applied achitectural structures: s concrete tructural...

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APPLIED ACHITECTURAL STRUCTURES:

STRUCTURAL ANALYSIS AND SYSTEMS

ARCH 631

DR. ANNE NICHOLS

FALL 2015

ten

reinforced concrete

constructionReinforced Concrete Construction 1Lecture 10

Applied Architectural StructuresARCH 631

lecture

http://nisee.berkeley.edu/godden

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Concrete

• columns

• beams

• slabs

• domes

• footings

Reinforced Concrete Construction 2Lecture 10

Architectural Structures IIIARCH 631


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Concrete Construction

• cast-in-place

• tilt-up

• prestressing

• post-tensioning




F2007abnReinforced Concrete Construction 4Lecture 9


Concrete Materials

• low strength to weight ratio

• relatively inexpensive

– Portland cement

– aggregate

– water

2

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Concrete Materials

• reinforcement

– deformed bars

– prestressing strand

– stirrups

– development length

– anchorage

– splices






Concrete Materials

• fire resistance– most fire-resistive structural material

– low rate of penetration

– retains strength ifexposure not too long

• stable to 900 – 1200 °Finternally

• loses 50% after that

– no toxic fumes

– cover necessary to protect steel



Concrete Beams

• types

– reinforced

– precast

– prestressed

• shapes

– rectangular, I

– T, double T’s, bulb T’s

– box

– spandrel



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Concrete Beams

• deformation

– camber (elastic)

• hogging

• sagging

– shrinkage strain

• 200-400 x 10-6

• about 2-3 years

– creep strain

• 2~3 times elastic strain

• about 2-3 years

3

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Concrete Beams

• shear

– vertical

– horizontal

– combination:

• tensile stresses

at 45°

• bearing

– crushing



Copyright © Kirk Martini



Concrete Beam Design

• composite of concrete and steel

• American Concrete Institute (ACI)

– design for failure

– strength design (LRFD)• service loads x load factors

• concrete holds no tension

• failure criteria is yield of reinforcement

• failure capacity x reduction factor

• factored loads < reduced capacity

– concrete strength = f’c

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• plane sections remain plane

• stress distribution changes

1

1 1

E yf E

Rε= = − 2

2 2

E yf E

Rε= = −



Behavior of Composite Members



Transformation of Material

• n is the ratio of E’s

• effectively widens a material to get same stress distribution

2

1

En

E=

4

F2007abnReinforced Concrete Construction 13

Lecture 9

Architectural Structures III

ARCH 631

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

E

E

En ==

1

2

dtransforme

cI

Myf −=

dtransforme

sI

Mynf −=



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Reinforced Concrete - stress/strain



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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

– force ductile failure



Location of n.a.

• ignore concrete below n.a.

• transform steel

• same area moments, solve for x

( ) 02

s

xbx nA d x⋅ − − =

5

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( )( )

( ) 02 2

fff f f w s

x hhb h x x h b nA d x

− − + − − − =



T sections

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

f f

bw

bw

hf hf

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ACI Load Combinations*

• 1.4D

• 1.2D + 1.6L + 0.5(Lr or S or R)

• 1.2D + 1.6(Lr or S or R) + (1.0L or 0.5W)

• 1.2D + 1.0W + 1.0L + 0.5(Lr or S or R)

• 1.2D + 1.0E + 1.0L + 0.2S

• 0.9D + 1.0W

• 0.9D + 1.0E

Reinforced Concrete Construction 18

Lecture 10

Applied Architectural Structures

ARCH 631

*can also use old ACI factors



Reinforcement

• deformed steel bars (rebar)

– Grade 40, Fy = 40 ksi

– Grade 60, Fy = 60 ksi - most common

– Grade 75, Fy = 75 ksi

– US customary in # of 1/8” φ

• longitudinally placed

– bottom

– top for compression reinforcement

– spliced, hooked, terminated...

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1cβ


Lecture 9


ARCH 631

Reinforced Concrete Design

• stress distribution in bending

Wang & Salmon, Chapter 3

b

As

a/2

TTNA

CCx a=

0.85f’c

actual stress Whitney stressblock

dh

6

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• C = 0.85 fćba

• T = Asfy• where

– fć = concrete compressive strength

– a = height of stress block

– β1 = factor based on fć– c = location to the n.a.

– b = width of stress block

– fy = steel yield strength

– As = area of steel reinforcementReinforced Concrete Construction 21

Lecture 9


ARCH 631

Force Equations

a/2

T

a=

0.85f’c

C1cβ

1

40000.85 (0.05) 0.65

1000

cf

β′ −

= − ≥



• T = C

• Mn = T(d-a/2)

– d = depth to the steel n.a.

• with As

– a = Asfy

– Mu ≤ φMn φ = 0.9 for flexure*

– φMn = φT(d-a/2) = φ Asfy (d-a/2)

Equilibrium

0.85f’cb

a=

β1c

a/2

T

C

0.85f’c

d

0.25* 0.65 ( ) 0.65

(0.005 )t y

y

φ ε εε

= + − ≥−

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sA

bdρ =



• over-reinforced

– steel won’t yield

• under-reinforced

– steel will yield

• reinforcement ratio

–

– use as a design estimate to find As, b, d

– max ρ is found with εsteel ≥ 0.004 (not ρbal)

– *with εsteel ≥ 0.005, φ = 0.9

Over and Under-reinforcement

www.thfrc.gov


Lecture 9


ARCH 631

As for a given Section

• several methods

– guess a and iterate

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

2.

