twenty two

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lecture

twenty two

concrete construction:

materials & beams

http:// nisee.berkeley.edu/godden

A

RCHITECTURAL

S

TRUCTURES

:

F

ORM,

B

EHAVIOR, AND

D

ESIGN

A

RCH 331

HÜDAVERDİ TOZAN

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Concrete Beams 2 Architectural Structures

Concrete Beam Design

• composite of concrete and steel

• American Concrete Institute (ACI)

– design for maximum stresses

– limit state design

• service loads x load factors

• concrete holds no tension

• failure criteria is yield of reinforcement

• failure capacity x reduction factor

• factored loads < reduced capacity

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Concrete Construction

• cast-in-place

• tilt-up

• prestressing

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Concrete Beams

• types

– reinforced

– precast

– prestressed

• shapes

– rectangular, I

– T, double T’s, bulb T’s

– box

– spandrel

(5)

Concrete Beams

• shear

– vertical

– horizontal

– combination:

• tensile stresses

at 45



• bearing

– crushing

http://urban.arch.virginia.edu

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S2013abn Concrete Beams 6 Lecture 22 Architectural Structures ARCH 331

Concrete

• low strength to weight ratio

• relatively inexpensive

– Portland cement

• types I - V

– aggregate

• course & fine

– water

– admixtures

• air entraining

(7)

Concrete

• hydration

– chemical reaction

– workability

– water to cement ratio

– mix design

• fire resistant

• cover for steel

• creep &

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Concrete

• placement (not pouring!)

• vibrating

• screeding

• floating

• troweling

• curing

• finishing

(9)

Reinforcement

• deformed steel bars (rebar)

– Grade 40, F

_y

= 40 ksi

– Grade 60, F

_y

= 60 ksi - most common

– Grade 75, F

_y

= 75 ksi

– US customary in # of 1/8”



• longitudinally placed

– bottom

– top for compression

reinforcement

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Reinforcement

• prestressing strand

• post-tensioning

• stirrups

• detailing

– development length

– anchorage

– splices

http:// nisee.berkeley.edu/godden

(11)

Composite Beams

• concrete

– in compression

• steel

– in tension

• shear studs

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Behavior of Composite Members

• plane sections remain plane

• stress distribution changes





E

y

E

f

₁



₁





1 



E

y

E

f

₂



₂





2

(13)

Transformation of Material

• n is the ratio of E’s

• effectively widens a material to get

same stress distribution

1

2 E

E

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Stresses in Composite Section

• with a section

transformed to one

material, new I

– stresses in that

material are

determined as usual

– stresses in the other

material need to be

adjusted by n

concrete

steel

E

n



1

2 d

transforme

c

I

My

f





d

transforme

s

I

Myn

f





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Reinforced Concrete Analysis

• for stress calculations

– steel is transformed to concrete

– concrete is in compression above n.a. and

represented by an equivalent stress block

– concrete takes no tension

– steel takes tension

(17)

Location of n.a.

• ignore concrete below n.a.

• transform steel

• same area moments, solve for x

0 )

(









x

nA

d

x

bx

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T sections

• n.a. equation is different if n.a. below

flange

f

b

w

b

w

h

f

h

f



 



(

)

0

2

2 



















 

x

h

x

h

b

x

h

nA

d

x

h

b

f

_f

_w

f

_s

f

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ACI Load Combinations*

• 1.4D

• 1.2D + 1.6L + 0.5(L

_r

or S or R)

• 1.2D + 1.6(L

_r

or S or R) + (1.0L or 0.5W)

• 1.2D + 1.0W + 1.0L + 0.5(L

_r

or S or R)

• 1.2D + 1.0E + 1.0L + 0.2S

• 0.9D + 1.0W

• 0.9D + 1.0E

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Reinforced Concrete Design

• stress distribution in bending

Wang & Salmon, Chapter 3

b

A

_s

a/2

T

NA

C

x

a=



₁

x

0.85 f’

_c

actual stress

Whitney stress

block

d

(21)

Force Equations

• C = 0.85 f´

_c

ba

• T = A

_s

f

_y

• where

– f´

_c

= concrete compressive

strength

– a = height of stress block

–



₁

= factor based on f´

_c

– x = location to the n.a.

– b = width of stress block

– f

= steel yield strength

a/2

T

a=



₁

x

0.85 f’

_c

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• T = C

• M

_n

= T(d-a/2)

– d = depth to the steel n.a.

• with A

_s

– a =

– M

_u





M

_n



= 0.9 for flexure

–



M

_n

=



T(d-a/2) =



A

_s

f

_y

(d-a/2)

Equilibrium

a/2

T

C

a=



₁

x

0.85 f’

_c

d

b

f

A

c

y

s



85 .

0

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• over-reinforced

– steel won’t yield

• under-reinforced

– steel will yield

• reinforcement ratio

–

– use as a design estimate to find A

_s

,b,d

– max



is found with





0.004 (not



)

Over and Under-reinforcement

bd

A

ρ



s

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A

_s

for a Given Section

• several methods

– guess a and iterate

1. guess a (less than n.a.)

2. 3. solve for a from M

_u

=



A

_s

f

_y

(d-a/2)

4. repeat from 2. until a from 3. matches a in 2.

y

c

s

f

ba

f

.

A



0

85 

















y

s

u

f

A

M

d

a



2

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A

_s

for a Given Section (cont)

• chart method

– Wang & Salmon Fig. 3.8.1 R

_n

vs.



1. calculate

2. find curve for f’

_c

and f

_y

to get



3. calculate A

_s

and a

• simplify by setting h = 1.1d

2 bd

M

R

_n



n

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Reinforcement

• min for crack control

• required

• not less than

• A

_s-max

:

• typical cover

– 1.5 in, 3 in with soil

• bar spacing

)

(

3 bd

f

A

y

c

s





)

bd

(

f

A

y

s

200 

cover

spacing

)

d

.

(

a





₁

0

375

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Annunciation Greek Orthodox Church

• Wright, 1956

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Annunciation Greek Orthodox Church

• Wright, 1956

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Cylindrical Shells

• can resist tension

• shape adds “depth”

• not vaults

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Kimball Museum, Kahn 1972

• outer shell edges

(33)

Kimball Museum, Kahn 1972

• skylights at peak

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