vertical forces review reinforced concrete design

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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


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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Definitions

2

Load(lbs)Stress psi

Area(in. )

Strength MaximumStress

Deformation(in.)Strain

OriginalLength(in.)

AreaLoad

Leng

th

D


Characteristics of Concrete

• Basic concept

– Strong in compression

– Weak in tensionFactor of 10!

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Load-Deformation Characteristics for Concrete


Rate of Loading

Faster loading rate

Slower loading rate

ASTM loading rate

6

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Styrofoam Beam – Loaded

Tension

Compression


Plain Concrete Beam (no rebar)

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Plain Concrete BeamLoad

Sudden Fracture


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


Deformed Reinforcement (ASTM)

0.002 Strain,

fy

Stress, fs

y = fy/Es+ 0.002

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Reinforced Concrete Beam

Tension

Steel Reinforcement

Tension

Load

Load

Concrete

Concrete

Steel Reinforcement


Load-Deformation for Plain and Reinforced Beams

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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 '


Strength Design Methodof ACI 318

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Strength Requirements

• Design strength ≥ required strength

– Design strength = strength reduction factor () ×nominal strength

– Required strength = load factor × service load effects


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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Material Strength Reduction Factors ()

Tension-controlled 0.90

Compression-controlledTied 0.65Spiral 0.75

Shear and torsion 0.75

Bearing on concrete 0.65


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)


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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Strength Requirements

Mn ≥ Mu

Pn ≥ Pu

Vn = (Vc + Vs) ≥ Vu

Tn ≥ Tu

Design Strength ≥ Required Strength


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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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


Design of Beams and One‐Way Slabs

Note: bw = 12 in. for one-way slabs

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Flexural Requirements


Nominal Flexural Strength

• Required

– Cross section

– Reinforcement h

bw

As

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Design Assumptions

• Static equilibrium

– External loads and moments and internal stresses

• Compatibility of strains

– Plane section before bending remains plane after bending


Strain Compatibility

h

bw

As

0.003

Strain

Neutral axisc

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Design Assumptions• Tensile strength of concrete in flexural calculations = 0


Stress Block Factor, 1

0.85

0.65

4000 8000

fc’

1

05.01000

4000f85.0

'c

1

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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)


Flexural Behavior

• Strength reduction factor, Φ depends on member behavior

Curvature

Bending moment

Tension controlled

Balanced condition

Compression controlled

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Strength Reduction Factor, Strength reduction factor, ϕ, depends on

maximum net tensile strain, εt,

at nominal strength, Mn

dt

t

0.003


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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Design For Flexure

• Nominal flexural strength

Mn = As fy (d – )

Mn = bd fy (d – )0.5d fy0.85fc

0.5As fy0.85fc b


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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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


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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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


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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Question

The MOST important variable affecting the flexural strength is:

a. concrete strength

b. beam depth

c. beam width

d. area of rebar


Development and Splices of Reinforcement

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Rebar Development

Why the ribs?


Figure R12.10.2

12.11.3

12.2.1 or 12.11.2

Positive M

Negative M

Rebar Development

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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.


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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Shear Requirements


Shear and Stirrups

Crack (saw cut)

Stirrup (bungee cord through drilled hole in beam)

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Shear and Stirrups

After concrete cracks, shear load supported by stirrups (stretching bungee cord)

Held together by stirrup


Shear and StirrupsUnderstanding Strut and Tie

Actually a Strut and Tie System52

Tension in Stirrup (tie)Compression in member (strut)

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Diagonal Tension Due to Shear

Stirrups

Section


Shear Strength

h

bw

Av at s

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Critical Section for Shear Design

Distributed loads

d Critical section


Critical Section for Shear Design

d

Critical section

Point load within dist. d

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Shear Design Strength

• Design strength = nominal strength

= 0.75

Vn = Vc + Vs


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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Shear Strength Provided by Concrete

• Members subject to shear and flexure

(lower bound without axial tension)

Vc = 2 bwdcf'


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


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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Minimum Shear Reinforcement

Av, min = 0.75 bws

fyt

50bws fyt

cf'


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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Shear Design – Summary


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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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


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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• Step 1: Geometry

d = 15.5 in. =

18‐1.5‐0.5‐0.5

h = 18”

b = 18 inches


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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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



• 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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• Summary

d = 15.5in. =

18‐1.5‐0.5‐0.5

h = 18in.

b = 18”

4#8

5#8

#3 at 6 in.


