Transcript Slide 1

PCI
th
6
Edition
Flexural Component Design
Presentation Outline
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What’s new to ACI 318
Gravity Loads
Load Effects
Concrete Stress Distribution
Nominal Flexural Strength
Flexural Strength Reduction Factors
Shear Strength
Torsion
Serviceability Requirements
New to ACI 318 – 02
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Load Combinations
Stress limits
Member Classification
Strength Reduction factor is a function of
reinforcement strain
• Minimum shear reinforcement requirements
• Torsion Design Method
Load Combinations
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U = 1.4 (D + F)
U = 1.2 (D + F + T) + 1.6 (L + H) + 0.5 (Lr or S or R)
U = 1.2D + 1.6 (Lr or S or R) + (1.0L or 0.8W)
U = 1.2D + 1.6W + 1.0L + 0.5(Lr or S or R)
U = 1.2D + 1.0E + 1.0L + 0.2S
U= 0.9D + 1.6W + 1.6H
U= 0.9D + 1.0E + 1.6H
Comparison of Load Combinations
• U=1.2D + 1.6 L
• U= 1.4D + 1.7L
If L=.75D
Ratio
2002
1999



 0.90
1.4D  1.7 .75D 
1.2D  1.6 .75D
i.e. a 10% reduction in required strength
Classifications
• No Bottom Tensile Stress Limits
• Classify Members Strength Reduction
Factor
– Tension-Controlled
– Transition
– Compression Controlled
• Three Tensile Stress Classifications
– Class U – Un-cracked
– Class T – Transition
– Class C – Cracked
Copied from ACI 318 2002, ACI 318-02 table R18.3.3
Class C Members
• Stress Analysis Based on Cracked Section
Properties
• No Compression Stress limit
• No Tension Stress limit
• Increase awareness on serviceability
– Crack Control
– Displacements
– Side Skin Reinforcement
Minimum Shear Reinforcing
1999
200
150
Av fy
bw s
100
50
2002
0
0
5,000
10,000
15,000
Concrete Strength, f'c, psi
20,000
System Loads
• Gravity Load Systems
–
–
–
–
Beams
Columns
Floor Member – Double Tees, Hollow Core
Spandrels
• Tributary Area
– Floor members, actual top area
– Beams and spandrels
• Load distribution
– Load path
– Floor members  spandrels or beams  Columns
Live Load Reduction
• Live Loads can be reduced based on:


15


L  L o  0.25 

K LL  A t 
Where:
KLL = 1
Lo = Unreduced live load and
At = tributary area
Live Load Reduction
• Or the alternative floor reduction shall not
exceed
R  r  (A t  150)
or

D
R  23.1  1  
Lo 

Where:
R = % reduction ≤ 40%
r = .08
Member Shear and Moment
• Shear and moments on members can be
found using statics methods and beam tables
from Chapter 11
Strength Design
• Strength design is based using the rectangular stress
block
• The stress in the prestressing steel at nominal
strength, fps, can be determined by strain
compatibility or by an approximate empirical equation
• For elements with compression reinforcement, the
nominal strength can be calculated by assuming that
the compression reinforcement yields. Then verified.
• The designer will normally choose a section and
reinforcement and then determine if it meets the
basic design strength requirement:
Mn  Mu
Concrete Stress Distribution
• Parabolic distribution
• Equivalent rectangular distribution
Stress Block Theory
• Stress-Strain
relationship
    E(f ' )
f’c=6,000 psi
c
–
E(f '
)
c
is not constant
f’c=3,000 psi
Stress Block Theory
• Stress-Strain relationship
– Stress-strain can be modeled by:
fc 
Where
and
 
