= 266 N 45 \u00b4 10 3 m 3 4 69 \u00b4 10 9 N\/m 2 10 \u00b4 10 3 m 5 \u00b4 10 3 m 3 = 7.0 \u00b4 10 5 m = 7.0 \u00b4 10 2 mm 2.5 \u00b4 10 3 in 12.43 A circular specimen of MgO

= 266 n 45 ´ 10 3 m 3 4 69 ´ 10 9 n/m 2 10 ´ 10 3

This preview shows page 54 - 58 out of 66 pages.

=(266 N)(45´10-3m)3(4)(69´109N/m2)(10´10-3m)(5´10-3m)3= 7.0 ´10-5m = 7.0 ´10-2mm (2.5 ´10-3in.)
12.43 A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 425 N (95.5 lbf), the flexural strength is 105 MPa (15,000 psi), and the separation between load points is 50 mm (2.0 in.).SolutionWe are asked to calculate the maximum radius of a circular specimen of MgO that is loaded using three-point bending. Solving for Rfrom Equation 12.7b R=FfLsfspé ë ê ê ù û ú ú 1/3which, when substituting the parameters stipulated in the problem statement, yields R=(425 N)(50´10-3m)(105´106N /m2)(p)é ë ê ù û ú 1/3= 4.0 ´10-3m = 4.0 mm (0.16 in.)
12.44 A three-point bending test was performed on an aluminum oxide specimen having a circular cross VHFWLRQ RI UDGLXVPP LQ WKHVSHFLPHQ IUDFWXUHG DWD ORDG RI950 N (215 lbf) when the distance between the support points was 50 mm (2.0 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of 12 mm (0.47 in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is 40 mm (1.6 in.)?SolutionFor this problem, the load is given at which a circular specimen of aluminum oxide fractures when subjected to a three-point bending test; we are then are asked to determine the load at which a specimen of the same material having a square cross-section fractures. It is first necessary to compute the flexural strength of the aluminum oxide, Equation 12.7b, and then, using this value, we may calculate the value of Ffin Equation 12.7a. From Equation 12.7b sfs=FfLpR3=(950 N)(50´10-3m)(p)(3.5´10-3m)3=352 ´106N/m2=352 MPa(50,000 psi)Now, solving for Fffrom Equation 12.7a, realizing that b= d= 12 mm, yields Ff=2sfsd33L=(2)(352´106N/m2)(12´10-3m)3(3)(40´10-3m)=10,100 N(2165 lbf)
12.45 (a)A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 390 MPa (56,600 psi). If the specimen radius is 2.5 mm (0.10 in.) and the support point separation distance is 30 mm (1.2 in.), predict whether or not you would expect the specimen to fracture when a load of 620 N (140 lbf) is applied. Justify your prediction. (b)Would you be 100% certain of the prediction in part (a)? Why or why not?Solution(a) This portion of the problem asks that we determine whether or not a cylindrical specimen of aluminum oxide having a flexural strength of 390 MPa (56,600 psi) and a radius of 2.5 mm will fracture when subjected to a load of 620 N in a three-point bending test; the support point separation is given as 30 mm. Using Equation 12.7b we will calculate the value of s; if this value is greater than sfs(390 MPa), then fracture is expected to occur. Employment of Equation 12.7b yields s=FLpR3=(620 N)(30´10-3m)(p)(2.5´10-3m)3=379 ´106N/m2=379 MPa (53,500 psi)Since this value is less than the given value of sfs(390 MPa), then fracture is not predicted.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture