Determine the LRFD design strength of the A36 plates shown T

Determine the LRFD design strength of the A36 plates shown. The PL 1/2x6 in. is fillet welded to transfer the tensile load, both transverse and longitudinal as shown. Include block shear strength in the calculations. See attached sketch.

Solution

(a) Strength based on yielding in the gross section
t Pn = t Fy Ag
= 0.9 (36 ksi) (6\" x 0.75\") = 145.8 Kips

(b) Strength based on tensile fracture
Ae = U x An
AISC Specification TABLE D3.1, Case 1, U= 1
Ae = (1) (6\" x 0.75\") = 4.5 Sq in.

t Pn = t Fu Ae
= (0.75) (58 ksi) (4.5 sqin) = 195.75 Kips.

t Pn = 195.75 Kips.

(c) Strength based on block shear failure
Design Block Shear Strength = Rn where =0.75

Rn = 0.6 Fu Anv + Ubs Fu Ant <= 0.6 Fy Agv + Ubs Fu Ant
Agv = (2 x 8\")(0.75\") = 12 sq in
Anv = 12 sq in
Ant = 6\" x 0.75\" = 4.5 sq in
Ubs = 1.0 for uniform tension stress
Rn = 0.6(58 ksi)(12 sq in) + (1)(58 ksi)(4.5) = 678.6 kips
Rn = 0.6(36 ksi)(12 sq in) + (1)(58 ksi)(4.5) = 520.2 Kips CONTROLS

Rn = 0.75x 520.2 = 390.15 kips

Based on a, b and c design strength is controled by a

Design strength is 145.8 Kips