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Force required to strip an M10 thread

  1. Feb 5, 2014 #1
    Hi all,
    im building a bench vice for college and I need to do a report on It also. the problem im having is working out the force required to strip the threads off an M10 x 1.5 metric thread. both the thread and nut are to be made from mild steel. any info that I could find online has only imperial data.
    if someone could please shed some light on this subject for me I would be highly obliged
  2. jcsd
  3. Feb 5, 2014 #2


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    Under tension there will be two potential modes of failure.
    The mode that will occur first is dependent on the length of the nut.

    1. Plastic failure of the bolt core.
    An M10x1.5 bolt has an effective tensile stress area of 58mm2.
    For mild steel expect 400 N/mm2
    58mm * 400 N/mm2 = 23200. N

    2. Removal of the helical thread from the nut or bolt.
    The thread shear area along the pitch-line per millimetre of the bolt length M10x1.5 is 4.0mm2.
    The standards for metal threads specify sufficient nut length.
  4. Feb 6, 2014 #3
    many thanks that really helps me. so I would then use the formula shear stress= load/area? so in that case mild steel has a shear strength of 80kN/mm2. So the load required to strip the thread would be 80,000=load/(4xL) where L is length of the nut?

    the object of the report is to prove that the nut length is sufficient enough to strip the threads rather that shear the bolt core.
  5. Feb 6, 2014 #4


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    Now you have confused me.
    To shear across the bolt core is a different thing to stretching the bolt core which is again different to stripping or shearing thread from the nut or bolt.

    A bolt is not usually sheared across the core. It is used to hold two materials together so that friction prevents relative shear movement between those materials. If the bolt stretches and deforms plastically, then the clamping effect is lost and the bolt may then be cut across the core. We rely on the bolt remaining elastic and torque the nut to place the bolt at about 65% of its yield point.

    Where did you get that 80kN/mm2 from?
    1 N/mm2 = 1,000,000. N/m2 = 1 MPa

    In steel the shear yield strength is usually taken as 0.58 * the tensile yield strength.
    Tensile strength of mild steel is usually 400 N/mm2, so shear strength will be about 240 N/mm2.

    For thread removal from an ISO 10x1.5 MS bolt we have 4mm2 per mm of nut length.
    which gives 4mm2 * 240 N/mm2 = 960. N per mm of nut length.

    The bolt tensile strength was computed earlier as 23,200 N.
    Then the thread in the nut must be 23,200 / 960 = 24.2 mm long to avoid stripping.
    Since the nut is only about 10 mm long it is probably safe to assume the thread will strip before the bolt yields.
  6. Feb 6, 2014 #5


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    Staff: Mentor

    For schoolwork questions you must do the bulk of the work yourself. Thread moved to the Homework Help forums...
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