Question

1. Consider a pair of 3/8 in welds that are each 10 in long. Determine the strength if these two welds if:
   1. They are loaded parallel to their length
   2. They are loaded perpendicular to their length

Do not consider simplifying assumptions

2. Consider an L5x3x1/2 supporting 45 kip dead load and 30 kip live load. It is being welded to an ASTM A36 PL5/8x8. Design and proportion the weld to make this a simple welded connection. The leg of the weld should be the average of the minimum and maximum size.  the 5 in leg of the angle is connected to the plate.

Answer 1

The strength of the welds remain the same whether they are load parallel to the lengths or perpendicular to the lengths.

Design strength = $\phi$t_e (0.6 F_EXX) L= 0.75 x (0.707 x 3/8) x (0.6 x 70 ) x (10+10) = 167.03 kips

: GIVEN DATA Member Details Grade of Steel Member designation Gross area of section Thickness of member, Centroidal distance, A 36 L5 x 3 x 1/2 2 A,-3.75 in t0.500 in 0.75 in Connection Details Member connected to one/two legs Grade of electrode Welding Process one leg E 70 (assumed) (assumed) SMAW Loads Service dead load Service live load D45 kips L = 30 kips CALCULATIONS For grade of steel A 36 Fy-36 ksi Fu 58 ksi Yield stress, Ultimate stress, Factored design load, T1.2 D 1.6 102.0 kips DESIGN OF WELD a) Selection of weld size and computation of strength 3 in 16 Min. size fillet weld Max. size fillet weld 2 - 7 in 16 16 Therefore, use 5 in size fillet weld with E 16 70 electrode Effective throat thickness, te0.707 x 5 16 0.221 in The design strength 0.75 x 0.221 x 0.6 x 70 = 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 -13.1 kips/in O.K Determination of lengths of weld to be used in the connection

Let F,, F3 be the forces due to longitudinal welds along outstand leg and connected leg respectively. Let F2 be the strength of weld along From moment equilibrium about back of the angle (at F) F1= 102.0 x 0.75 = 15.2 kips F102.0 15.2 Therefore, 86.8 kips Hence, the lengths of longitudinal welds are 15.2 6.96 LwF1 2.2 in Use 3 in LF86.812.5 in e 13 in R 6.96 2.5 2.5 0.75 FINAL DESIGN 16 513E 70(Typ.) 16