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SE EXAM REVIEW COURSE —March 2017
NCSEA Structural Engineering Exam Review Course
Vertical Forces ReviewReinforced Concrete Design
Spring 2017
Presented by Lawrence Novak, SE, F.ACI, F.SEI, CERT, LEED®AP
Senior Director of Structural Engineering & Codes
SE EXAM REVIEW COURSE —March 2017
Learning Objectives
• Structural Concrete Design by ACI 318‐11– Behavior of Materials for Reinforced Concrete
– Strength Design Methodology
– Analysis and Design of elements• Beams and one‐way slabs
• Columns
• Two‐way slabs
• Pre‐ and post‐tensioned structures
– Structural Detailing
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SE EXAM REVIEW COURSE —March 2017
Definitions
2
Load(lbs)Stress psi
Area(in. )
Strength MaximumStress
Deformation(in.)Strain
OriginalLength(in.)
AreaLoad
Leng
th
D
SE EXAM REVIEW COURSE —March 2017
Characteristics of Concrete
• Basic concept
– Strong in compression
– Weak in tensionFactor of 10!
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SE EXAM REVIEW COURSE —March 2017
Load-Deformation Characteristics for Concrete
SE EXAM REVIEW COURSE —March 2017
Rate of Loading
Faster loading rate
Slower loading rate
ASTM loading rate
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SE EXAM REVIEW COURSE —March 2017
Styrofoam Beam – Loaded
Tension
Compression
SE EXAM REVIEW COURSE —March 2017
Plain Concrete Beam (no rebar)
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Plain Concrete BeamLoad
Sudden Fracture
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Reinforced Concrete Beam
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Deformed Reinforcement
fy
Strain, 1
Es fs = Ess
fs = fy
Neglect in Design
Idealized stress-strain
Stress, fs
SE EXAM REVIEW COURSE —March 2017
Deformed Reinforcement (ASTM)
0.002 Strain,
fy
Stress, fs
y = fy/Es+ 0.002
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SE EXAM REVIEW COURSE —March 2017
Reinforced Concrete Beam
Tension
Steel Reinforcement
Tension
Load
Load
Concrete
Concrete
Steel Reinforcement
SE EXAM REVIEW COURSE —March 2017
Load-Deformation for Plain and Reinforced Beams
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SE EXAM REVIEW COURSE —March 2017
Modulus of Elasticity
• For wc = 90 to 160 lb/ft3
• For normal‐weight concrete (at 145 lb/ft3)
Ec = 57,000
Ec = wc1.533 cf '
cf '
SE EXAM REVIEW COURSE —March 2017
Strength Design Methodof ACI 318
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SE EXAM REVIEW COURSE —March 2017
Strength Requirements
• Design strength ≥ required strength
– Design strength = strength reduction factor () ×nominal strength
– Required strength = load factor × service load effects
SE EXAM REVIEW COURSE —March 2017
Strength Reduction Factor ()
• Understrength of a member
• Inaccuracies in the design equations
• Degree of ductility
• Importance of the member in the structure
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SE EXAM REVIEW COURSE —March 2017
Material Strength Reduction Factors ()
Tension-controlled 0.90
Compression-controlledTied 0.65Spiral 0.75
Shear and torsion 0.75
Bearing on concrete 0.65
SE EXAM REVIEW COURSE —March 2017
Design Strength
• Mn = design flexural strength
• Pn = design axial strength at given ecc.
