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Shear stress

  1. Oct 12, 2007 #1
    1. The problem statement, all variables and given/known data
    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 heres 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


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 13, 2007 #2

    Astronuc

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    Staff: Mentor

    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.
     
  4. Oct 13, 2007 #3
    thanks i will let you no how i get on
     
  5. Oct 13, 2007 #4
    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
     
  6. Oct 13, 2007 #5

    Astronuc

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    Staff: Mentor

    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.
     
  7. Oct 13, 2007 #6
    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 im lost thanks again
     
  8. Oct 13, 2007 #7

    Astronuc

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    Staff: Mentor

    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 Materials Carry Load.htm

    Stress is load P divided by area A or [itex]\sigma[/itex] = P/A or F/A. Conventially [itex]\sigma[/itex] refers to the normal stress, i.e. the stress vector is parallel with the normal to the surface of interest. Also, [itex]\tau[/itex] = 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: Oct 14, 2007
  9. Oct 14, 2007 #8
    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
     
  10. Oct 14, 2007 #9

    Astronuc

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    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.
     
  11. Oct 14, 2007 #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 im lost!!!!!
     
    Last edited: Oct 15, 2007
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