Transcript Slide 1

PCI 6th Edition
Connection Design
Presentation Outline
• Structural Steel Design
• Limit State Weld Analysis
• Strut – Tie Analysis for Concrete
Corbels
• Anchor Bolts
• Connection Examples
Changes
• 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 Design
• Focus on AISC LRFD 3rd Edition
– Flexural Strength
– Shear Strength
– Torsional Strength
– Combined Interaction
• Limit State Methods are carried through
examples
Structural Steel Details
• Built-up Members
• Torsional Strength
• Beam Seats
Steel Strength Design
• Flexure
fMp = f·Fy·Zs
Where:
fMp =
Fy =
Zs =
Flexural Design Strength
Yield Strength of Material
Plastic Section Modulus
Steel Strength Design
• Shear
fVn = f(0.6·Fy)·Aw
Where:
fVp = Shear Design Strength
Aw = Area subject to shear
Steel Strength Design
• Torsion (Solid Sections)
fTn = f(0.6·Fy)·a·h·t2
Where:
fTp =
a =
h =
t =
Torsional Design Strength
Torsional constant
Height of section
Thickness
Torsional Properties
• Torsional Constant, a
• Rectangular Sections
Steel Strength Design
• Torsion (Hollow Sections)
fTn = 2·f(0.6·Fy)·Ᾱ·t
Where:
fTp = Torsional Design Strength
Ᾱ = Area enclosed by centerline of walls
t = Wall thickness
Torsional Properties
• Hollow Sections
Ᾱ = w·d
Combined Loading Stress
• Normal Stress
fn 
P Mc M
,
,
A I S
• Bending Shear Stress
fv
bending
• Torsion Shear Stress
fv
torsion
VQ V

,
It A
Tc T
T

,
,
2
J aht 2At
Combined Loading
• Stresses are added based on direction
• Stress Limits based on Mohr’s circle analysis
– Normal Stress Limits
ffun  f  fy
f  0.90
– Shear Stress Limits
ffuv  f  0.60fy

f  0.90

Built-Up Section Example
Example
F
x
0
TC 0
A tF y A cFy  0
At  Ac
Determine Neutral Axis Location, y
Tension Area
Compression Area
A t  4in  y
A c  2  3 in 1in   3 in  y   4in
 8

8
A c  2.25  4  y
4  y  2.25  4  y
Tension = Compression
y
2.25
 0.281 in
8
Define Plastic Section Modulus, Zp
Either Tension or Compression Area x
Distance between the Tension /
Compression Areas Centroids

Zp  A t H  y t  y c

Determine Centroid Locations
• Tension
y 0.281
yt  
 0.14 in
2
2
• Compression
yc
__
Ay


A
 0.683 in
Calculate Zp


Z  4  y H  y  y 
 4  0.2811.375  0.14  0.683
Zp  A t H  y t  y c
p
Zp
t
Zp  0.62 in3
c
Beam Seats
• Stiffened Bearing
– Triangular
– Non-Triangular
Triangular Stiffeners
• Design Strength
fVn=f·Fy·z·b·t
Where:
fVn = Stiffener design
strength
f = Strength reduction
factor = 0.9
b = Stiffener projection
t
= Stiffener thickness
z = Stiffener shape factor
Stiffener Shape Factor
b
0.75 
 2.0
a
2
 b
 b
 b
z  1.39  2.2    1.27    0.25  
 a
 a
 a
3
Thickness Limitation
b 250

t
Fy
Triangular 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 Factor
b
8

 0.8  0.75 and  1.0
a 10
2
 b
 b
 b
z  1.39  2.2    1.27    0.25  
 a
 a
 a
 
3
   0.25 0.8
z  1.39  2.2 0.8  1.27 0.8
z  0.315
2
3
Thickness Limitation
b 250

t
Fy
8
 21.3
0.375
250
36
21.3  41.7
 41.7
Design Strength
fVn  f  Fy  z  b  t


 
fVn  0.9 36 ksi 0.315 8 in 0.375 in
fVn  28.9 kips

Weld 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)
• Stress at each point
calculated by mechanics of
materials principals
fx 
fy 
fz 
fr 
Px
Aw
Py
Aw
Pz
Aw