• Vn = design shear strength = (Vc + Vs)
• Tn = design torsional moment strength
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Load Types
• Dead (D)
• Live (L)
• Roof live (Lr)
• Snow (S)
• Rain (R)
• Wind (W)
• Seismic (E)
• Soil (H)
• Fluid (F)
• Temperature, creep, shrinkage (T)
SE EXAM REVIEW COURSE —March 2017
Primary Load Combinations
1. U = 1.4 D
2. U = 1.2 D + 1.6 L + 0.5(Lr or S or R)
3. U = 1.2 D + 1.6(Lr or S or R) +(1.0 L or 0.5 W)
4. U = 1.2 D + 1.0 W + 1.0 L + 0.5(Lr or S or R)
5. U = 1.2 D + 1.0 E + 1.0 L + 0.2 S
6. U = 0.9 D + 1.0 W
7. U = 0.9 D + 1.0 E
(see ACI 318‐11 Section 9.2 for additional information)
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SE EXAM REVIEW COURSE —March 2017
Strength Requirements
Mn ≥ Mu
Pn ≥ Pu
Vn = (Vc + Vs) ≥ Vu
Tn ≥ Tu
Design Strength ≥ Required Strength
SE EXAM REVIEW COURSE —March 2017
Question
If the service uniform dead load is 100 psf and the uniform live load is 50 psf, what is the maximum ultimate uniform load?
a. 140 psf
b. 150 psf
c. 200 psf
d. 225 psf
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SE EXAM REVIEW COURSE —March 2017
Question
If the service uniform dead load is 100 psf and the uniform live load is 50 psf, what is the maximum ultimate uniform load?
a. 140 psf
b. 150 psf
c. 200 psf 1.2 DL + 1.6 LL
d. 225 psf
SE EXAM REVIEW COURSE —March 2017
Design of Beams and One‐Way Slabs
Note: bw = 12 in. for one-way slabs
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SE EXAM REVIEW COURSE —March 2017
Flexural Requirements
SE EXAM REVIEW COURSE —March 2017
Nominal Flexural Strength
• Required
– Cross section
– Reinforcement h
bw
As
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SE EXAM REVIEW COURSE —March 2017
Design Assumptions
• Static equilibrium
– External loads and moments and internal stresses
• Compatibility of strains
– Plane section before bending remains plane after bending
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Strain Compatibility
h
bw
As
0.003
Strain
Neutral axisc
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Design Assumptions• Tensile strength of concrete in flexural calculations = 0
SE EXAM REVIEW COURSE —March 2017
Stress Block Factor, 1
0.85
0.65
4000 8000
fc’
1
05.01000
4000f85.0
'c
1
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SE EXAM REVIEW COURSE —March 2017
Nominal Flexural Strength
Mn = (C or T) (d – ) = As fy (d – )2a a
2
Mn = As fy (d – )0.5As fy0.85fc b
(Applies when tension steel yields)
SE EXAM REVIEW COURSE —March 2017
Flexural Behavior
• Strength reduction factor, Φ depends on member behavior
Curvature
Bending moment
Tension controlled
Balanced condition
Compression controlled
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SE EXAM REVIEW COURSE —March 2017
Strength Reduction Factor, Strength reduction factor, ϕ, depends on
maximum net tensile strain, εt,
at nominal strength, Mn
dt
t
0.003
SE EXAM REVIEW COURSE —March 2017
Strength Reduction Factor,
0.65
t = 0.002 t = 0.005
Transition
0.90
= 0.65 + (t – 0.002)(250/3)
Compression-controlled
Tension-controlled
Net Tensile Strain
Spiral
Other
0.75
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SE EXAM REVIEW COURSE —March 2017
Design For Flexure
• Nominal flexural strength
Mn = As fy (d – )
Mn = bd fy (d – )0.5d fy0.85fc
0.5As fy0.85fc b
SE EXAM REVIEW COURSE —March 2017
Design For Flexure
• Nominal strength coefficient, Rn
Rn = fy (1 – ), Mn = Mu /=
0.85fc
fy1 – 1 –
2Rn
0.85fc
0.5fy0.85fc
Mn
bd2
One Layer of Tension Reinforcement
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SE EXAM REVIEW COURSE —March 2017
Design For Flexure
• For f’c = 4,000 psi, fy = 60 ksi, and = 1.25%
Then the previous equation reduces to
Where
– Mu is in kip‐ft
– As is in in2
– D is in inches
– Usable where is less than 1.5%
As = Mu4d
Simplified Approach
SE EXAM REVIEW COURSE —March 2017
0
400
800
1200
0.00 0.01 0.02 0.03
bd/A s
2n
n bdM
R
t = 0.004
t = 0.005
fc’ = 5000 psi
fc’ = 6000 psi
fc’ = 4000 psi
fc’ = 3000 psi
Design Strength
