原子結構與鍵結
一、(a)什麼是同位素？
(b)為何元素的原子量不是整數？引證兩個理由
二、引證原子質量和原子量間的差異
三、給出下列離子之電子組態:
Fe2  , Fe3 , Cu  , Ba 2  , Br  , S 2 
四、溴化銫(CsBr)主要是離子鍵結，那 Cs  和 Br  離子所具有的電子結構與那兩
個惰性氣體相同？
五、關於電子組態，為何在週期表中 VIIA 族中的所有元素都具有共同點？
六、決定下列之電子組態是惰性氣體、鹵素、鹼金屬、鹼土金屬或過渡金屬，證
明你的選擇是正確的。
(a)1s22s22p63s23p63d24s2
(b)1s22s22p63s23p6
(c)1s22s22p5
(d)1s22s22p63s2
(e)1s22s22p63s23p63d24s2
(f)1s22s22p63s23p64s1
七、(a)在週期表中對稀土系列的元素而言，什麼電子副層是填滿的？
(b)對錒系而言什麼電子副層是填滿的？
結晶結構
一、證實 BBC 的原子堆積因子是 0.68
二、計算鈀原子的半徑，已知 Pd 具有 FCC 之晶體結構，密度為 12.0 g/cm3，且
原子量是 106.4 g/mol。
三、錫的單位晶胞具有正方對稱，其晶格常數 a 和 b 分別是 0.583 和 0.318nm。
若其密度、原子量和原子半徑分別是 7.30 g/cm3，1198.69g/mol 和 0.151nm，
計算原子堆積因子。
_
四、畫一正方單位晶胞，並在胞內劃出 [1 21] 方向和(210)平面。
五、示於下列立方單位晶胞中的方向，決定其指數
六、決定出下列單位晶胞中所示平面的米勒指數。
七、計算並比較 FCC 中[100]，[110]和[111]方向之線密度。
八、計算並比較 FCC 中(100)和(111)平面之平面密度。
九、金屬銥具有 FCC 晶體結構，若當使用具有波長為 0.1542nm 單光 x-輻射時，
(220)組平面之繞射角發在 69.22  (第一階反射)，計算(a)此組平面之平面間距;
和(b)銥原子之原子半徑。
十、下圖所示是具有 FCC 晶體結構的銅的 x-光繞射圖案之前四個波峰;若使用具
有波長為 0.1542nm 之單色 x-輻射。
(a)標定(即繪出 h, k, l 指數)每一波峰。
(b)決定出每一波峰之平面間距。
(c)對每一波峰而言，決定 Cu 的原子半徑
材料缺陷
一、計算鐵在 850  C 時，每立方米中空位的數目，空位的形成能是 1.08ev/atom，
此外，Fe 的密度和原子量分別是 7.65 g/cm3 和 55.85g/mol。
二、對 BCC 和 FCC 晶體結構而言，其有兩種不同型式的間隙位置，在每一情況
中，較大的間隙位置通常被填入雜質原子。對 FCC 而言，此較大的間隙位
置位於單位晶胞中每一個邊的中心;其被稱為八面體間隙的位置。另一方
面，BCC 中較大的間隙位置位於 0，
1
1
， -----亦即，位於{100}面，且處
2
4
在此面上兩單位晶胞邊線間的中點和其他兩單位晶胞邊線間距的四分之一
處;其被稱為四面體間隙位置。對 FCC 和 BCC 晶體結構而言，利用母原子
半徑尺來計算兩晶體結構中這些位置能剛好放入之雜質原子的半徑  。
三、30 wt% Zn 和 70 wt% Cu 所組成的合金，其原子百分比的成分是什麼？
四、含 99.7 lbm 銅、102 lbm 鋅和 21 lbm 鉛的合金，其原子百分比的組成成分為
何？
五、將上題中原子百分比換成重量百分比。
六、鉬和鎢形成一置換式的固溶體，計算加入多少重量百分比的鉬於鎢中，才會
產生每立方厘米含有 1.0 x 1022 Mo 原子的合金，純 Mo 和 W 的密度分別是
10.22 和 19.30g/cm3。
七、下列的每一個堆疊順序發現於 FCC 金屬中，舉例說明其存在的平面缺陷型
式為
(a)……ABCABCBACBA……
(b)……ABCABCBCABC…….在，複製堆疊順序且使用垂直虛線指出平面缺
陷的位置。
八、(a)對一 4 號 ASTM 晶粒大小而言，在放大倍率為 100 的照片上，每立平方
英吋將會有多少晶粒？
(b)估算下圖中顯微照片的晶粒號碼，假設放大 100 倍。
擴散理論
一、1.5mm 厚的鋼片在 1200  C 時兩邊都有氮氣氛圍，且已達到穩態擴散條件
。氮在鋼中於此溫度的擴散係數是 6 x 10-11m2/s，擴散通量是
1.2 x 10-7kg/m2-s。同時，亦知道在鋼片高壓表面的氮濃度是 4kg/m3。距高壓
邊多少距離處的濃度將會是 2.0kg/m3？假設為線性濃度分佈。
二、一鐵碳合金起初含有 0.20 wt% C，決定出在距表面 2mm 處達到 0.45 wt% 碳
濃度所需的時間。其表面濃度維持在 1.30 wt% C，且在 1000  C 下處理，使
用下表中  -Fe 之擴散數據。
三、對一鐵合金而言，已定出在滲碳熱處理 10h 期間將使距表面 2.5mm 處之碳
濃度升到 0.45 wt%，估算同一鋼種在相同滲碳溫度下距表面 5.0 mm 處達到
相同濃度所需的時間。
四、下圖所示的為擴散係數的對數值(以 10 為底)對絕對溫度的數來作圖，對鐵
在鉻中的擴散決定出活化能和指數前之係數。
五、(a)計算於 500  C 時銅在鋁中的擴散係數。
(b)在 600  C 時需要多久所產生的擴散結果(利用在某一特定點的濃度相同)
與 500  C 時 10h 相同？
六、一起初含有 0.20 wt% C 的 FCC 鐵碳合金於高溫下進行滲碳，滲碳氣氛使表
面碳濃度維持在 1.0 wt%。若在 49.5h 後於表面 4.0mm 處之碳濃度是 0.35
wt%，決定處理時之溫度。
材料之機械性質
一、一具有 107 GPa(15.5x106psi)的彈性模數和原有面積為 13.8mm(0.15in)的鈦合
金圓柱型試片，當所施加之拉伸負荷為 2000N(450 lbf)時，僅產生彈性變形，
若最大的允許伸長量是 0.42mm(0.0165in)，計算變形前試片的最大長度。
二、一鋼棒長 100mm(4.0 in)且具有正方形之橫截面積，正方形邊長為 20mm
(0.8 in)，此棒被 89000N(20000 lbf)之荷重拉伸，且伸長量為 0.1mm
(4.0x10-2 in)。假設變形是完全彈性的，計算鋼的彈性模數。
三、對青銅合金而言，塑性變形開始的應力是 275 MPa(40000 psi)，彈性模數為
115 GPa (16.7x106psi)
(a)若試片的橫截面積為 325mm2(0.5 in.2)，無塑性變形下，此試片能承受的
最大荷重為何？
(b)若試片的原長度是 115 mm(4.5 in)，不產生塑性變形情況下其可伸長之最
大長度為多少？
四、一圓柱棒長 380 mm(1.50 in)，直徑 10 mm(0.4 in)，承受拉荷重當承受
24500N(5500 pbf)的負荷時，若此棒不產生塑性變形和伸長量不大於
0.9 mm(0.035 in)，列於下面的四種金屬或合金的那一種是可能的候選者？
證明你的選擇。
材料
彈性模數(psi)
降伏強度(psi)
拉伸強度(psi)
鋁合金
黃銅合金
6
10x10
14.6x106
37000
50000
61000
61000
銅
鋼
16x106
30x106
30000
65000
40000
80000
五、一鋁合金圓柱試片的直徑為 0.505 in(12.8 mm)，標距長度為 2.00 in(50.800mm)
承受拉應力，使用下表之荷重—伸長特性，完成問題(a)到(f)。
(a)將數據劃成工程應力對工程應變圖。
(b)計算彈性模數。
(c)決定 0.002 偏位應變的降伏強度。
(d)決定此合金的拉伸強度。
(e)大概的延性是多少？以百分比伸長率來表示？
(f)計算彈性能模數。
荷
lbr
2850
5710
8560
11400
重
N
12700
25400
38100
50800
長
in.
2.001
2.002
2.003
2.004
度
mm
50.825
50.851
50.876
50.902
17100
76200
2.006
50.952
20000
20800
23000
24200
26800
28800
33650
89100
92700
102500
107800
119400
128300
149700
重
2.008
2.010
2.015
2.020
2.030
2.040
2.080
51.003
51.054
51.181
51.308
51.562
51.816
52.832
lbr
35750
N
159500
in.
2.120
mm
53.848
36000
35850
34050
160400
159500
151500
2.140
2.160
2.200
54.356
54.864
55.880
荷
長
度
28000
124700
2.230
56.642
破
裂
六、一圓柱型金屬試片具有原有直徑為 12.8mm(0.505 in)、標距長度為 50.80
mm(2.000 in)、承受拉力直到斷裂為止、破裂點的直徑是 8.13mm(0.320 in)、
標距長度是 74.17 mm(2.920 in)。利用百分斷面收縮率和百分伸長率來計算
延性。
七、(a)直徑為 10 mm 的勃氏硬度壓痕器，當使用 500kg 荷重時，在鋼合金上產
生 1.62 mm 的壓痕，計算此材料之 HB。
(b)當使用 500kg 荷重時，產生 450HB 的硬度，其壓痕之直徑為多少？
相圖
一、對成分為 74 wt% Zn – 26 wt% Cu 的合金，列出在下列溫度時出現的相和它
們的成分 : 850  C ， 750  C ， 680  C ， 600  C 和 500  C 。
二、成分 70 wt% 鎳 - 30 wt% 銅的銅-鎳合金由 1300  C ( 2370  F )的溫度慢慢加
熱。
(a) 在什麼溫度時液相首先形成？
(b) 此液相的成分為何？
(c) 在何溫度時此合金產生完全熔融？
(d) 在完全熔融前最後剩下的固體成分為何？
三、90 wt% 銀-10 wt% 銅的合金被加熱到β+液相範圍內的溫度，如果液相成分
是 85 wt% 銀，決定(a)合金溫度？(b)β相的成分;以及(c)兩相的質量分率。
四、成分 25 wt%銀-75 wt% 銅的銅-銀合金在 775  C ( 1425  F )回答下列問題
(a)決定α和β相的質量分率
(b)決定初晶α和共晶顯微組成物的質量分率。
(c)決定共晶α的質量分率。
五、鉛-錫合金在 180  C ( 355  F )的顯微結構包含初晶β和共晶結構。如果此兩個
顯微組成物的質量分率分別為 0.57 和 0.43，決定合金的成分。
六、對 85 wt% 鉛-15 wt% 鎂的合金而言，繪出在很慢冷卻速率冷到下列溫度所
觀察的顯微結構的示意圖: 600  C ( 1110  F )， 500  C ( 930  F )，
270  C ( 520  F )和 200  C ( 390  F ) 標示所有相並指出它們大約的成分。
七、下圖係鈦 – 銅相圖，圖中只有單相區被標示。定出共晶、共析、包晶以及
共軛相變態的溫度 – 成分點。也寫出在冷卻時的反應式。
八、(a)亞共析鋼和過共析鋼之間的差別是什麼？
(b)在亞共析鋼內，有共析和初析肥粒鐵存在。解釋它們之間的差異。在它
們二者中的碳濃度各是多少？
九、計算含 C 量 0.25 wt% 的鐵 – 碳合金其初析肥粒鐵和波來鐵的質量分率。
十、計算含碳量 0.43 wt% 的鐵 – 碳合金其共析肥粒鐵的質量分率。
相變態與熱處理
一、以熱處理和顯微組織發展的觀點，鐵 – 鐵碳化物相圖的兩個主要限制是什
麼？
二、(a)簡單描述過熱和過冷現象。
(b)為什麼它們會存在？
三、使用共析成分之鐵 – 碳合金的恆溫變態圖(下圖)，說明一小試片受下列時
間 – 溫度處理時最後顯微結構之本質(以出現之顯微組成物和每種大約百
分比表示)。在每一種情況下假設試片開始在 760  C ( 1400  F )並在此溫度保
持足夠長時間以達到完全且均質的沃斯田鐵結構。
(a)快速冷卻到 700  C ( 1290  F )，保持 104 秒，然後淬冷到室溫。
(b)將(a)部份之試片重新加熱至 700  C ( 1290  F )保持 20 小時。