2  f ''c  (  )
1  (  )2
1.71  f 'c
Ec
f ''c  .9  f 'c
:strain at max. stress
:max stress
Stress Block Theory
• The Whitney stress block is a simplified
stress distribution that shares the same
centroid and total force as the real stress
distribution
=
Equivalent Stress Block – b1 Definition
a  b1c
b1  1.05  05 
b1 = 0.85
when f’c < 3,000 psi
b1 = 0.65
when f’c > 8,000 psi
f 'c
1, 000psi
Design Strength
• Mild Reinforcement – Non - Prestressed
• Prestress Reinforcement
Strength Design Flowchart
• Figure 4.2.1.2
page 4-9
• Non-Prestressed
Path
• Prestressed Path
Non-Prestressed Members
• Find depth of compression block
Depth of Compression Block
a
A s  fy  A 's  f 'y
.85  f 'c  b
Where:
As is the area of tension steel
A’s is the area of compression steel
fy is the mild steel yield strength
Assumes
compression
steel yields
Flanged Sections
• Checked to verify that the compression block is truly
rectangular
Compression Block Area
• If compression block is rectangular, the flanged
section can be designed as a rectangular beam
=
=
A comp  a  b
Compression Block Area
• If the compression block is not rectangular (a> hf),
=
A f  (b  b w )  hf
To find “a”
A w  A comp  A f
a
Aw
bw
Determine Neutral Axis
• From statics and strain compatibility
c  a / b
Check Compression Steel
c  3 d '
• Verify that compression steel has reached yield using
strain compatibility
Compression Comments
• By strain compatibility, compression steel yields if:
c  3  d'
• If compression steel has not yielded, calculation for “a”
must be revised by substituting actual stress for yield
stress
• Non prestressed members should always be tension
controlled, therefore c / dt < 0.375
• Add compression reinforcement to create tesnion
controlled secions
Moment Capacity
• 2 equations
– rectangular stress block in the flange section
– rectangular stress block in flange and stem
section
Strength Design Flowchart
Figure 4.2.1.2
page 4-9
Non- Prestressed
Path
Prestressed Path
This portion of the
flowchart is dedicated to
determining the stress in
the prestress
reinforcement
Stress in Strand
fse - stress in the strand after losses
fpu - is the ultimate strength of the strand
fps - stress in the strand at nominal strength
Stress in Strand
• Typically the jacking force is 65% or
greater
• The short term losses at midspan are
about 10% or less
• The long term losses at midspan are
about 20% or less
fse  0.5  fpu
Stress in Strand
• Nearly all prestressed concrete is bonded
Stress in Strand
• Prestressed Bonded reinforcement

gp 
fpu
d
rp 
fps  fpu  1 

  '

b1 
f 'c dp




 

gp = factor for type of prestressing strand, see ACI 18.0
= .55 for fpy/fpu not less than .80
= .45 for fpy/fpu not less than .85
= .28 for fpy/fpu not less than .90 (Low Relaxation Strand)
rp = prestressing reinforcement ratio
Determine Compression Block
Compression Block Height
Assumes compression
steel yields
a
A ps  fps  A s  fy  A 's  f 'y
.85  f 'c  b
Prestress component
Where
Aps - area of prestressing steel
fps - prestressing steel strength
Flange Sections Check
Compression Steel Check
c  3 d '
• Verify that compression steel has reached yield using
strain compatibility
Moment Capacity
• 2 Equations
– rectangular stress block in flange section
– rectangular stress block in flange and stem
section
Flexural Strength Reduction Factor
• Based on primary reinforcement strain
• Strain is an indication of failure
mechanism
• Three Regions
Member Classification
• On figure 4.2.1.2
Compression Controlled
  < 0.002 at extreme
steel tension fiber or
• c/dt > 0.600
 = 0.70 with spiral ties
 = 0.65 with stirrups
Tension Controlled
  > 0.005 at extreme
steel tension fiber, or
• c/dt < 0.375
 = 0.90 with spiral ties
or stirrups
Transition Zone
• 0.002 <  < 0.005 at extreme
steel tension fiber, or
• 0.375 < c/dt < 0.6
 = 0.57 + 67() or
 = 0.48 + 83() with spiral
ties
 = 0.37 + 0.20/(c/dt) or
 = 0.23 + 0.25/(c/dt) with
stirrups
Strand Slip Regions
• ACI Section 9.3.2.7
‘where the strand embedment length is
less than the development length’
 =0.75
Limits of Reinforcement
• To prevent failure immediately upon cracking,
Minimum As is determined by:
A s,min 
3  f 'c
fy
 bw  d 
200  b w  d
fy
• As,min is allowed to be waived if tensile
reinforcement is 1/3 greater than required by
analysis
Limits of Reinforcement
• The flexural member must also have adequate
reinforcement to resist the cracking moment
– Where
M  1.2M
n
cr
 P Pe

