Charlie (Woo Yong) Choi 22609575 E45 6.7) In normal motion, the load exerted on the hip joint is 2.5 times body weight. (a) Calculate the corresponding stress (in MPa) on an artificial hip implant with a cross-‐ sectional area of 5.64cm2 in a patient weighing 150lbf. (b) Calculate the corresponding strain if the implant is made of Ti-‐6Al-‐4V, which has an elastic modulus of 124GPa. Solution a) Let’s first convert all units to the same one so we can work with them. 𝑁 1𝑃𝑎 = 1 ! 𝑚 1𝑙𝑏𝑠! = 4.448222𝑁 5.64𝑐𝑚! = 5.64 ∗ 10!! 𝑚! Plugging these values into the stress equation: 𝑃 2 ∗ 𝑏𝑜𝑑𝑦 𝑤𝑒𝑖𝑔ℎ𝑡 2.5 ∗ 150 ∗ 4.448222 𝜎= = = ≈ 𝟐. 𝟗𝟓𝟕𝟓𝟗𝑴𝑷𝒂 𝐴 𝜋𝑟 ! 5.64 ∗ 10!! b) Using the stress-‐strain equation with elastic modulus, we can find the strain: 𝜎 = 𝐸 ∗ 𝜖 𝜎 2.95759𝑀𝑃𝑎 𝜖= = ≈ 𝟐. 𝟑𝟖𝟓 ∗ 𝟏𝟎!𝟓 𝐸 124000𝑀𝑃𝑎 6.8) Repeat Problem 6.7 for the case of an athlete who undergoes a hip implant. The same alloy is used, but because the athlete weighs 200lbf a larger implant is required (with a cross-‐sectional area of 6.90cm2). Also, consider the situation in which the athlete expends his maximum effort exerting a load of five times his body weight. Solution 𝑃 2.5 ∗ 200 ∗ 4.448222 𝜎!.! !"#$% = = ≈ 𝟑. 𝟐𝟐𝟑𝟑𝑴𝑷𝒂 𝐴 6.9 ∗ 10!! 𝑃 5 ∗ 200 ∗ 4.44822 𝜎! !"#$% = = ≈ 𝟔. 𝟒𝟒𝟔𝟕𝑴𝑷𝒂 𝐴 6.9 ∗ 10!! 𝜎 3.2233𝑀𝑃𝑎 𝜖!.! !"#$% = = ≈ 𝟐. 𝟓𝟗𝟗𝟒 ∗ 𝟏𝟎!𝟓 𝐸 124000𝑀𝑃𝑎 𝜎 6.3367𝑀𝑃𝑎 𝜖! !"#$% = = ≈ 𝟓. 𝟏𝟗𝟖𝟗 ∗ 𝟏𝟎!𝟓 𝐸 124000𝑀𝑃𝑎 6.9) Suppose that you were asked to select a material for a spherical pressure vessel to be used in an aerospace application. The stress in the vessel wall is: 𝜎=
𝑝𝑟 2𝑡
where p is the internal pressure, r the outer radius of the sphere, and t the wall thickness. The mass of the vessel is: 𝑚 = 4𝜋𝑟 ! 𝑡𝜌 where r is the material density. The operating stress of the vessel will always be
𝜎≤
Charlie (Woo Yong) Choi 22609575 E45
𝑌. 𝑆 𝑆
where S is a safety factor. (a) show that the minimum mass of the pressure vessel will be 𝑚 = 2𝑆𝜋𝑝𝑟 !
𝜌 𝑌. 𝑆
(b) given Table 6.1 and the following data, select the alloy that will produce the lightest vessel. Alloy
𝜌 (
𝑀𝑔 ) 𝑚!
Cost ($/kg)
1040 carbon steel
7.80
0.63
304 stainless steel
7.80
3.70
3003-‐H14 aluminum
2.73
3.00
Ti-‐5Al-‐2.5Sn
4.46
15.00
(c) Given 6.1 and the data in the preceding table, select the alloy that will produce the minimum cost vessel. Solution ! a) Solving 𝑚 = 4𝜋𝑟 ! 𝑡𝜌 for t, t = !!! ! !. Plugging into expression for 𝜎, 𝑝𝑟 𝑝𝑟 4𝜋𝑟 ! 𝜌 2𝜋𝑝𝑟 ! 𝜌 𝜎= = = 2𝑡 2𝑚 𝑚 Then we plug this into operating stress requirement and rearrange: 𝑌. 𝑆 𝑆 2𝜋𝑝𝑟 ! 𝜌 𝑌. 𝑆 ≤ 𝑚 𝑆 𝟐𝑺𝝅𝒑𝒓𝟑 𝝆 𝑚= 𝒀. 𝑺 𝜎≤
b) Alloy
Y.S (MPa)
𝜌 𝑌. 𝑆
𝜌 (kg/m3)
𝜌 ∗ 𝑐𝑜𝑠𝑡 𝑌. 𝑆
1040 carbon steel
600
7800
13
8.19
304 stainless steel
205
7800
38.049
140.78
3003-‐H14 aluminum
145
2730
Charlie (Woo Yong) Choi 22609575 E45 18.828 56.484
Ti-‐5Al-‐2.5Sn
827
4460
5.3930
80.895
!
From looking at the !.! data, if all other variables such as radius, safety factor, !
pressure are kept constant and equal, the one with the lowest !.! value will require lowest minimum mass. Thus Ti-‐5Al-‐2.5Sn would require least mass. !
!
c) Since !.! of each material acts as a ratio of their masses, we can multiply !.! with the cost to see which of the materials have the lowest product. The one with the lowest product will be the cheapest. It turns out 1040 carbon steel has the lowest product, thus will be the cheapest. 6.16) A single crystal Al2O3 rod (precisely 6mm diameter x 50mm long) is used to apply loads to small samples in a high-‐precision dilatometer (a length-‐measuring device). Calculate the resulting rod dimensions if the crystal is subjected to a 25-‐kN axial compression load. Solution E = 380*103MPa
𝜈 = 0.26 𝑟! = 0.003𝑚
Assume change in r is minimal during compression, thus r is constant. !
𝜎=!=!
!"###! !.!!"! !
𝑧! = 0.05𝑚 𝜎 = 𝐸 ∗ 𝜖! 𝜎 ∆𝑧 𝜖! = = 0.002326 = 𝐸 𝑧! ∆𝑧 = 0.1163𝑚𝑚 𝑜𝑟 − 0.1163𝑚𝑚 𝑠𝑖𝑛𝑐𝑒 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝜖! 𝜈 = − = 0.26 𝜖! ∆𝑟 𝜖! = 0.00060476 = 𝑟! Since 𝜖! is very small, our assumption is reasonable. ∆𝑟 = 0.001814𝑚𝑚 Resulting rod dimensions: 0.04988m x 0.006003628m
