Finding Minimum Number of Spot-Welds for Steel Lap Joint

AI Thread Summary
To determine the minimum number of spot welds needed for a steel lap joint under tensile stress, one must calculate the load based on the tensile stress and the cross-sectional area of the joint. The load is then divided by the area of the spot welds to ensure that the shear stress in each weld does not exceed the safe limit of 35 MN/m2. The calculations indicate that approximately seven spot welds are required to maintain safety under the given conditions. It is crucial to round up the number of welds to ensure that the local stress remains below the critical shear stress. Accurate calculations and understanding of shear and tensile stress are essential for solving this problem effectively.
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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


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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:
 
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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.
 
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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!
 
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