1. CONNECTION DESIGN
Connections must be designed at the strength limit state
Average of the factored force effect at the connection and the force effect in the member at the same point
At least 75% of the force effect in the member
End connections for diaphragms, cross-frames, lateral bracing for straight flexural members - designed for factored member loads
Connections should be symmetrical about member axis
At least two bolts or equivalent weld per connection
Members connected so that their gravity axes intersect at a point
Eccentric connections should be avoided
End connections for floorbeams and girders
Two angles with thickness > in.
Made with high strength bolts
If welded account for bending moment in design

2. BOLTED CONNECTIONS Slip-critical and bearing type bolted connections.
Connections should be designed to be slip-critical where:
stress reversal, heavy impact loads, severe vibration
joint slippage would be detrimental to the serviceability of the structure
Joints that must be designed to be slip-critical include
Joints subject to fatigue loading or significant load reversal.
Joints with oversized holes or slotted holes
Joints where welds and bolts sharing in transmitting load
Joints in axial tension or combined axial tension and shear
Bearing-type bolted connections can be designed for joints subjected to compression or joints for bracing members

3. SLIP-CRITICAL BOLTED CONNECTION
Slip-critical bolted connections can fail in two ways: (a) slip at the connection; (b) bearing failure of the connection
Slip-critical connection must be designed to: (a) resist slip at load Service II; and (b) resist bearing / shear at strength limit states

4. SLIP-CRITICAL BOLTED CONNECTION
Slip-critical bolted connections can be installed with such a degree of tightness  large tensile forces in the bolt  clamp the connected plates together
Applied Shear force resisted by friction

5. SLIP-CRITICAL BOLTED CONNECTION
Slip-critical connections can resist the shear force using friction.
If the applied shear force is less than the friction that develops between the two surfaces, then no slip will occur between them
Nominal slip resistance of a bolt in a slip-critical connection
Rn = Kh Ks Ns Pt
Where, Pt = minimum required bolt tension specified in Table 1
Kh = hole factor specified in Table 1
Ks = surface condition factor specified in Table 3

6. SLIP-CRITICAL BOLTED CONNECTION
Values of Kh
Faying surfaces
Unpainted clean mill scale, and blast-cleaned surfaces with Class A coating
Unpainted blast-cleaned surfaces with Class B coating
Hot-dip galvanized surfaces roughened by wire brushing – Class C
Values of Pt
Values of Ks

7. SLIP-CRITICAL CONNECTION
Connection subjected to tensile force (Tu), which reduces clamping
Nominal slip resistance should be reduced by (1- Tu/Pt)
Slip is not a catastrophic failure limit-state because slip-critical bolted connections behave as bearing type connections after slip.
Slip-critical bolted connections are further designed as bearing-type bolted connection for the applicable factored strength limit state.

8. BEARING CONNECTION
In a bearing-type connection, bolts are subjected to shear and the connecting / connected plates are subjected to bearing stresses :

9. BEARING CONNECTION
Bearing type connection can fail in several failure modes
Shear failure of the bolts
Excessive bearing deformation at the bolt holes in the connected parts
Edge tearing or fracture of the connected plate
Tearing or fracture of the connected plate between two bolt holes
Failure of member being connected due to fracture or block shear or ...

10. BEARING CONNECTION Nominal shear resistance of a bolt
Threads excluded: Rn = 0.48 Ab Fub Ns
Threads included: Rn = 0.38 Ab Fub Ns
Where, Ab = area of the bolt corresponding to the nominal diameter
Fub = 120 ksi for A325 bolts with diameters 0.5 through 1.0 in.
Fub = 105 ksi for A325 bolts with diameters through 1.5 in.
Fub = 150 ksi for A490 bolts.
Ns = number of shear planes
Resistance factor for bolts in shear = fs = 0.80
Equations above - valid for joints with length < less than 50.0 in.
If the length is greater than 50 in., then the values from the equations have to be multiplied by 0.8

11. BEARING CONNECTION Rn = 1.2 Lc t Fu
Effective bearing area of a bolt = the bolt diameter multiplied by the thickness of the connected material on which it bears
Bearing resistance for standard, oversize, or short-slotted holes in any direction, and long-slotted holes parallel to the bearing force:
For bolts spaced with clear distance between holes greater than or equal to 3.0 d
and for bolts with a clear end distance greater than or equal to 2.0 d
Rn = 2.4 d t Fu
For bolts spaced with clear distance between holes less than 3.0 d
and for bolts with clear end distances less than 2.0 d
Rn = 1.2 Lc t Fu
Where, d = nominal bolt diameter
Lc= clear distance between holes or between the hole and the end of the member in
the direction of applied bearing force
Fu = tensile strength of the connected material
The resistance factor fbb for material in bearing due to bolts = 0.80

12. BEARING CONNECTION SPACING REQUIREMENTS
Minimum spacing between centers of bolts in standard holes shall not be less than three times the diameter of the bolt
For sealing against penetration of moisture in joints, the spacing on a single line adjacent to the free edge shall satisfy s ≤ ( t) ≤ 7.0
Minimum edge distances

13. BOLTED CONNECTION
Example 1 Design a slip-critical splice for a tension member. For the Service II load combination, the member is subjected to a tension load of 200 kips. For the strength limit state, the member is subjected to a maximum tension load of 300 kips.
The tension member is a W8 x 28 section made from M270-Gr. 50 steel. Use A325 bolts to design the slip-critical splice.
Step I. Service and factored loads
Service Load = 200 kips.
Factored design load = 300 kips
Tension member is W8 x 28 section made from M270 Gr.50. The tension splice must be slip critical (i.e., it must not slip) at service loads.

