0
12 in
y y + +
+ − =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Class formula using doubly reinforced rectangular section
behavior.
( ) ( )s s s s2
2
2 1 2 2 1 20
5.94 76.05 0
n A nA n A d nA dy y
b b
y y
′ ′ ′− + − + + − =
+ − =2 5.94 76.05 0
5.94 3406.25 in.
2
y y
y
+ − =
− ±= =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Calculate moment of inertia.
( ) ( ) ( ) ( )
( )( )
2 23
s scr +
3
11
3
112 in 6.25 in
I by n A y d nA d y′ ′= + − − + −
= ( )( )
( )( )( )
( )( )( )
22
22
4
12 in 6.25 in3
6.2 1.80 in 2.5 in 6.25 in
7.2 3.4 in 17.5 in 6.25 in
4230 in
=
+ −
+ −
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Weighted Icr
( ) ( )
( ) ( )cr cr1 cr2cr mid
4 4 4
0.7 0.15
0.7 4330 in 0.15 4230 in 4230 in
I I I I= + +
= + +( ) ( )4
0.7 4330 in 0.15 4230 in 4230 in
4300 in
= + +
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Instantaneous Dead and Live Load Deflection.
( )( )( ) ( )
( ) ( )
4 34
D
DL inst 4
1.00 k/ft 35 ft 12 in/ft
384 384 4030 ksi 4300 in
0.390 inches
l
EI
ω∆ = =
=
( ) ( )LL inst DL inst
0.390 inches
0.62 k/ft*
1.00 k/ft
0.242 inches
=
∆ = ∆
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Long term Deflection at the midspan
Dead Load (Duration > 5 years)
( ) ( ) ( )( )( )
2
s
w
3 0.31 in0.00221
2 2 12 in 17.5 in
A tt
b dρ′ = = =
Dead Load (Duration > 5 years)
( )
( ) ( ) ( )DLDL L.T. DL inst
2.01.80
1 50 1 50 0.00221
1.8 0.390 in 0.702 in
ξλ
ρ
λ
= = =′+ +
∆ = ∆ = =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Long term Deflection use the midspan information
Live Load (40 % sustained 6 months)
( ) ( ) ( )( )( )
2
s
w
3 0.31 in0.00221
2 2 12 in 17.5 in
A tt
b dρ ′ = = =
Live Load (40 % sustained 6 months)
( )
( ) ( ) ( )( )LLLL L.T. LL inst
1.21.08
1 50 1 50 0.00221
1.08 0.242 in 0.40
0.105 in
ξλ
ρ
λ
= = =′+ +
∆ = ∆ =
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Total Deflection after Installation of Glass Partition Wall.
( ) ( ) ( ) ( )total DL inst LL inst DL L.T. LL L.T.
0.390 in 0.242 in 0.702 in 0.105 in
= 1.44 in
∆ = ∆ + ∆ + ∆ + ∆
= + + +
∆ = ∆ − ∆ ( )
( )
after attachment total DL inst
permissible
1.44 in 0.105 in 1.33 in
480
35 ft 12 in/ft0.875 in < 1.33 in NO GOO
480
l
∆ = ∆ − ∆
= − =
∆ =
= = ( )D!
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Check whether modifying Icr to Ie will give an acceptable
deflection:
( )
3 3
cr crg cre midspan
a a
1M M
I I IM M
= + −
( )
( )
3
4
3
4
4
618 k-in16900 in
1380 k-in
618 k-in1 4330 in
1380 k-in
5460 in
=
+ −
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Check whether modifying Icr to Ie will give an acceptable
deflection:
( ) ( )
( )
3
4
e support
3
4
1610 k-in16900 in
2000 k-in
1610 k-in1 4230 in
I =
+ − ( )4
1 4230 in2000 k-in
10800 in
+ −
=
( ) ( ) ( )
( ) ( )e1 e2e avg e mid
4 4 4
4
0.7 0.15
0.7 5460 in 0.15 10800 in 10800 in
7060 in
I I I I= + +
= + +
=
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Floor Beam meets the ACI Code Maximum permissible
Deflection Criteria. Adjust deflections:
( )
4
DL inst 4
4
4300 in0.390 in 0.238 in
7060 in
4300 in0.242 in 0.147 in
∆ = =
∆ = = ( )
( )
( )
LL inst 4
4
DL L.T. 4
4
LL L.T. 4
4300 in0.242 in 0.147 in
7060 in
4300 in0.702 in 0.428 in
7060 in
4300 in0.105 in 0.064 in
7060 in
∆ = =
∆ = =
∆ = =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Adjust deflections:
( ) ( ) ( ) ( )total DL inst LL inst DL L.T. LL L.T.
0.238 in. 0.147 in. 0.428 in. 0.064 in.
0.877 in.
∆ = ∆ + ∆ + ∆ + ∆
= + + +
=
( )
( )
after attachment total DL inst
permissible
0.877 in.
0.877 in. 0.105 in.
0.772 in. 0.875 in. OKAY!
=
∆ = ∆ − ∆
= −
= < ∆ =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Part II: Check crack width @ midspan
3s cz f d A=
( )( )c s
2
e s
22e
2.5 in
A 2 2 2.5 in 12 in 60 in
60 in15 in
# bars 4
d d
d b
AA
= =
= = =
= = =
Serviceability Load Deflections Serviceability Load Deflections -- ExampleExample
Assume
( ) ( )s y0.6 0.6 60 ksi 36 ksi ACI 10.6.4f f= = =
( ) ( )( )( )
2336 ksi 2.5 in 15 inz =
= ( )120 k/in < 175 k/in OK=
For interior exposure, the crack width @ midspan
is acceptable.