PCI 6th EditionPCI 6th Edition

Connection Design

Presentation OutlinePresentation Outline

• Structural Steel Design• Limit State Weld Analysis• Strut – Tie Analysis for Concrete

Corbels• Anchor Bolts• Connection Examples

ChangesChanges

• New method to design headed studs (Headed Concrete Anchors - HCA)

• Revised welding section– Stainless Materials– Limit State procedure presented

• Revised Design Aids (moved to Chapter 11)• Structural Steel Design Section

– Flexure, Shear, Torsion, Combined Loading– Stiffened Beam seats

• Strut – Tie methodology is introduced• Complete Connection Examples

Structural Steel DesignStructural Steel Design

• Focus on AISC LRFD 3rd Edition– Flexural Strength– Shear Strength– Torsional Strength– Combined Interaction

• Limit State Methods are carried through examples

Structural Steel DetailsStructural Steel Details

• Built-up Members

• Torsional Strength

• Beam Seats

Steel Strength DesignSteel Strength Design

• Flexure

Mp = ·Fy·Zs

Where:

Mp = Flexural Design Strength

Fy = Yield Strength of Material

Zs = Plastic Section Modulus

Steel Strength DesignSteel Strength Design

• Shear

Vn = (0.6·Fy)·Aw

Where:

Vp = Shear Design Strength

Aw = Area subject to shear

Steel Strength DesignSteel Strength Design

• Torsion (Solid Sections)

Tn = (0.6·Fy)··h·t2

Where:

Tp = Torsional Design Strength

= Torsional constant

h = Height of section

t = Thickness

Torsional PropertiesTorsional Properties

• Torsional Constant, • Rectangular Sections

Steel Strength DesignSteel Strength Design

• Torsion (Hollow Sections)

Tn = 2·(0.6·Fy)·Ᾱ·t

Where:

Tp = Torsional Design Strength

Ᾱ = Area enclosed by centerline of walls

t = Wall thickness

Torsional PropertiesTorsional Properties

• Hollow Sections

Ᾱ = w·d

Combined Loading StressCombined Loading Stress

• Normal Stress

• Bending Shear Stress

• Torsion Shear Stress

fv

torsion

Tc

J,

T

ht2,

T

2At

fn

P

A,Mc

I,M

S

fv

bending

VQ

It,V

A

Combined LoadingCombined Loading

• Stresses are added based on direction• Stress Limits based on Mohr’s circle analysis

– Normal Stress Limits

– Shear Stress Limits

fuv

0.60fy

0.90

fun

fy

0.90

Built-Up Section ExampleBuilt-Up Section Example

ExampleExample

Fx 0

T C 0

AtF

y A

cF

y0

AtA

c

Determine Neutral Axis Location, yDetermine Neutral Axis Location, y

Tension Area Compression Area

Tension = Compression

At4iny

Ac

238

in1in 38

in y

4in

Ac

2.25 4y

4y 2.25 4y

y 2.25

80.281 in

Define Plastic Section Modulus, ZpDefine Plastic Section Modulus, Zp

Either Tension or Compression Area x Distance between the Tension / Compression Areas Centroids

Z

pA

tH y

t y

c

Determine Centroid LocationsDetermine Centroid Locations

• Tension

• Compression

y

t

y

2

0.281

20.14 in

y

c

A y__

A

0.683 in

Calculate ZpCalculate Zp

Zp

At

H yt

yc

Zp

4y H yt

yc

Zp

40.281 1.375 0.14 0.683 Z

p0.62 in3

Beam SeatsBeam Seats

• Stiffened Bearing– Triangular– Non-Triangular

Triangular StiffenersTriangular Stiffeners

• Design Strength

Vn=·Fy·z·b·t

Where:Vn = Stiffener design

strength= Strength reduction

factor = 0.9b = Stiffener projectiont = Stiffener thicknessz = Stiffener shape factor

Stiffener Shape FactorStiffener Shape Factor

0.75 b

a 2.0

z 1.39 2.2b

a

1.27

b

a

2

0.25b

a

3

Thickness LimitationThickness Limitation

b

t

250

Fy

Triangular Stiffener ExampleTriangular Stiffener Example

Given:A stiffened seat connection shown at right. Stiffener thickness, ts = 3/8 in.

Fy = 36 ksi

Problem:Determine the design shear resistance of the stiffener.

