Material Teknik (Engineering Materials)
1
Buku Pustaka • Materials Science and Engineering, An introduction, William D. Callister Jr, Wiley, 2004 • Ilmu dan Teknologi Bahan, Lawrence H. Van Vlack (terjemahan), Erlangga, 1995 • Pengetahuan Bahan, Tata Surdia dan Shinroku Saito, Pradnya Paramita, 1995 • Principle of Materials Science and Engineering, William F. Smith, Mc Graw Hill, 1996 2
Pokok Bahasan • • • • • • • • • • •
Pendahuluan Struktur dan ikatan atom Struktur dan cacat kristal Sifat mekanik Diagram fasa Proses anil dan perlakuan panas Logam besi Logam bukan besi Keramik Polimer Komposit 3
Material • Material adalah sesuatu yang disusun/dibuat oleh bahan. • Material digunakan untuk transportasi hingga makanan. • Ilmu material/bahan merupakan pengetahuan dasar tentang struktur, sifatsifat dan pengolahan bahan. 4
Jenis Material • Logam Kuat, ulet, mudah dibentuk dan bersifat penghantar panas dan listrik yang baik • Keramik Keras, getas dan penghantar panas dan listrik yang buruk • Polimer kerapatan rendah, penghantar panas dan listrik buruk dan mudah dibentuk • Komposit merupakan ganbungan dari dua bahan atau lebih yang masingmasing sifat tetap 5
Logam
6
Keramik
7
Polimer
8
Komposit
9
Struktur dan Ikatan Atom Material Teknik
10
Pendahuluan • Atom terdiri dari elektron dan inti atom • Inti atom disusun oleh proton dan neutron • Elektron mengelilingi inti atom dalam orbitnya masingmasing • Massa elektron 9,109 x 10-28 g dan bermuatan –1,602 x 1019 C • Massa proton 1,673 x 10-24 g dan bermuatan 1,602 x 10-19 C • Massa neutron 1,675 x 10-24 g dan tidak bermuatan • Massa atom terpusat pada inti atom • Jumlah elektron dan proton sama, sedangkan neutron neutral, maka atom menjadi neutral 11
Model atom Bohr
12
Konfiguration elektron unsur
*
l
No.
Element
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
K 1
L 2
M 3
N 4
O 5
P 6
Q 7
s
sp
spd
spd f
spd f
spd f
s
1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
1 2 2 2 2 1 2 2 2 2 1* 2 21 22 23 24 25 26
1 2 3 4 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
1 2 3 4 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6
1 2 3 5* 5 6 7 8 10 10 10 10 10 10 10 10
13
Tabel Periodik
14
Elektronegatip dari Unsur
15
Ikatan Atom Ionik
16
Ikatan Atom Kovalen
17
Ikatan Atom Logam
18
Ikatan Atom Hidrogen
19
Bilangan Koordinasi utk Ikatan Atom
20
Struktur dan Cacat Kristal Material Teknik
21
Pendahuluan z • Kristal adalah susunan atom-atom secara teratur dan kontinu pada arah tiga dimensi • Satuan sel adalah susunan terkecil dari kristal • Parameter kisi struktur kristal
c
x
y a
b
– Panjang sisi a, b, c – Sudut antara sumbu 22
Sistem Kristal Parameter kisi diklasifikasikan dalam tujuh sistem kristal dan empat belas kisi kristal • Arah kristal z dinyatakan sebagai vektor dalam [uvw] • uvw merupakan [111] bilangan bulat y c • Himpunan arah <111> [100] a [110] terdiri dari [111], b [111], [111], [111], x [111], [111], [111], [111] 23
Menentukan Indeks Miller Arah Kristal •
Prosedur menentukan arah kristal x y z Proyeksi a/2 b 0 Proyeksi (dlm a, b, c) ½ 1 0 Reduksi 1 2 0 Penentuan [120]
z
c
x
Proyeksi pd sb x: a/2
Proyeksi pd sb y: b
a y
b
24
Bidang Kristal • Dinyatakan dengan (hkl) • hkl merupakan bilangan bulat
z
Bid (110) mengacu titik asal O
y
c
a b
x
Bid. (110) ekivalen
z
Bid (111) mengacu titik asal O
c x
a b
y
Bid. (111) ekivalen 25
Menentukan Indeks Miller Bidang Kristal •
Prosedur menentukan bidang kristal x y z Perpotongan ~a -b c/2 Perpotongan (dlm a, b dan c) ~ -1 ½ Resiprokal 0 -1 2 Penentuan (012)
z
z’
c
a y b
x
x’
bid.(012)
26
14 kisi kristal
27
Kristal Kubik Berpusat Muka •
•
Faktor tumpukan padat = total volum bola / total volum satuan sel = Vs/Vc = 4x(4/3 r3)/16r32 = 0,74 Kerapatan = A / VcNA = (4x63,5) / (162x (1,28x10 -8)x(6,02x 1023)) g/cm3 = 8,89 g/cm3.
28
Kristal Kubik Berpusat Bidang
29
Kristal Heksagonal Tumpukan Padat
30
Cacat Kristal •
Cacat Kristal – Cacat titik • Kekosongan • Pengotor – Pengotor Intersisi – Pengotor Subtitusi – Cacat garis (dislokasi) • Dislokasi garis • Dislokasi ulir – Cacat bidang • Batas butir • Permukaan – Cacat volum 31
Cacat Titik
32
Dislokasi Garis
33
Dislokasi Ulir
34
Batas Butir
35
Permukaan
36
Inklusi
37
Sifat Mekanik Material Teknik
38
Sifat Mekanik • •
•
Material dalam pengunanya dikenakan gaya atau beban. Karena itu perlu diketahuo kharater material agar deformasi yg terjadi tidak berlebihan dan tidak terjadi kerusakan atau patah Karakter material tergantung pada: – Komposisi kimia – Struktur mikro – Sifat material: sifat mekanik, sifat fisik dan sifat kimia
Gaya/beban
Material
39
Sifat mekanik • Kekuatan (strength): ukuran besar gaya yang diperlukan utk mematahkan atau merusak suatu bahan • Kekuatan luluh (yield strength): kekuatan bahan terhadap deformasi awal • Kekuatan tarik (Tensile strength): kekuatan maksimun yang dapat menerima beban. • Keuletan (ductility): berhubungan dengan besar regangan sebelum perpatahan 40
Sifat Mekanik • Kekerasan (hardness): ketahanan bahan terhadap penetrasi pada permukaannya • Ketangguhan (toughness): jumlah energi yang mampu diserap bahan sampai terjadi perpatahan • Mulur (creep) • Kelelahan (fatique): ketahanan bahan terhadap pembebanan dinamik • Patahan (failure) 41
Konsep tegangan (stress) dan regangan (strain) F F
• Pembebanan statik: – Tarik – Kompressi – Geser F
F F Beban kompressi Beban tarik F Beban geser
42
Uji tarik Standar sampel untuk uji tarik 2¼’ 0,505’ 2’
¾’ R 3/8’
• Tegangan teknik, = F/Ao (N/m2=Pa) • Regangan teknik, = (li-lo)/lo • Tegangan geser, = F/Ao 43
Deformasi elastis • Pada pembebanan rendah dalam uji tarik, hubungan antara tegangan dan regangan linier
Beban dihilangkan Teg. Modulus elastis Pembebanan Reg.
44
Mesin uji tarik (Tensile Test)
45
Deformasi elastis • Hubungan tsb masih dalam daerah deformasi elastis dan dinyatakan dengan • Hubungan diatas dikenal sebagai Hukum Hooke • Deformasi yang mempunyai hubungan tegangan dan regangan linier (proporsional) disebut sebagai deformasi elastis 46
Paduan Modulus elastis Modulus geser Ratio logam (104 MPa) (104 MPa) Poisson Al 6,9 2,6 0,33 Cu-Zn 10,1 3,7 0,35 Cu 11,0 4,6 0,35 Mg 4,5 1,7 0,29 Ni 20,7 7,6 0,31 Baja 20,7 8,3 0,27 Ti 10,7 4,5 0,36 47 W 40 7 16 0 0 28
• Hubungan tegangan geser dan regangan geser dinyatakan dengan = G • Dengan = teg.geser = reg.geser G = modulus geser 48
Sifat elastis material • Ketika uji tarik dilakukan pada suatu logam, perpanjangan pada arah beban, yg dinyatakan dlm regangan z mengakibatkan terjadinya regangan kompressi pada x sb-x dan y pada sb-y • Bila beban pada arah sb-z uniaxial, maka x = y . Ratio regangan lateral & axial dikenal sebagai ratio Poisson
Z
z y x
Z
49
= x/y • Harga selalu positip, karena tanda x dan y berlawanan. • Hubungan modulus Young dengan modulus geser dinyatalan dengan E = 2 G (1 + ) • Biasanya <0,5 dan utk logam umumnya G = 0,4 E 50
Deformasi plastis • Utk material logam, umumnya deformso elastis terjadi < 0,005 regangan • Regangan > 0,005 terjadi deformasi plastis (deformasi permanen)
ys Teg.
0,002
Reg.
Titik luluh atas
ys Teg.
Titik Luluh bawah
Reg.
51
Deformasi elastis • Ikatan atom atau molekul putus: atom atau molekul berpindah tdk kembali pada posisinya bila tegangan dihilangkan • Padatan kristal: proses slip padatan amorphous (bukan kristal). Mekanisme aliran viscous
52
Perilaku uji tarik • Titik luluh: transisi elastis & platis • Kekuatan: kekuatan tarik: kekuatan maksimum • Dari kekuatan maksimum hingga titik terjadinya patah, diameter sampel uji tarik mengecil (necking)
53
Keuletan (ductility) • Keuletan: derajat deformasi plastis hingga terjadinya patah • Keuletan dinyatakan dengan – Presentasi elongasi, %El. = (lf-lo)/lo x 100% – Presentasi reduksi area, %AR = (Ao-Af)/Ao x 100% 54
Ketangguan (Toughness) B B’
Teg.
A
C
Reg.
C’
• Perbedaan antara kurva tegangan dan regangan hasil uji tarik utk material yang getas dan ulet • ABC : ketangguhan material getas • AB’C’ : ketangguhan material ulet
55
Logam Au Al Cu Fe Ni Ti Mo
Kekuatan luluh (MPa) 28 69 130 138 240 565
Kekuatan tarik (MPa) 130 69 200 262 480 330 655
Keuletan %El. 45 45 45 45 40 30 35
56
Tegangan dan regangan sebenarnya • Pada daerah necking, luas tampang lintang sampel uji material Teg. • Tegangan sebenarnya T = F/Ai • Regangan sebenarnya T = ln li/lo
sebenarnya teknik
Reg.
57
Bila volum sampel uji tidak berubah, maka Aili = Aolo
• Hubungan tegangan teknik dengan tegangan sebenarnya T = (1 + ) • Hubungan regangan teknik dengan regangan sebenarnya T = ln (1+ )
58
Uji Kekerasan (Hardness Test)
59
Uji Mulur (Creep Test)
60
Uji Kelelahan (Fatique Test)
61
Patahan (Failure)
62
Diagram Fasa Material Teknik
63
Pendahuluan • Sifat mekanik bahan salah satunya ditentukan oleh struktur mikro • Utk mengetahui struktur mikro, perlu mengetahui fasa diagram • Diagram fasa digunakan utk peleburan, pengecoran, kristalisasi dll • Komponen: logam murni dan/atau senyawa penyusun paduan • Cth. Kuningan, Cu sebagai unsur pelarut dan Zn sebagai unsur yang dilarutkan. • Batas kelarutan merupakan konsentrasi atom maksimum yang dapat dilarutkan oleh pelarut utk membentuk larutan padat (solid solution). Contoh Gula dalam air. 64
• Fasa adalah bagian homogen dari sistem yg mempunyai kharakteristik fisik & kimia yg uniform • Contoh fasa , material murni, larutan padat, larutan cair dan gas. • Material yg mempunyai dua atau lebih struktur disebut polimorfik • Jumlah fasa yg ada & bagiannya dlm material merupakan struktur mikro. 65
• Diagram kesetimbangan fasa merupakan diagram yang menampilkan struktur mikro atau struktur fasa dari paduan tertentu • Diagram kesetimbangan fasa menampilkan hubungan antara suhu dan komposisi serta jumlah fasa-fasa dalam keadaan setimbang.
