reinforced concrete construction

APPLIED ACHITECTURAL STRUCTURES: STRUCTURAL ANALYSIS AND SYSTEMS

Concrete

ARCH 631

• • • • •

DR. ANNE NICHOLS FALL 2013 lecture

ten

reinforced concrete construction Reinforced Concrete Construction 1 Lecture 10

Applied Architectural Structures ARCH 631

columns beams slabs domes footings

http://nisee.berkeley.edu/godden

http://nisee.berkeley.edu/godden

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Reinforced Concrete Construction 2 Lecture 10

Architectural Structures III ARCH 631

Concrete Construction

Concrete Materials

• • • •

• low strength to weight ratio • relatively inexpensive

cast-in-place tilt-up prestressing post-tensioning


http://nisee.berkeley.edu/godden Architectural Structures III ARCH 631

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– Portland cement – aggregate – water

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Concrete Materials

Concrete Materials

• reinforcement

• fire resistance – most fire-resistive structural material – low rate of penetration – retains strength if exposure not too long

– deformed bars – prestressing strand – stirrups – development length – anchorage – splices

• stable to 900 – 1200 F internally • loses 50% after that

– no toxic fumes – cover necessary to protect steel http://nisee.berkeley.edu/godden



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Concrete Beams

Concrete Beams

• types

• deformation – camber (elastic)

– reinforced – precast – prestressed

• hogging • sagging

– shrinkage strain

• shapes

• 200-400 x 10-6 • about 2-3 years

– rectangular, I – T, double T’s, bulb T’s – box – spandrel Reinforced Concrete Construction 7 Lecture 9

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– creep strain • 2~3 times elastic strain • about 2-3 years F2007abn



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Concrete Beams

Concrete Beam Design

• shear

• composite of concrete and steel • American Concrete Institute (ACI)

– vertical – horizontal – combination:

– design for failure – strength design (LRFD)

• tensile stresses at 45

Copyright © Kirk Martini

• bearing – crushing

• • • • •

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 Reinforced Concrete Construction 9 Lecture 10


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Behavior of Composite Members

Transformation of Material

• plane sections remain plane • stress distribution changes

• n is the ratio of E’s

n

E2 E1

• effectively widens a material to get same stress distribution

f1  E1   Reinforced Concrete Construction 11 Lecture 9

E1 y



f 2  E 2  


E2 y



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Stresses in Composite Section • with a section transformed to one material, new I

n

– stresses in that material are determined as usual – stresses in the other material need to be adjusted by n Reinforced Concrete Construction 13 Lecture 9

Reinforced Concrete - stress/strain

E E2  steel E1 Econcrete

fc   fs  

My I transformed Myn I transformed


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Reinforced Concrete Analysis

Location of n.a.

• for stress calculations

• ignore concrete below n.a. • transform steel • same area moments, solve for x

– 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

bx  Reinforced Concrete Construction 15 Lecture 9


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x  nA s ( d  x )  0 2 Architectural Structures III ARCH 631

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T sections

ACI Load Combinations*

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

• • • • • • •

f

f

hf

hf bw

bw

bf hf  x  

hf

   x  h b  x  hf   nA ( d  x )  0 f w s 2  2



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 *can also use old ACI factors

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Reinforcement

Reinforced Concrete Design

• deformed steel bars (rebar)

• stress distribution in bending

– Grade 40, Fy = 40 ksi – Grade 60, Fy = 60 ksi - most common – Grade 75, Fy = 75 ksi – US customary in # of 1/8” 

b d

h As

NA T actual stress

– bottom – top for compression reinforcement – spliced, hooked, terminated... Architectural Structures III ARCH 631

C

x

• longitudinally placed


0.85f’c a= 1 x

a/2 C

T Whitney stress block

Wang & Salmon, Chapter 3 F2007abn



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Force Equations • C = 0.85 f´cba • T = Asfy • where

Equilibrium 0.85f’c a/2 C

a= 1 x

0.85f’c

– d = depth to the steel n.a. T

– f´c = concrete compressive strength – a = height of stress block – b = width of stress block – fy = steel yield strength – As = area of steel reinforcement Reinforced Concrete Construction 21 Lecture 9

• T=C • Mn = T(d-a/2) • with As

T

0.85f’cb

– Mu  Mn  = 0.9 for flexure – Mn = T(d-a/2) =  Asfy (d-a/2) F2007abn



Over and Under-reinforcement

As for a given Section

• over-reinforced

• several methods

• under-reinforced

b

1. guess a (less than n.a.) 0.85 f cba 2. www.thfrc.gov

• reinforcement ratio A – ρ s bd

0.85f’c

As 

fy

 Mu a  2 d   As f y 

– max  is found with steel  0.004 (not bal) F2007abn

   

a/2

a

C

d

h As



3. solve for a from M u  As f y d  a

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


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– guess a and iterate

– steel won’t yield


d

– a = Asfy


– steel will yield

a/2 C

a=  1x

Asf y

2



4. repeat from 2. until a from 3. matches a in 2. Reinforced Concrete Construction 24 Lecture 9


