425 serviceability chp 6 s11

8/10/2019 425 Serviceability Chp 6 S11

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Chp-6:Lecture Goals

Serviceability

Deflection calculation

Deflection example


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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/oroccupancy of structures


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;


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:Time dependent Factor

:Amplification factor


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Solution:


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See Example 2.2

Ma: Max.Service

Load Moment


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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 materialproperties are fc= 4 ksi and fy= 60 ksi.

Determine Igt, Icr, Mcr(+), Mcr(-), and compare to the NA of

the beam.


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The components of the beam

2 2

s

2 2

s

c c

2 0.6 in 1.2 in

4 0.6 in 2.4 in

1 k57000 57000 40001000 lb

3605 ksi

A

A

E f


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The compute the n value and the centroid, I uncracked

n area (in ) n*area (in ) yi(in) yi*n*area (in2) I (in ) d (in) d *n*area(in )

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 ksi

8.043605 ksi

E

n E


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The compute the centroid and I uncracked

3

i i i2

i i

2 4 4

i 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


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The compute the centroid and I for a cracked doublyreinforced 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


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The compute the centroid for a cracked doubly reinforced

beam.

2

4.624 72.664 0y y

2

4.624 4.624 4 72.664

2

6.52 in.

y


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The compute the moment of inertia for a cracked doubly

reinforced beam.

2s2

s

3

cr 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


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The critical ratio of moment of inertia

4

cr

4g

cr g

5575.22 in

0.34616090.7 in

0.35

I

I

I I


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Find the components of the beam

c c

s

s s s

2

s 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

cc c

f Ec c

C A f f c

c


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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


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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 0.002073.44 in.

3.44 in.

y 6.52 in.

c


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

The maximum tension stress in tension is

r c7.5 7.5 4000474.3 psi 0.4743 ksi

f f

My

I

s


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Calculate the Deflections

(1) Instantaneous (immediate) deflections

(2) Sustained load deflection

Instantaneous Deflections

due to dead loads( unfactored) , live, etc.


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Calculate the DeflectionsInstantaneous Deflections

Equations for calculating Dinstfor common cases


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Sustained Load Deflections

Creep causes an increase

in concrete strainCurvature

increases

Compression steel

present

Increase in compressive

strains cause increase in

stress in compression

reinforcement (reducescreep strain in concrete)Helps limit thiseffect.


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Sustain load deflection = l Di

Instantaneous deflection

l

501 ACI 9.5.2.5

bd

As

at midspan for simple and continuous beams

at support for cantilever beams


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For dead and live loads

total DL inst LL inst

DL L.T. LL L.T.

D D D

D D

DL and LL may have different factors for LT ( long

term ) Dcalculations

instDLtotal

componentsN/S

ofattachmentaftertotal DDD


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The appropriate value of Icmust be used to calculate

D at each load stage.

instDLinstLL

instDL

DDD Some percentage of DL (if given)

Full DL and LL


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Serviceabil i ty Load Deflections - Example

fc= 5 ksi

fy= 60 ksi


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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 IICheck the ACI Code crack width provisions at midspan of the

beam.


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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


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The moment of inertia3

2

8151 in14.45 in 14.5 in

564 in

i i

i

y Ay

A

3 2

g

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


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The moment capacity

r c

c c

4

r g

cr -

t -

4

r 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-iny 14.5 in 1000 lbs/kip

f f

f

f I

f I


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Determine bending moments due to initial load (0.85DL) 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


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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 /1187.6 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


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Moment at C will be set equal to Mafor simplicity, asgiven in the problem statement.

A cr -

g

B cr +

cr

1050 k-in M

618 k-in Use I @ midspan

M

M


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Assume Rectangular Section Behavior and calculate the

areas of steel and ratio of Modulus of Elasticity

s

c

2 2

s

2 2

s

29000 ksi7.2

4030 ksi3#5 3 0.31 in 0.93 in

4#7 4 0.6 in 2.40 in

En

EA

A

S f


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Calculate the center of the T-beam

s s s s2

2 2

2

2 2

2

2 1 2 2 1 2 0

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 d y yb b

y y

y y

y


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The centroid is located at the As< 4.5 in. = tf Userectangular section behavior

2 23

s scr +

3

22

22

4

11

3

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

I by n A y d nA d y


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The moment of inertia at midspan

3 3

cr cr g cre midspan

a a

3

4

3

4

4

1

618 k-in16900 in

723 k-in

618 k-in1 4330 in723 k-in

12200 in

M MI I I

M M


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Calculate average effective moment of inertia, Ie(avg)forinterior span (for 0.85 DL) For beam with two ends

continuous and use Igfor 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

S i bil i t L d D fl ti E l


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Calculate instantaneous deflection due to 0.85 DL:

Use the deflection equation for a fixed-fixed beam but usethe 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

D



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

22

D L n

A

22

D L n

B

1.62 k/ft 33.67 ft

11 11

167 k-ft 2000 k-in1.62 k/ft 33.67 ft

16 16

115 k-ft 1380 k-in

lM

lM


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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-incracking at midspan

M M M

M M


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Assume beam is fully cracked under full DL + LL,therefore I= Icr(do not calculate Iefor now).

Icr

for supports

2 2

s

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

S i bi l i t L d D fl ti E l


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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 in 012 in

n A nA n A d nA d y y

b b

y y

Ser iceabil i t Load Deflections E ample


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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

n A nA n A d nA d y y

b b

y y

y


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Calculate moment of inertia.

2 23

s scr +

3

22

22

4

11

3

1 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


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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


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Instantaneous Dead and Live Load Deflection.

4 34

D

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/ft

0.242 inches

l

EI

D

D D


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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 0.002211.8 0.390 in 0.702 in

l

l

D D

Serviceabil i ty Load Deflections Example


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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

l

l

D D



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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

480

35 ft 12 in/ft0.875 in < 1.33 in NO GOO

480

l

D D D D D

D D D

D

D!



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Check whether modifying Icr

to Iewill give an acceptable

deflection:

3 3

cr cr g cre midspan

a a

3

4

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



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Check whether modifying Icrto Iewill give an acceptable

deflection:

3

4

e support

3

4

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



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Floor 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 in7060 in

4300 in0.702 in 0.428 in

7060 in

4300 in0.105 in 0.064 in

7060 in

D

D

D

D



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Adjust 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!

D D D D D

D D D

D



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Part II: Check crack width @ midspan

3s cz f d A

c s

2

e s2

2e

2.5 in

A 2 2 2.5 in 12 in 60 in

60 in15 in

# bars 4

d d

d b

AA



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Assume

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.


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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

425 serviceability chp 6 s11

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