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The critical stress intensity (KIC) for a material for a component of a design is
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- A bronze alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to fracture. Stress and strain data obtained during the test are shown in Figure. Determine: 400 - Upper scale a. the modulus of elasticity. 320 b. the proportional limit. Lower scale c. the ultimate strength. 240 d. the yield strength (0.20% offset). 160 e. the modulus of resilience. f. the true fracture stress if the 80 final diameter of the specimen at the location of 0.100 0.005 0.300 0.015 0.200 0.400 the fracture was 10.5 mm. 0.010 0.020 g. The Poisson's ratio. Strain (mm/mm) Stress (MPa)arrow_forward2. Another cylindrical component is madeof Enfennering ceramic Al203 but with different dimensions. Here, l=30 cm and the diameter is 4 cm. Assume the same Weibull modulus of 9. Calculate the level of the tensile strength for the following probability of failures: a. Pr (V) = 0.1 b. Pr (Vo) = 0.01 c. What is the survival probability and the failure probability of this component if a stress of 200 MPa is applied?arrow_forwardThe table below shows the deformation data that resulted from applying a pure tensile load to a brass alloy rod with an initial length of 30 mm and a diameter of 10 mm. This rod was subjected to necking and beyond necking deformation. The data at fracture is shown in the last row. Applied Load (N) Length (mm) Diameter (mm) 75,100 33.8 68, 250 34.9 50,200 35.7 6.7 Solve for the TRUE STRESS/ES and TRUE STRAINS. 8.3 7.8arrow_forward
- 4. The maximum stress a human tendon can withstand is estimated to be 1200 MPa. If you were to test it for rupture what load cell you should use (25ON or 5kN) what problems may occur if you select an incorrect load cell. 5. The figure below is a J-shaped (or concave upward) stress-strain curve. What does a J-shaped stress stain curve indicate about a material's response to stress and tendency to yield? Name two materials that have J-shaped stress strain curves.arrow_forwardPart A The stress-strain diagram for a steel alloy having an original diameter of 0.40 in. and a gage length of 9 in. is shown in the figure below. (Figure 1) Determine the modulus of elasticity for the material. Express your answer to three significant figures and include appropriate units. HA ? Eapprox = Value Units Submit Request Answer Figure < 1 of 1 Part B a (ksi) 80 Determine the load on the specimen that causes yielding. 70 Express your answer to three significant figures and include appropriate units. 60 50 40 HA 30 20 Py = Value Units 10 € (in./in.) 0.04 0,08 0.12 0,16 0,20 0,24 0,28 O 0.0005 0.001 0.0015 0.002 0.0025 0.0030.0035 Request Answer Submitarrow_forwardExample: Convert the change in length data in Table 3-2 to engineering stress and strain and plot a stress-strain curve Homework- help Table 3-2 The results of a tensile test of a 0.505 in. diameter aluminum alloy test bar, initial length (1o) = 2 in. Calculated LTO Load (Ib) Change in Length (in.) Stress (psi) Strain (in./in.) 0.000 1000 0.001 0.0005 4,993 14,978 24,963 34,948 37,445 39,442 39,941 39,691 37,944 3000 0.003 0.0015 5000 0.005 0.0025 7000 0.007 0.0035 7500 0.030 0.0150 7900 0.080 0.0400 8000 (maximum load) 0.120 0.0600 7950 0.160 0.0800 7600 (fracture) 0.205 0.1025arrow_forward
- A structural beam is built from steel with a yield strength of 345 MPa. The structure has a minimum service temperature of 5 oC, where the fracture toughness of the steel KIC = 60 MPam. What is the minimum surface crack lengths that must be detected by quality control to ensure that the full yield strength of the steel is available in the beam? Assume geometric parameter Y = 1 and plain strain conditions. Answer in millimeters.arrow_forwardQuestion 6 What is the calculated fracture toughness, K, for the perspex specimen in MPa root m? Give your answer to 2 decimal places. P = 250 N, crack length = 5.5 mm W= 13.3 mm, B = 5.45 mm Note that the span length (s) is 38mm. Take f(a/W) as 0.78arrow_forward1. A bronze alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to fracture. Stress and strain data obtained during the test are shown in Figure. Determine: 400 Upper scale a. the modulus of elasticity. b. the proportional limit. c. the ultimate strength. 320 Lower scale 240- d. the yield strength (0.20% offset). 160 e. the modulus of resilience. f. the true fracture stress if the 80 final diameter of the specimen at the location of the fracture was 10.5 mm. 0.100 0.200 0.300 0.400 0.005 0.010 0.015 0.020 g. The Poisson's ratio. Strain (mm/mm) Stress (MPa)arrow_forward
- A material with an Ultimate Tensile Strength (UTS) of 350 Mpa is loaded cyclically about mean stress of 70 MPa. If the stress range that will cause fatigue fracture in 100 000 cycles under zero mean stress is (+,- )60 MPa, what stress range (in MPa) about the mean of 70 MPa will give the same lifearrow_forwardQuestion 2 A machine component is made of Aluminium alloy having an ultimate tensile stress Ou = 350MPa, a fracture toughness Kic= 40 MPa√m and a geometry factor Y=1.12. The component is subject to cyclic tensile stress of 200 MPa and cyclic compressive stress of 70 MPa. Before the operation, using an ultrasonic non-destructive testing technique, it was found that there is an edge crack of length 2mm. The crack growth C(AK)m, where is in m/cycle, AK ΔσΥ νπα , C ~ 2.2×10-12 da rate is given by da dN dN and m=3.2. Find the critical crack length. = =arrow_forwardA tensile test for a copper specimen has been performed and the following data are obtained. - Percentage of Elongation = 69 % - Percentage of Reduction in Area = 39 % - Final length after fracture = 35.5 mm - Final Diameter after fracture = 3.6 mm & - Ultimate stress = 396 MPa SOLUTION: iv) Initial Diameter (in mm) = v) Ultimate Load (in N) =arrow_forward
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