8 Connection Design
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Transcript of 8 Connection Design
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?