Mz y

Mz x

Mx y
Ip
Ip
I xx

My x
I yy
fx 2  fy 2  fz2
Elastic Vector Method – (EVM)
• Weld Area ( Aw ) based on effective throat
• For a fillet weld:
Aw 
a
lw
2
Where:
a = Weld Size
lw = Total length of weld
Instantaneous Center Method (ICM)
• Deformation Compatibility Solution
• Rotation about an Instantaneous Center
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 Strength
• An elements capacity within the weld group is based
on the product of 3 functions.
– Strength
– Angular Orientation
– Deformation Compatibility
Rn 
j
f  g h
Strength, f
f  0.6  FEXX  A w
Aw - Weld area based on effective throat
Angular Orientation, g
Weld capacity increases as the angle of the force
and weld axis approach 90o
Rj  R  g

g  1.0  0.5 sin 
3
2
Deformation Compatibility, h


r
u
u


rcritical
rcritical
1.9  0.9

h
0.32
0.32

0.209   2
a
0.209   2
a




r



0.3

Where the ultimate element deformation u is:

 u  1.087   6

0.64
a  0.17a
Element Force

R n  0.6FEXX A w
j



 1.0  0.5 sin 

3
2



0.3 



r
r


u

u

 

rcritical
rcritical
1.9  0.9
 

0.32
0.32

 0.209   2
a
0.209   2
a 

 









Where: r and  are functions of the unknown location of
the instantaneous center, x and y
Equations of Statics
F
y
M
IC
0
Number of
Elements

R n  Pn  0
yj
j1
0
Number of
Elements

j1


R n rj  Pn e  r0  0
j
Tabulated Solution
• AISC LRFD 3rd Edition, Tables 8-5 to 8-12
fVn = C·C1· D·l
Where:
D =
C =
C1 =
l =
number of 16ths of weld size
tabulated value, includes f
electrode strength factor
weld length
Comparison of Methods
• Page 6-47:
Corbel 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 Steps
Step 1 – Determine maximum allowable shear
Step 2 – Determine tension steel by cantilever
Step 3 – Calculate effective shear friction coeff.
Step 4 – Determine tension steel by shear
friction
Step 5 – Compare results against minimum
Step 6 – Calculate shear steel requirements
Cantilever Beam Method
• Primary Tension Reinforcement
• Greater of Equation A or B
  a
 h 
Eq. A
 Vu    Nu   
 d  
  d 

1   2Vu 

Eq. B A s 
  Nu 
ffy   3 e 

1
As 
ffy
• Tension steel development is critical both in the column and
in the corbel
Cantilever Beam Method
• Shear Steel
A h  0.5  A s  A n 
• Steel distribution is
within 2/3 of d
Cantilever 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 Steps
Step 1 – Determine of bearing area of
plate
A pl 
Vu
f  0.85  f`c
f  0.75
Strut – Tie Analysis Steps
Step 2 – Select statically determinate
truss
AC I provides
guidelines for truss
angles, struts, etc.
Strut – Tie Analysis Steps
Step 3 – Determine of forces in the truss
members
Method of Joints or
Method of Sections
Strut – Tie Analysis Steps
Step 4 – Design of tension ties
As 
Fnt
ffy
f  0.75
Strut – Tie Analysis Steps
Step 5 – Design of critical nodal zone
fcu  0.85   n f`c
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
Strut – Tie Analysis Steps
Step 6 – Check compressive strut limits
based on Strut Shape
The design compressive
strength of a strut without
compressive reinforcement
fFns = f·fcu·Ac
where:
f = 0.75
Ac = width of corbel
× width of strut
Strut – Tie Analysis Steps Compression
Strut Strength
• From ACI 318-02, Section A.3.2:
fcu  0.85  s  f`c
Where:
s – function of strut shape / location
= 0.60l, bottle 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
Strut – Tie Analysis Steps
Step 7 – Consider detailing to ensure
design technique
Corbel 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)
h = 14 in
d = 13 in.
a = ¾ lp = 6 in.
From Table 4.3.6.1
2
Vumax  1000  l A
cr

     196 kips  80 kips
1000 12 14 14
1000
Step 2CBM – Tension Steel
• Cantilever Action
1
As 
ffy
  a
 h 
1
 Vu    Nu    
 d   .75 60
  d 
 1.18 in2
  6
 14  
80    15   
 13  
  13 
 
Step 3CBM – Effective Shear Friction Coefficient
    
1000  l  b  h   1000 1 14 14 1.4
e 

Vu
80
 3.43  3.4
Use  e  3.4
Step 4CBM – Tension Steel
• Shear Friction
 
   

