APPENDIX 9B

The estimation of the length of tensile and compressive stress zone behind a crack tip in a short rod fracture toughness specimen when the crack approaches the back face of the specimen (more accurate calculation)

The 1984-1987 calculation of the Appendix 9 of the Licentiate thesis ignored the weight of the specimen and made two simplifications: (i) the tensile strength to the maximum compressive stress ratio was much less than 1; and (ii) the ligament length to the specimen diameter ratio was much less than 1. Here is a more accurate calculation without the three assumptions with the computer aided numerical solution of the equation group using a Microsoft spreadsheet. We have the following two equations of equilibrium of force and moment:

     F - Mg + cseD/2 = bstD/2   
                                                              (9-2B)  

     Fw - Mgw/2 + (cseD/2)(c/3) = (bstD/2){b/3 + (1+st/3se)c}
                                                              (9-3B)
The two unknowns are b and se. The mass of half a specimen was M=0.63kg. Let a = se/st and A = w-a, the above equations are written as: 
     (A/2)a2 - [b/2 - (F-Mg)/(Dst)]a - [b/2 - (F-Mg)/(Dst)] = 0
                                                              (9-4B)  

     (F - Mg/2)[w/(Dst)] + (a3/6)[A/(1+a)]2 = (b2/6) + (b/2)(1/3 + a)[A/(1+a)]  
                                                              (9-5B)
The solution of (9-4B) is as follows: 
     a = (B/A){1 + [1 + 2A/B]1/2 }
                                                              (9-6B)  

          with B = [b/2 - (F-Mg)/(Dst)]
where the plus sign makes sense. Substituting a of (9-6B) into (9-5B), we have an equation of the unkown variable b. This equation is solved by trials of b values starting from the previous known solution of Appendix 9, that is 8.12 and 12.83 + 0.55 mm. For the parameters in Table 3.10, with A = (w-a) = 0.12 and 0.89 mm, and F = 100 and 280 N respectively, we obtain b = 8.0 and 12.85 mm that are in perfect agreement with the previous results. Notice that the specimen mass has negligible effect, the accurate initial values are 8.12 and 12.92 mm as compared to the previous 8.12 and 12.83 mm that had cumulative rounding errors due to manual calculator calculation. Thus, Appendix 9A and B have shown a two step problem solution method for complex problems: simplified initial solution and accurate solution.

Microsoft Excel Computer Spreadsheet Software

Alternate Javascript program:

LEFT side of (9-5B) = (F - Mg/2)[w/(Dst)] + (a3/6)[A/(1+a)]2
RIGHT side of (9-5B) = (b2/6) + (b/2)(1/3 + a)[A/(1+a)]

b (mm):    A = w-a (mm):    F (N):   
The reader may change the above three parameter values to make the balance to approach 0.
(LEFT - RIGHT) (mm^2):
Footnote Javascript code: 
   
    function CalBalance()
    {
    var Mg = 6.18, w=105.0, D=71.4, St=13;
    var A = this.form1.ligament.value;
    var b = this.form1.trial.value;
            this.form1.trial.focus();

    var F = this.form1.force.value;

    var B = b/2-(F-Mg)/(D*St);
    var alpha = (B*(1+Math.sqrt(1+2*A/B)))/A;
    var LEFT = (F-Mg/2)*w/(D*St)+(1/6)*(alpha*alpha*alpha)*(A/(1+alpha))*(A/(1+alpha));
    var RIGHT = b*b/6+(b/2)*(1/3+alpha)*(A/(1+alpha));
    var bal = LEFT - RIGHT;
    this.form1.balance.value = bal;
    //alert("A=" + A);
    }