(c)快速冷至 600  C ( 1110  F )，保持 4 秒，再快速冷至 450  C ( 840  F )，保持
10 秒，然後淬冷至室溫。
(d)快速冷至 400  C ( 750  F )保持 2 秒，然後淬冷至室溫。
(e)快速冷至 400  C ( 750  F )保持 20 秒，然後淬冷至室溫。
(f)快冷至 400  C ( 750  F )，保持 200 秒，然後淬冷至室溫。
(g)快冷至 575  C ( 1065  F )保持 20 秒，再快冷至 350  C ( 660  F )，保持 100
秒，然後淬冷至室溫。
(h)快速冷至 250  C ( 480  F )保持 100 秒，然後在水中淬至室溫。重新加熱到
315  C ( 600  F )1 小時然後慢冷至室溫。
四、使用 0.45 wt% 碳鋼合金之恆溫變態圖(下圖)，決定一小試片受下述時間 –
溫度處理的最後顯微結構(以出現的顯微組成物表示)。在每一種情況假設試
片開始在 845  C ( 1550  F )，而它在溫度保持足夠久使其達到完全和均質的沃
斯田鐵結構。
(a)快速冷至 250  C ( 480  F )，保持 103 秒，然後淬冷至室溫。
(b)快速冷至 700  C ( 1290  F )，保持 30 秒，然後淬冷至室溫。
(c)快速冷至 400  C ( 750  F )，保持 500 秒，然後淬冷至室溫。
(d)快速冷至 700  C ( 1290  F )，保持在此溫度 105 秒，然後淬冷至室溫。
(e)快速冷至 650  C ( 1200  F )，在此溫度保持 3 秒，快速冷至 400  C ( 750  F )，
保持 10 秒，然後淬冷至室溫。
(f)快速冷至 450  C ( 840  F )，保持 10 秒，然後淬冷至室溫。
(g)快速冷至 625  C ( 1155  F )，保持 1 秒，然後淬冷至室溫。
(h)快速冷至 625  C ( 1155  F )，然後在此溫度保持 10 秒，再快速冷至
400  C ( 750  F )，在此溫度保持 5 秒，然後淬冷至室溫。
五、複印 0.45 wt% 碳之鐵 – 碳合金的恆溫變態圖(上圖)，然後在圖上描繪並標明
產生下列顯微結構的時間 – 溫度路徑 :
(a)42% 初析肥粒鐵和 58%粗波來鐵
(b)50% 細波來鐵和 50% 變韌鐵
(c)100% 麻田散鐵
(d)50% 麻田散鐵和 50% 沃斯田鐵
六、將下列鐵 – 碳合金以及顯微結構由硬至軟排序之:
(a)含 c 量 0.25 wt% 之球化鐵 ;
(b)含 c 量 0.25 wt% 之粗波來鐵 ;
(c)含 c 量 0.6 wt% 之細波來鐵 ;
(d)含 c 量 0.6 wt% 之粗波來鐵。
驗證這些排序。
七、請指出下列鐵 – 碳合金在完全退火期間與每種合金沃斯田鐵化的可能溫度範
圍 : (a)0.25 wt% C : (b)0.45 wt% C (c)0.85 wt% C (d)1.10 wt% C。
八、球化熱處理的目的為何？那類合金常用這種處理。
九、簡單解釋硬度和硬化能的差異。
十、合金元素(除碳外)的出現對硬化能曲線的形狀有何影響？簡單解釋這些影響。
十一、自然和人工時效過程主要差異為何？
陶瓷材料
一、對於陶瓷化合物，什麼是成份離子的兩種用來決定晶體結構的特性？
二、証明配位數 4 時，陽離子和陰離子最小半徑比值是 0.225。
三、証明配為數為 6 時，陽離子和陰離子最小半徑比為 0.414。提示:用 NaCl 結晶
構造，並假設陰離子和陽離子沿立方體邊和橫過面的對角線時正好相接觸。
四、証實配位數為 8 時，陽離子和陰離子最小半徑比值等於 0.732。
五、引述一個理由為什麼陶瓷材料通常較金屬硬但較金屬脆。
六、引述玻璃的兩種優良性質。
七、鈉和石灰以鈉灰(Na2CO3)及石灰石(CaCO3)的形式加入一批玻璃內，加熱期
間，這兩成份會分解釋出二氧化碳(CO2)，最後產物是鈉和石灰，試計算鈉灰
和石灰石必須加入到 125 lbm 的石英(SiO2) 中產生一玻璃成分為 75wt%
SiO2，15 wt% Na2O 和 10 wt% CaO 的 重量。
八、(a)何謂去玻璃質化？
(b)引述兩種會改善去玻璃質化的性質及兩種會損害的。
九、簡單解釋為什麼玻璃陶瓷會產生不透明。
十、引述兩種黏土礦物在製程時的優良性質。
高分子材料
一、同質多晶現象與同質異構現象之差別。
二、對下列高分子描繪體結構 : (a)聚氟乙烯(b)聚三氟氯乙烯(c)聚氯乙烯。
三、對下列計算其體的分子量(a)聚氯乙烯(b)聚對苯二甲酸乙二酯(c)聚碳酸酯(d)
聚二甲基矽氧烷。
四、兩種尼龍 6，6 材料的密度和對應的結晶度如下:
ρ(g/cm2)
結晶度(%)
1.188
1.152
67.3
43.7
(a)計算尼龍 6，6 全部結晶及全部非晶質時之密度。
(b)決定試樣具 55.4% 結晶度時之密度。
五、簡單解釋下列如何影響半結晶高分子抗拉模數及為何 :
(a) 分子量 (b)結晶度 (c)拉伸變形 (d)未變形材料的退火 (e)拉伸材料的退火
六、下列高分子名稱中何者適合用於製造裝熱咖啡的杯子 : 聚乙烯、聚丙烯、聚
氯乙烯、PET 聚酯和聚碳酸酯。為什麼？
七、如下表所列高分子中，何者最適合當作製冰盒？為什麼？
八、(a)根據加熱時的機械特性 ; 和
(b) 依據可能的分子結構，比較熱塑性和熱固性高分子。
九、某些聚酯可能是熱塑性或熱固性，為此建議一種理由。
複合材料
一、引述介於大顆粒及分散強化顆粒增強複合物之間其強化機構間之一般性差異。
二、引述析出硬化和分散強化間的一個相似處和兩個相異處。
三、估計一水泥物中含 85 vol% TiC 顆粒於鎳基材中之最大和最小導熱值。假設
TiC 和 Ni 之熱傳導分別為 27 W/m-K 和 69 W/m-K。
四、(a)在一複合材料中什麼是基材和分散相間之差異？
(b)纖維增強複合物，對照其基材和分散相之機械性質。
五、想要生產一具縱向拉伸強度為 750MPa(109000 psi)之排列的碳纖維一環氧基複
合物。試計算纖維體積分率之需求，若
(1)平均纖維直徑和長度分別為 1.2 x10-2 mm(4.7 x 10-4 in) 和 1 mm (0.04 in)
(2)纖維破斷強度為 5000 MPa(725000 psi)
(3)纖維基材鍵強為 25 MPa (3625 psi)
(4)在破斷時基材應力為 10 MPa (1450 psi)
六、簡述拉出成形、絲織和預浸生產製造製程 ; 說明每一種之優缺點。
解答
原子結構與鍵結
一、<Sol>
(a) When two or more atoms of an element have different atomic masses, each is termed an
isotope.
(b) The atomic weights of the elements ordinarily are not integers because: (1) the
12
atomic masses of the atoms generally are not integers (except for C), and (2) the atomic
weight is taken as the weighted average of the atomic masses of an atom's naturally
occurring isotopes.
二、<Sol>
Atomic mass is the mass of an individual atom, whereas atomic weight is the
average (weighted) of the atomic masses of an atom's naturally occurring
isotopes.
三、<Sol>
The electron configurations of the ions are determined using Table 2.2.
2+
2 2 6 2 6 6
Fe - 1s 2s 2p 3s 3p 3d
3+
2 2 6 2 6 5
Fe - 1s 2s 2p 3s 3p 3d
+
2 2 6 2 6 10
Cu - 1s 2s 2p 3s 3p 3d
2+
2 2 6 2 6 10 2 6 10 2 6
Ba - 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p
2 2 6 2 6 10 2 6
Br - 1s 2s 2p 3s 3p 3d 4s 4p
22 2 6 2 6
S - 1s 2s 2p 3s 3p
四、<Sol>
+
The Cs ion is just a cesium atom that has lost one electron; therefore, it has an electron
configuration the same as xenon (Figure 2.6).
The Br ion is a bromine atom that has acquired one extra electron; therefore, it
has an electron configuration the same as krypton.
五、<Sol>
Each of the elements in Group VIIA has five p electrons.
六、<Sol>
2 2 6 2 6 7 2
(a) The 1s 2s 2p 3s 3p 3d 4s electron configuration is that of a transition metal because
of an incomplete d subshell.
2 2 6 2 6
(b) The 1s 2s 2p 3s 3p electron configuration is that of an inert gas because of filled
3s and 3p subshells.
2 2 5
(c) The 1s 2s 2p electron configuration is that of a halogen because it is one electron
deficient from having a filled L shell.
2 2 6 2
(d) The 1s 2s 2p 3s electron configuration is that of an alkaline earth metal because
of two s electrons.
2 2 6 2 6 2 2
(e) The 1s 2s 2p 3s 3p 3d 4s electron configuration is that of a transition metal
because of an incomplete d subshell.
2 2 6 2 6 1
(f) The 1s 2s 2p 3s 3p 4s electron configuration is that of an alkali metal because of
a single s electron.
七、<Sol>
(a) The 4f subshell is being filled for the rare earth series of elements.
(b) The 5f subshell is being filled for the actinide series of elements.
結晶結構
一、<Sol>
We are asked to show that the atomic packing factor for BCC is 0.68. The atomic packing factor
is defined as the ratio of sphere volume to the total unit cell volume, or
V
S
APF = V
C
Since there are two spheres associated with each unit cell for BCC
4R
V = 2(sphere volume) = 2 3
S

