1. PCI 6th Edition
Connection Design

2. Presentation Outline Structural Steel Design Limit State Weld Analysis
Strut – Tie Analysis for Concrete Corbels
Anchor Bolts
Connection Examples

3. 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

4. Structural Steel Design
Focus on AISC LRFD 3rd Edition
Flexural Strength
Shear Strength
Torsional Strength
Combined Interaction
Limit State Methods are carried through examples

5. Structural Steel Details
Built-up Members
Torsional Strength
Beam Seats

6. Steel Strength Design fMp = f·Fy·Zs Flexure Where:
fMp = Flexural Design Strength
Fy = Yield Strength of Material
Zs = Plastic Section Modulus

7. Steel Strength Design Shear fVn = f(0.6·Fy)·Aw Where:
fVp = Shear Design Strength
Aw = Area subject to shear

8. Steel Strength Design Torsion (Solid Sections) fTn = f(0.6·Fy)·a·h·t2
Where:
fTp = Torsional Design Strength
a = Torsional constant
h = Height of section
t = Thickness

9. Torsional Properties
Torsional Constant, a
Rectangular Sections

10. 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

11. Torsional Properties
Hollow Sections
Ᾱ = w·d

12. Combined Loading Stress
Normal Stress
Bending Shear Stress
Torsion Shear Stress

13. Combined Loading Stresses are added based on direction
Stress Limits based on Mohr’s circle analysis
Normal Stress Limits
Shear Stress Limits

14. Built-Up Section Example
For equilibrium the Tension force T must equal the compression force C.

15. Example
On space bar the Area equation are equated.

16. Determine Neutral Axis Location, y
Tension Area Compression Area
Tension = Compression
On space bar the compression area equation is displayed the solution for y

17. Define Plastic Section Modulus, Zp
Either Tension or Compression Area x Distance between the Tension / Compression Areas Centroids
On space bar the compression area equation is displayed the solution for y

18. Determine Centroid Locations
Tension
Compression

19. Calculate Zp

20. Beam Seats
Stiffened Bearing
Triangular
Non-Triangular

21. 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

22. Stiffener Shape Factor

23. Thickness Limitation

24. 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.

25. Shape Factor

26. Thickness Limitation

27. Design Strength

28. Weld Analysis Elastic Procedure Limit State (LRFD) Design introduced
Comparison of in-plane “C” shape
Elastic Vector Method - EVM
Instantaneous Center Method – ICM

29. Elastic Vector Method – (EVM)
Stress at each point calculated by mechanics of materials principals
The weld group has the following properties using a unit thickness:
Aw = total area of weld group
Ixx = moment of inertia about X-centroid axis
Iyy = moment of inertia about Y-centroid axis
Ip = polar moment of inertia Ixx + Iyy
X = dimension of weld groups centroid in X direction
Y = dimension of weld groups centroid in Y direction
The state of stress at each point is calculated with the following assumptions:
• Isotropic elastic material.
• Plane sections remain plane.

30. Elastic Vector Method – (EVM)
Weld Area ( Aw ) based on effective throat
For a fillet weld:
Where:
a = Weld Size
lw = Total length of weld

31. Instantaneous Center Method (ICM)
Deformation Compatibility Solution
Rotation about an Instantaneous Center

32. 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

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

34. Strength, f
Aw - Weld area based on effective throat

35. Angular Orientation, g
Weld capacity increases as the angle of the force and weld axis approach 90o

36. Deformation Compatibility, h
Where the ultimate element deformation Du is:

37. Element Force
Where: r and q are functions of the unknown location of the instantaneous center, x and y

38. Equations of Statics

39. Tabulated Solution AISC LRFD 3rd Edition, Tables 8-5 to 8-12
fVn = C·C1· D·l
Where:
D = number of 16ths of weld size
C = tabulated value, includes f
C1 = electrode strength factor
l = weld length

40. Comparison of Methods
Page 6-47:

41. Corbel Design Cantilever Beam Method Strut – Tie Design Method
Design comparison
Results comparison of Cantilever Method to Strut – Tie Method
Embedded Steel Sections

42. 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

43. Cantilever Beam Method
Primary Tension Reinforcement
Greater of Equation A or B
Tension steel development is critical both in the column and in the corbel

44. Cantilever Beam Method
Shear Steel
Steel distribution is within 2/3 of d

45. 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

46. Strut – Tie Analysis Steps
Step 1 – Determine of bearing area of plate

47. Strut – Tie Analysis Steps
Step 2 – Select statically determinate truss
AC I provides guidelines for truss angles, struts, etc.