Shape FactorShape Factor

b

a

8

10 0.8 0.75 and 1.0

z 1.39 2.2b

a

1.27

b

a

2

0.25b

a

3

z 1.39 2.2 0.8 1.27 0.8 2 0.25 0.8 3

z 0.315

Thickness LimitationThickness Limitation

b

t

250

Fy

8

0.37521.3

250

3641.7

21.3 41.7

Design StrengthDesign Strength

Vn

Fy

zbt

Vn

0.9 36 ksi 0.315 8 in 0.375 in V

n28.9 kips

Weld AnalysisWeld Analysis

• Elastic Procedure

• Limit State (LRFD) Design introduced

• Comparison of in-plane “C” shape– Elastic Vector Method - EVM– Instantaneous Center Method – ICM

Elastic Vector Method – (EVM)Elastic Vector Method – (EVM)

• Stress at each point calculated by mechanics of materials principals

fx

P

x

Aw

M

zy

Ip

fy

P

y

Aw

M

zx

Ip

fz

P

z

Aw

M

xy

Ixx

M

yx

Iyy

fr

fx

2 fy

2 fz

2

Elastic Vector Method – (EVM)Elastic Vector Method – (EVM)

• Weld Area ( Aw ) based on effective throat• For a fillet weld:

Where:a = Weld Size

lw = Total length of weld

A

w

a

2lw

Instantaneous Center Method (ICM)Instantaneous Center Method (ICM)

• Deformation Compatibility Solution• Rotation about an Instantaneous Center

Instantaneous Center Method (ICM)Instantaneous Center Method (ICM)

• Increased capacity– More weld regions achieve ultimate strength– Utilizes element vs. load orientation

• General solution form is a nonlinear integral• Solution techniques

– Discrete Element Method– Tabular Method

ICM Nominal StrengthICM Nominal Strength

• An elements capacity within the weld group is based on the product of 3 functions. – Strength– Angular Orientation– Deformation Compatibility

R

nj

g h f

Strength, fStrength, f

f 0.6FEXX

Aw

Aw - Weld area based on effective throat

Angular Orientation, gAngular Orientation, g

Weld capacity increases as the angle of the force and weld axis approach 90o

Rj

R g

g 1.0 0.5sin 32

Deformation Compatibility, hDeformation Compatibility, h

h

u

r

rcritical

0.209 2 0.32a

1.9 0.9

u

r

rcritical

0.209 2 0.32a

0.3

u

1.087 6 0.64a 0.17a

Where the ultimate element deformation u is:

Element ForceElement Force

Where: r and are functions of the unknown location of the instantaneous center, x and y

Rn

j

0.6FEXX

Aw 1.0 0.5sin 3

2

u

r

rcritical

0.209 2 0.32a

1.9 0.9

u

r

rcritical

0.209 2 0.32a

0.3

Equations of StaticsEquations of Statics

Fy 0 R

nyjj1

Number of Elements

Pn

0

MIC 0 R

njrj

j1

Number of Elements

Pn

e r0 0

Tabulated SolutionTabulated Solution

• AISC LRFD 3rd Edition, Tables 8-5 to 8-12Vn = C·C1· D·l

Where:D = number of 16ths of weld sizeC = tabulated value, includes C1 = electrode strength factorl = weld length

Comparison of MethodsComparison of Methods

• Page 6-47:

Corbel DesignCorbel Design

• Cantilever Beam Method

• Strut – Tie Design Method

• Design comparison– Results comparison of Cantilever

Method to Strut – Tie Method

• Embedded Steel Sections

Cantilever Beam Method StepsCantilever Beam Method Steps

Step 1 – Determine maximum allowable shearStep 2 – Determine tension steel by cantileverStep 3 – Calculate effective shear friction coeff.Step 4 – Determine tension steel by shear

frictionStep 5 – Compare results against minimumStep 6 – Calculate shear steel requirements

Cantilever Beam MethodCantilever Beam Method

• Primary Tension Reinforcement • Greater of Equation A or B

• Tension steel development is critical both in the column and in the corbel

Eq. A As

1

fy

Vu

a

d

N

u

h

d

Eq. B As

1

fy

2Vu

3e

N

u

Cantilever Beam MethodCantilever Beam Method

• Shear Steel

• Steel distribution is within 2/3 of d

A

h0.5 A

s A

n

Cantilever Beam Method StepsCantilever Beam Method Steps

Step 1 – Determine bearing area of plate

Step 2 – Select statically determinate truss

Step 3 – Calculate truss forces

Step 4 – Design tension ties

Step 5 – Design Critical nodes

Step 6 – Design compression struts

Step 7 – Detail Accordingly

Strut – Tie Analysis StepsStrut – Tie Analysis Steps

Step 1 – Determine of bearing area of plate

Apl

V

u

0.85fc̀

0.75

Strut – Tie Analysis StepsStrut – Tie Analysis Steps

Step 2 – Select statically determinate truss

AC I provides guidelines for truss angles, struts, etc.