3
3
 = 8R
3

3
Also, the unit cell has cubic symmetry, that is V = a . But a depends on R according to
C
Equation (3.3), and
V =
C
4R
 3
3
=
3
64R
3 3
Thus,
APF =
3
8R /3
= 0.68
3
64R /3 3
二、<Sol>
We are asked to determine the radius of an iridium atom, given that Ir has an FCC crystal
structure. For FCC, n = 4 atoms/unit cell, and V = 16R
C
3
2 [Equation (3.4)]. Now,
nA
Ir
= V N
C A
And solving for R from the above two expressions yields
 nAIr 
R= 

16NA 2
1/3
(4 atoms/unit cell)(192.2 g/mol)

= 

( 2)(16)(22.4 g/cm3)(6.023 x 1023 atoms/mol)
= 1.36 x 10
-8
1/3
cm = 0.136 nm
三、<Sol>
In order to determine the APF for Sn, we need to compute both the unit cell volume (V ) which
C
2
is just the a c product, as well as the total sphere volume (V ) which is just the product of
S
the volume of a single sphere and the number of spheres in the unit cell (n). The value of
n may be calculated from Equation (3.5) as
V N
C A
n= A
Sn
=
2
-24
23
(7.30)(5.83) (3.18)(x 10 )(6.023 x 10 )
118.69
= 4.00 atoms/unit cell
Therefore
4 3
(4)3R 
V


S
APF = V
=
2
C
(a) (c)
4
3
(4)3()(0.151) 


2
(0.583) (0.318)
= 0.534
四、<Sol>
_
(a) This portion of the problem calls for us to draw a [121 ] direction within an orthorhombic
unit cell (a ≠ b ≠ c,  =  =  = 90). Such a unit cell with its origin positioned at point O is
shown below. We first move along the +x-axis a units (from point O to point A), then
parallel to the +y-axis 2b units (from point A to point B). Finally, we proceed parallel to
_
the z-axis -c units (from point B to point C). The [121 ] direction is the vector from the
origin (point O) to point C as shown.
z
a
90
c
O
90
90
A
x
y
B
b
C
(b) We are now asked to draw a (210) plane within an orthorhombic unit cell. First
remove the three indices from the parentheses, and take their reciprocals--i.e., 1/2, 1, and
. This means that the plane intercepts the x-axis at a/2, the y-axis at b, and parallels the
z-axis.
The plane that satisfies these requirements has been drawn within the
orthorhombic unit cell below.
z
a
c
y
x
b
五、<Sol>
__
Direction A is a [011 ] direction, which determination is summarized as follows. We first of all
position the origin of the coordinate system at the tail of the direction vector; then in terms
of this new coordinate system
x
y
z
Projections
0a
-b
-c
0
-1
-1
Projections in terms of a, b,
and c
Reduction to integers
not necessary
__
[011 ]
Enclosure
_
Direction B is a [2 10] direction, which determination is summarized as follows.
We first of all position the origin of the coordinate system at the tail of the direction vector;
then in terms of this new coordinate system
x
y
z
-a
b
2
0c
and c
-1
1
2
0
Reduction to integers
-2
1
0
Projections
Projections in terms of a, b,
_
[2 10]
Enclosure
Direction C is a [112] direction, which determination is summarized as follows.
We first of all position the origin of the coordinate system at the tail of the direction vector;
then in terms of this new coordinate system
x
y
z
a
2
b
2
c
and c
1
2
1
2
1
Reduction to integers
1
1
2
Projections
Projections in terms of a, b,
Enclosure
[112]
_
Direction D is a [112 ] direction, which determination is summarized as follows.
We first of all position the origin of the coordinate system at the tail of the direction vector;
then in terms of this new coordinate system
Projections
Projections in terms of a, b,
x
y
z
a
2
b
2
-c
and c
1
2
1
2
-1
Reduction to integers
1
1
-2
_
[112 ]
Enclosure
六、<Sol>
For plane A since the plane passes through the origin of the coordinate system as shown, we
will move the origin of the coordinate system one unit cell distance to the right along the y
_
axis; thus, this is a (32 4) plane, as summarized below.
x
y
z
2a
3
-b
c
2
2
3
-1
1
2
Reciprocals of intercepts
3
2
-1
2
Reduction
3
-2
4
Intercepts
Intercepts in terms of a, b,
and c
_
(32 4)
Enclosure
For plane B we will leave the origin at the unit cell as shown; this is a (221) plane,
as summarized below.
Intercepts
x
y
z
a
2
b
2
c
1
2
1
2
1
2
2
1
Intercepts in terms of a, b,
and c
Reciprocals of intercepts
Enclosure
(221)
七、<Sol>
For FCC the linear density of the [100] direction is computed as follows:
The linear density, LD, is defined by the ratio
L
c
LD = L
l
where L is the line length within the unit cell along the [100] direction, and L is line length
l
c
passing through intersection circles. Now, L is just the unit cell edge length, a which, for
l
FCC is related to the atomic radius R according to a = 2R 2 [Equation (3.1)]. Also for
this situation, L = 2R and therefore
c
LD =
2R
= 0.71
2R 2
For the [110] direction, L = L = 4R and therefore,
l
c
4R
LD = 4R = 1.0
For the [111] direction L = 2R, whereas L = 2R 6 , therefore
c
l
LD =
2R
= 0.41
2R 6
八、<Sol>
Planar density, PD, is defined as
A
c
PD = A
p
where A is the total plane area within the unit cell and A is the circle plane area within
p
c
this same plane. For the (100) plane in FCC, in terms of the atomic radius, R, and the unit
cell edge length a
2
2
2
A = a = (2R 2) = 8R
p
Also, upon examination of that portion of the (100) plane within a single unit cell, it may be
noted that there reside 2 equivalent atoms--one from the center atom, and one-fourth of
each of the four corner atoms. Therefore,
A = (2)R
c
2
Hence
PD =
2
2R
2 = 0.79
8R
That portion of a (111) plane that passes through a FCC unit cell forms a triangle
as shown below.
2R 3
R
4R
In terms of the atomic radius R, the length of the triangle base is 4R, whereas the height is
2R 3 . Therefore, the area of this triangle, which is just A is
p
1
2
A = 2(4R)(2R 3) = 4R 3
p
Now it becomes necessary to determine the number of equivalent atoms residing within
this plane. One-sixth of each corner atom and one-half of each middle atom belong
belongs to the unit cell. Therefore, since there are 3 corner and 3 middle atoms, there is
an equivalent of 2 atoms within the unit cell. Hence
2
A = 2(R )
c
PD =
and
2
2R
= 0.91
2
4R 3
九、<Sol>
(a) From the data given in the problem, and realizing that 69.22 = 2, the interplanar spacing
for the (220) set of planes may be computed using Equation (3.9) as
n
(1)(0.1542 nm)
d
=
=
= 0.1357 nm
220 2 sin 
69.22
(2)sin
2 