S

bc
Mcr  Sbc   
 fr   Mnc 
 1
 A Sb

 Sb

Section after composite
has been applied,
including prestress
forces
Correction for
initial stresses on
non-composite,
prior to topping
placement
Critical Sections
Horizontal Shear
• ACI requires that the interface between
the composite and non-composite, be
intentionally roughened, clean and free of
laitance
• Experience and tests have shown that
normal methods used for finishing precast
components qualifies as “intentionally
roughened”
Horizontal Shear, Fh Positive Moment Region
• Based on the force transferred in topping (page 4-53)
Horizontal Shear, Fh Negative Moment Region
• Based on the force transferred in topping (page 4-53)
Unreinforced Horizontal Shear
Fh    80  b v  l vh
Where
 – 0.75
bv – width of shear area
lvh - length of the member subject to shear, 1/2 the
span for simply supported members
Reinforced Horizontal Shear
Fh    (260  0.6  rv  fy )    b v  l
A cs 
Fh
  m e  fy
Where
 – 0.75
rv - shear reinforcement ratio
Acs - Area of shear reinforcement
me - Effective shear friction coefficient
vh
Shear Friction Coefficient
me 
1000    A cr  m
Vu
Fh
Shear Resistance by Non-Prestressed Concrete
• Shear strength for
non-prestressed
sections
Vc  2  f 'c  b w  d
Prestress Concrete Shear Capacity

Vu  d 
Vc   0.6  f 'c  700
  bw  d
Mu 

Where:
Vu  d
Mu
•
•
•
•
1
ACI Eq 11-9
Effective prestress must be 0.4fpu
Accounts for shear combined with moment
May be used unless more detail is required
Prestress Concrete Shear Capacity
• Concrete shear strength is minimum is
Vc  2  f 'c  b w  d
• Maximum allowed shear resistance from
concrete is:
Vc  5  f 'c  b w  d
Shear Capacity, Prestressed
• Resistance by concrete when diagonal cracking is a
result of combined shear and moment
Vi  Mcr
Vci  0.6  f 'c  b w  d  Vd 
Mmax
Where:
Vi and Mmax - factored
externally applied loads
e.g. no self weight
Vd - is un-factored dead
load shear
Shear Capacity, Prestressed
• Resistance by concrete when diagonal cracking is a
result of principal tensile stress in the web is in excess of
cracking stress.


Vcw  3.5  f 'c  0.3  fpc  bw  d  Vp
Where:
Vp = the vertical
component of effective
prestress force (harped
or draped strand only)
Vcmax
• Shear capacity is the minimum of Vc, or if a
detailed analysis is used the minimum of Vci
or Vcw
Shear Steel
If:
Vu  Vc
Then:
v s  Vn  Vc
or v s 
Vu

 Vc
Shear Steel Minimum Requirements
• Non-prestressed members
A v  0.75  f 'c 
bw  s
fy
 50  b w
s
fy
• Prestressed members
Av 
A ps  fpu  s
80  fy  d

d
bw
Remember
both legs of a stirrup count for Av
Torsion
• Current ACI
– Based on compact sections
– Greater degree of fixity than PC can provide
• Provision for alternate solution
– Zia, Paul and Hsu, T.C., “Design for Torsion and
Shear in Prestressed Concrete,” Preprint 3424,
American Society of Civil Engineers, October,
1978. Reprinted in revised form in PCI JOURNAL,
V. 49, No. 3, May-June 2004.
Torsion
For members loaded two sides, such as inverted
tee beams, find the worst case condition with
full load on one side, and dead load on the
other
1.0D
1.2D+1.6L
Torsion
• In order to neglect Torsion
Tu  Tu(min)
Where:
Tu(min) – minimum torsional strength provided
by concrete
Minimum Torsional Strength
Tu(min)