48. Strut – Tie Analysis Steps
Step 3 – Determine of forces in the truss members
Method of Joints or Method of Sections

49. Strut – Tie Analysis Steps
Step 4 – Design of tension ties

50. Strut – 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

51. 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

52. Strut – Tie Analysis Steps Compression Strut Strength
From ACI , Section A.3.2:
Where:
bs – 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

53. Strut – Tie Analysis Steps
Step 7 – Consider detailing to ensure design technique

54. Corbel Example Given: Problem: 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

55. Step 1CBM – Cantilever Beam Method (CBM)
h = 14 in
d = 13 in.
a = ¾ lp = 6 in.
From Table

56. Step 2CBM – Tension Steel
Cantilever Action

57. Step 3CBM – Effective Shear Friction Coefficient

58. Step 4CBM – Tension Steel
Shear Friction

59. As based on cantilever action governs
Step 5CBM – As minimum
As based on cantilever action governs
As = 1.18 in2

60. Use (2) #3 ties = (4) (0.11 in2) = 0.44 in2
Step 6CBM – Shear Steel
Use (2) #3 ties = (4) (0.11 in2) = 0.44 in2
Spaced in top 2/3 (13) = 8 ½ in

61. 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

62. Step 2ST – Truss Geometry
tan qR=Nu / Vu = (15)/(80) = 0.19
l1 = (h - d) tanqR + aw + (hc - cc)
= ( )(0.19) ( )
= in.
l2 = (hc - cc) – ws/2
= ( ) - ws/2
= ws/2

63. 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]

64. Step 2ST – Truss Geometry
Maximum compressive stress at the nodal zone p (anchors one tie, βn = 0.8)
fcu = 0.85·bn·f`c = 0.85(0.8)(5)= 3.4 ksi
An = area of the nodal zone
= b·ws = 14ws

65. Step 2ST – Determine ws , l2 From Eq. 2 and 3
0.014Nc Nc = 0
Nc = 175 kips
ws = 0.28Nc = (0.28)(175) = 4.9in
l2 = ws
= (4.9) = 9.3

66. Step 3ST – Solve for Strut and Tie Forces
Solving the truss ‘mnop’ by statics, the member forces are:
Strut op = kips (c)
Tie no = kips (t)
Strut np = kips (c)
Tie mp = kips (t)
Tie mn = kips (t)

67. Step 4ST – Critical Tension Requirements
For top tension tie ‘no’
Tie no = 68.2 kips (t)
Provide 2 – #8 = 1.58 in2 at the top

68. 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 , 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.

69. 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

70. Step 6ST – Strut Width
Width of strut ‘np’

71. Step 6ST – Compression Strut Strength
From ACI , Section A.3.2:
Where - bottle shaped strut, bs = 0.60l
161 kips ≥ kips OK

72. Step 7ST – Surface Reinforcement
Since the lowest value of bs was used, surface reinforcement is not required based on ACI 318 Appendix A

73. Example Conclusion
Cantilever Beam Method
Strut-and-Tie Method

74. Embedded Steel Sections

75. Concrete and Rebar Nominal Design Strengths
Concrete Capacity

76. Concrete and Rebar Nominal Design Strengths
Additional Tension Compression Reinforcement Capacity

77. Corbel Capacity
Reinforced Concrete

78. Steel 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
f = 0.9

79. Steel Section Nominal Design Strengths
Shear
where:
h, t = depth and thickness of steel web
f = 0.9

80. Anchor Bolt Design
ACI , Appendix D, procedures for the strength of anchorages are applicable for anchor bolts in tension.

81. Strength Reduction Factor
Function of supplied confinement reinforcement
f = 0.75 with reinforcement
f = 0.70 with out reinforcement
It is highly recommended that such confinement be provided around all concrete controlled headed stud connections.

82. 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

83. Hooked Anchor Bolts No = 126·f`c·eh·do·Ccrp Where:
eh = hook projection ≥ 3do
do = bolt diameter
Ccrp = cracking factor (Section )

84. Column Base Plate Design
Column Structural Integrity requirements 200Ag

85. Completed Connection Examples
Examples Based
Applied Loads
Component Capacity
Design of all components
Embeds
Erection Material
Welds
Design for specific load paths

86. Completed Connection Examples
Cladding “Push / Pull”
Wall to Wall Shear
Wall Tension
Diaphragm to Wall Shear

102. Questions?