(b) In order to compute the atomic radius we must first determine the lattice parameter, a
using Equation (3.10), and then R from Equation (3.1) since Ir has a FCC crystal structure.
Therefore,
a=d
220
2
2
2
(2) + (2) + (0) = (0.1357 nm)( 8 ) = 0.3838 nm
And
R=
a
2 2
=
0.3838 nm
= 0.1357 nm
2 2
十、<Sol>
A material in which atomic bonding is predominantly ionic in nature is less likely to form a
noncrystalline solid upon solidification than a covalent material because covalent bonds
are directional whereas ionic bonds are nondirectional; it is more difficult for the atoms in
a covalent material to assume positions giving rise to an ordered structure.
材料缺陷
一、<Sol>
Determination of the number of vacancies per cubic meter in iron at 850C (1123 K) requires
the utilization of Equations (4.1) and (4.2) as follows:
N 
 QV
 QV
A Fe
N = N exp - kT  = A
exp - kT 
V




Fe
=
23
3
(6.023 x 10 atoms/mol)(7.65 g/cm )
exp 55.85 g/mol
1.08 eV/atom


 (8.62 x 10-5 eV/atom-K)(1123 K)
18
-3
24 -3
= 1.18 x 10 cm = 1.18 x 10 m
二、<Sol>
In the drawing below is shown the atoms on the (100) face of a FCC unit cell; the interstitial
site is at the center of the edge.
R
2r
R
a
The diameter of an atom that will just fit into this site (2r) is just the difference between that
unit cell edge length (a) and the radii of the two host atoms that are located on either side
of the site (R); that is
2r = a - 2R
However, for FCC a is related to R according to Equation (3.1) as a = 2R 2 ; therefore,
solving for r gives
r=
a - 2R
2R 2 - 2R
=
= 0.41R
2
2
A (100) face of a BCC unit cell is shown below.
R
R
+r
a/4
a/2
The interstitial atom that just fits into this interstitial site is shown by the small circle. It is
situated in the plane of this (100) face, midway between the two vertical unit cell edges,
and one quarter of the distance between the bottom and top cell edges. From the right
triangle that is defined by the three arrows we may write
a
2
2
However, from Equation (3.3), a =
 4R 
2 3
a 2
2
+ 4 = (R + r)
 
4R
, and, therefore, the above equation takes the form
3
2
+
 4R 
4 3
2
2
2
= R + 2Rr + r
After rearrangement the following quadratic equation results:
2
2
r + 2Rr - 0.667R = 0
And upon solving for r, r = 0.291R.
Thus, for a host atom of radius R, the size of an interstitial site for FCC is
approximately 1.4 times that for BCC.
三、<Sol>
In order to compute composition, in atom percent, of a 30 wt% Zn-70 wt% Cu alloy, we employ
Equation (4.6) as
C A
Zn Cu
C ' = C A
x 100
+C A
Zn
Zn Cu
Cu Zn
(30)(63.55 g/mol)
= (30)(63.55 g/mol) + (70)(65.39 g/mol) x 100
= 29.4 at%
C A
Cu Zn
C ' = C A
x 100
+C A
Cu
Zn Cu
Cu Zn
(70)(65.39 g/mol)
= (30)(63.55 g/mol) + (70)(65.39 g/mol) x 100
= 70.6 at%
四、<Sol>
In this problem we are asked to determine the concentrations, in atom percent, of the
Cu-Zn-Pb alloy. It is first necessary to convert the amounts of Cu, Zn, and Pb into grams.
m ' = (99.7 lb )(453.6 g/lb ) = 45224 g
Cu
m
m
m ' = (102 lb )(453.6 g/lb ) = 46267 g
Zn
m
m
m ' = (2.1 lb )(453.6 g/lb ) = 953 g
Pb
m
m
These masses must next be converted into moles, as
m '
Cu
45224 g
n
= A
= 63.55 g/mol = 711.6 mol
mCu
Cu
46267 g
n
=
= 707.6 mol
mZn 65.39 g/mol
953 g
n
=
= 4.6 mol
mPb 207.2 g/mol
Now, employment of a modified form of Equation (4.5)
n
mCu
C ' = n
x 100
+
n
+n
Cu
mCu
mZn
mPb
711.6 mol
= 711.6 mol + 707.6 mol + 4.6 mol x 100 = 50.0 at%
707.6 mol
C ' = 711.6 mol + 707.6 mol + 4.6 mol x 100 = 49.7 at%
Zn
4.6 mol
C ' = 711.6 mol + 707.6 mol + 4.6 mol x 100 = 0.3 at%
Pb
五、<Sol>
This problem calls for a conversion of composition in atom percent to composition in weight
percent. The composition in atom percent for Problem 4.11 is 50 at% Cu, 49.7 at% Zn,
and 0.3 at% Pb. Modification of Equation (4.7) to take into account a three-component
alloy leads to the following
C
C ' A
Cu Cu
= C ' A
x 100
+C ' A +C ' A
Cu
Cu Cu
Zn Zn
Pb Pb
50(63.55 g/mol)
= 50(63.55 g/mol) + 49.7(65.39 g/mol) + 0.3(207.2 g/mol) x 100
= 49.0 wt%
C
C ' A
Zn Zn
= C ' A
x 100
+C ' A +C ' A
Zn
Cu Cu
Zn Zn
Pb Pb
49.7(65.39 g/mol)
= 50(63.55 g/mol) + 49.7(65.39 g/mol) + 0.3(207.2 g/mol) x 100
= 50.1 wt%
C
C ' A
Pb Pb
= C ' A
x 100
+C ' A +C ' A
Pb
Cu Cu
Zn Zn
Pb Pb
0.3(207.2 g/mol)
= 50(63.55 g/mol) + 49.7(65.39 g/mol) + 0.3(207.2 g/mol) x 100
1.0 wt%
六、<Sol>
This problem asks us to determine the weight percent of Mo that must be added to W such that
22
the resultant alloy will contain 10
Mo atoms per cubic centimeter. To solve this
problem, employment of Equation (4.18) is necessary, using the following values:
22
3
N =N
= 10 atoms/cm
1
Mo
 =
= 10.22 g/cm
1
Mo
 =  = 19.30 g/cm
2
W
3
3
A =A
= 95.94 g/mol
1
Mo
A = A = 183.85 g/mol
2
W
Thus
C
Mo
=
100
N 

A W
W
1+N A
Mo Mo Mo
100
=
3
3
(6.023 x 20 atoms/mole)(19.30 g/cm ) 19.30 g/cm 


22
3
(10 atoms/cm )(95.94 g/mol)
10.22 g/cm3
23
1+
= 8.91 wt%
七、<Sol>
(a) The interfacial defect that exists for this stacking sequence is a twin boundary, which
occurs at the following position
ABCABCBACBA
The stacking sequence on one side of this position is mirrored on the other side.
(b) The interfacial defect that exists within this FCC stacking sequence is a stacking
fault, which occurs over the region indicated
ABCABCBCABC
For this region, the stacking sequence is HCP.
八、<Sol>
(a) We are asked for the number of grains per square inch (N) at a magnification of 100X,
and for an ASTM grain size of 4. From Equation (4.16), n = 4, and
(n - 1)
N=2
(4 - 1)
=2
3
=2 =8
(b) This problem calls for an estimation of the grain size number (n) for the micrograph
shown in Figure 4.12b. By observation, the number of grains per square inch (N) ranges
between eight and twelve. Now, rearranging Equation (4.16) so that n becomes the
dependent variable yields
log N
n = log 2 + 1
For N = 8
log 8
n = log 2 + 1 = 4.0
Whereas, for N = 12
log 12
n = log 2 + 1 = 4.6
Thus, the ASTM grain size number will lie between 4.0 and 4.6.
擴散理論
一、<Sol>
3
We are asked to determine the position at which the nitrogen concentration is 2 kg/m . This
problem is solved by using Equation (5.3) in the form
C -C
A
B
J=-D x -x
A B
If we take C
A
3
to be the point at which the concentration of nitrogen is 4 kg/m , then it
becomes necessary to solve for x , as
B
x =x +D
B
A
CA - CB


 J 
Assume x is zero at the surface, in which case
A
x = 0 + (6 x 10
B
-11
(4 kg/m3 - 2 kg/m3)
2
m /s) 

