twenty one service loads x load factors concrete holds no tension - - PowerPoint PPT Presentation

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S2009abn Concrete Beams 1 Lecture 21 Elements of Architectural Structures ARCH 614

ELEMENTS OF ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN

ARCH 614

• DR. ANNE NICHOLS

SPRING 2019 lecture

twenty one

concrete construction:

materials & beams

http:// nisee.berkeley.edu/godden S2007abn Concrete Beams 2 Lecture 21 Elements of Architectural Structures ARCH 614

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

– concrete strength = f’c

S2007abn Concrete Beams 3 Lecture 21 Elements of Architectural Structures ARCH 614

Concrete Construction

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

http://nisee.berkeley.edu/godden Concrete Beams 4 Lecture 21 Elements of Architectural Structures ARCH 614 S2007abn

Concrete

• low strength to weight ratio
• relatively inexpensive

– Portland cement – aggregate – water

• hydration
• fire resistant
• creep & shrink

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Concrete Beams 5 Lecture 21 Elements of Architectural Structures ARCH 614 S2007abn

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” 

• longitudinally placed

– bottom – top for compression reinforcement – spliced, hooked, terminated...

S2007abn

1 1 1

E y f E R    

2 2 2

E y f E R    

Concrete Beams 6 Lecture 21 Elements of Architectural Structures ARCH 614

Behavior of Composite Members

• plane sections remain plane
• stress distribution changes

Concrete Beams 7 Lecture 21 Elements of Architectural Structures ARCH 614 S2007abn

Transformation of Material

• n is the ratio of E’s
• effectively widens a material to get

same stress distribution

1 2

E E n 

Concrete Beams 8 Lecture 21 Elements of Architectural Structures ARCH 614 S2007abn

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 E n  

1 2 d transforme c

I My f  

d transforme s

I Myn f  

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S2007abn Concrete Beams 9 Lecture 21 Elements of Architectural Structures ARCH 614

Reinforced Concrete - stress/strain

Concrete Beams 10 Lecture 21 Elements of Architectural Structures ARCH 614 S2007abn

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

Concrete Beams 11 Lecture 21 Elements of Architectural Structures ARCH 614 S2007abn

Location of n.a.

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

) ( 2     x d nA x bx

s

Concrete Beams 12 Lecture 21 Elements of Architectural Structures ARCH 614 S2007abn

T sections

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

flange

f f

bw bw hf hf

   

) ( 2 2              x d nA h x b h x h x h b

s f w f f f f

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S2007abn Concrete Beams 13 Lecture 21 Elements of Architectural Structures ARCH 614

ACI Load Combinations*

*can also use old ACI factors

• 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

S2007abn Concrete Beams 14 Lecture 21 Elements of Architectural Structures ARCH 614

Reinforced Concrete Design

• stress distribution in bending

Wang & Salmon, Chapter 3

b As a/2 T T NA C C x a= 1c 0.85f’c actual stress Whitney stress block d h

S2007abn Concrete Beams 15 Lecture 21 Elements of Architectural Structures ARCH 614

Force Equations

• C = 0.85 f cba
• T = Asfy
• 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 – fy = steel yield strength – As = area of steel reinforcement a/2 T 0.85f’c C a=1c

S2007abn Concrete Beams 16 Lecture 21 Elements of Architectural Structures ARCH 614

• T = C
• Mn = T(d-a/2)

– d = depth to the steel n.a.

• with As

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

Equilibrium

a/2 T C 0.85f’c d

b f f A

c y s

 85 .

a=1x

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S2009abn Concrete Beams 17 Lecture 21 Elements of Architectural Structures ARCH 614

• over-reinforced

– steel won’t yield

• under-reinforced

– steel will yield

• reinforcement ratio

– – use as a design estimate to find As,b,d – max  is found with steel  0.004 (not bal)

Over and Under-reinforcement

bd A ρ

s



http://people.bath.ac.uk/abstji/concrete_video/virtual_lab.htm Concrete Beams 18 Lecture 21 Elements of Architectural Structures ARCH 614 S2007abn

As for a Given Section

• several methods

– guess a and iterate

• 1. guess a (less than n.a.)

2.

• 3. solve for a from Mu =  Asfy (d-a/2)
• 4. repeat from 2. until a from 3. matches a in 2.

y c s

f ba f . A   85          

y s u

f A M d a  2

Concrete Beams 19 Lecture 21 Elements of Architectural Structures ARCH 614 S2007abn

As for a Given Section (cont)

• chart method

– Wang & Salmon Fig. 3.8.1 Rn vs. 

• 1. calculate
• 2. find curve for f’c and fy to get 
• 3. calculate As and a
• simplify by setting h = 1.1d

2

bd M R

n n 

S2007abn Concrete Beams 20 Lecture 21 Elements of Architectural Structures ARCH 614

Reinforcement

• min for crack control
• required
• not less than
• As-max :
• typical cover

– 1.5 in, 3 in with soil

• bar spacing

) ( 3 bd f f A

y c s

  ) ( 200 bd f A

y s 

cover spacing

) 375 . (

1

d a  

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Concrete Beams 21 Lecture 21 Elements of Architectural Structures ARCH 614 S2009abn

Approximate Depths