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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
http://nisee.berkeley.edu/godden
F2009abn
Concrete Construction
• cast-in-place
• tilt-up
• prestressing
• post-tensioning
Reinforced Concrete Construction 3Lecture 10
Architectural Structures IIIARCH 631
http://nisee.berkeley.edu/godden
F2007abnReinforced Concrete Construction 4Lecture 9
Architectural Structures IIIARCH 631
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
Reinforced Concrete Construction 5Lecture 10
Architectural Structures IIIARCH 631
http://nisee.berkeley.edu/godden
F2007abnReinforced Concrete Construction 6Lecture 9
Architectural Structures IIIARCH 631
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
F2007abnReinforced Concrete Construction 7Lecture 9
Architectural Structures IIIARCH 631
Concrete Beams
• types
– reinforced
– precast
– prestressed
• shapes
– rectangular, I
– T, double T’s, bulb T’s
– box
– spandrel
Reinforced Concrete Construction 8Lecture 9
Architectural Structures IIIARCH 631
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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
Reinforced Concrete Construction 9Lecture 10
Architectural Structures IIIARCH 631
Copyright © Kirk Martini
F2007abnReinforced Concrete Construction 10Lecture 9
Architectural Structures IIIARCH 631
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= = −
Reinforced Concrete Construction 11Lecture 9
Architectural Structures IIIARCH 631
Behavior of Composite Members
F2007abnReinforced Concrete Construction 12Lecture 9
Architectural Structures IIIARCH 631
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 −=
Reinforced Concrete Construction 14Lecture 9
Architectural Structures IIIARCH 631
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Reinforced Concrete - stress/strain
Reinforced Concrete Construction 15Lecture 9
Architectural Structures IIIARCH 631
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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
F2007abnReinforced Concrete Construction 16Lecture 9
Architectural Structures IIIARCH 631
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
− − + − − − =
Reinforced Concrete Construction 17Lecture 9
Applied Architectural StructuresARCH 631
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
F2007abnReinforced Concrete Construction 19Lecture 9
Applied Architectural StructuresARCH 631
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
Reinforced Concrete Construction 20
Lecture 9
Architectural Structures III
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 f´cba
• T = Asfy• where
– f´c = concrete compressive strength
– a = height of stress block
– β1 = factor based on f´c– 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
Architectural Structures III
ARCH 631
Force Equations
a/2
T
a=
0.85f’c
C1cβ
1
40000.85 (0.05) 0.65
1000
cf
β′ −
= − ≥
F2008abnReinforced Concrete Construction 22Lecture 10
Architectural Structures IIIARCH 631
• 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 =
Reinforced Concrete Construction 23Lecture 9
Architectural Structures IIIARCH 631
• 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
F2007abnReinforced Concrete Construction 24
Lecture 9
Architectural Structures III
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
F2007abnReinforced Concrete Construction 25
Lecture 9
Architectural Structures III
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 =
Reinforced Concrete Construction 26Lecture 9
Architectural Structures IIIARCH 631
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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 =
Reinforced Concrete Construction 27Lecture 9
Architectural Structures IIIARCH 631
ACI Shear Values
λ for lightweight materials
F2007abnReinforced Concrete Construction 28
Lecture 9
Architectural Structures III
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
Reinforced Concrete Construction 29
Lecture 10
Applied Architectural Structures
ARCH 631
0.75
F2007abnReinforced Concrete Construction 30
Lecture 9
Architectural Structures III
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
Reinforced Concrete Construction 31Lecture 9
Architectural Structures IIIARCH 631
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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
Reinforced Concrete Construction 32Lecture 9
Architectural Structures IIIARCH 631
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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
Reinforced Concrete Construction 33Lecture 9
Architectural Structures IIIARCH 631
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Prestressed Concrete
• pretensioned
– reinforcement bonded
• post-tensioned
