Question

The stainless steel and polystyrene rulers in Example 5.2 are loaded as cantilevers of length 250 mm. Both have sharp transverse scratches of depth 0.2 mm, near the base of the cantilever on the tensile side. For this surface crack geometry, assume the factor Y in equation (8.4) is 1.1. The fracture toughness K1c of steel and polystyrene are 80 and 1 MPa.m", respectively. Calculate the stress needed to cause fast fracture in each case. Will the cantilevers fail by yielding or by fast fracture? (b) A compact tension specimen, as in Figure 8.4, is manufactured from an Al alloy with dimensions w = 50 mm and b = 20 mm, with a sharp pre-crack of length c = 10 mm. The maximum load capacity available on the test machine is 50 kN. Is the machine capacity sufficient to fracture the alloy, if its fracture toughness is 25 MPa.m"2?

Answer 1

fast fracture for * llac The Stress at So, the ruler will stainless Steel 6- 1x 10 80x105 XX2X10 4 6 2.9 Gpa Ruler will Yield before it fractures Polystyrening Fracture stress 6 1-1 XT7X24154 6 = 36mpa It is less than Yield Strength- fracture before it Yields Scanned by TapScanner

Stress Intensity Intensity Factor Kic= 1064x F' XXC bxw Kic = 25 Mpa rm Failure load TI _3 -3 25X10 X 20X10 X 50X10 VT Xox 10 3 F = 86 KN Test Machine is Not Strong Scanned by TapScanner