Strut – Tie Analysis StepsStrut – Tie Analysis Steps

Step 3 – Determine of forces in the truss members

Method of Joints or Method of Sections

Strut – Tie Analysis StepsStrut – Tie Analysis Steps

Step 4 – Design of tension ties

As

F

nt

fy

0.75

Strut – Tie Analysis StepsStrut – Tie Analysis Steps

Step 5 – Design of critical nodal zone

where:βn = 1.0 in nodal zones bounded

by structure or bearing areas= 0.8 in nodal zones

anchoring one tie = 0.6 in nodal zones

anchoring two or more ties

fcu 0.85nf

c̀

Strut – Tie Analysis StepsStrut – Tie Analysis Steps

Step 6 – Check compressive strut limits based on Strut Shape

The design compressive strength of a strut without compressive reinforcement

Fns = ·fcu·Ac

where: = 0.75

Ac = width of corbel × width of strut

Strut – Tie Analysis Steps Compression Strut Strength

Strut – Tie Analysis Steps Compression Strut Strength

• From ACI 318-02, Section A.3.2:

Where:

s – function of strut shape / location= 0.60bottle shaped strut= 0.75, when reinforcement is provided= 1.0, uniform cross section= 0.4, in tension regions of members= 0.6, for all other cases

fcu 0.85sf

c̀

Strut – Tie Analysis StepsStrut – Tie Analysis Steps

Step 7 – Consider detailing to ensure design technique

Corbel ExampleCorbel Example

Given:Vu = 80 kips

Nu = 15 kips

fy = Grade 60

f′c = 5000 psi Bearing area – 12 x 6 in.

Problem:Find corbel depth and reinforcement based on Cantilever

Beam and Strut – Tie methods

Step 1CBM – Cantilever Beam Method (CBM)Step 1CBM – Cantilever Beam Method (CBM)

h = 14 in

d = 13 in.

a = ¾ lp = 6 in.

From Table 4.3.6.1

V

umax10002A

cr

1000 12 14 14 1000

196 kips 80 kips

Step 2CBM – Tension SteelStep 2CBM – Tension Steel

• Cantilever Action

As

1

fy

Vu

a

d

N

u

h

d

1

.75 60 806

13

15

14

13

1.18 in2

Step 3CBM – Effective Shear Friction CoefficientStep 3CBM – Effective Shear Friction Coefficient

e

1000 bh

Vu

1000 114 14 1.4

80

3.43 3.4

Use e

3.4

Step 4CBM – Tension SteelStep 4CBM – Tension Steel

• Shear Friction

As

1

fy

2Vu

3e

N

u

1

0.75 60 2 80 3 3.4

15

0.68 in2

Step 5CBM – As minimumStep 5CBM – As minimum

As,min

0.4bdf

c̀

fy

0.4 14 13 5

60

0.61 in2

As based on cantilever action governs

As = 1.18 in2

Step 6CBM – Shear SteelStep 6CBM – Shear Steel

Ah

0.5 As

An

0.5 1.18 15

0.75 60

0.42 in

Use (2) #3 ties = (4) (0.11 in2) = 0.44 in2

Spaced in top 2/3 (13) = 8 ½ in

Step 1ST – Strut - Tie Solution (ST)Step 1ST – Strut - Tie Solution (ST)

Determination of bearing plate size and protection for the corner against spalling

Required plate area:

Use 12 by 6 in. plate, area = 72 in2 > 25.1 in2

Abearing

V

u

0.85f`c

80

0.75 0.85f`c

25.1 in2

Step 2ST – Truss GeometryStep 2ST – Truss Geometry

tan R=Nu / Vu = (15)/(80) = 0.19

l1 = (h - d) tanR + aw + (hc - cc) = (14 - 13)(0.19) + 6 + (14 - 2.25) = 17.94 in.

l2 = (hc - cc) – ws/2

= (14 - 2.25) - ws/2

= 11.75 - ws/2

Step 2ST – Truss GeometryStep 2ST – Truss Geometry

Find ws

Determine compressive force,

Nc, at Node ‘p’:

∑Mm = 0

Vu·l1+Nu·d – Nc·l2=0 [Eq. 1]

(80)(17.94) + (15)(13) – Nc(11.75 – 0.5ws) = 0

[Eq. 2]

Step 2ST – Truss GeometryStep 2ST – Truss Geometry

• Maximum compressive stress at the nodal zone p (anchors one tie, βn = 0.8)

fcu = 0.85·n·f`c = 0.85(0.8)(5)= 3.4 ksi

An = area of the nodal zone

= b·ws = 14ws

Step 2ST – Determine ws , l2Step 2ST – Determine ws , l2

• From Eq. 2 and 3

0.014Nc2 - 11.75Nc - 1630 = 0

Nc = 175 kips

ws = 0.28Nc = (0.28)(175) = 4.9in

l2 = 11.75 - 0.5 ws = 11.75 - 0.5(4.9) = 9.3

Step 3ST – Solve for Strut and Tie ForcesStep 3ST – Solve for Strut and Tie Forces