 1.2 x 10-7 kg/m2-s 
-3
= 1 x 10 m = 1 mm
二、<Sol>
We are asked to compute the diffusion time required for a specific nonsteady-state diffusion
situation. It is first necessary to use Equation (5.5).
C -C
x
o
 x 
C - C = 1 - erf 2 Dt
s
o
wherein, C = 0.45, C = 0.20, C = 1.30, and x = 2 mm = 2 x 10
x
o
s
-3
C -C
x
o
0.45 - 0.20
 x 
C - C = 1.30 - 0.20 = 0.2273 = 1 - erf 2 Dt
s
o
or
erf
 x  = 1 - 0.2273 = 0.7727
2 Dt
By linear interpolation from Table 5.1
z
erf(z)
0.85
0.7707
z
0.7727
0.90
0.7970
z - 0.850
0.7727 - 0.7707
0.900 - 0.850 = 0.7970 - 0.7707
From which
z = 0.854 =
x
2 Dt
Now, from Table 5.2, at 1000C (1273 K)
D = (2.3 x 10
-5
148000 J/mol
2
m /s) exp - (8.31J/mol-K)(1273 K)


= 1.93 x 10
-11
2
m /s
m. Thus,
Thus,
0.854 =
(
(2)
-3
2 x 10 m
-11 2
1.93 x 10
m /s (t)
)
Solving for t yields
4
t = 7.1 x 10 s = 19.7 h
三、<Sol>
This problem calls for an estimate of the time necessary to achieve a carbon concentration of
0.45 wt% at a point 5 mm from the surface. From Equation (5.6b),
2
x
Dt = constant
But since the temperature is constant, so also is D constant, and
2
x
t = constant
or
x
2
1
t
1
x
2
2
= t
2
Thus,
(2.5 mm)
10 h
2
=
(5.0 mm)
t
2
2
from which
t = 40 h
2
四、<Sol>
This problem asks us to determine the values of Q and D for the diffusion of Fe in Cr from
d
o
the plot of log D versus 1/T. According to Equation (5.9b) the slope of this plot is equal to
- Q /2.3R (rather than - Q /R since we are using log D rather than ln D) and the intercept
d
d
at 1/T = 0 gives the value of log D . The slope is equal to
o
slope =
(log D)
=
1
T
 
Taking 1/T and 1/T as 0.65 x 10
1
2
-3
log D - log D
1
2
1
1
T -T
1
2
and 0.60 x 10
-3 -1
K , respectively, then the values of
log D and log D are -15.60 and -14.74, respectively. Therefore,
1
2
Q = - 2.3 R
d
(log D)
1
T
 
= - (2.3)(8.31 J/mol-K)
-15.60 - (-14.74)




(0.65 x 10-3 - 0.60 x 10-3) K-1
= 329,000 J/mol
Rather than trying to make a graphical extrapolation to determine D , a more accurate
o
value is obtained analytically using Equation (5.9b) taking a specific value of both D and T
-15 2
(from 1/T) from the plot given in the problem; for example, D = 1.0 x 10
m /s at T =
-3
1626 K (1/T = 0.615 x 10 ). Therefore
Qd
D = D exp RT
o
 
= 1.0 x 10
-15
329000 J/mol
2
m /s exp (8.31 J/mol-K)(1626 K)


= 3.75 x 10
-5
2
m /s
五、<Sol>
(a) We are asked to calculate the diffusion coefficient for Cu in Al at 500C. Using the data
in Table 5.2
 Qd
D = D exp - RT
o


= (6.5 x 10
-5
136000 J/mol
2
m /s) exp - (8.31 J/mol-K)(500 + 273 K)


= 4.15 x 10
-14
2
m /s
(b) This portion of the problem calls for the time required at 600C to produce the same
diffusion result as for 10 h at 500C. Equation (5.7) is employed as
(Dt)
500
= (Dt)
600
Now, from Equation (5.8)
D
600
= (6.5 x 10
-5
136000 J/mol
2
m /s) exp - (8.31 J/mol-K)(600 + 273 K)


= 4.69 x 10
-13
2
m /s
Thus,
(Dt)
500
t
= D
600
600
=
-14
2
m /s)(10 h)
= 0.88 h
-13 2
(4.69 x 10
m /s)
(4.15 x 10
六、<Sol>
This problem asks us to compute the temperature at which a nonsteady-state 49.5 h diffusion
anneal was carried out in order to give a carbon concentration of 0.35 wt% C in FCC Fe at
a position 4.0 mm below the surface. From Equation (5.5)
C -C
x
o
0.35 - 0.20
 x 
C - C = 1.0 - 0.20 = 0.1875 = 1 - erf 2 Dt
s
o
Or
erf
 x  = 0.8125
2 Dt
Now it becomes necessary to, using the data in Table 5.1 and linear interpolation, to
determine the value of x/2 Dt
z
erf (z)
0.90
0.7970
y
0.8125
0.95
0.8209
y - 0.90
0.8125 - 0.7970
0.95 - 0.90 = 0.8209 - 0.7970
From which
y = 0.9324
Thus,
x
= 0.9324
2 Dt
And since t = 49.5 h and x = 4.0 mm
D=
x
2
(4t)(0.9324)
2
-3 2 2
(4.0 x 10 ) m
-11 2
= (4)(178,200 s)(0.869) = 2.58 x 10
m /s
Now, in order to solve for the temperature at which D has the above value, we must
employ Equation (5.9a); solving for T yields
Q
d
T = R(ln D - ln D)
o
From Table 5.2, D
o
and Q
d
for the diffusion of C in FCC Fe are 2.3 x 10
-5
2
m /s and
148,000 J/mol, respectively. Therefore
T=
148000 J/mol
-5
-11
(8.31 J/mol-K) ln(2.3 x 10 ) - ln(2.58 x 10 )
[
]
= 1300 K = 1027C
Design Problems
材料之機械性質
一、<Sol>
We are asked to compute the maximum length of a cylindrical titanium alloy specimen that is
deformed elastically in tension. For a cylindrical specimen
do 2
A = π 2 
o
 
where d
is the original diameter.
o
solving for l leads to
o
Combining Equations (6.1), (6.2), and (6.5) and
2
Eπd l
o
l =
4F
o
=
9
2
-3 2
-3
(107 x 10 N/m )(π)(3.8 x 10 m) (0.42 x 10 m)
(4)(2000 N)
= 0.25 m = 250 mm (10 in.)
二、<Sol>
This problem asks us to compute the elastic modulus of steel. For a square cross-section, A
o
= b
2
, where b is the edge length. Combining Equations (6.1), (6.2), and (6.5) and
o
o
solving for E, leads to
E=
Fl
-3
o
(89000 N)(100 x 10 m)
=
2
-3 2
-3
b l
(20 x 10 m) (0.10 x 10 m)
o
9
2
6
= 223 x 10 N/m = 223 GPa (31.3 x 10 psi)
三、<Sol> (a) This portion of the problem calls for a determination of the maximum load that
can be applied without plastic deformation (F ). Taking the yield strength to be 275 MPa,
y
and employment of Equation (6.1) leads to
6
2
-6 2
F =  A = (275 x 10 N/m )(325 x 10 m )
y
y o
= 89,375 N (20,000 lb )
f
(b)
The maximum length to which the sample may be deformed without plastic
deformation is determined from Equations (6.2) and (6.5) as

l = l 1 + E
i o

= (115 mm)1 +

275 MPa 
 = 115.28 mm (4.51 in.)
3
115 x 10 MPa
Or
l = l - l = 115.28 mm - 115.00 mm = 0.28 mm (0.01 in.)
i o
四、<Sol>
This problem asks that we ascertain which of four metal alloys will not 1) experience plastic
deformation, and 2) elongate more than 0.9 mm when a tensile load is applied. It is first
necessary to compute the stress using Equation (6.1); a material to be used for this
application must necessarily have a yield strength greater than this value. Thus,
F
= A =
o
24500 N
= 312 MPa
10 x 10-3 m2


π
2


Of the metal alloys listed, only brass and steel have yield strengths greater than this
stress.
Next, we must compute the elongation produced in both brass and steel using
Equations (6.2) and (6.5) in order to determine whether or not this elongation is less than
0.9 mm. For brass
l
o
(312 MPa)(380 mm)
∆l = E =
= 1.19 mm
3
100 x 10 MPa
Thus, brass is not a candidate. However, for steel
l
o
(312 MPa)(380 mm)
∆l = E =
= 0.57 mm
3
207 x 10 MPa
Therefore, of these four alloys, only steel satisfies the stipulated criteria.
五、<Sol>
This problem calls for us to make a stress-strain plot for aluminum, given its tensile load-length
data, and then to determine some of its mechanical characteristics.
(a)
The data are plotted below on two plots:
the first corresponds to the entire
stress-strain curve, while for the second, the curve extends just beyond the elastic region
of deformation.
400
Stress (MPa)
300
200
100
0
0.00
0.10
Strain
Stress (MPa)
300
200
100
0
0.000
0.002
0.004
0.006
0.008
0.010
0.012
Strain
(b) The elastic modulus is the slope in the linear elastic region as
E=
∆
200 MPa - 0 MPa
3
6
=
= 62.5 x 10 MPa = 62.5 GPa (9.1 x 10 psi)
0.0032 - 0
∆
(c) For the yield strength, the 0.002 strain offset line is drawn dashed. It intersects the
stress-strain curve at approximately 285 MPa (41,000 psi ).
(d) The tensile strength is approximately 370 MPa (54,000 psi), corresponding to the
maximum stress on the complete stress-strain plot.
(e) The ductility, in percent elongation, is just the plastic strain at fracture, multiplied by
one-hundred. The total fracture strain at fracture is 0.165; subtracting out the elastic
strain (which is about 0.005) leaves a plastic strain of 0.160. Thus, the ductility is about
16%EL.
(f) From Equation (6.14), the modulus of resilience is just
2

y
U = 2E
r
which, using data computed in the problem yields a value of
2
(285 MPa)
5
3
U =
= 6.5 x 10 J/m
r (2)(62.5 x 103 MPa)
3
(93.8 in.-lb /in. )
f
六、<Sol>
This problem calls for ductility in both percent reduction in area and percent elongation.
Percent reduction in a area is computed using Equation (6.12) as
%RA =
in which d
o
do2 df2
π 2  - π 2 
 
 
d
2
 o
π 2 
 
x 100
and d are, respectively, the original and fracture cross-sectional areas.
f
Thus,
2
12.8 mm2
6.60 mm
π
 2 
 2 
x 100 = 73.4%
12.8 mm2