t max
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SE EXAM REVIEW COURSE —March 2017
Minimum Reinforcement for Flexural Members
As,min = 3( fc/fy)bwd
≥ 200bwd/fy(Governs when fc < 4444 psi )
= 0.33% for f’c = 4 ksi and fy = 60 ksi
SE EXAM REVIEW COURSE —March 2017
Question
The MOST important variable affecting the flexural strength is:
a. concrete strength
b. beam depth
c. beam width
d. area of rebar
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SE EXAM REVIEW COURSE —March 2017
Question
The MOST important variable affecting the flexural strength is:
a. concrete strength
b. beam depth
c. beam width
d. area of rebar
SE EXAM REVIEW COURSE —March 2017
Development and Splices of Reinforcement
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SE EXAM REVIEW COURSE —March 2017
Rebar Development
Why the ribs?
SE EXAM REVIEW COURSE —March 2017
Figure R12.10.2
12.11.3
12.2.1 or 12.11.2
Positive M
Negative M
Rebar Development
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SE EXAM REVIEW COURSE —March 2017
Splices of Deformed Bars and Wires in Tension
• Lap length 12 in. for Classes A and BClass A 1.0dClass B 1.3d
• d computed from ACI 318 section 12.2, but without– 12 in. min. of 12.2, and– modification factor for excess reinforcement.
SE EXAM REVIEW COURSE —March 2017
Reinforcing Bar Details
1 inch
db
1.33(max. agg. size)
No. 3 stirrup with No. 6 bars and smallerNo. 4 stirrup with No. 7 thru No. 11 bars
1½ inchclear (7.7.1)
cc
db
Min. clear space =
ds
r = ¾ inch for No. 3 stirrups
1 inch for No. 4 stirrups
cs
bw
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SE EXAM REVIEW COURSE —March 2017
Shear Requirements
SE EXAM REVIEW COURSE —March 2017
Shear and Stirrups
Crack (saw cut)
Stirrup (bungee cord through drilled hole in beam)
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SE EXAM REVIEW COURSE —March 2017
Shear and Stirrups
After concrete cracks, shear load supported by stirrups (stretching bungee cord)
Held together by stirrup
SE EXAM REVIEW COURSE —March 2017
Shear and StirrupsUnderstanding Strut and Tie
Actually a Strut and Tie System52
Tension in Stirrup (tie)Compression in member (strut)
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SE EXAM REVIEW COURSE —March 2017
Diagonal Tension Due to Shear
Stirrups
Section
SE EXAM REVIEW COURSE —March 2017
Shear Strength
h
bw
Av at s
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Critical Section for Shear Design
Distributed loads
d Critical section
SE EXAM REVIEW COURSE —March 2017
Critical Section for Shear Design
d
Critical section
Point load within dist. d
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SE EXAM REVIEW COURSE —March 2017
Shear Design Strength
• Design strength = nominal strength
= 0.75
Vn = Vc + Vs
SE EXAM REVIEW COURSE —March 2017
Shear Design Strength• Vn = (Vc + Vs) ≥ Vu
• Vs = Vu –VcVn = nominal shear strength
Vc = nominal shear strength provided by concrete
Vs = nominal shear strength provided by shear reinforcement
Vu = factored shear force at section
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SE EXAM REVIEW COURSE —March 2017
Shear Strength Provided by Concrete
• Members subject to shear and flexure
(lower bound without axial tension)
Vc = 2 bwdcf'
SE EXAM REVIEW COURSE —March 2017
Shear Strength Provided by Shear Reinforcement
• Members with shear reinforcement perpendicular to axis of member
Vs =Avfytd
s d
Avfyt
45o
ds
V
Avfyt
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Ab
sAv = 2 × Ab for two legs
Shear Reinforcement
SE EXAM REVIEW COURSE —March 2017
Shear Reinforcement
• If Vu ≤ (Vc/2), no shear reinforcement is required
• If Vu > (Vc/2), at least minimum shear reinforcement must be provided (with exceptions)
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SE EXAM REVIEW COURSE —March 2017
Minimum Shear Reinforcement
Av, min = 0.75 bws
fyt
50bws fyt
cf'
SE EXAM REVIEW COURSE —March 2017
Maximum Stirrup Spacing
– For Vu – Vc = Vs < 4 (i.e., low shear)
s < d/2 or 24 in.