2
   0.5     f 'c   x  y   g


Where:
x and y - are short and long
side, respectively of a
component rectangle
g  is the prestress factor
Prestress Factor, g
• For Prestressed Members
g  1  10
fpc
f`c
Where:
fpc – level of prestress after losses
Maximum Torsional Strength
• Avoid compression failures due to over
reinforcing
1
 K    f`  x y
3
2
Tn(max) 
t
 K V 
t
t
1 

30

C

T

t
u
Tn(max) 
Where:
c
2
Tu


fpc 
bw  d
K t  g 12  10  C t 

f`c 
 x2 y
Maximum Shear Strength
• Avoid compression failures due to over
reinforcing
Vn(max) 
10    f`c  b w  d
 30  C  T 
t
u
1 

 K t  Vt 
Vn(max) 
Vu

2
Torsion/Shear Relationship
• Determine the torsion carried by the concrete
Tc 
T 'c
T' T 
1  c u
 V 'c Vu 
2
Where:
T’c and V’c - concrete resistance under pure
torsion and shear respectively
Tc and Vc - portions of the concrete resistance
of torsion and shear
Torsion/Shear Relationship
• Determine the shear carried by the concrete
V 'c
Vc 
2
 V' V 
1  c u
 T 'c Tu 
Torsion Steel Design
• Provide stirrups for torsion
moment - in addition to
shear
T

u
  Tc   s
 

At 
 t  x1  y1  fy
 t  0.66  0.33  y1 x1  1.5
Where
x and y - short and long
dimensions of the closed
stirrup
Torsion Steel Design
• Minimum area of closed stirrups is
limited by
A
v
 2A t

min
 50 
bw  s
fy
 (g )  200 
2
bw  s
fy
Longitudinal Torsion Steel
• Provide longitudinal steel for
torsion based on equation
Al 
2  A t  (x1  y1 )
s
or








Tu
2  At 
 400  x 

Al  


 x 1  y1



f
V
s 
u
 y
T 

u




3

C
t


• Whichever greater


Longitudinal Steel limits


 400  x
Al  
 fy


The
2  At






T
2

A


u
t

  x 1  y1


V
s 
T  u 

 u 3  C 
t


factor in
s
the second equation need not exceed
12  fpc  50  b
50  b w 
w
 1 
 

fy
f`c 
fy

Detailing Requirements, Stirrups
• 135 degree hooks are required unless sufficient
cover is supplied
• The 135 degree stirrup hooks are to be anchored
around a longitudinal bar
• Torsion steel is in addition to shear steel
Detailing Requirements, Longitudinal Steel
• Placement of the bars should be around the
perimeter
• Spacing should spaced at no more than 12 inches
• Longitudinal torsion steel must be in addition to
required flexural steel (note at ends flexural demand
reduces)
• Prestressing strand is permitted (@ 60ksi)
• The critical section is at the end of simply supported
members, therefore U-bars may be required to meet
bar development requirements
Serviceability Requirements
• Three classifications for prestressed
components
– Class U: Uncracked
– Class T: Transition
– Class C: Cracked
 t  7.5 f 'c
Stress
7.5 f 'c   t  12 f 'c
 t  12 f 'c
Uncracked Section
• Table 4.2.2.1 (Page 4.24)
• Easiest computation
• Use traditional mechanics
of materials methods to
determine stresses, gross
section and deflection.
• No crack control or side
skin reinforcement
requirements
Transition Section
• Table 4.2.2.1 (Page 4.24)
• Use traditional mechanics
of materials methods to
determine stresses only.
• Use bilinear cracked
section to determine
deflection
• No crack control or side
skin reinforcement
requirements
Cracked Section
• Table 4.2.2.1 (Page 4.24)
• Iterative process
• Use bilinear cracked
section to determine
deflection and to
determine member
stresses
• Must use crack control
steel per ACI 10.6.4
modified by ACI 18.4.4.1
and ACI 10.6.7
Cracked Section Stress Calculation
• Class C member require stress to be
check using a Cracked Transformed
Section
• The reinforcement spacing
requirements must be adhered to
Cracked Transformed
Section Property Calculation Steps
Step 1 – Determine if section is cracked
Step 2 – Estimate Decompression Force in Strand
Step 3 – Estimate Decompression Force in mild
reinforcement (if any)
Step 4 – Create an equivalent force in topping if present
Step 5 – Calculate transformed section of all elements
and modular ratios
Step 6 – Iterate the location of the neutral axis until the
normal stress at this level is zero
Step 7 – Check Results with a a moment and force
equilibrium set of equations
Steel Stress
• fdc – decompression stress
stress in the strand when the
surrounding concrete stress is zero –
Conservative to use, fse (stress after
losses) when no additional mild steel is
present.
Simple Example
Page 4-31
Deflection Calculation –
Bilinear Cracked Section
• Deflection before the
member has cracked is
calculated using the
gross (uncracked)
moment of inertia, Ig
• Additional deflection
after cracking is
calculated using the
moment of inertia of the
cracked section Icr
Effective Moment of Inertia
• Alternative method
3
3
M 

