425-Serviceability-Chp-6-S11.ppt

Chp-6:Lecture GoalsChp-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:Time dependent Factor

:Amplification factor

ζ

Solution:

See Example 2.2

Ma: Max. ServiceLoad Moment

Reinforced Concrete Sections - ExampleReinforced 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.


The components of the beam

2 2s

2 2s

c c

2 0.6 in 1.2 in

4 0.6 in 2.4 in

1 k57000 57000 4000

1000 lb

3605 ksi

A

A

E f

Reinforced Concrete Sections - Example Reinforced Concrete Sections - Example

The compute the n value and the centroid, I uncracked

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

s

c

29000 ksi8.04

3605 ksi

En

E


The compute the centroid and I uncracked

3i i i

2i i

2 4 4i i i i

4

3840.38 in12.26 in.

313.34 in

I 13824 in 2266.7 in

16090.7 in

y n Ay

n A

I d n A

Reinforced Concrete Sections - Example Reinforced Concrete Sections - Example

The compute the centroid and I for a cracked doubly reinforced beam.

0

212212 ssss2

b

dnAAny

b

nAAny

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

26.52 in.

y

Reinforced Concrete Sections - Example Reinforced Concrete Sections - Example The compute the moment of inertia for a cracked doubly reinforced beam.

2s

2s

3cr 1

3

1ydnAdyAnybI

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 - Example Reinforced Concrete Sections - Example The critical ratio of moment of inertia

4cr

4g

cr g

5575.22 in0.346

16090.7 in

0.35

I

I

I I

Reinforced Concrete Sections - ExampleReinforced Concrete Sections - ExampleFind the components of the beam

c c

s

s s s

2s s s c

0.85 0.85 4 ksi 12 in. 0.85 34.68

2.5 in. 0.00750.003 0.003

0.0075 217.529000 0.003 87

217.50.85 1.2 in 87

261100.32

C f ba c c

c

c c

f Ec c

C A f fc

c


Find the components of the beam

2

c s

2

2

2.4 in 60 ksi 144 k

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.

T

T C C

c c cc

c

The neutral axis


The strain of the steel

Note: At service loads, beams are assumed to act elastically.

s

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

3.44 in.

3.44 in.

y 6.52 in.

c


Using a linearly varying and = E along the NA is the centroid of the area for an elastic center

The maximum tension stress in tension is

r c7.5 7.5 4000

474.3 psi 0.4743 ksi

f f

My

I


The uncracked moments for the beam

4

rcr

4

rcr

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.

My IM

I y

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 strain

Curvature increases

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 = i

Instantaneous deflection

501 ACI 9.5.2.5

bd

As at midspan for simple and continuous beams

at support for cantilever beams


= time dependent factor for sustained load

5 years or more 12 months 6 months 3 months

1.4 1.2 1.0

2.0

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 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.

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 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 - ExampleServiceability Load Deflections - Example

fc = 5 ksi

fy = 60 ksi


Part IDetermine 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 IICheck the ACI Code crack width provisions at midspan of the beam.


Deflection before glass partition is installed (85 % of DL)

e

w

35ft 12 in/ft105 in

4 48t 2 8 4.5 in 2 12 in = 84 in

s = 10 ft = 120 in.

lb

b


Compute the centroid and gross moment of inertia, Ig.

b h Area yi A i * yiF lange 84 4.5 378 17.75 6709.5W eb 15.5 12 186 7.75 1441.5

564 8151


The moment of inertia3

2

8151 in14.45 in 14.5 in

564 ini i

i

y Ay

A

3 2g

3 22

3 22

4 4

1

12

184 in 4.5 in 378 in 17.75 in - 14.45 in

121

12 in 15.5 in 186 in 7.75 in - 14.45 in12

16,950 in 16900 in

I bh Ad


The moment capacity

r c

c c

4r g

cr -t -

4r g

cr +t +

7.5 7.5 5000 530 psi

E 57000 57000 5000 /1000 lbs=4030 ksi

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 f

f

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

2

B D n

2

0.85 /11

0.85*1.00k/ft* 33.67 ft /11

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

M w l


Moment at C will be set equal to Ma for simplicity, as given in the problem statement.