– For Vu – Vc = Vs > 4 (i.e., high shear)
s < d/4 or 12 in.
C wf ' b d
C wf ' b d
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SE EXAM REVIEW COURSE —March 2017
Shear Design – Summary
SE EXAM REVIEW COURSE —March 2017
Question
For uniform loading, the critical section for beam shear is located at:
a. face of support
b. a distance d/2 from the face of the support
c. a distance d from the face of the support
d. all sections are critical and need be checked
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SE EXAM REVIEW COURSE —March 2017
Question
For uniform loading, the critical section for beam shear is located at:
a. face of support
b. a distance d/2 from the face of the support
c. a distance d from the face of the support
d. all sections are critical and need be checked
SE EXAM REVIEW COURSE —March 2017
R/C Beam Design Example
6 in
. S
lab
Design 18 in.× 18 in. beam
d =18 in. – .5 in. – 0.5 in. –0.5 in.= 15.5 in.
f’c = 4ksi, fy = 60ksi
30 ft 0 in. typical (centerline)
Live load = 50 psf
SDL = 10 psf 15 f
t-
0 in
.15
ft-
0 in
.
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SE EXAM REVIEW COURSE —March 2017
R/C Beam Design Example
• Step 1: Geometry
d = 15.5 in. =
18‐1.5‐0.5‐0.5
h = 18”
b = 18 inches
SE EXAM REVIEW COURSE —March 2017
R/C Beam Design Example• Step 2: Determine dead load
– Dead load slab = 6”/(12’’/ft)*150pcf = 75 psf– Dead load beam = 1.5ft*(1.5ft‐0.5ft)*150pcf/15’ = 15 psf– Superimposed dead load (given) = 10 psf
Total dead load = 100 psf
• Step 3: Determine ultimate uniform load
– wult = 1.2 DL+1.6 LL = 1.2(100 psf) + 1.6(50 psf) = 200 psf
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SE EXAM REVIEW COURSE —March 2017
R/C Beam Design Example• Step 4: Determine ultimate moments and longitudinal steel (top and
bottom) by simplified approach (use ACI coefficients for moments)
– End moment at face of support = wult Ln2/10
= 200psf × 15ft × (30ft – 1.5ft)2 / 10 = 244 kip‐ft = Mu‐
As = Mu / (4d) = 244 / (4 × 15.5) = 4.0 in2 use 5#8 top
( = 1.4% < 1.5% OK for simplified approach)
– Midspan positive moment = wult Ln2/16
= 200psf × 15ft × (30ft – 1.5ft)2 / 16 = 152 kip‐ft = Mu+
As = Mu / (4d) = 152 / (4 × 15.5) = 2.5 in2 use 4#8 bot
SE EXAM REVIEW COURSE —March 2017
R/C Beam Design Example
• Step 5: Shear design
– Ultimate shear at distance d from support
• Vu = 200 psf × 15ft × {30ft/2 – [1.5ft+15.5in./(12in./ft)]} = 39 kips
– Determine Vc (concrete component)
• Vc = 2 = 2 ( ) × 18in. × 15.5in. = 35 kips
– Determine required Vs (steel component)
• Vs req = Vu/ – Vc = 39 kips / 0.75 – 35 kips = 17 kips
– Determine required Av/s for stirrups
• Av/s = Vs/(fy d) = 17 kips / (60 ksi × 15.5in.) = 0.018 in2/in
– Select stirrups
• Use #3 at 6 inches (Av/s provided = 2 × 0.11/6 = 0.037 in2/in OK)
4 , 0 0 0 C wf ' b d
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SE EXAM REVIEW COURSE —March 2017
R/C Beam Design Example
• Summary
d = 15.5in. =
18‐1.5‐0.5‐0.5
h = 18in.
