Figure 1: A 3-D ball-and-spring model of the tin wire. Figure 2: The top view and side view of a simple cubic lattice.

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Question (5) Bond stiffness of solid tin. A certain tin wire (white tin, as opposed to gray tin) has a length of.5 m and a square rectangular cross section that is ( mm x mm).

1 Question (5) Bond stiffness of solid tin. A certain tin wire (white tin, as opposed to gray tin) has a length of.5 m and a square rectangular cross section that is ( mm x mm). With a tension of 200 N applied to the wire, it stretches 6 mm. Use a simple ball-and-spring model of a solid, and assume that tin s crystal lattice is simple cubic, with atoms essentially stacked next to and on top of one another, like inflated basketballs stacked in a box. From these measurements, approximate the bond stiffness of tin. The density of white tin is 7365 kg/m 3. Its atomic weight is 8.7 g/mol. Figure : A 3-D ball-and-spring model of the tin wire. Figure 2: The top view and side view of a simple cubic lattice. Solution There are numerous springs in series and in parallel. We can compute the effective spring stiffness of springs in series using k eff,series = k + k 2 +

2 and we can compute the effective spring stiffness of spring in parallel using k eff,parallel = k + k 2 + Therefore, our plan is to: (a) Determine the stiffness of the entire wire, by modeling it as a single spring. (b) Determine the number of parallel strands (or chains) are in the wire and find the stiffness each strand. (c) Determine the number of springs connected in series in a strand and find the stiffness of each spring. Model the wire as a single spring. Figure 3: A model of the wire as a single spring. The stiffness of the wire is found by Hooke s law. The tension applied to the wire is 200 N and the wire stretches 6 mm. F on spring = k wire s 200 N = k(0.006 m) k = 200 N m k = N/m We need to know how many strands we have. Look at the top view of the wire. The number of strands is N strands = N side N side where N side is the number of atoms along the width of the wire. By examining the wire, you can see that the width of the wire is

3 Figure 4: Each atom in the top view represents a strand of atoms of length L. w = N side d where d is the diameter of an atom. The diameter of an atom is found from the density of tin, Avogadro s number, and the atomic mass of tin. Use dimensional analysis. The portion of the volume of the entire solid that is taken up by one atom is d 3 where d is the diameter of the atom. The total volume of the solid is the volume per atom times the number of atoms. Use this to solve for the volume of a box taken up by one atom. = N atoms atom atom = N atoms The quantity on the right side of the above equation is the volume per atom. Use the mass density of aluminum, its molar mass, and Avogadro s number to get the volume per atom. Simply cancel units to give the correct units. = volume mole mass N atoms mass atoms mole = M ρ N A where N A is Avogadro s number and M is the molar mass of aluminum. Substituting the density and molar mass and Avodadro s number gives N atoms = ( m 3 ) ( mol ) ( ) 0.87 kg 7365 kg atoms mol = m 3 /atom The volume of the box filled by one atom is atom = d 3 since each side of the box has a length equal to d the diameter of the atom. Take the cubed root to get the diameter.

4 = d 3 d = m 3 /atom = m Now, solve for N side, w = N side d N side = w d N side = 0.00 m m N side = atoms The total number of strands is N strands = N side N side = ( atoms) 2 = strands All of these strands are in parallel. The effective stiffness of parallel springs is k eff,parallel = k + k 2 and we already know the effective stiffness of all of the strands. Substituting this, we can find the stiffness of an individual strand. k eff,parallel = k + k 2 k eff,parallel = N strands k strand N/m = ( strands)k strand k strand = N/m strands k strand = N/m Now that we know the stiffness of an individual strand, we an find the stiffness of an individual spring. First, we need the to know the number of springs in the strand. This is approximately equal to the number of atoms along the length of the wire. (It s actually one less, but because there are so many atoms, a large number minus is still a large number.) The number of atoms along the length of the wire found by

Figure 5: The number of springs, in the side view, is actually less than the number of atoms. L = N length d N length = L d N length = 5.02 0 9 atoms 5.

5 Figure 5: The number of springs, in the side view, is actually less than the number of atoms. L = N length d N length = L d N length = atoms springs The effective spring stiffness of springs in series is given by k eff,series = k + k 2 + where the effective stiffness is the stiffness of a strand. = + + k eff,series k k 2 = N length k strand k k k strand = N length k = N length k strand k = ( springs)( N/m) k = 4.9 N/m k 5 N/m So, the bond stiffness of tin to two significant figures is 5 N/m. This is of similar order of magnitude to other metals, such as copper (27 N/m) for example.

1/7/14. Measuring Matter. How can you convert among the count, mass, and volume of something? Apples can be measured in three different ways.

1/7/14. Measuring Matter. How can you convert among the count, mass, and volume of something? Apples can be measured in three different ways. Chapter 10 Chemical Quantities 102 Mole-Mass and Mole-Volume Relationships 103 Percent Composition and Chemical Formulas 1 Measuring Matter Measuring Matter How can you convert among the count, mass, and