Finding Minimum Number of Spot-Welds for Steel Lap Joint

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Discussion Overview

The discussion revolves around calculating the minimum number of spot welds required for a steel lap joint subjected to tensile stress. Participants explore the application of shear stress concepts and relevant equations to solve a homework problem involving specific dimensions and material properties.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a homework problem involving two strips of steel and asks for guidance on how to approach it, specifically regarding shear stress and tensile stress.
  • Another participant suggests using tensile stress and cross-sectional area to determine the load, and then dividing this load by the total area of the spot welds to find the number of welds needed.
  • A participant calculates the load based on the safe shearing stress and the area of the plate, arriving at a number of spot welds but expresses uncertainty about the correctness of their calculations.
  • Further clarification is provided that the number of spot welds must ensure that the shear stress in any spot weld does not exceed the safe shear stress of 35 MN/m².
  • Another participant emphasizes the need to divide the load by the total area of the N spot welds to maintain the shear stress below the critical value.
  • One participant struggles with the calculations and seeks assistance in determining how to find the number of spot welds based on their derived stress values.
  • Another participant advises rounding up the number of welds since the number must be a whole number, reiterating the importance of keeping local stress below the critical stress.

Areas of Agreement / Disagreement

Participants generally agree on the approach of using tensile and shear stress calculations to determine the number of spot welds needed. However, there is uncertainty regarding specific calculations and the interpretation of results, leading to some confusion and lack of consensus on the final answer.

Contextual Notes

Participants express uncertainty about specific mathematical steps and the dependence on correct interpretations of stress and area calculations. There are unresolved issues regarding the accuracy of derived values and the application of rounding rules for the number of welds.

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Homework Statement


please help i have just been introduced to shear stress and can do the basics i have just got a problem with this homework question and i am after some pointers as how to tackle it here's the question TWO STRIPS OF STEEL,100*2.5MM,ARE SPOT WELDED TOGETHER TO FORM A LAP JOINT, THE SPOT-WELDS BEING SINGLE SHEAR. THE SPOT WELD HAVE A DIAMETER OF 4MM AND THE SAFE SHEARING STRESS ON SPOT-WELD IS 35MN/m2. EVALUATE THE MINIMUM NUMBER OF COMPLETE SPOT-WELDS NEEDED IF THE JOINT IS SUBJECTED TO A TENSILE STRESS OF 20MN/m2 hpe you can help cheers


Homework Equations





The Attempt at a Solution

 
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Use the tensile stress and cross-sectional area to determine the load.

Then take the load and divide by the total area of the spot welds = N*Aspot.

The shear stress is just the load divided by the area supporting that load.
 
thanks i will let you no how i get on
 
IS THIS RIGHT = LOAD = 35MN/m2 X AREA OF PLATE (250X10-6) =8.75X10 3 THEN DIVIDE THE LOAD 20X10 6/AREA OF SPOT WELD(3.14X10-3)=6.369X10 9 =7 SPOT WELDS THANKS FOR THE HELP
 
The load is related to the Tensile Stress 20MN/m2. The cross-sectional area is correct.

One must find the number of spot welds, such that the shear stress in one spot weld is less than the safe shear stress of 35 MN/m2.
 
ok double checked load 20MN/m2 x 250x10-6 = 5x10 3 then 35x10 6 / 5 x103 = 7 x 10 3 if this is not right I am lost thanks again
 
Ok. With the load, one must divide the load by the total area of the N spot welds in order for the shear stress in any spot weld to be less than 35 MN/m2.

See if this reference helps.

http://em-ntserver.unl.edu/NEGAHBAN/Em325/01-How-Materials-carry-load/How%20Materials%20Carry%20Load.htm

Stress is load P divided by area A or \sigma = P/A or F/A. Conventially \sigma refers to the normal stress, i.e. the stress vector is parallel with the normal to the surface of interest. Also, \tau = P/A or F/A, but here the load or force is perpendicular to the normal of the surface, i.e. the load/force is parallel with the surface.

So in the problem at hand, determine P/A, where A = N As, whereAs is the area of one spot weld.

You're getting there. :smile:
 
Last edited by a moderator:
i have found p/a or f/a =5 x10 3/1.25x10-5(area of one spot weld)=400 x 10 6 how do i find how many spot welds now as i have tried to divide the answer by 35 MN/m2 your help is much appreiciated
 
So take the total stress and divide by critical stress/spot weld. The number of welds N, must be a whole number, and one has to round up regardless, since the local stress (in any given spot weld) must be less than the critical stress.
 
  • #10
do i divide 5 x 10 3/1.25 x 10 - 5 as i have the answer on the question sheet as 7 but this formula =400 10 6 I am lost!
 
Last edited:

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