π
 2 
π
%RA =
While, for percent elongation, use Equation (6.11) as
%EL =
=
lf - lo
 l  x 100
 o 
72.14 mm - 50.80 mm
x 100 = 42%
50.80 mm
七、<Sol>
(a) We are asked to compute the Brinell hardness for the given indentation. It is necessary to
use the equation in Table 6.4 for HB, where P = 500 kg, d = 1.62 mm, and D = 10 mm.
Thus, the Brinell hardness is computed as
2P
HB =
[
πD D -
2 2
D -d
]
(2)(500 kg)
=
[
(π)(10 mm) 10 mm -
2
2
(10 mm) - (1.62 mm)
]
= 241
(b) This part of the problem calls for us to determine the indentation diameter d which will
yield a 450 HB when P = 500 kg. Solving for d from this equation in Table 6.4 gives
d=
=
2
2P
2
D - D - (HB)πD


(2)(500 kg)
2
(10 mm) - 10 mm - (450)(π)(10 mm)


2
= 1.19 mm
相圖
一、<Sol>
This problem asks us to determine the phases present and their concentrations at several
temperatures, as an alloy of composition 74 wt% Zn- 26 wt% Cu is cooled. From Figure
9.17:
At 850C, a liquid phase is present; C = 74 wt% Zn-26 wt% Cu
L
At 750C,  and liquid phases are present; C = 76 wt% Zn-24 wt% Cu; C =

L
68 wt% Zn-32 wt% Cu
At 680C,  and liquid phases are present; C = 74 wt% Zn-26 wt% Cu; C =

L
82 wt% Zn-18 wt% Cu
At 600C, the  phase is present; C = 74 wt% Zn-26 wt% Cu

At 500C,  and  phases are present; C = 69 wt% Zn-31 wt% Cu; C = 78


wt% Zn-22 wt% Cu
二、<Sol>
Upon heating a copper-nickel alloy of composition 70 wt% Ni-30 wt% Cu from 1300C and
utilizing Figure 9.2a:
(a) The first liquid forms at the temperature at which a vertical line at this composition
intersects the -( + L) phase boundary--i.e., about 1350C;
(b) The composition of this liquid phase corresponds to the intersection with the ( + L)-L
phase boundary, of a tie line constructed across the  + L phase region at 1350C--i.e., 59
wt% Ni;
(c) Complete melting of the alloy occurs at the intersection of this same vertical line at 70
wt% Ni with the ( + L)-L phase boundary--i.e., about 1380C;
(d) The composition of the last solid remaining prior to complete melting corresponds to
the intersection with -( + L) phase boundary, of the tie line constructed across the  + L
phase region at 1380C--i.e., about 78 wt% Ni.
三、<Sol>
(a) In order to determine the temperature of a 90 wt% Ag-10 wt% Cu alloy for which  and
liquid phases are present with the liquid phase of composition 85 wt% Ag, we need to
construct a tie line across the  + L phase region of Figure 9.6 that intersects the liquidus
line at 85 wt% Ag; this is possible at about 850C.
(b)
The composition of the  phase at this temperature is determined from the
intersection of this same tie line with solidus line, which corresponds to about 95 wt% Ag.
(c) The mass fractions of the two phases are determined using the lever rule, Equations
(9.1) and (9.2) with C = 90 wt% Ag, C = 85 wt% Ag, and C = 95 wt% Ag, as
o
L

C -C
o
L
90 - 85
W = C - C = 95 - 85 = 0.50


L
C -C

o
95 - 90
W = C - C = 95 - 85 = 0.50
L

L
四、<Sol>
(a) This portion of the problem asks that we determine the mass fractions of  and  phases
for an 25 wt% Ag-75 wt% Cu alloy (at 775C). In order to do this it is necessary to employ
the lever rule using a tie line that extends entirely across the  +  phase field (Figure 9.6),
as follows:
C -C

o
91.2 - 25
W = C - C = 91.2 - 8.0 = 0.796



C -C
o

25 - 8.0
W = C -C
= 91.2 - 8.0 = 0.204



(b)
Now it is necessary to determine the mass fractions of primary  and eutectic
microconstituents for this same alloy. This requires us to utilize the lever rule and a tie
line that extends from the maximum solubility of Ag in the  phase at 775C (i.e., 8.0 wt%
Ag) to the eutectic composition (71.9 wt% Ag). Thus
W
C
-C
eutectic
o
71.9 - 25
= C
= 71.9 - 8.0 = 0.734
-C
'
eutectic

C -C
o

25 - 8.0
W = C
= 71.9 - 8.0 = 0.266
-C
e
eutectic

(c) And, finally, we are asked to compute the mass fraction of eutectic , W
. This
e
quantity is simply the difference between the mass fractions of total  and primary  as
W
e
= W - W = 0.796 - 0.734 = 0.062

'
五、<Sol>
This problem asks that we determine the composition of a Pb-Sn alloy at 180C given that W
'
= 0.57 and W = 0.43. Since there is a primary  microconstituent present, then we know
e
that the alloy composition, C is between 61.9 and 97.8 wt% Sn (Figure 9.7).
o
Furthermore, this figure also indicates that C = 97.8 wt% Sn and C
= 61.9 wt% Sn.

eutectic
Applying the appropriate lever rule expression for W
'
W
C -C
C - 61.9
o
eutectic
o
= C -C
= 97.8 - 61.9 = 0.57
'

eutectic
and solving for C yields C = 82.4 wt% Sn.
o
o
六、<Sol>
Schematic sketches of the microstructures that would be observed for an 85 wt% Pb-15 wt%
Mg alloy at temperatures of 600C, 500C, 270C, and 200C are shown below. The
phase compositions are also indicated.
七、<Sol>
In this problem we are asked to specify temperature-composition points for all eutectics,
eutectoids, peritectics, and congruent phase transformations for a portion of the
titanium-copper phase diagram.
There is one eutectic on this phase diagram, which exists at about 51 wt% Cu-49
wt% Ti and 960C. Its reaction upon cooling is
L  Ti Cu + TiCu
2
There is one eutectoid for this system. It exists at about 7.5 wt% Cu-92.5 wt% Ti
and 790C. This reaction upon cooling is
  + Ti Cu
2
There is one peritectic on this phase diagram. It exists at about 40 wt% Cu-60
wt% Ti and 1005C. The reaction upon cooling is
 + L  Ti Cu
2
There is a single congruent melting point that exists at about 57.5 wt% Cu-42.5
wt% Ti and 982C. The reaction upon cooling is
L  TiCu
八、<Sol>
(a) A hypoeutectoid steel has a carbon concentration less than the eutectoid; on the other
hand, a hypereutectoid steel has a carbon content greater than the eutectoid.
(b) For a hypoeutectoid steel, the proeutectoid ferrite is a microconstituent that formed
above the eutectoid temperature. The eutectoid ferrite is one of the constituents of
pearlite that formed at a temperature below the eutectoid. The carbon concentration for
both ferrites is 0.022 wt% C.
九、<Sol>
The mass fractions of proeutectoid ferrite and pearlite that form in a 0.25 wt% C iron-carbon
alloy are considered in this problem. From Equation (9.20)
W =
p
C ' - 0.022
o
0.25 - 0.022
=
= 0.31
0.74
0.74
And, from Equation (9.21)
W
'
=
0.76 - C '
o
0.76 - 0.25
=
= 0.69
0.74
0.74
十、<Sol>
This problem asks that we compute the mass fraction of eutectoid ferrite in an iron-carbon alloy
that contains 0.43 wt% C. In order to solve this problem it is necessary to compute mass
fractions of total and proeutectoid ferrites, and then to subtract the latter from the former.
To calculate the mass fraction of total ferrite, it is necessary to use the lever rule and a tie
line that extends across the entire  + Fe C phase field as
3
C
-C
o
6.70 - 0.43
W = C
= 6.70 - 0.022 = 0.939
-C

Fe C

3
Fe3C
Now, for the mass fraction of proeutectoid ferrite we use Equation (9.21)
W
'
=
0.76 - C '
o
0.76 - 0.43
=
= 0.446
0.74
0.74
And, finally, the mass fraction of eutectoid ferrite W
W
''
''
is just
= W - W = 0.939 - 0.446 = 0.493