b = 18”
4#8
5#8
#3 at 6 in.
SE EXAM REVIEW COURSE —March 2017
Progressive Collapse vs. Structural Integrity
• Design against progressive collapse requires extensive analysis and design assuming loss of one member at a time
• ACI 318 Structural Integrity Requirements Call for Minor Changes in Detailing of Reinforcement—no analysis or design needed
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Ronan Point(1968)
Explosion on 18th
floorof 22 floors
Example
SE EXAM REVIEW COURSE —March 2017
Explosion in lower story
Local damage
Example
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R7.13 – What is Structural Integrity?
• It is the intent of the structural integrity provisions to improve the redundancy and ductility of structures, and thereby reduce the risk of failure or collapse of parts or all of a building due to damage occurring to a relatively small area of a building.
• The overall integrity of the structure can be substantially enhanced by minor changes in the detailing of reinforcement.
• The intent is to avoid “house of cards” type collapses.
SE EXAM REVIEW COURSE —March 2017
Conven onal Flexural Reinforcement―3‐Span Beam
Flexural Reinforcement
Bending Moment DiagramGreater of:
‐ d‐ 12 db‐ Span/16
6 in.(12.11.1)
(12.12.3)
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Perimeter Beams and Other Beams
Largest of A+s1/4 or A+
s2/4 (Min. 2 Bars)
• Continuous
• Class B tension splice
• Mechanical or welded splice
A+s1
A+s2
A+s1/4
SE EXAM REVIEW COURSE —March 2017
Perimeter Beams
Largest of A–s1/6 or A–
s2/6 (Min. 2 Bars)
• Continuous
• Class B tension splice
• Mechanical or welded splice
A-s1 A-
s2
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SE EXAM REVIEW COURSE —March 2017
Perimeter Beams
or
Min. 135o Hook
Note: Size and spacing of stirrups is based on shear and/or torsion demand.
SE EXAM REVIEW COURSE —March 2017
Perimeter Beams
OR
Unacceptable OK
Crosstie
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SE EXAM REVIEW COURSE —March 2017
Columns
SE EXAM REVIEW COURSE —March 2017
Design Axial Strength
Pn,max ≥ Pn ≥ Pu
Spiral
Pn,max = 0.85[0.85fc(Ag – Ast) + fyAst ]
Tied
Pn,max = 0.80[0.85fc(Ag – Ast) + fyAst ]
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SE EXAM REVIEW COURSE —March 2017
Pc
Axial Strength – Concrete Contribution
Concrete
Max. Pc = 0.85fcAc = 0.85fc(Ag – Ast)
L
P
SE EXAM REVIEW COURSE —March 2017
L
P
Steel
Ps
Axial Strength – Steel Contribution
Max. Ps = fyAst
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SE EXAM REVIEW COURSE —March 2017
L
P = Po
Po = Pc + Ps = 0.85fc(Ag – Ast) + fy Ast
Axial Strength
SE EXAM REVIEW COURSE —March 2017
Pn,max = Constant × Po
Constant = 0.85 (Spiral)
= 0.80 (Tied)
Nominal Axial Strength – (Pn,max )
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SE EXAM REVIEW COURSE —March 2017
Strength Reduction Factor
• Compression‐controlled section
– Tied = 0.65
– Spiral = 0.75
Design axial strength = Pn,max
SE EXAM REVIEW COURSE —March 2017
Preliminary Column Sizing• Reinforcement ratios in the range of 1–2% are usually the most economical.
