Question 1 Copper-rich copper beryllium alloys are precipitation hardenable. After consulting the portion of the phase diagram given below, answer the following questions: (a) Specify the range of compositions over which these alloys may be precipitation hardened. (b) Briefly describe the heat-treatment procedures (in terms of temperatures) that would be used to precipitation harden an alloy having a composition of your choosing, yet lying within the range given for part (a). This problem is concerned with the precipitation-hardening of copper-rich Cu-Be alloys. It is necessary for us to use the Cu-Be phase diagram, which is shown above. (a) The range of compositions over which these alloys may be precipitation hardened is between approximately 0.2 wt% Be (the maximum solubility of Be in Cu at about 300 C) and 2.7 wt% Be (the maximum solubility of Be in Cu at 866 C). (b) The heat treatment procedure, of course, will depend on the composition chosen. First of all, the solution heat treatment must be carried out at a temperature within the α phase region, after which, the specimen is quenched to room temperature. Finally, the precipitation heat treatment is conducted at a temperature within the α +γ 2 phase region. For example, for a 1.5 wt% Be-98.5 wt% Cu alloy, the solution heat treating temperature must be between about 600 C and 900 C, while the precipitation heat treatment would be below 600 C, and probably above 300 C. Below 300 C, diffusion rates are low, and heat treatment times would be relatively long.
Question 2 In an aligned and continuous glass fiber-reinforced nylon 6,6 composite, the fibers are to carry 94% of a load applied in the longitudinal direction. (a) Using the data provided in the table below, determine the volume fraction of fibers that will be required. (b) What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 30 MPa. a) Given some data for an aligned and continuous glass-fiber-reinforced nylon 6,6 composite, we are asked to compute the volume fraction of fibers that are required such that the fibers carry 94% of load applied in the longitudinal direction. From Equation 16.11 Now, using values for F f and F m from the problem statement And, solving for V f yields, V f 0.393 b) We are now asked for the tensile strength of this composite. From Equation 16.17, Since values for σ f (3400 MPa) and σ m ' (30 MPa) are given in the problem statement.
Question 3 Look at the following Figure, and perform the following tasks: (a) Obtain approximate values of fracture toughness K IC for AISI 4340 steel heat treated to yield strengths of 1000 and 1800MPa. (b) For each of these yield strengths, calculate the transition (or critical) crack length a t, and comment on the significance of the values obtained. Figure 1 Fracture toughness vs. yield strength for AISI 4340 steel quenched and tempered to various strength levels. (Adapted from an illustration courtesy of W. G. Clark, Jr., Westinghouse Science and Technology Ctr., Pittsburgh, PA.)
Question 4 Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa m. It has been determined that fracture results at a stress of 365 MPa when the maximum internal crack length is 2.5 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 4.0 mm. This problem asks us to determine the stress level at which a wing component on an aircraft will fracture for a given fracture toughness (40 MPa m ) and maximum internal crack length (4.0 mm), given that fracture occurs for the same component using the same alloy at one stress level (365 MPa) and another internal crack length (2.5 mm). It first becomes necessary to solve for the parameter Y for the conditions under which fracture occurred using Equation 8.5. Therefore, Y K Ic s pa 40 MPa m (365 MPa) (p) 2.5 10-3 m Ł 2 ł 1.75 Now we will solve for σ c using Equation 8.6 as s c Y 40 MPa m pa 3 4 10 - m (1.75) ( p) Ł 2 ł K Ic 288 MPa
Question 5 A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 77.0 MPa m and a yield strength of 1400 MPa. The flaw size resolution limit of the flaw detection apparatus is 4.0 mm. If the design stress is one half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection. This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection, given the limit of the flaw detection apparatus (4.0 mm), the value of K Ic ( 77MPa m ), the design stress (σ y /2 in which σ y 1400 MPa), and Y 1.0. We first need to compute the value of a c using Equation 8.7; thus 2 1 K Ic ac p Ł Ys ł Ø ø 1 Œ 77 MPa m œ Œ œ pœ 1400 MPa (1.0) œ Œ 2 œ º Ł łß 2 0.0039 m3.9 mm Therefore, the critical flaw is not subject to detection since this value of a c (3.9 mm) is less than the 4.0 mm resolution limit.
Question 6 A spherical pressure vessel is made of ASTM A517-F steel and operates at room temperature. The inner diameter is 1.5 m, the wall thickness is 10 mm, and the maximum pressure is 6MPa. Is the leak-before-break condition met? For ASTM A517-F steel, K II ii 187 MMM m. This problem asks that we check if leak-before-break condition is met, i.e. whether the vessel will leak before it ruptures or not. For the leak-before-break condition to be satisfied, the critical crack length must be greater than the wall thickness of the pressure vessel. So, we have to find the critical crack length given the fracture toughness value of the material (K II ii 187 MMM m) and the internal pressure (P is 6MPa). First, we need to calculate what is the maximum stress on the vessel wall is caused by this internal pressure. So, The maximum stress in the pressure vessel wall is: σ mmm σ t P r 2 t (6 MMM)(750 mm) 2 (10mm) 225 MMM Combining this value with K II, the critical crack length can be calculated as: a c 1 2 π K II σ t 220 mm This value of the critical crack length far exceeds the thickness (t 10 mm). Therefore the pressure vessel will leak before break.