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Modulus of rupture formula

  1. Mar 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove the formula of modulus of rupture using the third point loading test.
    R = Pl/bd^2 P: load at fracture
    l: length
    b: width
    d: depth
    2. Relevant equations

    3. The attempt at a solution
    I don't know how to start. PLease help me.
  2. jcsd
  3. Mar 2, 2007 #2


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    Homework Helper

    Use the definition of modulus of rupture - hint, it's a stress at some point in the beam.

    Set up the shear force and bending moment diagram for the loading in a 3-point test, and find the stress from the usual beam equations.
  4. Oct 12, 2011 #3
    The Modulus of Rupture formula can be derrived simply by using basic statics and strength of material equations.

    Let's start with a beam of length L loaded at midspan with a force F and is simply supported at the 2 ends of the beam. The reaction forces are P1 and P2. The cross section of the beam is rectangle and has a width of b (dimension into the paper), and a depth (height or thickness of the cross section) of d. This loading configuration represents a simple 3 point bending test.

    The beam will fail (rupture) at the point of maximum stress where it exceeds the ultimate strength of the material. In the 3 point bending test, the beam will fail at the point of maximum bending stress (maximum outer fiber stress) that exceeds the ultimate material strength. Maximum bending stress in the beam is determined by the famous formula of Mc/I, where M is the bending moment, c is the distance from the neutral axis to the outer fiber, and I is the moment of inertia of the beam cross section.

    It should be clear that the maximum bending moment is located at the midpoint of the beam (for a beam loaded at midspan) which can be determined by P1 x L/2 or P2 x L/2, where L/2 is the length from the support to the beam midspan. P1 = P2 = F/2 for a midspan loaded beam. For a symmetric cross section such as a rectangle, c is just d/2 (the distance from the center of the cross section to the outer most fiber). The moment of inertia I for a rectangle cross section is bd^3/12.

    Therefore, putting everything together: Mc/I = (P1 x L/2) x (d/2) / (bd^3/12) = (F/2 x L/2) x (d/2) / (bd^3/12) which simplifies to: 3FL/(2bd^2).
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