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SE EXAM REVIEW COURSE —March 2017
Nonslender Square Tied Columns
Pn(max) = 0.80Ag[0.85fc’ + ρg(fy – 0.85fc’)]
ρg = Ast/Ag, %
Pn(max) = Pu
(kips)
= 0.65fc
’ = 4000 psify = 60,000 psi
SE EXAM REVIEW COURSE —March 2017
Nonslender Tied Columns
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SE EXAM REVIEW COURSE —March 2017
Short Tied Column Axial Design
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
0.5% 0.6% 0.7% 0.8% 0.9% 1.0% 1.1% 1.2% 1.3% 1.4% 1.5%
Pu / Ag (ksi)
= As / Ag
Simplified Column Capacity (short tied column, fy = 60 ksi)
f’c = 6,000 psi
f’c = 5,000 psi
f’c = 4,000 psi concrete
Per ACI 10.8.4
(not allowed for special moment frames for seismic)
SE EXAM REVIEW COURSE —March 2017
Short Tied Column Axial Design
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
0.5% 0.6% 0.7% 0.8% 0.9% 1.0% 1.1% 1.2% 1.3% 1.4% 1.5%
Pu / Ag (ksi)
= As / Ag
Simplified Column Capacity (short tied column, fy = 60 ksi)
f’c = 6 ksi
f’c = 5 ksi
f’c = 4 ksi concrete
1% steel is generally optimum At 1% steel, Column Capacity (Pu / Ag) is approx. = f’c / 2
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SE EXAM REVIEW COURSE —March 2017
Circular Ties for Circular Columns7.10.5.4 —Where longitudinal bars are located around the perimeter of a circle, a complete circular tie shall be permitted.
The ends of the circular tie shall overlap by not less than 6” and terminate with standard hooks that engage a longitudinal column bar.
Overlaps at ends of adjacent circular ties shall be staggered around the perimeter enclosing the longitudinal bars.
This image cannot currently be displayed.
SE EXAM REVIEW COURSE —March 2017
Combined Flexural and Axial LoadThis image cannot currently be displayed.
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Interaction Diagram
h
b
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SE EXAM REVIEW COURSE —March 2017
Interaction Diagram
Moment, Mn
Axi
al L
oad,
Pn
– Pure Compression
Po = 0.85fc(Ag – Ast) + fy AstThis image cannot currently be displayed.
cu = 0.0031
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SE EXAM REVIEW COURSE —March 2017
Interaction Diagram
Moment, Mn
Axi
al L
oad,
Pn 2
This image cannot currently be displayed.
cu = 0.003
SE EXAM REVIEW COURSE —March 2017
Interaction Diagram
Moment, Mn
Axi
al L
oad,
Pn
3
This image cannot currently be displayed.
cu = 0.003
t = 0
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SE EXAM REVIEW COURSE —March 2017
Interaction Diagram
Moment, Mn
Axi
al L
oad,
Pn
4
This image cannot currently be displayed.
cu = 0.003
t = y
SE EXAM REVIEW COURSE —March 2017
Moment, Mn
Axi
al L
oad,
Pn
5
This image cannot currently be displayed.
cu = 0.003
t > y
Interaction Diagram
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SE EXAM REVIEW COURSE —March 2017
Moment, Mn
Axi
al L
oad,
Pn
6
This image cannot currently be displayed.
cu = 0.003
t >> y
Interaction Diagram
SE EXAM REVIEW COURSE —March 2017
Simplified Interaction DiagramThis image cannot currently be displayed.
