serviceability • deflection calculation • deflection …€¢ deflection calculation •...

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.


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


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

= = =

= + = +

=

∑∑ ∑


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

++

+− =

+ − =


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− + +

=

=


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 =

+ −

+ −

=


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

′ = = −

′= = − = −

′= − = −

= −


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


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 =

=


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


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

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.


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


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


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

∆−∆=∆


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


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.


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

≤ = =

≤ + = +

≤


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


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 = +

= +

+ +

= ≅

∑


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

= = =

= = =


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


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

= ⇒

=

= −

= ⇒


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

= →

=

= →


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

+ − =

+ − =

− ±= =


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

=

+ −

+ −

=


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

=

+ −

=


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

=

+ +

=


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∆ = =

=


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+= =

= = −


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

= = − > = −

⇒

= > =

⇒


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

= +

=

=

′ = =

′ =


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


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 + +

+ − =


Class formula using doubly reinforced rectangular section

behavior.

( ) ( )s s s s2

2

2 1 2 2 1 20

5.94 76.05 0


b b

y y

′ ′ ′− + − + + − =

+ − =2 5.94 76.05 0

5.94 3406.25 in.

2

y y

y

+ − =

− ±= =


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

=

+ −

+ −

=


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

= + +

=


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

=

∆ = ∆

=


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

= = =′+ +

∆ = ∆ = =


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

= = =′+ +

∆ = ∆ =

=


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!


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

=

+ −

=


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= + +

= + +

=


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

∆ = =

∆ = =

∆ = =


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!

=

∆ = ∆ − ∆

= −

= < ∆ =


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

= =

= = =

= = =


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.

serviceability • deflection calculation • deflection …€¢ deflection calculation •...

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