66
Diagram Cu-Ni • L = larutan cair homogen yang mengandung Cu dan Ni • A = larutan padat subtitusi yang terdiri dari Cu dan Ni, yang mempunyai struktur FCC
67
Diagram Cu-Ni • Jumlah persentasi cair (Wl) = S/(R+S)x100% • Jumlah persentasi a (W) = R/(R+S)x100%
68
Sistem binary eutektik •
•
•
•
• •
Batas kelarutan atom Ag pada fasa dan atom Cu pada fasa tergantung pada suhu Pada 780C, Fasa dapat melarutkan atom Ag hingga 7,9%berat dan Fasa dapat melarutkan atom Cu hingga 8,8%berat Daerah fasa padat: fasa , fasa +, dan fasa , yang dibatasi oleh garis solidus AB, BC, AB, BG, dan FG, GH. Daerah fasa padat + cair: fasa + cair, dan fasa + cair, yang dibatasi oleh garis solidus Daerah fasa cair terletak diatas garis liquidus AE dan FE Reaksi Cair padat() + padat () pada titik E disebut reaksi Eutektik.
A F G
H
E
B
C
69
Diagram Fasa Pb-Sn •
Reaksi eutektik Cair (61,9%Sn) (19,2%Sn)+(97,6%Sn)
70
Diagram Fasa Cu-Zn
71
Diagram Fasa Fe-Fe3C •
•
• •
•
Besi- (ferrit); Struktur BCC, dapat melarutkan C maks. 0,022% pada 727C. Besi- (austenit); struktur FCC, dapat melarutkan C hingga 2,11% pada 1148C. Besi- (ferrit); struktur BCC Besi Karbida (sementit); struktur BCT, dapat melarutkan C hingga 6,7%0 Pearlit; lamel-lamel besi- dan besi karbida 72
Reaksi pada Diagram Fasa Fe-C • Reaksi eutektik pada titik 4,3%C, 1148C L (2,11%C) + Fe3C(6,7%C) • Reaksi eutektoid pada titik 0,77%C, 727C (0,77%C) (0,022%C) + Fe3C(6,7%C) • Reaksi peritektik
73
Pengaruh unsur pada Suhu Eutektoid dan Komposisi Eutektoid • Unsur pembentuk besi-: Mn & Ni • Unsur pembentuk besi-: Ti, Mo, Si & W
74
Diagram Fasa Al-Si •
•
•
Paduan hipoeutektik AlSi mengandung Si <12,6% Paduan eutektik Al-Si mengandung Si sekitar 12,6% Paduan hipereutektik Al-Si mengandung Si >12,6%
75
Proses Anil & Perlakuan Panas Material Teknik
76
Pendahuluan • Proses anil merupakan proses perlakuan panas suatu bahan melalui pemanasan pada suhu cukup tinggi dan waktu yang lama, diikuti pendinginan perlahan-lahan • Anil – Bahan: Gelas – Tujuan: menghilangkan tegangan sisa & menghindari terjadinya retakan panas – Prosedur: suhu pemanasan mendekati suhu transisi gelas dan pendinginan perlahan-lahan – Perubahan strukturmikro: tidak ada 77
• Menghilangkan Tegangan – Bahan: semua logam, khususnya baja – Tujuan: menghilangkan tegangan sisa – Prosedur: Pemanasan sampai 600C utk baja selama beberapa jam – Perubahan strukturmikro: tidak ada • Rekristalisasi – Bahan: logam yang mengalami pengerjaan dingin – Tujuan: pelunakan dengan meniadakan pengerasan regangan – Prosedur: Pemanasan antara 0,3 dan 0,6 titik lebur logam – Perubahan strukturmikro: butir baru 78
Anil Sempurna • Bahan: baja • Tujuan: Pelunakan sebelum pemesinan • Prosedur: austenisasi 2-30C • Perubahan strukturmikro: pearlit kasat
900 C
800 700
normalisasi anil 0,77%C +Fe3C
79
Speroidisasi – Bahan: baja karbon tinggi, seperti bantalan peluru – Tujuan: meningkatkan ketangguhan baja – Prosedur: dipanaskan pada suhu eutektoid (~700C) untuk 1-2 jam – Perubahan strukturmikro: speroidit
80
Laku Mampu Tempa (Malleabilisasi) • Bahan: besi cor • Tujuan: besi cor lebih ulet • Prosedur: – anil dibawah suhu eutektoid (<750C) Fe3C 3Fe() + C(garfit) Dan terbentuk besi mampu tempa ferritik – Anil diatas suhu eutektoid (>750C) Fe3C 3Fe() + C(garfit) Dan terbentuk besi mampu tempa austenitik • Perubahan strukturmikro: terbentuknya gumpalan grafit. 81
Normalisasi terdiri dari homogenisasi dan normalisasi •
•
Homogenisasi – Bahan: logam cair – Tujuan: menyeragamkan komposisi bahan – Prosedur: pemanasan pada suhu setinggi mungkin asalkan logam tidak mencair dan tidak menumbuhkan butir – Perubahan strukturmikro: homogenitas lebih baik, mendekati diagram fasa Normalisasi – Bahan: baja – Tujuan: membentuk strukturmikro dengan butir halus & seragam – Prosedur: austenisasi 50-60C, disusul dengan pendinginan udara – Perubahan strukturmikro: pearlit halus dan sedikit besi- praeutektoid 82
Anil
83
Recovery, Rekristalisasi, Pertumbuhan Butir
84
Proses Presipitasi • Pengerasan presipitasi dilakukan dengan memanaskan logam hingga unsur pemadu larut, kemudian celup cepat, dan dipanaskan kembali pada suhu relatip rendah
85
Diagram Transformasi-Isotermal
86
Diagram Transformasi-Isotermal untuk Baja Eutektoid
87
Logam Besi Material Teknik
88
Logam besi • • • • •
Baja karbon Baja paduan Baja pekakas & dies Baja tahan karat Besi tuang
89
Baja karbon • Menurut kadungan C – Baja karbon rendah: C<0,3%, utk baut, mur, lembaran, pelat, tabung, pipa, komponen mesin berkekuatan rendah – Baja karbon menengah: 0,3%
Klasifikasi baja menurut AISI & SAE
91
Baja seri 1045 utk yoke ball • 1045 termasuk seri 10xx atau seri baja karbon • Angka 45 merupakan kandungan karbon = 45/100 % = 0,45%
92
