][ STEEL STRUCTURES ][

Problem:
A bar of A 36 steel, 
3 x 7/16" in cross section
is to be welded with E 70 XX electrodes
to the back of a channel
so that the full tensile strength of the bar
may be developed.
Specify the weld to be used.

A. Given:
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1. bar of A 36 steel, 
2. 3 x 7/16" in cross section
3. E 70 XX electrodes

B. Asked:
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What size is the weld?
	thickness
	length
	type

C. Graph:
======= 

D. Calculation:
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1. Area of Bar:
		Area = 3 x 7/16"
		Area = 3 x 0.4375
		Area = 1.3125 sq. in.
		Area = 1.313 sq. in.

2. Allowable Unit tensile strength of A 36 steel
		P& A 266 Table 8.1

Allowable tensile stress
			 = 0.60 x Fy
		 	= 0.60 x 36 ksi
		 	= 21.6
		 	= 22 Ksi	

3. Allowable tensile strength of bar:
	 =  Ft x A 	  
	 =  22 ksi x  1.313 square inch
	 =  28.886 kips
	 =  28.9 kips


4. Tensile strength of weld:
	The weld must be of ample dimensions
	to resist a force equivalent to
	the allowable tensile strength of the bar
	in our case the magnitude is (28.9 kips).

5. Selection of Weld:
	P&A 370:
	The maximum size of a fillet weld applied
	to a square edge of a plate or section that 
	is 1/4" or more in the thickness should be
	1/16" less than the nominal thickness of the
	edge.
	Along edges of material less than 1/4" thick,
	the maximum size may be equal to the thickness of the material.

	Plate Thickness		 	Weld Size			
	plate >=1/4"			thickness - 1/16"
	plate < 1/4"			thickness 		
	7/16" - 1/16 = 6/16 = 3/8" 
	3/8" WELD has been preselected

6. Allowable working Strength 
	P&A 371   Table 12.1

	Allowable working strength = 5.6 kips/in.

7. Required calculated length of the weld:
		=load/allowable working strength
		= 28.9 kips / 5.6 kips/in
		= 5.16 in.

8. Other Weld requirements 
	P&A 371
	a. For starting and stopping the arc,
		approximately 1/4" should added
		to the length of the fillet welds.

	P&A 372
	b. box weld
			return = 2 x weld size
			return = 2 x 3/8"
					= 6/8" 
					= 3/4" return

	P&A 380
	c. minimum length of welds equals 4 times
		the weld size 4 x 3/8 = 12/8 = 1.5 inches
 

	P&A 372
	d. weld is greater than 1/4" therefor requires
			double pass weld

	 

Length = calculated length + start/stop + 
		    2 returns

Length 	= L + 1/4" + 3/4" + 3/4"
            	= 5.16 " + 0.25" + 0.75" + 0.75" 
            	= 6.91 
	= 7"

E. SOLUTION:
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9. Layout of Weld:
	thickness = 3/8"
	length = 7"
	returns, etc.
	The position of the weld with respect to the
	bar has several options, three of which are 
	shown in the figure below.

utilize the welding symbols 
to specify the weld

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