th
PCI 6 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
φMp = φ·Fy·Zs Where: φMp = Flexural Design Strength Fy = Yield Strength of Material Zs = Plastic Section Modulus
Steel Strength Design • Shear
φVn = φ(0.6·Fy)·Aw Where: φVp = Shear Design Strength Aw = Area subject to shear
Steel Strength Design • Torsion (Solid Sections)
φTn = φ(0.6·Fy)·α·h·t2 Where: φTp = α = h = t =
Torsional Design Strength Torsional constant Height of section Thickness
Torsional Properties
• Torsional Constant, α • Rectangular Sections
Steel Strength Design • Torsion (Hollow Sections)
φTn = 2·φ(0.6·Fy)·Ᾱ·t Where: φTp = 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
P Mc M fn = , , A I S
• Bending Shear Stress fv
bending
• Torsion Shear Stress fv
torsion
VQ V = , It A
Tc T T = , , 2 J αht 2At
Combined Loading • Stresses are added based on direction • Stress Limits based on Mohr’s circle analysis – Normal Stress Limits φfun = φ ⋅ fy φ = 0.90
– Shear Stress Limits φfuv = φ ⋅ 0.60fy
(
φ = 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 Tension = Compression
A c = 2 ⋅ 3 in ⋅1in + 3 in − y ⋅ 4in 8 8 A c = 2.25 − 4 ⋅ y 4 ⋅ y = 2.25 − 4 ⋅ y 2.25 y= = 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
φVn=φ·Fy·z·b·t Where: φVn = Stiffener design strength φ = 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 φVn = φ ⋅ Fy ⋅ z ⋅ b ⋅ t
(
)(
)( )(
φVn = 0.9 36 ksi 0.315 8 in 0.375 in φVn = 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 Ip Mz x Ip Mx y 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 2
lw
Where: a = Weld Size lw = Total length of weld
Instantaneous Center Method (ICM) â&#x20AC;˘ Deformation Compatibility Solution â&#x20AC;˘ 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 â&#x2039;&#x2026; g
()
g = 1.0 + 0.5 sin θ
3
2
Deformation Compatibility, h r ∆u ∆u rcritical rcritical 1.9 − 0.9 h= −0.32 −0.32 0.209 θ + 2 a 0.209 θ + 2 a r
(
)
(
0.3
)
Where the ultimate element deformation ∆u is:
(
∆u = 1.087 θ + 6
)
−0.64
a ≤ 0.17a
Element Force
{
R n = 0.6FEXX A w j
}
()
⋅ 1.0 + 0.5 sin θ
3
2
⋅
0.3 r r ∆u ∆ u rcritical rcritical 1.9 − 0.9 −0.32 −0.32 0.209 θ + 2 a 0.209 θ + 2 a
(
)
(
)
Where: r and θ 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 φVn = C·C1· D·l Where: D = C = C1 = l =
number of 16ths of weld size tabulated value, includes φ electrode strength factor weld length
Comparison of Methods â&#x20AC;˘ 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 φfy 3µ e 1 As = φfy
• 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 φ ⋅ 0.85 ⋅ f`c
φ = 0.75
Strut â&#x20AC;&#x201C; Tie Analysis Steps Step 2 â&#x20AC;&#x201C; Select statically determinate truss AC I provides guidelines for truss angles, struts, etc.
Strut â&#x20AC;&#x201C; Tie Analysis Steps Step 3 â&#x20AC;&#x201C; 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 φfy
φ = 0.75
Strut – Tie Analysis Steps Step 5 – Design of critical nodal zone fcu = 0.85 ⋅ β 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 φFns = φ·fcu·Ac where: φ = 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 ⋅ βs ⋅ f`c Where: βs – function of strut shape / location = 0.60λ, 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 â&#x20AC;&#x201C; Tie Analysis Steps
Step 7 â&#x20AC;&#x201C; 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 ⋅ λ A
cr
=
( )( )( ) = 196 kips ≥ 80 kips
1000 12 14 14 1000
Step 2CBM – Tension Steel
• Cantilever Action 1 As = φfy
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 ⋅ λ ⋅ b ⋅ h ⋅ µ 1000 1 14 14 1.4 µe = = Vu 80 = 3.43 ≥ 3.4 Use µ e = 3.4
Step 4CBM – Tension Steel
• Shear Friction
( ) ( ) ( )
1 2Vu 1 As = + N = u φfy 3µ 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 φfy 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 â&#x20AC;&#x201C; Strut - Tie Solution (ST) Determination of bearing plate size and protection for the corner against spalling Required plate area:
A bearing =
(
Vu
Ď&#x2020; 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 θR=Nu / Vu = (15)/(80) = 0.19 l1 = (h - d) tanθR + 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·l 1+Nu·d – Nc·l 2=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·βn·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 φfy
=
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 ⋅ β s ⋅ f`c Where - bottle shaped strut, βs = 0.60λ
()
fcu = 0.85 ⋅ 0.6 1 ⋅ 5 = 2.55 ksi
(
) ( )(
)
φFns = φ ⋅ 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 βs 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
• Additional Tension Compression Reinforcement Capacity 2 ⋅ A s ⋅ fy
Vr =
6e 1+
4.8s
le
le
−1
Corbel Capacity
• Reinforced Concrete
(
φVn = φ Vc + VR φ = 0.75
)
Steel Section Nominal Design Strengths • Flexure - Based on maximum moment in section; occurs when shear in steel section = 0.0 φVn =
φ ⋅ Zs ⋅ fy a+
0.5 ⋅ Vu
0.85 ⋅ f`c ⋅b
Where: b = effective width on embed, 250 % x Actual φ = 0.9
Steel Section Nominal Design Strengths
• Shear
φVs = φ ⋅ 0.6fy ⋅ h ⋅ t where: h, t = depth and thickness of steel web φ = 0.9
Anchor Bolt Design
â&#x20AC;˘ 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 Ď&#x2020; = 0.75 with reinforcement Ď&#x2020; = 0.70 with out reinforcement
Headed Anchor Bolts No = Cbs·AN·Ccrb·Ψed,N Where: Ccrb = Cracked concrete factor, 1 uncracked, 0.8 Cracked AN = Projected surface area for a stud or group Ψed,N =Modification for edge distance Cbs = Breakout strength coefficient Cbs = 2.22 ⋅ λ ⋅
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 â&#x20AC;˘ 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?