Mcr 
cr
Ie  
 Ig  1 
 Icr
M
M
 a

a 
or based on stress
f f 
Mcr
 1   tl r 
Ma
 fl 
Where:
ftl = final stress
fl = stress due to live load
fr = modulus of rupture
Prestress Losses
• Prestressing losses
– Sources of total prestress loss (TL)
TL = ES + CR + SH + RE
– Elastic Shortening (SH)
– Creep (CR)
– Shrinkage (SH)
– Relaxation of tendons (RE)
Elastic Shortening
• Caused by the prestressed force in the precast member
ES  K es  Eps  fcir Eci
Where:
Kes = 1.0 for pre-tensioned members
Eps = modulus of elasticity of prestressing tendons (about
28,500 ksi)
Eci = modulus of elasticity of concrete at time prestress is
applied
fcir = net compressive stress in concrete at center of gravity of
prestressing force immediately after the prestress has been
applied to the concrete
fcir
 P P  e2  M  e
g
fcir  K cir   i  i


 A
Ig 
Ig
g
Where:
Pi = initial prestress force (after anchorage seating loss)
e = eccentricity of center of gravity of tendons with
respect to
center of gravity of concrete at the cross
section
considered
Mg = bending moment due to dead weight of prestressed
member and any other permanent loads
in place at time of
prestressing
Kcir = 0.9 for pretensioned members
Creep
• Creep (CR)
– Caused by stress in the concrete


CR  K cr  Eps Eci  fcir  fcds

Where:
Kcr = 2.0 normal weight concrete
= 1.6 sand-lightweight concrete
fcds = stress in concrete at center of gravity of
prestressing force due to all uperimposed
permanent dead loads that are applied to
the member after it has been prestressed
fcds
fcds 
Msd  e
Ig
Where:
Msd = moment due to all superimposed permanent
dead and sustained loads applied after prestressing
Shrinkage
• Volume change determined by section and
environment





SH  8.2  106  K sh  Eps  1  0.06  V S  100  R.H.
• Where:
Ksh = 1.0 for pretensioned members
V/S = volume-to-surface ratio
R.H. = average ambient relative humidity
from
map
Relative Humidity
Page 3-114 Figure 3.10.12
Relaxation
• Relaxation of prestressing tendons is based on the
strand properties


RE  K re  J  SH  CR  ES   C


Where:
Kre and J - Tabulated in the PCI handbook
C - Tabulated or by empirical equations in the PCI
handbook
Relaxation Table
• Values for Kre and J
for given strand
• Table 4.7.3.1
page 4-85
Relaxation Table Values for C
• fpi = initial stress in
prestress strand
• fpu = ultimate stress
for prestress strand
• Table 4.7.3.2
(Page 4-86)
Prestress Transfer Length
• Transfer length –
Length when the stress
in the strand is applied
to the concrete
• Transfer length is not
used to calculate
capacity


lt  fse 3  db
lt   fse 3  db
Prestress Development Length
• Development length length required to
develop ultimate strand
capacity
• Development length is
not used to calculate
stresses in the member

ld  lt  fps  fse




ld  fse 3  db  fps  fse

Beam Ledge Geometry
Beam Ledge Design
• For Concentrated loads where s > bt + hl, find
the lesser of:


Vn  3      f 'c  hl  2  bl  b  b t  hl 




Vn      f 'c  hl  2  bl  b  b t  hl  2  de 


Beam Ledge Design
• For Concentrated loads where s < bt + hl, find
the lesser of:


Vn  1.5      f 'c  hl  2  b l  b  b t  hl  s 




b  h 
l
Vn      f 'c  hl   b l  b   t
  de  s 


 2 


Beam Ledge Reinforcement
• For continuous loads or closely spaced concentrated
loads:
Vn  24    hl    f 'c
• Ledge reinforcement should be provided by 3 checks
– As, cantilevered bending of ledge
– Al, longitudinal bending of ledge
– Ash, shear of ledge
Beam Ledge Reinforcement
• Transverse (cantilever) bending reinforcement, As
• Uniformly spaced over width of 6hl on either side of the
bearing
• Not to exceed half the
distance to the next load
• Bar spacing should not
exceed the ledge depth,
hl, or 18 in
h 
 a
1 
A s    Vu     Nu   l  
fy 
 d
 d  
Nu
0.2 
Vdl
Longitudinal Ledge Reinforcement
• Placed in both the top and bottom of the ledge
portion of the beam:
200  bl  b  dl
Al 
fy
Where:
dl - is the depth of steel


U-bars or hooked bars may
be required to develop
reinforcement at the end
of the ledge
Hanger Reinforcement
• Required for attachment of the ledge to the web
• Distribution and spacing
of Ash reinforcement
should follow the same
guidelines as for As
A sh 
Vu
  fy

 m
Hanger (Shear) Ledge Reinforcement
• Ash is not additive to shear and torsion
reinforcement
• “m” is a modification factor which can be
derived, and is dependent on beam section
geometry. PCI 6th edition has design aids
on table 4.5.4.1
Dap Design
(1) Flexure (cantilever bending) and axial tension in the
extended end. Provide flexural reinforcement, Af, plus
axial tension reinforcement, An.
Dap Design
(2) Direct shear at the junction of the dap and the main body
of the member. Provide shear friction steel, composed of
Avf + Ah, plus axial tension reinforcement, An
Dap Design
(3) Diagonal tension emanating from the re-entrant
corner. Provide shear reinforcement, Ash
Dap Design
(4) Diagonal tension in the extended end. Provide shear
reinforcement composed of Ah and Av
Dap Design
(5) Diagonal tension in the undapped portion. This is
resisted by providing a full development length for As
beyond the potential crack.
Dap Reinforcement
5 Main Areas of Steel
• Tension - As
• Shear steel - Ah
• Diagonal cracking – Ash, A’sh
• Dap Shear Steel - Av
Tension Steel – As
• The horizontal reinforcement is determined in
a manner similar to that for column corbels:

h 
 a
A s  A f  An 
  Vu     Nu   l  
fy 
 d
 d  
1
and 0.2 
Nu
Vdl
Shear Steel – Ah
• The potential vertical crack (2) is resisted by a
combination of As and Ah
Ah 
2  Vu
3    fy  m e
 An
Shear Steel – Ah
• Note the development ld of Ah beyond the
assumed crack plane. Ah is usually a U-bar
such that the bar is developed in the dap
Diagonal Cracking Steel – Ash
• The reinforcement required to resist diagonal
tension cracking starting from the re-entrant
corner (3) can be calculated from:
A sh 
Vu
  fy
and   .75
Dap Shear Steel – Av
• Additional reinforcement for Crack (4) is
required in the extended end, such that:

Vn    A v  fy  Ah  fy  2  b  d    f 'c

Dap Shear Steel – Av
• At least one-half of the reinforcement
required in this area should be placed
vertically. Thus:
V

u
Av 
   2  b  d    f 'c 
2  fy  

1
Dap Limitations and Considerations
• Design Condition as a dap if any of the
following apply
– The depth of the recess exceeds 0.2H or 8 in.
– The width of the recess (lp) exceeds 12 in.
– For members less than 8 in. wide, less than onehalf of the main flexural reinforcement extends to
the end of the member above the dap
– For members 8 in. or more wide, less than onethird of the main flexural reinforcement extends to
the end of the member above the dap
Questions?