• Solving the truss ‘mnop’ by statics, the member forces are:

Strut op = 96.0 kips (c)

Tie no = 68.2 kips (t)

Strut np = 116.8 kips (c)

Tie mp = 14.9 kips (t)

Tie mn = 95.0 kips (t)

Step 4ST – Critical Tension RequirementsStep 4ST – Critical Tension Requirements

• For top tension tie ‘no’

Tie no = 68.2 kips (t)

Provide 2 – #8 = 1.58 in2 at the top

As

F

nt

fy

62

0.75 60 1.52in2

Step 5ST – Nodal ZonesStep 5ST – Nodal Zones

• The width `ws’’ of the nodal zone ‘p ’ has been chosen in Step 2 to satisfy the stress limit on this zone

• The stress at nodal zone ‘o ’ must be checked against the compressive force in strut ‘op ’ and the applied reaction, Vu

• From the compressive stress flow in struts of the corbel, Figure 6.8.2.1, it is obvious that the nodal zone ‘p ’ is under the maximum compressive stress due to force Nc.

• Nc is within the acceptable limit so all nodal zones are acceptable.

Step 6ST – Critical Compression RequirementsStep 6ST – Critical Compression Requirements

• Strut ‘np’ is the most critical strut at node ‘p’. The nominal compressive strength of a strut without compressive reinforcement

Fns = fcu·Ac

Where:

Ac = width of corbel × width of strut

Step 6ST – Strut WidthStep 6ST – Strut Width

• Width of strut ‘np’

Strut Width w

s

sin(54.4o)

4.9

sin(54.4o)

6.03 in

Step 6ST – Compression Strut StrengthStep 6ST – Compression Strut Strength

• From ACI 318-02, Section A.3.2:

Where - bottle shaped strut, s = 0.60

161 kips ≥ 116.8 kips OK

fcu 0.85sf

c̀

fcu

0.850.6 15 2.55 ksi

F

ns f

cuA

c0.75 2.55 14 6.03

161.5 kips

Step 7ST – Surface ReinforcementStep 7ST – Surface Reinforcement

• Since the lowest value of s was used, surface reinforcement is not required based on ACI 318 Appendix A

Example ConclusionExample Conclusion

Cantilever Beam Method Strut-and-Tie Method

Embedded Steel SectionsEmbedded Steel Sections

Concrete and Rebar Nominal Design StrengthsConcrete and Rebar Nominal Design Strengths

• Concrete Capacity

Vc

0.85f

c̀bl

e

1 3.6el

e

Concrete and Rebar Nominal Design StrengthsConcrete and Rebar Nominal Design Strengths

• Additional Tension Compression Reinforcement Capacity

Vr

2A

sf

y

1

6el

e

4.8sl

e

1

Corbel CapacityCorbel Capacity

• Reinforced Concrete

Vn

Vc

VR

0.75

Steel Section Nominal Design StrengthsSteel Section Nominal Design Strengths

• Flexure - Based on maximum moment in section; occurs when shear in steel section = 0.0

Where:b = effective width on embed, 250 % x Actual

= 0.9

Vn

Z

sf

y

a0.5V

u

0.85fc̀b

Steel Section Nominal Design StrengthsSteel Section Nominal Design Strengths

• Shear

where:h, t = depth and thickness of steel web

= 0.9

V

s 0.6f

yht

Anchor Bolt DesignAnchor Bolt Design

• ACI 318-2002, Appendix D, procedures for the strength of anchorages are applicable for anchor bolts in tension.

Strength Reduction FactorStrength Reduction Factor

Function of supplied confinement reinforcement

= 0.75 with reinforcement

= 0.70 with out reinforcement

Headed Anchor BoltsHeaded Anchor Bolts

No = Cbs·AN·Ccrb·ed,N

Cbs

2.22 f '

c

hef

3

Where:

Ccrb = Cracked concrete factor,

1 uncracked, 0.8 Cracked

AN = Projected surface area for a stud or group

ed,N =Modification for edge distance

Cbs = Breakout strength coefficient

Hooked Anchor BoltsHooked Anchor Bolts

Where:

eh = hook projection ≥ 3do

do = bolt diameter

Ccrp = cracking factor (Section 6.5.4.1)

No = 126·f`c·eh·do·Ccrp

Column Base Plate DesignColumn Base Plate Design

• Column Structural Integrity requirements 200Ag

Completed Connection ExamplesCompleted Connection Examples

• Examples Based– Applied Loads– Component Capacity

• Design of all components– Embeds– Erection Material– Welds

• Design for specific load paths

Completed Connection ExamplesCompleted Connection Examples

• Cladding “Push / Pull”

• Wall to Wall Shear

• Wall Tension

• Diaphragm to Wall Shear

Questions?Questions?