14. BOLTED CONNECTION Step II. Slip-critical splice connection
Slip resistance of one fully-tensioned slip-critical bolt = Rn = Kh Ks Ns Pt
f = 1.0 for slip-critical resistance evaluation
Assume bolt diameter = d = ¾ in. Therefore Pt = 28 kips from Table 1
Assume standard holes. Therefore Kh = 1.0
Assume Class A surface condition. Therefore Ks = 0.33
Therefore, fRn = 1.0 x 0.33 x 1 x 28 = 9.24 kips
Therefore, number of ¾ in. diameter bolts required for splice to be slip-critical at service loads = 200 / 9.24 =
Therefore, number of bolts required ≥ 22

15. BOLTED CONNECTION Step III: Layout of flange-plate splice connection
To be symmetric about centerline, need the number of bolts = multiple of 8.
Therefore, choose 24 fully tensioned 3/4 in. A325 bolts with layout above.
Slip-critical strength of the connection = 24 x 9.24 kips = kips
Minimum edge distance (Le) = 1 in. from Table 4.
Design edge distance Le = 1.25 in.
Minimum spacing = s = 3 x bolt diameter = 3 x ¾ = 2.25 in.
Design spacing = 2.5 in.

16. BOLTED CONNECTION Step IV: Connection strength at factored loads
The connection should be designed as a normal shear/bearing connection beyond this point for the factored load of 300 kips
Shear strength of high strength bolt = f Rn = 0.80 x 0.38 x Ab x Fub Ns
Equation given earlier for threads included in shear plane.
Ab = 3.14 x / 4 = in2
Fub = 120 ksi for A325 bolts with d < 1-1/8 in.
Ns= 1
Therefore, fRn = 16.1 kips
The shear strength of 24 bolts = 16.1 kips/bolt x 24 = kips

17. BOLTED CONNECTION
Bearing strength of 3/4 in. bolts at edge holes (Le = 1.25 in.)
fbb Rn = 0.80 x 1.2 Lc t Fu
Because the clear edge distance = 1.25 – (3/4 + 1/16)/2 = in. < 2 d
fbb Rn = 0.80 x 1.2 x x 65 kips x t = kips / in. thickness
Bearing strength of of 3/4 in. bolts at non-edge holes (s = 2.5)
fbb Rn = 0.80 x 2.4 d t Fu
Because the clear distance between holes = 2.5 – (3/4 + 1/16) = in. > 2d
fbb Rn = 0.80 x 2.4 x 0.75 x 65 kips x t = 93.6 kips / in. thickness
Bearing strength of bolt holes in flanges of wide flange section W8 x 28 (t = in.)
8 x x x 93.6 x = 892 kips

18. CONNECTION STRENGTH
Connection strength (fRn) > applied factored loads (gQ).
Therefore ok

19. WELDED CONNECTIONS Introduction
The shielded metal arc welding (SMAW) process for field welding.
Submerged metal arc welding (SAW) used for shop welding – automatic or semi-automatic process
Quality control of welded connections is particularly difficult because of defects below the surface, or even minor flaws at the surface, will escape visual detection.
Welders must be properly certified, and for critical work, special inspection techniques such as radiography or ultrasonic testing must be used.

20. WELDED CONNECTIONS
Two most common types of welds are the fillet and the groove weld.
lap joint – fillet welds placed in the corner formed by two plates
Tee joint – fillet welds placed at the intersection of two plates.
Groove welds – deposited in a gap or groove between two parts to be connected e.g., butt, tee, and corner joints with beveled (prepared) edges
Partial penetration groove welds can be made from one or both sides with or without edge preparation.

21. WELDED CONNECTIONS Design of fillet welded connections
Fillet welds are most common and used widely
Weld sizes are specified in 1/16 in. increments
Fillet welds are usually fail in shear, where the shear failure occurs along a plane through the throat of the weld
Shear stress in fillet weld of length L subjected to load P
fv =
·       

22. FILLET WELDED CONNECTIONS
The shear strength of the fillet weld = fe Fexx
Where, fe2 = 0.80
Fexx is the tensile strength of the weld electrode used in the welding process. It can be 60, 70, 80, 90, 100, 110, or 120 ksi. The corresponding electrodes are specified using the nomenclature E60XX, E70XX, E80XX, and so on.
Therefore, the shear strength of the fillet weld connection
fRn = fe2 x 0.60 Fexx x a Lw
Electrode strength should match the base metal strength
If yield stress (sy) of the base metal is  ksi, use E70XX electrode
If yield stress (sy) of the base metal is  ksi, use E80XX electrode
E70XX is the most popular electrode used for SMAW fillet welds
For E70XX, fRn = 0.80 x 0.60 x 70 x a Lw = a Lw kips