A cr -

g

B cr +

cr

1050 k-in <M

1610 k-in Use I @ supports

723 k-in>M

618 k-in Use I @ midspan

M

M

Serviceability Load Deflections - ExampleServiceability Load Deflections - ExampleAssume Rectangular Section Behavior and calculate the areas of steel and ratio of Modulus of Elasticity

s

c

2 2s

2 2s

29000 ksi7.2

4030 ksi

3#5 3 0.31 in 0.93 in

4 #7 4 0.6 in 2.40 in

En

E

A

A

Serviceability Load Deflections - ExampleServiceability Load Deflections - ExampleCalculate the center of the T-beam

s s s s2

2 2

2

2 2

2

2 1 2 2 1 20

2 6.2 0.93 in 2 7.2 2.4 in

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

n A nA n A d nA dy y

b b

y y

y y

y


The centroid is located at the A’s < 4.5 in. = tf Use rectangular section behavior

2 23s scr +

3

22

22

4

11

31

84 in 2.49 in3

6.2 0.93 in 2.5 in 2.49 in

7.2 2.4 in 17.5 in 2.49 in

4330 in

I by n A y d nA d y


The moment of inertia at midspan

3 3

cr crg cre midspan

a a

34

34

4

1

618 k-in16900 in

723 k-in

618 k-in1 4330 in

723 k-in

12200 in

M MI I I

M M


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 mid

4

4 4

4

0.7 0.15

0.7 12200 in

0.15 16900 in 16900 in

13600 in

I I I I


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


Calculate additional short-term Deflections (full DL & LL)

22D L n

A

22D L n

B

1.62 k/ft 33.67 ft

11 11167 k-ft 2000 k-in

1.62 k/ft 33.67 ft

16 16115 k-ft 1380 k-in

lM

lM


Calculate additional short-term Deflections (full DL & LL)

Let Mc = Ma = - 2000 k-in for simplicity see problem statement

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 2s

2

2 2s

7.20

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

n

A

d

A

d


Class formula using doubly reinforced rectangular section behavior.

s s s s2

2 2

2

2 2

2 1 2 2 1 20

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


b b

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

5.94 3406.25 in.

2


b b

y y

y


Calculate moment of inertia.

2 23s scr +

3

22

22

4

11

31

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

I by n A y d nA d y


Weighted Icr

cr cr1 cr2cr mid

4 4 4

4

0.7 0.15

0.7 4330 in 0.15 4230 in 4230 in

4300 in

I I I I


Instantaneous Dead and Live Load Deflection.

4 34D

DL inst 4

LL inst DL inst

1.00 k/ft 35 ft 12 in/ft

384 384 4030 ksi 4300 in

0.390 inches

0.62 k/ft*

1.00 k/ft0.242 inches

l

EI


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 d

DLDL L.T. DL inst

2.01.80

1 50 1 50 0.00221

1.8 0.390 in 0.702 in


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 d

LLLL L.T. LL inst

1.21.08

1 50 1 50 0.00221

1.08 0.242 in 0.40

0.105 in


Total Deflection after Installation of Glass Partition Wall.

total DL inst LL inst DL L.T. LL L.T.

after attachment total DL inst

permissible

0.390 in 0.242 in 0.702 in 0.105 in

= 1.44 in

1.44 in 0.105 in 1.33 in

48035 ft 12 in/ft

0.875 in < 1.33 in NO GOO480

l

D!


Check whether modifying Icr to Ie will give an acceptable deflection:

3 3

cr crg cre midspan

a a

34

34

4

1

618 k-in16900 in

1380 k-in

618 k-in1 4330 in

1380 k-in

5460 in

M MI I I

M M


Check whether modifying Icr to Ie will give an acceptable deflection:

34

e support

34

4

1610 k-in16900 in

2000 k-in

1610 k-in1 4230 in

2000 k-in

10800 in

I

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 - ExampleServiceability Load Deflections - ExampleFloor Beam meets the ACI Code Maximum permissible Deflection Criteria. Adjust deflections:

4

DL inst 4

4

LL inst 4

4

DL L.T. 4

4

LL L.T. 4

4300 in0.390 in 0.238 in

7060 in

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 - ExampleServiceability Load Deflections - ExampleAdjust deflections:

total DL inst LL inst DL L.T. LL L.T.

after attachment total DL inst

permissible

0.238 in. 0.147 in. 0.428 in. 0.064 in.

0.877 in.

0.877 in. 0.105 in.

0.772 in. 0.875 in. OKAY!

Serviceability Load Deflections - ExampleServiceability Load Deflections - ExamplePart II: Check crack width @ midspan

3s cz f d A

c s

2e 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 - ExampleServiceability Load Deflections - ExampleAssume

s y0.6 0.6 60 ksi 36 ksi ACI 10.6.4f f

2336 ksi 2.5 in 15 in

120 k/in < 175 k/in OK

z

For interior exposure, the crack width @ midspan is acceptable.

http://civilengineeringreview.com/book/theory-structures/beam-deflection-method-superposition

http://911review.org/WTC/concrete-core.html

http://iisee.kenken.go.jp/quakes/kocaeli/index.htm

Documents

425-Serviceability-Chp-6-S11.ppt