Question

3. The plates below are made of A36. There are two 1/4in fillet welds as shown. Assume a filler material of E70xx. What is the allowable load P (lb) that the joint can withstand? Assume a FOS of 3 for the base material and use basic AWS D1.1 assumptions for the weld allowable stresses (see Table 2.3 below). Please show your work. 1/2 in 1/2 in A in

Answer 1

A Size of the weled 1. h / M. a Base 1 1/2 20 Hp wilth of base and weld material, bahin Material properties Yield strength of 436 speel, y = 36, no psi yreld strengon of E70 XX yeler , oy = 70, ano psi Given that, fus of base matemal = 3 objective! To find the allowable load (P), that the joint can withstand. solution! The allowable strength of base matemal, Tallow ... Jallow = 36, 0o = 12,110 psi As per AWS D 1.1 apump in the shear strength of the effective weld area = 0.3 x tensile strength of hillur material. it013 X70, oro = 21,00 psl. Now, the effective area of weld, Aweld = 1414 x hxb >> A weld = 1.414 X A X4 B (Parallel filled weld) =1414 18,2 And the cls area of base material, a base / 2 x 4 = 2 in² Now, the manimum wood on the weld, p = A weld y z > P = 414 X 21, Uw = 29,694 lb But the mastmum load required for the base material is given by 1. P = A base X Tallent - Tatay = 29694 = 14847 psi, which is more than the allowable stress off base material (120000 psi) There fore, the manimum exord should be calentated from the resisting power of base makewal only. g.e p = Abase & Tallow = 2 x 12,000 = 24 oro eb Answer! Therefore, the auth wable lood that the joint can withstand is p = 24, erro pound car. .