3. solve for a from

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

b

As

a/2

Asfy

a

0.85f’c

dhC

y

cs

f

baf.A

′=

850

−=

ys

u

fA

Mda

φ2

( )2

adfA M ysu −= φ

7


Lecture 9


ARCH 631

As For Given Section (cont)

• chart method

– Wang & Salmon

Fig. 3.8.1 Rn vs. ρ

1. calculate

2. find curve for f’c and fy to get ρ

3. calculate As and a

• simplify by setting h = 1.1d

2bd

MR n

n =



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Shear in Concrete Beams

• flexure combines with shear to form diagonal cracks

• horizontal reinforcement doesn’t help

• stirrups = vertical reinforcement

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• Vu is at distance d from face of support

• shear capacity:– where bw means thickness of web at n.a.

• shear stress (beams)– = 0.75 for shear

f’c is in psi

• shear strength:– Vs is strength from stirrup reinforcement

c c wV 2 f b dφ φ λ ′=

c c2 fλ ′υ =



ACI Shear Values

λ for lightweight materials


Lecture 9


ARCH 631

Stirrup Reinforcement

• shear capacity:

– Av = area in all legs of stirrups

– s = spacing of stirrup

• may need stirrups when concrete has enough strength!

s

dfAV

yv

s =

8

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greater of w

yt

50 b s

f and

′c w

yt

0.75 f b s

f

− φ

φ

u c

yt

(V V )s

f d

φ

− φ

yt

u c

A f d

V V

vsmaller of

yt

w

A f

50 b

vand

′

v yt

c w

A f

0.75 f b

′φ = φ λ φ =c c wV 2 f b d 0.75 (ACI 11.3.1.1)

NOTE: section numbers are pre ACI 318-14

Required Stirrup Reinforcement

• spacing limits


Lecture 10

Applied Architectural Structures

ARCH 631

0.75


Lecture 9


ARCH 631

Concrete Deflections

• elastic range– I transformed

– Ec (with f’c in psi)• normal weight concrete (~ 145 lb/ft3)

• concrete between 90 and 155 lb/ft3

• cracked– I cracked

– E adjusted

cc fE ′= 000,57

ccc fwE ′= 335.1



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Deflection Limits

• relate to whether or not beam supports or is attached to a damageable non-structural element

• need to check service live load and long term deflection against these

supporting masonry – live + long term

supporting plaster – live

floor systems (typical) – live + long term

roof systems (typical) – live

L/480

L/360

L/240

L/180



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Prestressed Concrete

• impose a longitudinal force on a member in order to withstand more loading until the member reaches a tensile limit

9



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• pretensioned

– reinforcement bonded

• post-tensioned

– bonded or unbonded

– end bearing

• precast

– concrete premade in a position other than

its final position in the structure



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• high strength tendons

– grade 250

– grade 270



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g

tt Mc

A

Pf

I−−=

)(wM

g

bb

Mc

A

Pf

I+−=

t - top

b - bottom

c - distance to fiberIg - gross cross section inertia

• axial prestress (e=0)



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)(wM

• axial prestress (e≠0)

g

tt

g

t

g

tt Mc

r

ec

A

PMcPec

A

Pf

III−

−−=−+−=

21

g

bb

g

b

g

bb

Mc

r

ec

A

PMcPec

A

Pf

III+

+−=+−−=

21

)(A

rrememberI

=−

10



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cr f ′= 5.7f

B

A

0

Self weight

- areas of design importance

(wi)

CDesign load

(DL+LL)

D

EF

G



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Composite Beams

• concrete

– in compression

• steel

– in tension

• shear studs



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Continuous Beams

• reduced size

• reduced moments

• moments canreverse with loading patterns

• need top & bottom reinforcement

• sensitive to settlement

11



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Approximate Depths


Lecture 9


ARCH 631

Concrete Columns

• columns require

– ties or spiral reinforcement

to “confine” concrete

(#3 bars minimum)

– minimum amount of longitudinal steel (4 bars minimum)



Concrete Columns

• effective length in monolithic casts mustbe found with respect to stiffness of joint

• not slender when

*not braced



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Concrete Columns

12



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Concrete Columns

• Po – no bending

• φc = 0.65 for ties with Pn = 0.8Po

• φc = 0.70 for spirals with Pn = 0.85Po

• Pu ≤ φcPn

• nominal axial capacity:

– presumes steel yields

– concrete at ultimate stress

stystgco AfAAfP +−′= )(85.0



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Columns with Bending

• eccentric loads can cause moments

• moments can change shape and induce more deflection

(P- ∆)P

∆



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• for ultimate strength behavior, ultimate strains can’t be exceeded

– concrete 0.003

– steel

• P reduces with M

s

y

E

f



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• need to consider combined stresses

• plot interaction

diagram

1≤+o

n

o

n

M

M

P

P

13



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Concrete Floor Systems

• types & spanning direction



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• flexure design as T-beams (+/- M)

• increase of 10% Vc permitted

• one-way and two-way moments

• slabs need steel

• effective width is smallest of

– L/4

– bw + 16t

– center-to-center

of beams



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One-way

Joists

– standard

stems

– 2.5” to 4.5”

slab

– ~30” widths

– reusable forms

14



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One-way

Joists

– wide pans

– 5’, 6’ up

– light loads & long spans

– one-leg

stirrups



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Two-way

Joists

– domed pans

– 3’, 4’ & 5’



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Construction Supervision

• proper placement of allreinforcement

– welding

– splices

• mix design

– slump

– in-situ strength

• cast cylinders

• cylinder cores – if needed

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