Baja Paduan • Baja paduan rendah berkekuatan tinggi (high strength alloy steel) – C<0,30% – Strukturmikro: butir besi- halus, fasa kedua martensit & besi- – Produknya: pelat, balok, profil
• Baja fasa ganda (Dual- phase steel) – Strukturmikro: campuran besi- & martensit 93
Baja paduan rendah berkekuatan tinggi Kekuatan luluh 103 Psi
MPa
35
240
40
275
45
310
50
350
60
415
70
485
80
550
100
690
120
830
140
970
Komposis kimia
Deoksidasi
S = kualitas struktur
F = kill + kontrol S
X = paduan rendah
K = kill
W = weathering
O = bukan kill
D = fasa ganda
Cth. 50XF 50 kekuatan luluh 50x103 Psi X paduan rendah F kill + kontrol S
94
Baja tahan karat • Sifatnya tahan korosi, kekuatan & keuletan tinggi dan kandungan Cr tinggi • Kandungan lain : Ni, Mo, Cu, Ti, Si, Mg, Cb, Al, N dan S
95
Jenis baja tahan karat • Austenitik (seri 200 & 300) – Mengandung Cr, Ni dan Mg – Bersifat tidak magnit, tahan korosi – Utk peralatan dapur, fitting, konstruksi, peralatan transport, tungku, komponen penukar panas, linkungan kimia
• Ferritik (seri 400) – Mengandung Cr tinggi, hingga 27% – Bersifat magnit, tahan korosi – Utk peralatan dapur. 96
Jenis baja tahan karat • Martemsitik (seri 400 & 500) – Mengandung 18%Cr, tdk ada Ni – Bersifat magnit, berkekuatan tinggi, keras, tahan patah dan ulet – Utk peralatan bedah, instrument katup dan pegas
• Pengerasan presipitasi – Mengandung Cr, Ni, Cu, Al, Ti, & Mo – Bersifat tahan korosi, ulet & berkekuatan tinggi pada suhu tinggi – Utk komponen struktur pesawat & pesawat ruang angkasa 97
Jenis baja tahan karat • Struktur Duplek – Campuran austenit & ferrit – Utk komponen penukar panas & pembersih air
98
Besi cor • Besi tuang disusun oleh besi, 2,11-4,50% karbon dan 3,5% silikon • Kandungan Si mendekomposisi Fe3C menjadi Fe dan C (garfit)
99
Jenis besi cor • • • •
Besi cor kelabu Besi cor nodular (ulet) Besi cor tuang putih Besi cor malleable
100
Besi cor kelabu • Disusun oleh serpihan C (grafit) yang tersebar pada besi- • Bersifat keras & getas
101
Besi cor nodular (ulet) • C (grafit)nya berbentuk bulat (nodular) tersebar pada besi-. • Nodular terbentuk karena besi cor kelabu ditambahkan sedikit unsur magnesium dan cesium • Keras & ulet 102
Besi cor putih • Disusun oleh besi- dan besi karbida (Fe3C) • Terbentuk melalui pendinginan cepat • Getas, tahan pakai & sangat keras
103
Besi cor malleable • Disusun oleh besi- dan C (grafit) • Dibentuk dari besi cor putih yang dianil pada 800-900oC dalam atmosphere CO & CO2
104
Logam Bukan Besi Material Teknik
105
Pendahuluan •
•
•
Logam & paduan bukan besi – Logam biasa: Al, Cu, Mg – Logam/paduan tahan suhu tinggi: W, Ta, Mo Aplikasi utk – Ketahanan korosi – Konduktifitas panas $ listrik tinggi – Kerapatan rendah – Mudah dipabrikasi Cth. – Al utk pesawat terbang, peralatan masak – Cu utk kawat listrik, pipa air – Zn utk karburator – Ti utk sudu turbin mesinjet – Ta utk mesin roket 106
Alimunium Produk Wrough 1xxx 2xxx 3xxx 4xxx 5xxx 6xxx 7xxx 8xxx
Al murni: 99,00% Al+Cu Al+Mn Al+Si Al+Mg Al+Mg+Si Al+Zn Al+unsur lain 107
Alimunium Produk Cor 1xx.x 2xx.x 3xx.x 4xx.x 5xx.x 6xx.x 7xx.x 8xx.x
Al murni: 99,00% Al+Cu Al+Si, Cu, Mg Al+Si Al+Mg Tidak digunakan Al+Zn Al+Pb 108
Perlakuan utk produk aluminium wrough dan cor F
Hasil pabrikasi (pengerjaan dingin atau panas atau cor)
O
Proses anil (hasil pengerjaan dingin atau panas atau cor)
H
Pengerjaan regangan melalui pengerjaan dingin (utk produk wrough) Perlakuan panas
T
109
Magnesium & paduan magnesium • Logam terringan dan penyerap getaran yg baik • Aplikasi: – – – – – – –
Komponen pesawat & missil Mesin pengankat Pekakas Tangga Koper Sepeda Komponen ringan lainnya. 110
Paduan magnesium: produk wrough dan cor Paduan
Komposisi (%)
Kondisi
Pembentukk an Ekstrusi lembaran & pelat Ekstrusi & tempa Lembaran & pelat Ekstrusi & tempa 111
Al
Zn
Mn
AZ31B
3,0
1,0
0,2
F H24
AZ80A
8,5
0,5
0,2
T5
HK31A ZK60A
5,7
Zr
0,7
H24
0,55
T5
Penamaan paduan magnesium • • • •
Hurup 1&2 menyatakan unsur pemadu utama Angka 3&4 menyatakan % unsur pemadu utama Hurup 5 menyatakan standar paduan Hurup dan angka berikutnya menyatakan perlakuan panas Contoh. AZ91C-T6 A Al Z Zn 9 9%Al 1 1%Zn C Standar C T6 Perlakuan panas 112
Tembaga & paduan tembaga •
•
Sifat paduan tembaga: – Konduktifitas listrik dan panas tinggi – Tidak bersifat magnit – Tahan korosi Aplikasi – Komponen listrik dan elektronik – Pegas – Cartridge – Pipa – Penukar panas – Peralatan panas – Perhiasan, dll 113
Jenis paduan tembaga • • • • • •
Kuningan (Cu+Zn) Perunggu (Cu+Sn) Perunggu Al (Cu+Sn+Al) Perunggu Be (Cu+Sn+Be) Cu+Ni Cu+Ag 114
Nikel & paduan nikel •
• • • • • • •
Sifat paduan nikel – Kuat – Getas – Tahan korosi pada suhu tinggi Elemen pemadu nikel: Cr, Co, Mo dan Cu Paduan nikel base = superalloy Paduan nikel tembaga = monel Paduan nikel krom = inconel Paduan nikel krom molybdenum = hastelloy Paduan nikel kron besi = nichrome Paduan nikel besi = invar 115
