All solids are elastic3 and re ta in th eir volum e and shape. Solids are characterized by the long-range order in the arrangem ent of th e particles of which they are composed. These p articles are atom s, m olecules, or ions.
The c ry sta llin e stru c tu re of a solid is a re su lt of th e or dered arrangem ent of its particles. The physical properties of a cry stal depend on th e o rie n tatio n of th e sym m etry axis in th e cry stal (anisotropy).
U nder the action of e x te rn al forces, solids are deformed. The deform ations disappearing after the forces stop acting on
§ 7. Properties of Solids. D eform ations 69 a body are known as elastic. There are a large num ber of different deform ations, b u t extension (compression) and shear are im p o rtan t.
E xtension is characterized by an absolute deform ation A I: M = I — l0
and a rela tiv e deform ation (strain)
M echanical stress a is defined as the ra tio of th e in te rn al force em erging in a body as a resu lt of deform ation to the cross-sectional area of th e body: a — F/S. The u n it of m echanical stress is the pascal (Pa).
H ooke’s law establishes a re la tio n between elastic deform a tions and the in te rn al forces. The m echanical stress a is d irectly p roportional to the s tra in e:
7 17 AZ
o = k e , or o = E - r—,
where E is Y oung’s m odulus. Y oung’s m odulus is m easured in the same u n its as stress, viz. in pascals.
The elastic lim it is th e m axim um stress em erging in a m ate ria l for w hich H ooke’s law rem ains in force.
The safety factor is defined as th e ra tio of th e m axim um (u ltim ate) stress a u of a co n stru ctio n to th e adm issible stress a ad:
For an elastic deform ation, th e p o ten tial energy E p of a body is equal to th e work done during the deform ation (extension or compression) of a body:
£ P = - i r - = - l r ( A 0 2-
W hen a solid changes its s ta te of aggregation (when i t m elts), th e sep aratio n between th e particles in the cry stal la ttic e increases, and the la ttic e is destroyed. The p o ten tial energy of th e in te rac tio n between m olecules (particles) increases.
To m elt 1 kg of a solid a t its m elting point, a certain am ount of h eat X is requ ired , which is known as the specific
70 Ch. I. Fundam entals of Molecular P h ysics la te n t heat of fusion:
k = Qlm.
The specific la te n t heat of fusion is m easured in joules per kilogram (J/kg).
In order to m elt a c ry stallin e substance, heat m ust be spent to heat it to its m elting point and to convert it in to a liq u id :
Q = cm ( T m — T) -f- km. W orked Problems
Problem 36. D eterm ine the elongation of a copper rod h av ing a len gth of 6 m and a cross-sectional area of 0.4 cm2
under th e action of a force of 2 kN.
Given: I = 6 m is the length of the rod, S = 0.4 x 10” 4 m2 is its cross-sectional area, and F = 2 X 103 N is th e applied force. From tab les, we tak e Y oung’s m odulus for copper E = 130 x 109 Pa.
Find: the elongation A/ of the rod.
Solution. We sh all solve th e problem using H ooke’s law - y - = a , or - y - = , whence
a i FI XJ 2 x 103 N X 6 m 9 Q v i a .
E S ’ 130 X 109 Pa X 0 .4 X 10”4 m 2 “ X 0 m * Answer. The copper rod elongates by ap proxim ately 0.23 cm.
Problem 37. A chandelier of mass 250 kg is suspended on an alum inium rod w ith an u ltim a te stress of 0.11 G Pa. W h at m ust the cross-sectional area of the rod be for the safety factor to be 4? W h a t is the stra in in the rod?
Given: m = 250 kg is the mass of th e chandelier, n = 4 is the safety factor, and cru = 1.1 X 108 Pa is the u ltim a te stress. From tables, we tak e th e free fall acceleration g = 9.81 m /s2 and Y oung’s m odulus for alu m inium , E — 7 x
1 0 10 Pa.
Find: th e cross-sectional area S of the rod and the s tra in e. Solution. The deform ation of the rod is caused by the force of g ra v ity G = mg. The cross-sectional area of the rod is chosen depending on th e m echanical stress a em erging in the rod: a = G/S = m g I S , whence
§ 7. Properties of Solids. D eform ations 71 G iven th e safety factor n and the u ltim a te stress a u, we can determ ine the adm issible stress: n = a u/cr, a = o j n . F inally,
S = mg <*u ” ,
o _ 250 kg X 9.81 m /s 2 x 4 ^ 1.1 X 1 0 ® Pa — The strain in the rod is
a u _ 1 .1 X 10® Pa
8.9 X 10" 6 m2.
n E e== 4 x 7 X 1010 Pa
10
~ '\Answer. The cross-sectional area of the rod is 0.89 cm2
and the stra in is about 4 X 10~4.
Problem 38. How m uch h eat is required to convert 0.8 kg of ice a t —10°C into steam at
100°C? P lo t the graph of t = f (Q). Given: m = 0.8 kg is the mass of the ice, t x = —10°C is the in itia l tem p eratu re of the ice and ^2 = 1 0 0°C is th e tem p erature of the steam . From tab le s, we tak e the m eltin g point for ice, t0 = 0°C, the boiling point for w ater, t h = 100°C, the specific heat for ice, = 2090 J/(kg *K ), the specific h eat for w ater, cw = 4187 J /(k g -K ), the specific la te n t h eat of fusion for ice, A,=3.35 x 105 J/k g , and th e spe
cific la te n t h eat of v ap o rization of w ater, r = 2.26 x
106 J/k g .
Find: th e am ount of h eat Q required to convert the ice in to steam .
Solution. The required am ount of heat Q is determ ined by sum m ing the am ounts of heat Q = Qx + @2 + (?3 + (?4
(Fig. 13), where Qx is the h eat required to h eat th e ice to the m elting point,
Q, = (t0 — f,),
Qi = 2090 J/(kg- K) x 0.8 kg x 10 K = 16720 J - 16.72 k J ,
72 Ch. I. Fundam entals of M olecular P h ysics Q2 is th e am ount of h eat required to m elt the ice,
Q2 = km, Q2 = 3.35 x 105 J /k g x 0.8 kg - 2.6 x 105 J = 260 k J ,
Qs is th e am ount of heat required to h eat th e w ater obtained from th e ice to the boiling point,
Q3 = CwJ7l (t2 £q),
Q3 = 4187 J/(kg* K) x 0.8 kg x 100 K = 334960 J ~ 335 k J ,
and Q4 is th e am ount of h eat required to evaporate th e w ater Q4 = r m , Qt = 2.26 x 106 J /k g x 0.8 kg = 1.808
x 106 J ~ 1810 k J. The to ta l am ount of h eat is
Q = 16.72 k J + 260 kJ + 335 kJ + 1810 kJ ~ 2420 k J .
Answer. In order to convert 0.8 kg of ice into steam , an energy of 2420 kJ = 2.42 MJ is required.
Problem 39. A certain am ount of steam a t 100°C is in tro