'
相變態與熱處理
一、<Sol>
Two limitations of the iron-iron carbide phase diagram are:
1) The nonequilibrium martensite does not appear on the diagram; and
2) The diagram provides no indication as to the time-temperature relationships for the
formation of pearlite, bainite, and spheroidite, all of which are composed of the equilibrium
ferrite and cementite phases.
二、<Sol>
(a) Superheating and supercooling correspond, respectively, to heating or cooling above or
below a phase transition temperature without the occurrence of the transformation.
(b) They occur because right at the phase transition temperature, the driving force is not
sufficient to cause the transformation to occur. The driving force is enhanced during
superheating or supercooling.
三、<Sol>
This problem asks us to determine the nature of the final microstructure of an iron-carbon alloy
of eutectoid composition, that has been subjected to various isothermal heat treatments.
Figure 10.14 is used in these determinations.
(a) 50% coarse pearlite and 50% martensite
(b) 100% spheroidite
(c) 50% fine pearlite, 25% bainite (upper), and 25% martensite
(d) 100% martensite
(e) 40% bainite (upper) and 60% martensite
(f) 100% bainite (upper)
(g) 100% fine pearlite
(h) 100% tempered martensite
四、<Sol>
We are asked to determine which microconstituents are present in a 0.45 wt% C iron-carbon
alloy that has been subjected to various isothermal heat treatments.
(a) Martensite
(b) Proeutectoid ferrite and martensite
(c) Bainite
(d) Spheroidite
(e) Ferrite, medium pearlite, bainite, and martensite
(f) Bainite and martensite
(g) Proeutectoid ferrite, pearlite, and martensite
(h) Proeutectoid ferrite and fine pearlite
五、<Sol>
Below is shown an isothermal transformation diagram for a 0.45 wt% C iron-carbon alloy, with
time-temperature paths that will produce (a) 42% proeutectoid ferrite and 58% coarse
pearlite; (b) 50% fine pearlite and 50% bainite; (c) 100% martensite; and (d) 50%
martensite and 50% austenite.
六、<Sol>
This problem asks us to rank four iron-carbon alloys of specified composition and
microstructure according to tensile strength. This ranking is as follows:
0.6 wt% C, fine pearlite
0.6 wt% C, coarse pearlite
0.25 wt% C, coarse pearlite
0.25 wt% C, spheroidite
The 0.25 wt% C, coarse pearlite is stronger than the 0.25 wt% C, spheroidite since coarse
pearlite is stronger than spheroidite; the composition of the alloys is the same. The 0.6
wt% C, coarse pearlite is stronger than the 0.25 wt% C, coarse pearlite, since increasing
the carbon content increases the strength. Finally, the 0.6 wt% C, fine pearlite is stronger
than the 0.6 wt% C, coarse pearlite inasmuch as the strength of fine pearlite is greater
than coarse pearlite because of the many more ferrite-cementite phase boundaries in fine
pearlite.
七、<Sol>
We are asked for the temperature range over which several iron-carbon alloys should be
austenitized during a full-anneal heat treatment.
(a) For 0.25 wt% C, heat to between 845 and 870C (1555 and 1600F) since the A
3
temperature is 830C (1525F).
(b) For 0.45 wt% C, heat to between 790 and 815C (1450 and 1500F) since the A
3
temperature is 775C (1425F).
(c) For 0.85 wt% C, heat to between 742 and 767C (1368 and 1413F) since the A
1
temperature is 727C (1340F).
(d) For 1.10 wt% C, heat to between 742 and 767C (1368 and 1413F) since the A
1
temperature is 727C (1340F).
八、<Sol>
The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy
having a spheroiditic microstructure. It is normally used on medium- and high-carbon
steels, which, by virtue of carbon content, are relatively hard and strong.
九、<Sol>
Hardness is a measure of a material's resistance to localized surface deformation, whereas
hardenability is a measure of the depth to which a ferrous alloy may be hardened by the
formation of martensite. Hardenability is determined from hardness tests.
十、<Sol>
The presence of alloying elements (other than carbon) causes a much more gradual
decrease in hardness with position from the quenched end for a hardenability curve.
The reason for this effect is that alloying elements retard the formation of pearlitic
and bainitic structures which are not as hard as martensite.
十一、<Sol>
For precipitation hardening, natural aging is allowing the precipitation process to occur at the
ambient temperature; artificial aging is carried out at an elevated temperature.
Design Problems
陶瓷材料
一、<Sol>
The two characteristics of component ions that determine the crystal structure are: 1) the
magnitude of the electrical charge on each ion; and 2) the relative sizes of the cations and
anions.
二、<Sol>
In this problem we are asked to show that the minimum cation-to-anion radius ratio for a
coordination number of four is 0.225. If lines are drawn from the centers of the anions,
then a tetrahedron is formed. The tetrahedron may be inscribed within a cube as shown
below.
C
F
E
Anion (r
A
)
a
D
Cation
B
(r
C)
A
a
The spheres at the apexes of the tetrahedron are drawn at the corners of the cube, and
designated as positions A, B, C, and D. (These are reduced in size for the sake of clarity.)
The cation resides at the center of the cube, which is designated as point E. Let us now
express the cation and anion radii in terms of the cube edge length, designated as a. The
spheres located at positions A and B touch each other along the bottom face diagonal.
Thus,
__
AB = 2r
But
A
__
2
2
2
2
(AB ) = a + a = 2a
or
__
AB = a 2 = 2r
A
And
2r
a=
A
2
There will also be an anion located at the corner, point F (not drawn), and the cube
diagonal AEF will be related to the ionic radii as
___
AEF = 2(r + r )
A C
(The line AEF has not been drawn to avoid confusion.) From the triangle ABF
__
__
___
2
2
2
(AB ) + (FB ) = (AEF )
But,
__
2r
FB = a =
A
2
and
__
(AB ) = 2r
A
from above. Thus,
2
( A)
2r
+
2
2rA2
  = [2(rA + rC)]
 2
Solving for the r /r ratio leads to
C A
r
C
=
r
A
6-2
= 0.225
2
三、<Sol>
Below is shown one of the faces of the rock salt crystal structure in which anions and cations
just touch along the edges, and also the face diagonals.
r
A
r
C
H
G
F
From triangle FGH,
__
__
GF = 2r
A
and
__
FH = GH = r + r
A C
Since FGH is a right triangle
__
__
__
2
2
2
(GH ) + (FH ) = (FG )
or
(r A + r C )
2
(
+ r +r
A C
)
2
( A)
= 2r
2
which leads to
2r
r +r =
A C
A
2
Or, solving for r /r
C A
r
C
=
r
A
 2 - 1 = 0.414
 2 
四、<Sol> This problem asks us to show that the minimum cation-to-anion radius ratio for a
coordination number of 8 is 0.732. From the cubic unit cell shown below
y
x
2rA
the unit cell edge length is 2r , and from the base of the unit cell
A
2
2
2
2
x = (2r ) + (2r ) = 8r
A
A
A
Or
x = 2r
A
2
Now from the triangle that involves x, y, and the unit cell edge
2
2
2
2
x + (2r ) = y = (2r + 2r )
A
A
C
(2r
2
2
2
2 ) + 4r
= (2r + 2r )
A
A
C
A
Which reduces to
2r
A
(
3 - 1) = 2r
C
Or
r
C
=
r
A
3 - 1 = 0.732
五、<Sol>
Crystalline ceramics are harder yet more brittle than metals because they (ceramics) have
fewer slip systems, and, therefore, dislocation motion is highly restricted.
六、<Sol>
The two desirable characteristics of glasses are optical transparency and ease of
fabrication.
七、<Sol>
We are asked to compute the weight of soda ash and limestone that must be added to 100 lb
m
of SiO to yield a glass composition of 75 wt% SiO , 15 wt% Na O, and 10 wt% CaO.
2
2
2
Inasmuch as the concentration of SiO in the glass is 75wt%, the final weight of the glass
2
(m
) is just
glass
m
glass
=
100 lb
m
= 133.3 lb
0.75
m
Therefore, the weights of Na O (m
) and CaO (m
) are as follows:
2
Na2O
CaO
m
Na2O
=
15 wt% (133.3 lb ) = 20.0 lb
m
m
 100 
and
m
CaO
=
10 wt% (133.3 lb ) = 13.3 lb
m
m
 100 
In order to compute the weights of Na CO and CaCO we must employ molecular
2
3
3
weights, as
m
molecular wt. Na2CO3
= (20.0 lb ) molecular wt. Na O 
Na2CO3
m
2


