CHAPTER 2. NATURE OF MATERIALS

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1 Materials for Civil and Construction Engineers 4th Edition Mamlouk SOLUTIONS MANUAL Full clear download (no formatting errors) at: CHAPTER 2. NATURE OF MATERIALS 2.1. See Section See Section See Section See Section See Section See Section See Section See Section See Section If the atomic masses and radii are the same, then the material that crystalizes into a lattice with a higher APF will have a larger density. The FCC structure has a higher APF than the BCC structure For the face-center cubic crystal structure, number of equivalent whole atoms in each unit cell = 4 By inspection the diagonal of the face of a FCC unit cell = 4r Using Pythagorean theory: (4r) 2 = a 2 + a 2 16r 2 = 2 a 2 8r 2 = a 2 a 2 2r a. Number of equivalent whole atoms in each unit cell in the BCC lattice structure = 2 b. Volume of the sphere = (4/) r

2 Volume of atoms in the unit cell = 2 x (4/) r = (8/) r By inspection, the diagonal of the cube of a BCC unit cell = 4r = a 2 a 2 a 2 = a a = Length of each side of the unit cell = 4r

3 4 r c. Volume of the unit cell = volume of atoms in the unit cell APF = (8 / ).r = 0.68 total unit volume of the cell (4r / ) 2.1. For the BCC lattice structure: a Volume of the unit cell of iron = 4r r 4x0.124x = = 2.48 x m For the FCC lattice structure: a 2 2r Vol. of unit cell of aluminum = (2 2r) = (2 2x0.14) = nm = x10-29 m From Table 2., copper has an FCC lattice structure and r of nm Volume of the unit cell of copper =(2 2r) = (2 2x0.1278) = nm = 4.72 x10-29 m For the BCC lattice structure: a 4r Volume of the unit cell of iron = Density = r 4x0.124x = = 2.48 x m n = Number of equivalent atoms in the unit cell = 2 A = Atomic mass of the element = 55.9 g/mole NA= Avogadro s number = 6.02 x x x10 29 x6.02x10 2 = x 10 6 g/m = Mg/m 4r For the BCC lattice structure: a Vol. of the unit cell of molybdenum = 4 r 4 = x0.16x109 =.119 x m

4 = 2 x x 10 6 g/m = Mg/ m.119x10 29 x6.02x10 2

5 2.18. For the BCC lattice structure: a 4r Volume of the unit cell of the metal = = For the FCC lattice structure: a 2 4 r = 4 x0.128x109 = 2.58 x m 2 x x 10 6 g/m = 8.16 Mg/ m 2.58x10 29 x6.02x10 2 Volume of unit cell of the metal = (2 2r 2r) = (2 2 x0.12) = nm = 5.204x10-29 m = 4 x x 10 6 g/m = Mg/ m 5.204x10 29 x6.02x For the FCC lattice structure: a 2 Volume of unit cell of aluminum = ( a. 2r 2r) = (2 2x0.14) = nm = x10-29 m Density = For FCC lattice structure, n = 4 A = Atomic mass of the element = g/mole NA= Avogadro s number = 6.02 x x26.98 = x 10 6 g/m = Mg/m x10 29 x6.02x10 2 For FCC lattice structure, n = 4 Vc = 4x x10 6 x6.02x10 2 4x(4 / ).r APF = 0.74 = x10 29 r = x m r =0.128 x 10-9 m = nm = x m For FCC lattice structure, n = 4 4x40.08 Vc = = x m 1.55x10 6 x6.02x10 2

6 4x(4 / ).r b. APF = 0.74 = x10 28 r = x m r =0.196 x 10-9 m = nm See Section See Section See Section See Figure See Section mt = 100 g PB = 65 % PlB = 0 % PsB = 80 % From Equations 2.4 and 2.5, ml + ms = ml+ 80 ms = 65 x 100 Solving the two equations simultaneously, we get: m l = mass of the alloy which is in the liquid phase = 0 g m s = mass of the alloy which is in the solid phase = 70 g 2.0. mt = 100 g PB = 45 % PlB = 17 % PsB = 65 % From Equations 2.4 and 2.5, ml + ms = ml+ 65 ms = 45 x 100 Solving the two equations simultaneously, we get: m l = mass of the alloy which is in the liquid phase = g m s = mass of the alloy which is in the solid phase = 58.9 g

7 2.1. mt = 100 g PB = 60 % PlB = 25 % PsB = 70 % From Equations 2.4 and 2.5, ml + ms = ml+ 70 ms = 60 x 100 Solving the two equations simultaneously, we get: m l = mass of the alloy which is in the liquid phase = g m s = mass of the alloy which is in the solid phase = g 2.2. mt = 100 g PB = 40 % PlB = 20 % PsB = 50 % From Equations 2.4 and 2.5, ml + ms = ml+ 50 ms = 40 x 100 Solving the two equations simultaneously, we get: m l = mass of the alloy which is in the liquid phase =. g m s = mass of the alloy which is in the solid phase = g 2.. a. Spreading salt reduces the melting temperature of ice. For example, at a salt composition of 5%, ice starts to melt at -21 o C. When temperature increases more ice will melt. At a temperature of -5 o C, all ice will melt. b. -21 o C c. -21 o C 2.4. See Section See Section See Section 2.4. Materials for Civil and Construction Engineers 4th Edition Mamlouk SOLUTIONS MANUAL Full clear download (no formatting errors) at: materials for civil and construction engineers rd edition pdf free download materials for civil and construction engineers 4th edition pdf

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