weld
DESCRIPTION
shshssTRANSCRIPT
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Dr. M.E. Haque, P.E. (Weld Connections - Rev) Page 1 of 7
Electrodes should be selected to match the base metal.
Use E70XX electrodes with steels that have a yield stress less than 60 ksi.
Use E80XX electrodes with steels that have a yield stress of 60 ksi or 65 ksi.
Nominal load capacity of weld, Rn = Fw Aw = [ 0.707 w L ] Fw
Design Strength, Rn = Fw Aw = [ 0.707 w L ] Fw
Where = 0.75
Where Fw = nominal strength of the weld metal per unit area = 0.6 FEXX
Fw based on the angle of the load to the longitudinal axis of the weld ( ):
Fw = 0.60 FEXX [1.0 + 0.5 Sin 1.5 ]
Throat = w x Cos 450 = 0.707 w
w
w
Root
FILLET WELD
Shear Plane
Weld Connections
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Dr. M.E. Haque, P.E. (Weld Connections - Rev) Page 2 of 7
1/4 7 E70
Near side (arrow side)
1/4 7 E70 Other side
1/4 7 E70
Both sides 7 1/4
1/4 7 E70
Weld all around
1/4 7 E70
Field weld
Fillet weld symbols
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Dr. M.E. Haque, P.E. (Weld Connections - Rev) Page 3 of 7
EXAMPLE 1: Determine the strength of the following welds.
(a) Nominal Design Strength, Rn = Fw Aw = [ 0.707 w L ] Fw
Where = 0.75
Rn = 0.75 [ 0.707(3/4)(8)](0.6x70) = 133.62 kips
(b) Fw = 0.60 FEXX [1.0 + 0.5 Sin 1.5 ] = 0.60 FEXX (1.5) { Sin = Sin (900 ) = 1}
Rn = 133.62 x 1.5 = 200.4 kips
(c) Fw = 0.60 FEXX [1.0 + 0.5 Sin 1.5 ] = 0.60 FEXX (1.292456) { = 450 }
Rn = 133.62 x 1.292456 = 172.7 kips
8
(a) 3/4 weld, 8 long, loaded along the weld
8
(b) 3/4 weld, 8 long, loaded perpendicular to the weld
8
(c) 3/4 weld, 8 long, loaded at a 450 to the weld
450
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Dr. M.E. Haque, P.E. (Weld Connections - Rev) Page 4 of 7
Example 2: Determine the design strength for C-shaped welds. Use E70XX electrodes and a 3/16 Fillet weld.
For 3/16 weld, 2-10 long, loaded along the weld
Rw(L) = 0.707(3/16)(2x10)](0.6x70) = 111.35 kips
For 3/16 weld, 8 long, loaded perpendicular to the weld
Rw(T) = 0.707(3/16)(8)](0.6x70) = 44.54 Kips
Without considering the weld and load orientation, Weld Design Strength AISC Specification, Equation (J2-10a) (AISC Steel Manual 14th Ed.) Rn = 111.35 + 44.54 = 155.89 Kips
Weld Design Strength considering the added contribution of the transverse welds while reducing the contribution of the longitudinal welds, AISC Specification, Equation (J2-10b) (AISC Steel Manual 14th Ed.) Rn = (0.85)111.35 + (1.5)44.54 = 161.46 Kips
Select the largest one, Rn = (0.75)(161.46) = 121.10 Kips
E70 3/16
10
8
PL (A36 Steel)
PL (A36 Steel)
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Dr. M.E. Haque, P.E. (Weld Connections - Rev) Page 5 of 7
Example 3: Determine the LRFD design strength of the connection as shown in Figure. Follow the following steps:
(a) Strength based on yielding in the gross section (b) Strength based on tensile fracture (c) Strength based on block shear failure (d) Weld strength. (e) What is the design strength of this connection?
SOLUTION:
(a) Strength based on yielding in the gross section t Pn = t Fy Ag
= 0.9 (36 ksi) (8" x 0.5") = 129.6 Kips
(b) Strength based on tensile fracture
Ae = U x An
t Pn = t Fu Ae
AISC Specification TABLE D3.1, Case 1, U= 1
Ae = (1) (8" x 0.5") = 4 Sq in.
E70 5/16
10
8
PL (A36 Steel)
PL (A36 Steel)
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Dr. M.E. Haque, P.E. (Weld Connections - Rev) Page 6 of 7
t Pn = t Fu Ae
= (0.75) (58 ksi) (4 sqin) = 174 Kips.
(c) Strength based on block shear failure
Design Block Shear Strength = Rn where =0.75
Rn = 0.6 Fu Anv + Ubs Fu Ant
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Dr. M.E. Haque, P.E. (Weld Connections - Rev) Page 7 of 7
Tee Groove Weld
Backing bar Butt Groove Weld
Corner Groove Weld
Fig 1: Complete penetration groove welds
Fig 2: Partial penetration groove welds