Zone 1Zone 2
Zone 3
Axial load
Bending moment
1
2
3
4
5s = y
s = 0
0.003
0.003
s = y /2
0.003
Strain
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SE EXAM REVIEW COURSE —March 2017
Strength Reduction Factor,
0.65
t = 0.002 t = 0.005
Transition
0.90
= 0.65 + (t – 0.002)(250/3)
Compression-controlled
Tension-controlled
Net tensile strain
Spiral
Other
0.75
SE EXAM REVIEW COURSE —March 2017
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Nominal strengthPn, Mn
Design strengthPn, Mn
Po
Pn, max = 0.80Po
Compression-controlled
Balanced condition
Po
Pn, max
Interaction Diagram
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SE EXAM REVIEW COURSE —March 2017
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Pn, Mn
Pn, Mn
Compression-controlled
Tension-controlled = 0.90
Transition
Interaction Diagram
SE EXAM REVIEW COURSE —March 2017
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Nominal strength Pn, Mn
Design strength Pn, Mn
+ (Pu, Mu)OK
Interaction Diagram
+ (Pu, Mu)NG
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SE EXAM REVIEW COURSE —March 2017
Splices
Moment
Axi
al lo
ad
fs = 0
Zone 1
Zone 1: Compression
Zone 2
Zone 2: Class A or B
fs = 0.5fy
Zone 3: Class B
Zone 3fs = fy
fs 0.5fy
SE EXAM REVIEW COURSE —March 2017
Slender ColumnsThis image cannot currently be displayed.
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Slender Columns
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Question
What is the strength reduction factor, , for a tied column?
a. 0.65
b. 0.75
c. 0.90
d. Varies from 0.65 to 0.90 based on steel strain
SE EXAM REVIEW COURSE —March 2017
QuestionWhat is the strength reduction factor, , for a tied column?
a. 0.65
b. 0.75
c. 0.90
d. Varies from 0.65 to 0.90 based on steel strain
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Two‐Way Slabs
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SE EXAM REVIEW COURSE —March 2017
Two‐way Concrete Floor Systems
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Flat plate Flat slab
Waffle slab
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Two-way beam supported
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Drop PanelThis image cannot currently be displayed.
d d1
/6 /6 h/4
h
SE EXAM REVIEW COURSE —March 2017
Shear Requirements for Slabs and Footings
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Shear Strength for Slabs and Footings
• Critical shear sections
• Nominal shear strength of concrete
• Shear reinforcement
SE EXAM REVIEW COURSE —March 2017
Shear Strength
• Critical shear sections
– One‐way shear
– Two‐way shear
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One‐Way Shear1
2
Tributary area
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Critical section
d
Slab plan viewd = effective flexural depth of slab
SE EXAM REVIEW COURSE —March 2017
Vu = 2
Vu = qu × Tributary area
One‐Way Shear
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Two‐Way Shear – Tributary AreasThis image cannot currently be displayed.
SE EXAM REVIEW COURSE —March 2017
Two‐Way Shear1
2
Tributary area
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Critical perimeter, bo
d/2
Slab plan viewd = effective flexural depth of slab
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Shear Strength
• Select the maximum of three conditions
1. Maximum nominal shear strength, Vc
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SE EXAM REVIEW COURSE —March 2017
• Nominal shear strength, Vc
2. Reduction for rectangular columns
Vc = 2 + 4
= ratio of long to short column dimensions
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Shear Strength
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• Nominal shear strength, Vc
3. Check for column location
Vc =sdbo
+ 2
s = 40 for interior columns
= 30 for edge columns
= 20 for corner columns
(Think 10 x the # of supported sides)
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Shear Strength
SE EXAM REVIEW COURSE —March 2017
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Critical Perimeter
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Headed Shear Stud Reinforcement for SlabsThis image cannot currently be displayed.