Supperalloy • Tahan panas dan tahan suhu tinggi • Aplikasi: mesin jet, turbin gas, mesin roket, pekakas, dies, industri nuklir, kimia dan petrokimia • Jenis superalloy – Superalloy besi base: 32-67%Fe, 15-22%Cr, 9-38%Ni – Superalloy kobalt base: 35-65%Co, 19-30%Cr, 35%Ni – Superalloy nikel base: 38-76%Ni, 27%Cr, 20%Co. 116
Keramik Material Teknik
117
Keramik • Senyawa logam atau bukan logam yang mempunyai ikatan atom ionik dan kovalen • Ikatan ionik dan kovalen menyebabkan keramik mempunyai titik lebur tinggi dan bersifat isolator • Keramik terdiri dari – Keramik tradisional, disusun oleh tanah liat, silika dan feldspar. Cth. bata, ubin, genteng dan porselen – Keramik murni atau teknik, disusun oleh senyawa murni. 118
Struktur Kristal • Sebagian besar keramik diikat secara ionik dan hanya sedikit tang diikat secara kavalen • Ikatan ionik biasanya mempunyai diameter atom kation < atom anion, akibatnya atom kation selalu dikelilingi atom anion. • Jumlah atom tetangga terdekat (mengelilingi) atom tertentu dikenal sbg bilangan koordinasi (Coordination number). 119
Hub.bil.koordinasi dan perbandingan jari2atom kation-anion Bilangan koordinasi 2
Perbandingan jari-jari kationanion <0,155
3
0,115-0,225
4
0,225-0,414
6
0,414-0,732
8
0,723-1,0
Geometri koordinasi
120
Jari-jari kation dan anion Kation
Jari-jari ion (nm)
Anion
Jari-jari ion (nm)
Al 3+
0,053
Br -
0,196
Ba 2+
0,136
Cl -
0,181
Ca 2+
0,100
F-
0,133
Cs +
0,170
I-
0,220
Fe 2+
0,077
O 2-
0,140
Fe 3+
0,069
S 2-
0,184
K+
0,138
Mg 2+
0,072
Mn 2+
0,067
Na 2+
0,102
Ni 2+
0,069
Si 4+
0,040
Ti 2+
0,061
121
Struktur Kristal Tipe AX Cth.; NaCl, CsCl, ZnS dan intan • Struktur NaCl (Garam) – Bentuk kubik berpusat muka (FCC) – 1 atom kation Na+ dikelilingi 6 atom anion Cl- (BK 6) – Posisi atom kation Na+: ½½½, 00½, 0½0, ½00 – Posisi atom anion Cl-: 000, ½½0, ½0½, 0½½ – Cth seperti kristal garam: MgO, MnS, LiF dan FeO. – Perbadingan jari-jari atom kation dan anion = 0,102/0,181 = 0,56 122
Struktur kristal tipe AX • Struktur CsCl – Bentuk kubik sederhana (simple cubic) – 1 atom kation Cs+ dikelilingi 8 atom anion Cl- (BK 8) – Posisi atom kation Na+: ½½ – Posisi atom anion Cl-:000 – Perbandingan jari-jari aton kation dan anion = 0,170/0,181 = 0,94.
123
Struktur kristal tipe AX • Struktur ZnS – Bentuk Sphalerite – 1 atom kation Zn+ dikelilingi 4 atom anion S- (BK 4) – Posisi atom kation Zn+: ¾¾¾, ¼¼¾, ¼¾¼, ¾¼¼ – Posisi atom anion S-: 000, ½½0, ½0½, 0½½ – Cth seperti kristal ZnS: ZnTe, BeO dan SiO. – Perbandingan jari-jari atom kation dan anion = 0,060/0,174 = 0,344 124
Struktur kristal AX • Struktur intan – Bentuk sama seperti ZnS, tetapi seluruh atomnya diisi atom C. – Ikatan atomnya ikatan atom kovalen
Struktur kristal intan 125
Struktur kristal AmXp • Al2O3 (korundum) – Bentuk heksagonal tumpukan padat
Struktur kristal Al2O3 126
Struktur kristal AmBnXp • BaTiO3 – Bentuk kristal perouskite – Atom kation: Ba2+ dan Ti4+ – Atom anion: O2-
Struktur kristal perouskite 127
Polimer Material Teknik
128
Polymer Structures
TEM of spherulite structure in natural rubber(x30,000). • Chain-folded lamellar crystallites (white lines) ~10nm thick extend radially.
Polymer Structures ISSUES TO ADDRESS... What are the basic • Classification? • Monomers and chemical groups? • Nomenclature? • Polymerization methods? • Molecular Weight and Degree of Polymerization? • Molecular Structures? • Crystallinity? • Microstructural features?
Polymer Microstructure • Polymer = many mers mer H H H H H H C C C C C C H H H H H H
Polyethylene (PE)
mer H H H H H H C C C C C C H CH3 H CH3 H CH3
mer H H H H H H C C C C C C H Cl H Cl H Cl
Polyvinyl chloride (PVC)
Polypropylene (PP)
Adapted from Fig. 14.2, Callister 6e.
Polyethylene perspective of molecule
A zig-zag backbone structure with covalent bonds
Polymer Microstructure
• Covalent chain configurations and strength: More rigid
Van der Waals, H secondary
bonding
Linear
Branched
Cross-Linked
Network
Direction of increasing strength Adapted from Fig. 14.7, Callister 6e.
Common Examples - Textile fibers: polyester, nylon… - IC packaging materials. - Resists for photolithography/microfabrication. - Plastic bottles (polyethylene plastics). - Adhesives and epoxy. - High-strength/light-weight fibers: polyamides, polyurethanes, Kevlar… - Biopolymers: DNA, proteins, cellulose…
Common Classification • Thermoplastics: polymers that flow more easily when squeezed, pushed, stretched, etc. by a load (usually at elevated T). – Can be reheated to change shape.
• Thermosets: polymers that flow and can be molded initially but their shape becomes set upon curing. – Reheating will result in irreversible change or decomposition.