23. FILLET WELDED CONNECTIONS
The shear strength of the base metal must be considered:
f Rn = fv x 0.58 Ag Fy
where, fv = 1.0
Fy is the yield strength of the base metal and Ag is the gross area in shear
Strength of weld in shear Strength of base metal
= 0.80 x 0.60 x Fexx x x a x Lw = 1.0 x 0.58 x Fy x t x Lw
Smaller governs the strength of the weld

24. FILLET WELDED CONNECTIONS
Limitations on weld dimensions
Minimum size (amin)
Weld size need not exceed the thickness of the thinner part joined.
amin depends on the thickness of the thicker part joined
If the thickness of the thicker part joined (T) is less than or equal to ¾ in.  amin = ¼ in.
If T is greater than ¾ in.  amin = 5/16 in.
Maximum size (amax)
Maximum size of fillet weld along edges of connected parts
for material with thickness < 0.25 in., amax­ = thickness of the material
for plates with thickness  0.25 in., amax = thickness of material - 1/16 in.
Minimum length (Lw)
Minimum effective length of fillet weld = 4 x size of fillet weld
Effective length of fillet weld > 1.5 in.

25. FILLET WELDED CONNECTIONS
Weld terminations and end returns
End returns must not be provided around transverse stiffeners
Fillet welds that resist tensile forces not parallel to the weld axis or proportioned to withstand repeated stress shall not terminate at corners of parts or members
Where end returns can be made in the same plane, they shall be returned continuously, full size around the corner, for a length equal to twice the weld size (2a)

26. FILLET WELD DESIGN
Example 1 Design the fillet welded connection system for a double angle tension member 2L 5 x 3½ x 1/2 made from A36 steel to carry a factored ultimate load of 250 kips.
Step I. Design the welded connection
Considering only the thickness of the angles; amin = 1/4 in.
Considering only the thickness of the angles; amax = 1/2 - 1/16 in. = 7/16 in.
Design, a = 3/8 in. = in.
§   Shear strength of weld metal = f Rn = 0.80 x 0.60 x FEXX x x a x Lw
= 8.9 x Lw kips
Strength of the base metal in shear = f Rn = 1.0 x 0.58 x Fy x t x Lw
= Lw kips
§         Shear strength of weld metal governs, f Rn = 8.9 Lw kips

27. FILLET WELD DESIGN Design strength f Rn > 250 kips
Therefore, 8.9 Lw > 250 kips
Therefore, Lw > 28.1 in.
Design length of 3/8 in. E70XX fillet weld = 30.0 in.
Shear strength of fillet weld = 267 kips
Connection layout
Connection must be designed to minimize eccentricity of loading. Therefore, the center or gravity of the welded connection must coincide with the center of gravity of the member.

28. FILLET WELD DESIGN Connection layout
Connection must be designed to minimize eccentricity of loading.
The c.g. of the welded connection must coincide with c.g. of the member
Total length of weld required = 30 in.
Two angles assume each angle will have weld length of 15 in.

29. FILLET WELD DESIGN
The tension force Tu acts along the c.g. of the member, which is 1.65 in. from the top and 3.35 in. from the bottom (AISC manual).
Let, f be the strength of the fillet weld per unit length.
Therefore, fL1 + fL2 = Tu
And fL2 x fL1 x 1.65 = 0 - taking moments about the member c.g.
Therefore, L1 = 2.0 L2
But, L1 + L2 = 15.0 in.
Therefore, L1 = 10 in. and L2 = 5 in.
Design: L1 = 10.0 in. and L2 = 5.0 in.

30. FILLET WELD DESIGN Consider another layout fL1 + fL2 + 5f = Tu
fL2 x f x fL1 x 1.65 = 0 - Moment about member c.g.
Additionally, L1 + L2 + 5 = 15.0 in.
Therefore, L1 = 7.6 in. and L2 = 2.4 in.
Design: L1 = 8.0 in. and L2 = 3.0 in.

31. Groove Welded Connections
Connects structural members that are aligned in the same plane
Basic Types:
Complete joint penetration groove weld: transmits full load of the member they join and have the same strength as the base metal.
Partial penetration groove weld: Welds do not extend completely through the thickness of the pieces being joined.

32. Groove Welds Complete penetration groove welded connections
Tension and compression loaded
Factored resistance = factored resistance of base metal
Shear loaded on effective area  lesser of
Factored resistance of weld = 0.6 x fe1 x Fexx = 0.6 x 0.85 x Fexx
60% of factored resistance of base metal in tension
Partial penetration groove-welded connections
Tension or compression parallel to the weld axis and compression normal to effective area  factored resistance of the base metal
Tension normal to the effective area  lesser of
Factored resistance of the weld = 0.6 fe2 Fexx = 0.60 x 0.80 x Fexx
Factored resistance of the base metal
Shear loaded  lesser of
Factored resistance of base metal = 0.58 Fy

33. Groove Welds