1st Stud d/2 Stud Spacing
d/2 0.75d for prestressed slabs
and footings
Section A‐A
Studs with base rail
Av = cross‐sectional area of studs on any peripheral line
SE EXAM REVIEW COURSE —March 2017
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Headed Shear Stud Reinforcement for Slabs
Interior Column
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Two‐Way Slabs: Shear Stress at Critical Section (Shear Studs Compared to Stirrups)
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Stirrups (11.11.3) Shear Studs (11.11.5)This image cannot currently be displayed.
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SE EXAM REVIEW COURSE —March 2017
Question
For the two‐way punching shear check for a slab, the equation (4) represents:a. the upper bound on Vcb. the lower bound on Vcc. Vcd. none of the above
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Question
For the two‐way punching shear check for a slab, the equation (4 ) represents:a. the upper bound on Vcb. the lower bound on Vcc. Vcd. none of the above
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SE EXAM REVIEW COURSE —March 2017
Prestressed Concrete
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Prestressed Concrete: General Principles
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Tension
Compression
No tension
Area A
Prestressing tendon
Ff
FA
Prestress Stress due to loading
Combined stresses
SE EXAM REVIEW COURSE —March 2017
Methods of Prestressing Concrete Members
• PretensioningThis image cannot currently be displayed.
This image cannot currently be displayed.• Post-tensioning
Anchorage
BeamBonded tendon
Jack
Stressing bed Abutment
Jacking end anchor
Jack
Unbonded tendon in ductBeam
Dead end anchor
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Prestressed Concrete Beams
SE EXAM REVIEW COURSE —March 2017
Advantages of Post‐tensioned Structures
• Reduced structural depth for lower story heights and reduced dead load– Additional savings in labor and material for M/E/P, elevator, and cladding
• Long economical spans
• Wide flexibility and variation in design
• Reduced cracking
• However, additional inspections are required.
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Unbonded Tendons
• The prestressing strand is prevented from bonding and is free to move, relative to the surrounding concrete.
– 7‐wire strands (0.5 inch diameter)
– Sheathing
– Anchor
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SE EXAM REVIEW COURSE —March 2017
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Unbonded Tendons
• Prestressing force can only be transferred to the concrete through the anchorage
– Casting: 5 in. by 2¼ in. typical
– WedgesThis image cannot currently be displayed.
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Typical Tendon ProfileThis image cannot currently be displayed.
Continuous post‐tensioned beam
Drape Post‐tensioningtendon
Neutral axis
DL3CBA L1 L2
SE EXAM REVIEW COURSE —March 2017
Post‐Tensioned Slabs (Option 1)This image cannot currently be displayed.
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Post‐Tensioned Slabs (Option 2)
Not less than 300bwd/fy in each direction. Must be located inside column vertical bars.
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, Butd
SE EXAM REVIEW COURSE —March 2017
Structural Design Standards Relevant for Reinforced Concrete Design
In order of precedence of controlling requirements for forces and concrete design:• International Building Code:
IBC 2012 Edition
• Minimum Design Loads for Buildings and Other Structures:
ASCE 7‐10
• Building Code Requirements for Structural Concrete:
ACI 318‐11
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Recommended References and Additional Study Materials
• PCA Notes on ACI 318‐11 Building Code Requirements for Structural Concrete FREE PDF
• CRSI Design Handbook
• Design of Concrete Structures (McGraw Hill)
• Reinforced Concrete Mechanics and Design (Prentice Hall)
• PCA’s Simplified Design Manual FREE PDF
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SE EXAM REVIEW COURSE —March 2017
“Engineering does not tell men what they should want or why they want it.
Rather it recognizes a need and tries to meet it.”
‐ Hardy Cross, 1952This image cannot currently be displayed. This image cannot currently be displayed.
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NCSEA Structural Engineering Exam Review Course
Vertical Forces ReviewReinforced Concrete Design
Spring 2017
Presented by Lawrence Novak, SE, F.ACI, F.SEI, CERT, LEED®AP
Senior Director of Structural Engineering & Codes