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As For Given Section (cont)

Shear in Concrete Beams

• chart method

• flexure combines with shear to form diagonal cracks

– Wang & Salmon Fig. 3.8.1 Rn vs.  1. calculate Rn 

• horizontal reinforcement doesn’t help • stirrups = vertical reinforcement

Mn bd 2

2. find curve for f’c and fy to get  3. calculate As and a

• simplify by setting h = 1.1d Reinforced Concrete Construction 25 Lecture 9


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ACI Shear Values

Stirrup Reinforcement

• Vu is at distance d from face of support • shear capacity: Vc   c  bw d

• shear capacity:

– where bw means thickness of web at n.a.

• shear stress (beams)  = 0.75 for shear – c  2 f c

Vc   2 f c bw d • shear strength:

Vs 

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Av f y d s

– Av = area in all legs of stirrups – s = spacing of stirrup

f’c is in psi

• may need stirrups when concrete has enough strength!

Vu  Vc  Vs

– Vs is strength from stirrup reinforcement ACI 11.3, 11.12 Reinforced Concrete Construction 27 Lecture 9


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Required Stirrup Reinforcement

Concrete Deflections

• spacing limits

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

E c  57 ,000

f c

• concrete between 90 and 155 lb/ft3

E c  wc1.5 33 f c

• cracked – I cracked – E adjusted

0.75 0.75





Deflection Limits

Prestressed Concrete

• relate to whether or not beam supports or is attached to a damageable nonstructural element • need to check service live load and long term deflection against these

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

L/180 L/240 L/360 L/480 Reinforced Concrete Construction 31 Lecture 9

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roof systems (typical) – live floor systems (typical) – live + long term supporting plaster – live supporting masonry – live + long term Architectural Structures III ARCH 631

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• pretensioned

• high strength tendons

– reinforcement bonded

– grade 250 – grade 270

• post-tensioned – bonded or unbonded – end bearing

• precast – concrete premade in a position other than its final position in the structure Reinforced Concrete Construction 33 Lecture 9


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• axial prestress (e=0)

• axial prestress (e0)

M (w )

ft   fb  

P Mc t  A Ig P Mc b  A Ig


M (w )

t - top b - bottom c - distance to fiber Ig - gross cross section inertia


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P Pect Mc t P  ec  Mc t     1  2t   A Ig Ig A r  Ig P Pecb Mc b P  ecb  Mc b fb       1  2   A Ig Ig A r  Ig

ft  

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(remember r 

I ) A

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Prestressed Concrete G F E D

fr  7.5 fc

C Design load (DL+LL)

0 (wi)

B

Self weight

A - areas of design importance



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Composite Beams

Continuous Beams

• concrete

• reduced size • reduced moments

– in compression

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• steel – in tension

• moments can reverse with loading patterns • need top & bottom reinforcement • sensitive to settlement

• shear studs



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Approximate Depths

Concrete Columns • columns require – ties or spiral reinforcement to “confine” concrete (#3 bars minimum)

– minimum amount of longitudinal steel (#5 bars minimum: 4 with ties, 5 with spiral)



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Concrete Columns



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Concrete Columns

• effective length in monolithic casts must be found with respect to stiffness of joint • not slender when kLu  22 r



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Concrete Columns

Columns with Bending

• Po – no bending

• eccentric loads can cause moments • moments can change shape and induce more deflection (P-)

Po  0.85 f c( Ag  Ast )  f y Ast

• c = 0.65 for ties with Pn = 0.8Po • c = 0.70 for spirals with Pn = 0.85Po

P

• Pu  cPn



• nominal axial capacity: – presumes steel yields – concrete at ultimate stress Reinforced Concrete Construction 45 Lecture 9


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

• need to consider combined stresses

– concrete 0.003 – steel

fy

Pn M n  1 Po M o

Es

• P reduces with M


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• plot interaction diagram Architectural Structures III ARCH 631

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Concrete Floor Systems


• types & spanning direction



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One-way

• • • • •

Joists

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


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– standard stems – 2.5” to 4.5” slab – ~30” widths – reusable forms

– L/4 – bw + 16t – center-to-center of beams Reinforced Concrete Construction 51 Lecture 9


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One-way

Two-way

Joists

Joists

– wide pans – 5’, 6’ up – light loads & long spans – one-leg stirrups


– domed pans – 3’, 4’ & 5’


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Construction Supervision • proper placement of all reinforcement – welding – splices

• mix design – slump – in-situ strength • cast cylinders • cylinder cores – if needed Reinforced Concrete Construction 55 Lecture 9


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