• Other ways to classify polymers. – By chemical functionality (e.g. polyacrylates, polyamides, polyethers, polyeurethanes…). – Vinyl vs. non-vinyl polymers. – By polymerization methods (radical, anionic, cationic…). – Etc…
Common Chemical Functional Groups H
Ethylene (ethene) Propylene (propene)
H C C
H
H
H
H
=
C C H
C H H
H
1-butene 2-butene trans
Acetylene (ethyne) Saturated hydrocarbons (loose H to add atoms)
cis
H C C H
Unsaturated hydrocarbons (double and triple bonds)
Common Hydrocarbon Monomers Alcohols
Methyl alcohols
Ethers
Dimethyl Ether
Acids
Acetic acid
Aldehydes
Aromatic hydrocarbons
Formaldehyde
Phenol
Some Common Polymers Common backbone with substitutions Polyacrylonitrile (PAN)
H
H
C
C
H
C N
Vinyl polymers (one or more H’s of ethylene can be substituted) H
H C C
H
X
H
H
C
C
H
X
Nomenclature Monomer-based naming: poly________ Monomer name goes here
e.g. ethylene -> polyethylene if monomer name contains more than one word: poly(_____ ____) Monomer name in parentheses
e.g. acrylic acid -> poly(acrylic acid) Note: this may lead to polymers with different names but same structure. H H H H
…
C C C C
H H H H
…
H H H H
polyethylene
…
C C C C
…
H H H H
polymethylene
Polymerization Methods A. Free Radical Polymerization 1. Initiation H
R
H
R
C C
Free radical initiator (unpaired electron)
H
H
monomer
H
H
C
C
H
H
Radical transferred
R
R H
H
H C
sp2 carbons
C
C
H
H bonds bond
H
C H
H sp3 carbon
Polymerization Methods A. Free Radical Polymerization H
2. Propagation R
H
H
C
C
H
H
H
C C
H
R
C C H
H
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
R
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
H
H C H
R
H
H C H
C H
H
C
Both carbon atoms will change from sp2 to sp3.
Polymerization Methods A. Free Radical Polymerization 3. Termination
R
R
H
H
C
C
H
H
H
H
C
C
H
H
+
R
+
R
R
H
H
C
C
H
H
H
H
C
C R
H
H
R
H
H
H
H
C
C
C
C
H
H
H
H
R
Intentional or unintentional molecules/impurities can also terminate.
Polymerization Methods Loses water (condensation)
B. Stepwise polymerization O
O H2N
R
C
+ OH
H2N
R
C
O
OH
H2N
R
C
N H
R
C
OH
+
H
H N
H N H C C R
Various R groups… n
C. Other methods Anionic polymerization, cationic polymerization, coordination polymerization…
H
O
Proteins (polypeptides have similar composition) O
O
O R
+ (n-1)
C
n
H
O
H
Molecular Weights Not only are there different structures (molecular arrangements) …… but there can also be a distribution of molecular weights (i.e. number of monomers per polymer molecule).
16 mers
20 mers
10 mers
Average molecular weight =
20 16 10 M monomer 15.3M monomer 3
This is what is called number average molecular weight.
Number average molecular weight:
N j M j
Mn j
mo N j j j
N j
Mj = jmo mass of polymer chain with length j (mo = monomer molecular weight).
N j
j
Note:
Nj = # of polymer chains with length j
j
N j
N j M j Total weight
Total # of polymer chains
j
j
Weight average molecular weight:
N j M 2j
W j M j
Mw j
W j
j
In general:
j
N j M j
W j N jM j
j
N j M j 1
M j
N j M j j
If = 0 then
Mn
If = 1 then
Mw
Molecular Weight: Different Notations In Lecture Notes
N j M j Mn
j
N j j
N j M 2j
Mw j
N j M j j
In Callister Textbook
Mn x i Mi i
Ni xi N j
wi
j
Ni M i N j M j j
Mw w i Mi i
Molecular Weights Why do we care about weight average MW? -some properties are dependent on MW (larger MW polymer chains can contribute to overall properties more than smaller ones).
Distribution of polymer weights
Examples – Light scattering: larger molecules scatter more light than smaller ones. Infrared absorption properties: larger molecules have more side groups and light absorption (due to vibrational modes of side groups) varies linearly with number of side groups.
Polydispersity and Degree of Polymerization Polydispersity:
Mw 1 Mn
When polydispersity = 1, system is monodisperse.
Degree of Polymerization: Number avg degree of polymerization
Mn nn mo
Weight avg degree of polymerization
Mw nw mo
Example 1 Compute the number-average degree of polymerization for polypropylene, given that the number-average molecular weight is 1,000,000 g/mol.
C3 H6
What is “mer” of PP?
Mer molecular weight of PP is
mo=3AC+6AH =3(12.01 g/mol)+6(1.008 g/mol) = 42.08 g/mol
Number avg degree of polymerization
Mn 106 g / mol nn 23,700 mo 42.08g / mol
Example 2 (a, b, and c) A. Calculate the number and weight average degrees of polymerization and polydispersity for a polymer sample with the following distribution. Avg # of monomers/chain 10 100 500 1000 5000 50,000
M m nn n 0 mo m0
jN N j
j
j
j
jN N j
j
Relative abundance 5 25 50 30 10 5
j
j
5 *10 25 *100 50 * 500 30 *1000 10 * 5000 5 * 50000 2860.4 5 25 50 30 10 5
M 1 nw w m o mo
( jm ) N j N N ( jm ) jN 2
2
o
j
j
j
j
o
j
j
j
Note: m0 cancels in all these!
j
5 *10 2 25 *100 2 50 * 500 2 30 *1000 2 10 * 5000 2 5 * 50000 2 35, 800 5 *10 25 *100 50 * 500 30 *1000 10 * 5000 5 * 50000
Example 2 (cont.) B. If the polymer is PMMA, calculate number and weight average molecular weights. Mw if monomer is methylmethacrylate (5C, 2O, and 8H) So m0= 5(12)+2(16)+8(1)= 100 g/mol
CH3 | -CH2-C| CO2CH3
Mn nnmo 2860.4(100g / mol ) 286,040g / mol Mw nwmo 35,800(100g / mol ) 3,580,000g / mol Mw 3,580,000 ~ 12.52 Polydispersity: Mn 286,040
Example 2 (cont.) C. If we add polymer chains with avg # of monomers = 10 such that their relative abundance changes from 5 to 10, what are the new number and weight average degrees of polymerization and polydispersity? M nn = n = mo =
jN N j
j
j
Add 5 more monomers of length 10 ….
j
10 * 10 + 25 * 100 + 50 * 500 + 30 * 1000 + 10 * 5000 + 5 * 50000 = 2750 10 + 25 + 50 + 30 + 10 + 5
2 Mw j j N j nw 35,800 mo j jN j
Note: significant change in number average (3.8 %) but no change in weight average!
