Question

(1.5 points) Determine the size and length of weld required for the connection below to make the weld design strength and controlling tension design capacity roughly equal. Assume plates are A36 steel and weld material is E70 electrodes. Welds are fillet welds made by the SMAW process. PLļx 12

Answer 1

GIVEN DATA Member Details Grade of Steel A 36 Bar 1/2 x 8 Member designation 0.500 Thickness of member : PL 1/2 x 12 Plate designation 0.500 Connection Details Grade of electrode E 70 SMAW (assumed) Welding Process CALCULATIONS A 36, For grade of steel Fy 36 ksi Yield stress, Fu58 ksi Ultimate stress, DESIGN OF WELD Yielding in the gross section Design strength of gross section is obtained as: 0.9 x 58 x 4 208.8 kips Fracture in the net section An A Net area - 4 in Shear lag Factor Effective area 4.00in2 Nominal strength of net section is obtained as 0.75 x 58 x 4 174 kips Thus, design strength of the member is the smaller of the two values, that is 9.Tn: 174.0 kips Selection of weld size and computation of strern : 3 in Min. size fillet weld 16 Max. Size fillet weld 16 2 7 in 16

Therefore, use 5 in size fillet weld with E 70 electrode 16 0.707x 5 Effective throat thickness, 16 0.221 in The design strength 0.75 x 0.221 x 0.6 x70 6.96 kips/in This cannot exceed the shear rupture strength of the base metal 0.75 x 0.500 x 0.6 x 58 13.05 kips/in O.K b) Determination of lengths of weld to be used in the connection Total length of weld, Lw of fillet weld required is し,-174.0 6.96 25 in Hence, the length of weld on each side, 12.5 in Use 13 in weld E 70 (Тур) 16