What is the allowable capacity of two lines of 3/16\" fillet welds made with E60 electrodes? The lines are each 4-in long and the load is applied parallel to the weld. Solution yield strength of E60xx = 50 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 ] in our case = 0o Fw = 0.60 FEXX = 0.6*5= 30 ksi Design Strength, Rn = Fw Aw = 2x[0.75*[ 0.707 3/16* 4 ] 30] = 23.86 kips thanks.

1. What is the amount of total debits on the post-closing trial balance for this The following items
are taken from the financial statements of A Company for the year ending December 31, 2013:
Solution
1) Post closing trial balance Debit Cash 16 Accounts Receivable 26 Supplies 4
Equipment 191 237 2) Current Assets Account Receivable 11000 Cash 11000
Prepaid Expense 6000 28000