Polydispersity:
M w 3, 580, 000 = ~ 13 Mn 275000
Sequence isomerism For an asymmetric monomer T
H
+
T
H
e.g. poly(vinyl fluoride):
T
H
T
H
T
H
H
T
H
T
T
H
e.g. PMMA
H
F
H
H
F
H
H
H
C
C
C
C C
C
C
C
H3C H3C H3C O O O O O O O O H C H C H C H C
H
H
H
F H
H
H
F
C
C
C
C C
C
H
CH3 H
CH3H
CH3 H
H to T
T to T H to H
H3C
H to T
Random arrangement
H to T
C
C CH3
H to T
Exclusive H to T arrangement (Why?)
Polymer Molecular Configurations • Regularity and symmetry of side groups affect properties
Polymerize
Can it crystallize? Melting T?
• Stereoisomerism: (can add geometric isomerism too) Syndiotactic Alternating sides
Isotactic On one side
Atactic Randomly placed
- Conversion from one stereoisomerism to another is not possible by simple rotation about single chain bond; bonds must be severed first, then reformed!
Polymer Geometrical Isomerism • Regularity and symmetry of side groups affect properties
H
H
cis-structure
trans-structure
with R= CH3 to form rubber Cis-polyisoprene trans-polyisoprene -Conversion from one isomerism to another is not possible by simple rotation about chain bond because double-bond is too rigid! -See Figure 4.8 for taxonomy of polymer structures
Polymer Structural Isomerism Some polymers contain monomers with more than 1 reactive site e.g. isoprene
2 1
H2C
4
CH3 C
C H
CH2 3
trans-isoprene
trans-1,4-polyisoprene
CH3 C H2
C
C H
trans-1,2-polyisoprene
3,4-polyisoprene
H2 C
H2 C
C n
H3C
n
H2 C H C
CH H2C
Note: there are also cis-1,4- and cis-1,2-polyisoprene
C H2C
n
CH3
Polymer Microstructure • Covalent chain configurations and strength: More rigid
Van der Waals, H secondary
bonding
Linear
Branched
Cross-Linked
Network
Direction of increasing strength
Adapted from Fig. 14.7, Callister 6e.
Short branching
Long branching
Star branching
Dendrimers
CoPolymers • Random, Alternating, Blocked, and Grafted • Synthetic rubbers are often copolymers. e.g., automobile tires (SBR)
Styrene-Butadiene Rubber random polymer
Molecular Structure How do crosslinking and branching occur in polymerization? 1. Start with or add in monomers that have more than 2 sites that bond with other monomers, e.g. crosslinking polystyrene with divinyl benzene …
stryene
…
polystyrene
Control degree of + crosslinking by styrene-divinyl styrene benzene ratio divinyl benzene
…
…
crosslinked polystyrene
Monomers with trifunctional groups lead to network polymers.
Molecular Structure Branching in polyethylene (back-biting)
H2C CH2
R
H2 C
C H2
H2 C
Same as
C H2
H2 C
C H
H
H
H
H
H C
R
H
CH2
Radical moves to a different carbon
C
CH2
(H transfer)
C H H2
C
H
CH2
C
R
CH2
C H H 2
Polymerization continues from this carbon
Process is difficult to avoid and leads to (highly branched) low-density PE . When there is small degree of branching you get high-density PE.
Example 3 Nitrile rubber copolymer, co-poly(acrylonitrile-butadiene), has
Mn 106,740g / mol
nn 2000
Calculate the ratio of (# of acrylonitrile) to (# of butadiene). 3 C = 3 x 12.01 g/mol 3 H = 3 x 1.008 g/mol 1 N = 1 x 14.007 g/mol
4 C = 4 x 12.01 g/mol 6 H = 6 x 1.008 g/mol m0= 54.09 g/mol
m0= 53.06 g/mol 1,4-addition product
We need to use an avg. monomer MW:
mo
Mn 106,740 53.57g / mol 2000 nn
mo f1m1 f2m2 f1(m1 m2 ) m2 m m2 53.37 54.09 f1 0 0.7 m1 m2 53.06 54.09
f2 1 f1 0.3
f2 0.7 7 :3 f1 0.3
Vulcanization
See also sect. in Chpt. 8
• Crosslinking in elastomers is called vulcanization, and is achieved by irreversible chemical reaction, usually requiring high temperatures.
• Sulfur compounds are added to form chains that bond adjacent polymer backbone chains and crosslinks them. • Unvulcnaized rubber is soft and tacky an poorly resistant to wear. e.g., cis-isoprene
Double bonds
+ (m+n) S
Single bonds
(S)m (S)n
Stress-strain curves
Molecular Weight and Crystallinity • Molecular weight, Mw: Mass of a mole of chains. smaller Mw
larger Mw
• Tensile strength (TS): --often increases with Mw. --Why? Longer chains are entangled (anchored) better.
• % Crystallinity: % of material that is crystalline. --TS and E often increase with % crystallinity. --Annealing causes crystalline regions to grow. % crystallinity increases.
crystalline region amorphous region
Adapted from Fig. 14.11, Callister 6e.
Polymer Crystallinity polyethylene
• Some are amorphous. • Some are partially crystalline (semi-crystalline). • Why is it difficult to have a 100% crystalline polymer?
%crystallinity
c ( s a ) 100% s ( c a )
s = density of specimen in question a = density of totally amorphous polymer c = density of totally crystalline polymer
Volume fraction of crystalline component.