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


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.


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


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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Perimeter Beams

or

Min. 135o Hook

Note: Size and spacing of stirrups is based on shear and/or torsion demand.


Perimeter Beams

OR

Unacceptable OK

Crosstie

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Columns


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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Pc

Axial Strength – Concrete Contribution

Concrete

Max. Pc = 0.85fcAc = 0.85fc(Ag – Ast)

L

P


L

P

Steel

Ps

Axial Strength – Steel Contribution

Max. Ps = fyAst

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L

P = Po

Po = Pc + Ps = 0.85fc(Ag – Ast) + fy Ast

Axial Strength


Pn,max = Constant × Po

Constant = 0.85 (Spiral)

= 0.80 (Tied)

Nominal Axial Strength – (Pn,max )

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Strength Reduction Factor

• Compression‐controlled section

– Tied = 0.65

– Spiral = 0.75

Design axial strength = Pn,max


Preliminary Column Sizing• Reinforcement ratios in the range of 1–2% are usually the most economical.

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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


Nonslender Tied Columns

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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)


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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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.


Combined Flexural and Axial LoadThis image cannot currently be displayed.

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Interaction Diagram

h

b



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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Interaction Diagram

Moment, Mn

Axi

al L

oad,

Pn 2


cu = 0.003


Interaction Diagram

Moment, Mn

Axi

al L

oad,

Pn

3


cu = 0.003

t = 0

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Interaction Diagram

Moment, Mn

Axi

al L

oad,

Pn

4


cu = 0.003

t = y


Moment, Mn

Axi

al L

oad,

Pn

5


cu = 0.003

t > y

Interaction Diagram

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Moment, Mn

Axi

al L

oad,

Pn

6


cu = 0.003

t >> y

Interaction Diagram


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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Strength Reduction Factor,

0.65

t = 0.002 t = 0.005

Transition

0.90

= 0.65 + (t – 0.002)(250/3)


Tension-controlled

Net tensile strain

Spiral

Other

0.75



Nominal strengthPn, Mn

Design strengthPn, Mn

Po

Pn, max = 0.80Po


Balanced condition

Po

Pn, max

Interaction Diagram

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Pn, Mn

Pn, Mn


Tension-controlled = 0.90

Transition

Interaction Diagram



Nominal strength Pn, Mn

Design strength Pn, Mn

+ (Pu, Mu)OK

Interaction Diagram

+ (Pu, Mu)NG

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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


Slender ColumnsThis image cannot currently be displayed.


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


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

This image cannot currently be displayed. This image cannot currently be displayed.


Two‐way Concrete Floor Systems

This image cannot currently be displayed.This image cannot currently be displayed.


Flat plate Flat slab

Waffle slab


Two-way beam supported

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Drop PanelThis image cannot currently be displayed.

d d1

/6 /6 h/4

h


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


Shear Strength

• Critical shear sections

– One‐way shear

– Two‐way shear

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One‐Way Shear1

2

Tributary area


Critical section

d

Slab plan viewd = effective flexural depth of slab


Vu = 2

Vu = qu × Tributary area

One‐Way Shear


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Two‐Way Shear – Tributary AreasThis image cannot currently be displayed.


Two‐Way Shear1

2

Tributary area




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

Vc = 4This image cannot currently be displayed.


• Nominal shear strength, Vc

2. Reduction for rectangular columns

Vc = 2 + 4

= ratio of long to short column dimensions


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)



Shear Strength



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



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)


Stirrups (11.11.3) Shear Studs (11.11.5)This image cannot currently be displayed.




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



Prestressed Concrete



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Prestressed Concrete: General Principles


Tension

Compression

No tension

Area A

Prestressing tendon

Ff

FA

Prestress Stress due to loading

Combined stresses


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


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




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


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.


, Butd


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




“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

vertical forces review reinforced concrete design

Documents