– bonded or unbonded
– end bearing
• precast
– concrete premade in a position other than
its final position in the structure
Reinforced Concrete Construction 34Lecture 9
Architectural Structures IIIARCH 631
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Prestressed Concrete
• high strength tendons
– grade 250
– grade 270
Reinforced Concrete Construction 35Lecture 9
Architectural Structures IIIARCH 631
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Prestressed Concrete
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)
Reinforced Concrete Construction 36Lecture 9
Architectural Structures IIIARCH 631
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Prestressed Concrete
)(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
Reinforced Concrete Construction 37Lecture 9
Architectural Structures IIIARCH 631
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Prestressed Concrete
Reinforced Concrete Construction 38Lecture 9
Architectural Structures IIIARCH 631
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Prestressed Concrete
cr f ′= 5.7f
B
A
0
Self weight
- areas of design importance
(wi)
CDesign load
(DL+LL)
D
EF
G
Reinforced Concrete Construction 39Lecture 9
Architectural Structures IIIARCH 631
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Composite Beams
• concrete
– in compression
• steel
– in tension
• shear studs
Reinforced Concrete Construction 40Lecture 9
Architectural Structures IIIARCH 631
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Continuous Beams
• reduced size
• reduced moments
• moments canreverse with loading patterns
• need top & bottom reinforcement
• sensitive to settlement
11
Reinforced Concrete Construction 41Lecture 9
Architectural Structures IIIARCH 631
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Approximate Depths
F2007abnReinforced Concrete Construction 42
Lecture 9
Architectural Structures III
ARCH 631
Concrete Columns
• columns require
– ties or spiral reinforcement
to “confine” concrete
(#3 bars minimum)
– minimum amount of longitudinal steel (4 bars minimum)
F2007abnReinforced Concrete Construction 43Lecture 9
Architectural Structures IIIARCH 631
Concrete Columns
• effective length in monolithic casts mustbe found with respect to stiffness of joint
• not slender when
*not braced
Reinforced Concrete Construction 44Lecture 9
Architectural Structures IIIARCH 631
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Concrete Columns
12
Reinforced Concrete Construction 45Lecture 9
Architectural Structures IIIARCH 631
F2007abn
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
Reinforced Concrete Construction 46Lecture 9
Architectural Structures IIIARCH 631
F2007abn
Columns with Bending
• eccentric loads can cause moments
• moments can change shape and induce more deflection
(P- ∆)P
∆
Reinforced Concrete Construction 47Lecture 9
Architectural Structures IIIARCH 631
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Columns with Bending
• for ultimate strength behavior, ultimate strains can’t be exceeded
– concrete 0.003
– steel
• P reduces with M
s
y
E
f
Reinforced Concrete Construction 48Lecture 9
Architectural Structures IIIARCH 631
F2007abn
Columns with Bending
• need to consider combined stresses
• plot interaction
diagram
1≤+o
n
o
n
M
M
P
P
13
Reinforced Concrete Construction 49Lecture 9
Architectural Structures IIIARCH 631
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Concrete Floor Systems
• types & spanning direction
Reinforced Concrete Construction 50Lecture 9
Architectural Structures IIIARCH 631
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Concrete Floor Systems
Reinforced Concrete Construction 51Lecture 9
Architectural Structures IIIARCH 631
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Concrete Floor Systems
• 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
Reinforced Concrete Construction 52Lecture 9
Architectural Structures IIIARCH 631
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One-way
Joists
– standard
stems
– 2.5” to 4.5”
slab
– ~30” widths
– reusable forms
14
Reinforced Concrete Construction 53Lecture 9
Architectural Structures IIIARCH 631
F2007abn
One-way
Joists
– wide pans
– 5’, 6’ up
– light loads & long spans
– one-leg
stirrups
Reinforced Concrete Construction 54Lecture 9
Architectural Structures IIIARCH 631
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Two-way
Joists
– domed pans
– 3’, 4’ & 5’
Reinforced Concrete Construction 55Lecture 9
Architectural Structures IIIARCH 631
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Construction Supervision
• proper placement of allreinforcement
– welding
– splices
• mix design
– slump
– in-situ strength
• cast cylinders
• cylinder cores – if needed