CONNECTION DESIGN Connections must be designed at the

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 • End connections for floorbeams and girders – 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 – Two angles with thickness > 0. 375 in. – Made with high strength bolts – If welded account for bending moment in design Visit for more Learning Resources

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

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

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

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

SLIP CRITICAL BOLTED CONNECTION Values of Kh Values of Pt Values of Ks • 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

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.

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

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

BEARING CONNECTION • Nominal shear resistance of a bolt – Threads excluded: – Threads included: Rn = 0. 48 Ab Fub Ns Rn = 0. 38 Ab Fub Ns Where, Ab = area of the bolt corresponding to the nominal diameter Fub = 120 ksi for A 325 bolts with diameters 0. 5 through 1. 0 in. Fub = 105 ksi for A 325 bolts with diameters 1. 125 through 1. 5 in. Fub = 150 ksi for A 490 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

BEARING CONNECTION • 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

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 ≤ (4. 0 + 4. 0 t) ≤ 7. 0 – Minimum edge distances

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 W 8 x 28 section made from M 270 Gr. 50 steel. Use A 325 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 W 8 x 28 section made from M 270 Gr. 50. The tension splice must be slip critical (i. e. , it must not slip) at service loads.

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, f. Rn = 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 = 21. 64. • Therefore, number of bolts required ≥ 22

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. A 325 bolts with layout above. – Slip critical strength of the connection = 24 x 9. 24 kips = 221. 7 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.

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 0. 752 / 4 = 0. 442 in 2 – Fub = 120 ksi for A 325 bolts with d < 1 1/8 in. – Ns= 1 – Therefore, f. Rn = 16. 1 kips • The shear strength of 24 bolts = 16. 1 kips/bolt x 24 = 386. 9 kips

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 = 0. 84375 in. < 2 d – fbb Rn = 0. 80 x 1. 2 x 0. 84375 x 65 kips x t = 52. 65 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) = 1. 6875 in. > 2 d – 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 W 8 x 28 (t = 0. 465 in. ) – 8 x 52. 65 x 0. 465 +16 x 93. 6 x 0. 465 = 892 kips

CONNECTION STRENGTH • Connection strength (f. Rn) > applied factored loads (g. Q). – Therefore ok

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.

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.

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 =

FILLET WELDED CONNECTIONS • The shear strength of the fillet weld = fe 2 0. 60 Fexx – Where, fe 2 = 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 E 60 XX, E 70 XX, E 80 XX, and so on. • Therefore, the shear strength of the fillet weld connection – f. Rn = fe 2 x 0. 60 Fexx x 0. 707 a Lw • Electrode strength should match the base metal strength – If yield stress (sy) of the base metal is 60 65 ksi, use E 70 XX electrode – If yield stress (sy) of the base metal is 60 65 ksi, use E 80 XX electrode • E 70 XX is the most popular electrode used for SMAW fillet welds – For E 70 XX, f. Rn = 0. 80 x 0. 60 x 70 x 0. 707 a Lw = 0. 2375 a Lw kips

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 0. 707 x a x Lw = 1. 0 x 0. 58 x Fy x t x Lw Smaller governs the strength of the weld

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) • Minimum length (Lw) – 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 effective length of fillet weld = 4 x size of fillet weld – Effective length of fillet weld > 1. 5 in.

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 (2 a)

FILLET WELD DESIGN Example 1 Design the fillet welded connection system for a double angle tension member 2 L 5 x 3½ x 1/2 made from A 36 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. = 0. 375 in. § Shear strength of weld metal = f Rn = 0. 80 x 0. 60 x FEXX x 0. 707 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 = 10. 44 Lw kips § Shear strength of weld metal governs, f Rn = 8. 9 Lw kips

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. E 70 XX 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.

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.

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, f. L 1 + f. L 2 = Tu And f. L 2 x 3. 35 - f. L 1 x 1. 65 = 0 taking moments about the member c. g. – Therefore, L 1 = 2. 0 L 2 But, L 1 + L 2 = 15. 0 in. – Therefore, L 1 = 10 in. and L 2 = 5 in. Design: L 1 = 10. 0 in. and L 2 = 5. 0 in.

FILLET WELD DESIGN • Consider another layout f. L 1 + f. L 2 + 5 f = Tu f. L 2 x 3. 5 + 5 f x 0. 85 f. L 1 x 1. 65 = 0 Moment about member c. g. Additionally, L 1 + L 2 + 5 = 15. 0 in. Therefore, L 1 = 7. 6 in. and L 2 = 2. 4 in. Design: L 1 = 8. 0 in. and L 2 = 3. 0 in.

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.

Groove Welds Complete penetration groove welded connections • Tension and compression loaded • Shear loaded on effective area lesser of – Factored resistance of weld = 0. 6 x fe 1 x Fexx = 0. 6 x 0. 85 x Fexx – Factored resistance = factored resistance of base metal – 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 fe 2 Fexx = 0. 60 x 0. 80 x Fexx – Factored resistance of the base metal Shear loaded lesser of – Factored resistance of the weld = 0. 6 fe 2 Fexx = 0. 60 x 0. 80 x Fexx – Factored resistance of base metal = 0. 58 Fy

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