105.99 g/mol
= (20.0 lb ) 61.98 g/mol  = 34.2 lb
m
m

and
m
molecular wt. CaCO3
= (13.3 lb ) molecular wt. CaO 
CaCO3
m

100.09 g/mol
= (13.3 lb ) 56.08 g/mol  = 23.8 lb
m
m

八、<Sol>
(a)
Devitrification is the process whereby a glass material is caused to transform to a
crystalline solid, usually by a heat treatment.
(b) Two properties that may be improved by devitrification are 1) a lower coefficient of
thermal expansion, and 2) a higher thermal conductivity. Two properties that may be
impaired are 1) a loss of optical transparency, and 2) a lowering of mechanical strength
when stresses are introduced from volume changes that attend the transformation. In
some cases, however, strength may actually be improved.
九、<Sol>
Glass-ceramics may not be transparent because they are polycrystalline.
Light will
be scattered at grain boundaries in polycrystalline materials if the index of
refraction is anisotropic, and when those grains adjacent to the boundary have
different crystallographic orientations. This phenomenon is discussed in Section
22.10.
十、<Sol>
Two desirable characteristics of clay minerals relative to fabrication processes are 1) they
become hydroplastic (and therefore formable) when mixed with water; and 2) during
firing, clays melt over a range of temperatures, which allows some fusion and bonding of
the ware without complete melting and a loss of mechanical integrity and shape.
高分子材料
一、<Sol>
Polymorphism is when two or more crystal structures are possible for a material of given
composition. Isomerism is when two or more polymer molecules or mer units have the
same composition, but different atomic arrangements.
二、<Sol>
The mer structures called for are sketched below.
(a) Polyvinyl fluoride
H
H
C
C
H
F
(b) Polychlorotrifluoroethylene
F
F
C
C
F
Cl
H
H
C
C
H
OH
(c) Polyvinyl alcohol
三、<Sol>
Mer weights for several polymers are asked for in this problem.
(a) For polyvinyl chloride, each mer unit consists of two carbons, three hydrogens, and
one chlorine (Table 15.3). If A , A , and A represent the atomic weights of carbon,
C H
Cl
hydrogen, and chlorine, respectively, then
m = 2(A ) + 3(A ) + 1(A )
C
H
Cl
= (2)(12.01 g/mol) + (3)(1.008 g/mol) + 35.45 g/mole = 62.49 g/mol
(b) For polyethylene terephthalate, from Table 15.3, each mer unit has ten carbons, eight
hydrogens, and four oxygens. Thus,
m = 10(A ) + 8(A ) + 4(A )
C
H
O
= (10)(12.01 g/mol) + (8)(1.008 g/mol) + (4)(16.00 g/mol) = 192.16 g/mol
(c) For polycarbonate, from Table 15.3, each mer unit has sixteen carbons, fourteen
hydrogens, and three oxygens. Thus,
m = 16(A ) + 14(A ) + 3(A )
C
H
O
= (16)(12.01 g/mol) + (14)(1.008 g/mol) + (3)(16.00 g/mol)
= 254.27 g/mol
(d)
For polydimethylsiloxane, from Table 15.5, each mer unit has two carbons, six
hydrogens, one silicon, and one oxygen. Thus,
m = 2(A ) + 6(A ) + (A ) + (A )
C
H
Si
O
= (2)(12.01 g/mol) + (6)(1.008 g/mol) + (28.09 g/mol)
+ (16.00 g/mol) = 74.16 g/mol
四、<Sol>
(a) We are asked to compute the densities of totally crystalline and totally amorphous nylon
6,6 [ and  from Equation (15.10)]. From Equation (15.10) let C =
c
a
such that
C=
 ( -  )
c s a
 ( -  )
s c a
Rearrangement of this expression leads to
% crystallinity
,
100
 (C -  ) +   - C  = 0
c
s s
c a
s a
in which  and  are the variables for which solutions are to be found. Since two values
c
a
of  and C are specified in the problem, two equations may be constructed as follows:
s
 (C  -  ) +   - C   = 0
c 1 s1 s1
c a
1 s1 a
 (C  -  ) +   - C   = 0
c 2 s2 s2
c a
2 s2 a
In which 
s1
3
= 1.188 g/cm , 
s2
3
= 1.152 g/cm , C = 0.673, and C = 0.437. Solving the
1
2
above two equations leads to
  (C - C )
s1 s2 1
2
 =
a
C  -C 
1 s1
2 s2
=
3
3
(1.188 g/cm )(1.152 g/cm )(0.673 - 0.437)
3
3
3 = 1.091 g/cm
(0.673)(1.188 g/cm ) - (0.437)(1.152 g/cm )
And
  (C - C )
s1 s2 2
1
 =
c  (C - 1) -  (C - 1)
s2 2
s1 1
=
3
3
(1.188 g/cm )(1.152 g/cm )(0.437 - 0.673)
3
= 1.242 g/cm
3
3
(1.152 g/cm )(0.437 - 1.0) - (1.188 g/cm )(0.673 - 1.0)
(b) Now we are asked to determine the density of a specimen having 55.4% crystallinity.
Solving for  from Equation (15.10) and substitution for  and  which were computed
s
a
c
in part (a) yields
- 
c a
 =
s C( -  ) - 
c a
c
=
3
3
- (1.242 g/cm )(1.091 g/cm )
3
3
3
(0.554) 1.242 g/cm - 1.091 g/cm - 1.242 g/cm
(
)
= 1.170 g/cm
五、<Sol>
3
(a) The tensile modulus is not directly influenced by a polymer's molecular weight.
(b) Tensile modulus increases with increasing degree of crystallinity for semicrystalline
polymers. This is due to enhanced secondary interchain bonding which results from
adjacent aligned chain segments as percent crystallinity increases.
This enhanced
interchain bonding inhibits relative interchain motion.
(c) Deformation by drawing also increases the tensile modulus. The reason for this is
that drawing produces a highly oriented molecular structure, and a relatively high degree
of interchain secondary bonding.
(d)
When an undeformed semicrystalline polymer is annealed below its melting
temperature, the tensile modulus is increased.
(e) An drawn semicrystalline polymer that is annealed experiences a decrease in tensile
modulus as a result of a reduction in chain-induced crystallinity, and a reduction in
interchain bonding forces.
六、<Sol>
This question asks us to name which, of several polymers, would be suitable for the fabrication
of cups to contain hot coffee. At its glass transition temperature, an amorphous polymer
begins to soften. The maximum temperature of hot coffee is probably slightly below 100C
(212F).
Of the polymers listed, only polystyrene and polycarbonate have glass
transition temperatures of 100C or above (Table 16.2), and would be suitable for this
application.
七、<Sol>
In order for a polymer to be suited for use as an ice cube tray it must have a glass-transition
temperature below 0C. Of those polymers listed in Table 16.2 only low-density and
high-density polyethylene, PTFE, and polypropylene satisfy this criterion.
八、<Sol>
This question asks for comparisons of thermoplastic and thermosetting polymers.
(a) Thermoplastic polymers soften when heated and harden when cooled, whereas
thermosetting polymers, harden upon heating, while further heating will not lead to
softening.
(b) Thermoplastic polymers have linear and branched structures, while for thermosetting
polymers, the structures will normally be network or crosslinked.
九、<Sol>
Thermosetting polyesters will be crosslinked, while thermoplastic ones will have linear
structures without any appreciable crosslinking.
複合材料
一、<Sol>
The
major
difference
in
strengthening
mechanism
between
large-particle
and
dispersion-strengthened particle-reinforced composites is that for large-particle the
particle-matrix interactions are not treated on the molecular level, whereas, for
dispersion-strengthening these interactions are treated on the molecular level.
二、<Sol>
The similarity between precipitation hardening and dispersion strengthening is the
strengthening mechanism--i.e., the precipitates/particles effectively hinder dislocation
motion.
The two differences are: 1) the hardening/strengthening effect is not retained at
elevated temperatures for precipitation hardening--however, it is retained for dispersion
strengthening; and 2) the strength is developed by a heat treatment for precipitation
hardening--such is not the case for dispersion strengthening.
三、<Sol>
This problem asks for the maximum and minimum thermal conductivity values for a TiC-Co
cermet. Using a modified form of Equation (17.1) the maximum thermal conductivity k (u)
c
is calculated as
k (u) = k V + k V = k V
+k
V
c
m m
p p
Co Co
TiC TiC
= (69 W/m-K)(0.15) + (27 W/m-K)(0.85) = 33.3 W/m-K
The minimum thermal conductivity k (l) will be
c
k k
Co TiC
k (l) = V k
+V
k
c
Co TiC
TiC Co
(69 W/m-K)(27 W/m-K)
= (0.15)(27 W/m-K) + (0.85)(69 W/m-K)
= 29.7 W/m-K
四、<Sol>
(a) The matrix phase is a continuous phase that surrounds the noncontinuous dispersed
phase.
(b) In general, the matrix phase is relatively weak, has a low elastic modulus, but is quite
ductile. On the other hand, the fiber phase is normally quite strong, stiff, and brittle.
五、<Sol>
In this problem, for an aligned carbon fiber-epoxy matrix composite, we are given the desired
-2
longitudinal tensile strength (750 MPa), the average fiber diameter (1.2 x 10 mm), the
average fiber length (1.0 mm), the fiber fracture strength (5000 MPa), the fiber-matrix
bond strength (25 MPa), and the matrix stress at composite failure (10 MPa); and we are
asked to compute the volume fraction of fibers that is required. It is first necessary to
compute the value of the critical fiber length using Equation (17.3). If the fiber length is
much greater than l , then we may determine V using Equation (17.17), otherwise, use of
c
f
either Equation (17.18) or Equation (17.19) is necessary. Thus,
*d
-2
f
(5000 MPa)(1.2 x 10 mm)
l =
=
= 1.20 mm
2(25 MPa)
c 2
c
Inasmuch as l < l (1.0 mm < 1.20 mm), then use of Equation (17.19) is required.
c
Therefore,
l
c
 * = d V +  ' (1 - V )
c d'
f
m
f
750 MPa =
-3
(1.0 x 10 m)(25 MPa)
(V ) + (10 MPa)(1 - V )
-5
f
f
1.2 x 10 m
Solving this expression for V leads to V = 0.357.
f
f
六、<Sol>
Pultrusion, filament winding, and prepreg fabrication processes are described in Section 17.13.
For pultrusion, the advantages are: the process may be automated, production
rates are relatively high, a wide variety of shapes having constant cross-sections are
possible, and very long pieces may be produced. The chief disadvantage is that shapes
are limited to those having a constant cross-section.
For filament winding, the advantages are: the process may be automated, a
variety of winding patterns are possible, and a high degree of control over winding
uniformity and orientation is afforded.
The chief disadvantage is that the variety of
shapes is somewhat limited.
For prepreg production, the advantages are: resin does not need to be added to
the prepreg, the lay-up arrangement relative to the orientation of individual plies is variable,
and the lay-up process may be automated. The chief disadvantages of this technique are
that final curing is necessary after fabrication, and thermoset prepregs must be stored at
subambient temperatures to prevent complete curing.