1   2Vu 
1


As 

N


u
ffy   3 e 
 0.75 60
 0.68 in2
  2 80 


  15
  3 3.4 



Step 5CBM – As minimum
A s,min
f`c
  
5
 0.4  b  d
 0.4 14 13
ffy
60
 0.61 in2
As based on cantilever action governs
As = 1.18 in2
Step 6CBM – Shear Steel

15



A h  0.5  A s  A n   0.5 1.18 

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)
Determination of bearing plate size and protection
for the corner against spalling
Required plate area:
A bearing 

Vu
f 0.85f`c


80

0.75 0.85f`c

 25.1 in2
Use 12 by 6 in. plate, area = 72 in2 > 25.1 in2
Step 2ST – Truss Geometry
tan R=Nu / Vu = (15)/(80) = 0.19
l1 = (h - d) tanR + 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 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 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 , 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 Forces
• Solving the truss ‘mnop’ by
statics, the member forces
are:
Strut op
Tie no
Strut np
Tie mp
Tie mn
=
=
=
=
=
96.0 kips (c)
68.2 kips (t)
116.8 kips (c)
14.9 kips (t)
95.0 kips (t)
Step 4ST – Critical Tension Requirements
• For top tension tie ‘no’
Tie no = 68.2 kips (t)
As 
Fnt
ffy

62
 
0.75 60
 1.52in2
Provide 2 – #8 = 1.58 in2 at the top
Step 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 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 Width
• Width of strut ‘np’
Strut Width 
ws
o
sin(54.4 )
 6.03 in

4.9
sin(54.4 o )
Step 6ST – Compression Strut Strength
• From ACI 318-02, Section A.3.2:
fcu  0.85  s  f`c
Where - bottle shaped strut, s = 0.60l

fcu  0.85  0.6 1  5  2.55 ksi

  

fFns  f  f cu  Ac  0.75 2.55  14 6.03   161.5 kips
161 kips ≥ 116.8 kips
OK
Step 7ST – Surface Reinforcement
• Since the lowest value of s was used,
surface reinforcement is not required
based on ACI 318 Appendix A
Example Conclusion
Cantilever Beam Method
Strut-and-Tie Method
Embedded Steel Sections
Concrete and Rebar Nominal Design Strengths
• Concrete Capacity
Vc 
0.85  f`c b  l e
1  3.6e
le
Concrete and Rebar Nominal Design Strengths
• Additional Tension Compression
Reinforcement Capacity
2  A s  fy
Vr 
6e
1
le
4.8s
le
1
Corbel Capacity
• Reinforced Concrete

fVn  f Vc  VR
f  0.75

Steel Section Nominal Design Strengths
• Flexure - Based on maximum moment in
section; occurs when shear in steel section =
0.0
f  Zs  fy
fVn 
a
0.5  Vu
0.85  f`c b
Where:
b = effective width on embed, 250 % x Actual
f = 0.9
Steel Section Nominal Design Strengths
• Shear
fVs  f  0.6fy  h  t
where:
h, t = depth and thickness of steel web
f = 0.9
Anchor Bolt Design
• ACI 318-2002, Appendix D, procedures
for the strength of anchorages are
applicable for anchor bolts in tension.
Strength Reduction Factor
Function of supplied confinement reinforcement
f = 0.75 with reinforcement
f = 0.70 with out reinforcement
Headed Anchor Bolts
No = Cbs·AN·Ccrb·Yed,N
Where:
Ccrb = Cracked concrete factor,
1 uncracked, 0.8 Cracked
AN = Projected surface area for a stud or group
Yed,N =Modification for edge distance
Cbs = Breakout strength coefficient
Cbs  2.22  l 
f 'c
3
hef
Hooked Anchor Bolts
No = 126·f`c·eh·do·Ccrp
Where:
eh = hook projection ≥ 3do
do = bolt diameter
Ccrp = cracking factor (Section 6.5.4.1)
Column Base Plate Design
• Column Structural Integrity
requirements 200Ag
Completed Connection Examples
• Examples Based
– Applied Loads
– Component Capacity
• Design of all components
– Embeds
– Erection Material
– Welds
• Design for specific load paths
Completed Connection Examples
• Cladding “Push / Pull”
• Wall to Wall Shear
• Wall Tension
• Diaphragm to Wall Shear
Questions?