THE UNIVERSITY OF OKLAHOMA - COLLEGE OF ARCHITECTURE

Click here for the FULL GRAPHICS version

Problem:A bar of A 36 steel, 3 x 7/16" in cross sectionis to be welded with E 70 XX electrodesto the back of a channelso that the full tensile strength of the barmay be developed.Specify the weld to be used.

A. Given:=======1. bar of A 36 steel, 2. 3 x 7/16" in cross section3. E 70 XX electrodes

B. Asked:=======What size is the weld?

thicknesslengthtype

Dr. Gruenwald - Weld Calculations http://www.ou.edu/class/hgruenwald/teach/4333/4333l34t.htm

1 of 5 8/25/2010 7:11 AM

C. Graph:=======

D. Calculation:=========1. Area of Bar:

Area = 3 x 7/16"Area = 3 x 0.4375Area = 1.3125 sq. in.Area = 1.313 sq. in.

2. Allowable Unit tensile strength of A 36 steelP& 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 dimensionsto resist a force equivalent tothe allowable tensile strength of the barin our case the magnitude is (28.9 kips).

5. Selection of Weld:P&A 370:The maximum size of a fillet weld appliedto a square edge of a plate or section that is 1/4" or more in the thickness should be1/16" less than the nominal thickness of theedge.Along edges of material less than 1/4" thick,the maximum size may be equal to the thickness of the material.

Plate Thickness Weld Sizeplate >=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

Dr. Gruenwald - Weld Calculations http://www.ou.edu/class/hgruenwald/teach/4333/4333l34t.htm

2 of 5 8/25/2010 7:11 AM

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 371a. For starting and stopping the arc,

approximately 1/4" should addedto the length of the fillet welds.

P&A 372b. box weld

return = 2 x weld sizereturn = 2 x 3/8"

= 6/8" = 3/4" return

P&A 380c. minimum length of welds equals 4 times

the weld size 4 x 3/8 = 12/8 = 1.5 inches

P&A 372d. 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:===========9. Layout of Weld:

thickness = 3/8"length = 7"returns, etc.The position of the weld with respect to thebar has several options, three of which are shown in the figure below.

utilize the welding symbols to specify the weld

Dr. Gruenwald - Weld Calculations http://www.ou.edu/class/hgruenwald/teach/4333/4333l34t.htm

3 of 5 8/25/2010 7:11 AM

To hear the announcement click on the speaker ! sound

LECTURES

To Home Page

Timetable

Prof. Dr. Hermann Gruenwald

(mail comments to: HGRUENWALD@ou.edu)

College of Architecture

The University of Oklahoma

Dr. Gruenwald - Weld Calculations http://www.ou.edu/class/hgruenwald/teach/4333/4333l34t.htm

4 of 5 8/25/2010 7:11 AM

©Dr. Gruenwald 1996,1997

Dr. Gruenwald - Weld Calculations http://www.ou.edu/class/hgruenwald/teach/4333/4333l34t.htm

5 of 5 8/25/2010 7:11 AM