Mcrystalline %crystallinity 100% Mtotal
cVc 100% sVs
c fc 100% s
Mtotal Mcrystalline Mamophous Using definition of volume fractions: V Ms Mc Ma V fa a fc c Vs Vs sVs cVc aVa V V s c c a a c fc a fa c fc a (1 fc ) fc ( c a ) a Vs Vs
s a fc c a
Substituting in fc into the original definition:
c ( s a ) %crystallinity 100% s ( c a )
Polymer Crystallinity Degree of crystallinity depends on processing conditions (e.g. cooling rate) and chain configuration. Cooling rate: during crystallization upon cooling through MP, polymers become highly viscous. Requires sufficient time for random & entangled chains to become ordered in viscous liquid. Chemical groups and chain configuration: More Crystalline
Less Crystalline
Smaller/simper side groups
Larger/complex side groups
Linear
Highly branched Crosslinked, network
Isotactic or syndiotactic
Random
Semi-Crystalline Polymers Fringed micelle model: crystalline region embedded in amorphous region. A single chain of polymer may pass through several crystalline regions as well as intervening amorphous regions.
s a fc c a Crystalline volume fractions Important
Semi-Crystalline Polymers Chain-folded model: regularly shaped platelets (~10 – 20 nm thick) sometimes forming multilayers. Average chain length
>> platelet thickness.
Semi-Crystalline Polymers Spherulites: Spherical shape composed of aggregates of chain-folded crystallites.
Natural rubber
Cross-polarized light through spherulite structure of PE.
Diblock copolymers
Representative polymer-polymer phase behavior with different architectures:
F. Bates, Science 1991.
A) Phase separation with mixed LINEAR homopolymers. B) Mixed LINEAR homopolymers and DIBLOCK copolymer gives surfactant-like stabilized state. C) Covalent bond between blocks in DIBLOCK copolymer give microphase segregation.
Thermoplastics vs Thermosets • Thermoplastics: --little cross linking --ductile --soften w/heating --polyethylene (#2) polypropylene (#5) polycarbonate polystyrene (#6)
• Thermosets:
T mobile liquid
viscous liquid
crystalline solid
Callister, rubber Fig. 16.9 tough plastic
Tm Tg
partially crystalline solid
Molecular weight
--large cross linking (10 to 50% of mers) Adapted from Fig. 15.18, Callister 6e. --hard and brittle Tm: melting over wide range of T --do NOT soften w/heating depends upon history of sample --vulcanized rubber, epoxies, consequence of lamellar structure thicker lamellae, higher Tm. polyester resin, phenolic resin Tg: from rubbery to rigid as T lowers
Packing of Polymers • Packing of “spherical” atoms as in ionic and metallic crystals led to crystalline structures. • How polymers pack depend on many factors: • long or short, e.g. long (-CH2-)n. • stiff or flexible, e.g. bendy C-C sp3. • smooth or lumpy, e.g., HDPE. • regular or random • single or branched • slippery or sticky, e.g. C-H covalent (nonpolar) joined via vdW.
Analogy: Consider dried (uncooked) spaghetti (crystalline) vs. cooked and buttered spaghetti (amorphous). • pile of long “stiff” spaghetti forms a random arrangement. • cut into short pieces and they align easily.
Candle wax more crystalline than PE, even though same chemical nature.
What Are Expected Properties? • Would you expect melting of nylon 6,6 to be lower than PE? nylon
O H O H || | || 6,6 | N C N C C N C | | | | | 4 H 6 H H H H
+ + + + Hydrogen bonds + + O ||
O ||
H H | | N C N C C N C | | | | | H H 6 H 4 H H
a) b)
H
H
C C H H
polyethylene
+ + + bonds + Waals Van der + +
H
H
C C H H
What is the source of intermolecular cohesion in Nylon vs PE? How does the source of linking affect temperature?
With H-bonds vs vdW bonds, nylon is expected to have (and does) higher melting T.
What Are Expected Properties? Which polymer more likely to crystallize? Can it be decided? Linear syndiotactic polyvinyl chloride
Linear isotactic polystyrene
• For linear polymers, crystallization is more easily accomplished as chain alignment is not prevented. • Crystallization is not favored for polymers that are composed of chemically complex mer structures, e.g. polyisoprene. • Linear and syndiotactic polyvinyl chloride is more likely to crystallize. • The phenyl side-group for PS is bulkier than the Cl side-group for PVC. • Generally, syndiotactic and isotactic isomers are equally likely to crystallize.
What Are Expected Properties? Which polymer more likely to crystallize? Can it be decided? Networked Phenol-Formaldehyde (Bakelite)
Linear and highly crosslink cis-isoprene
+
H
+ H20
• Networked and highly crosslinked structures are near impossible to reorient to favorable alignment.
• Not possible to decide which might crystallize. Both not likely to do so.
What Are Expected Properties? Which polymer more likely to crystallize? Can it be decided? alternating Poly(Polystyrene-Ethylene) Copolymer
random poly(vinyl chloride-tetra-fluoroethylne) copolymer
• Alternating co-polymer more likely to crystallize than random ones, as they are always more easily crystallized as the chains can align more easily.
Detergents • Soap is a detergent based on animal or vegetable product, some contain petrochemicals
water
detergent grease
• What properties of soap molecules do you need to remove grease? • “green” end must be “hydrophilic”. Why? • Opposite end must be hydrocarbon. Why?
Water must be like oxygen (hoard electrons and promote H-bonding)
e.g., oxy-clean®
grease
Simple polymer: Elmers glue + Borax SLIME! Chemistry
Elmer’s glue is similar to “poly (vinyl alcohol)” with formula: OH
OH
OH
OH
OH
OH
OH
OH
OH
OH
OH
OH
OH
this is a SHORT, n=15 chain of poly(vinyl alcohol)
Borax is sodium tetraborate decahydrate (B4Na2O7 • 10 H2O). The borax actually dissolves to form boric acid, B(OH)3. This boric acid-borate solution is a buffer with a pH of about 9 (basic). Boric acid will accept a hydroxide OH- from water.
B(OH)3 + 2H2O B(OH)4- + H3O+ pH=9.2
Hydrolyzed molecule acts in a condensation reaction with PVA, crosslinking it.
OH
OH
Simple polymer: Elmer’s glue + Borax SLIME! Hydrolyzed molecule acts in a condensation reaction with PVA, crosslinking it. B(OH)3 + 2H2O B(OH)4- + H3O+ pH=9.2
Crosslinked
Crosslinking ties chains via weak non-covalent (hydrogen) bonds, so it flows slowly.
Range of Bonding and Elastic Properties Is “slime” a thermoset or thermoplastic, or neither?
Thermoset bonding • Covalent bonds form crosslinks
Slime?
• H-bonds form crosslinks
Stiffness increases Where is nylon?
Thermoplastic bonding • Induced dipolar bonds form crosslinks
Summary • Polymers are part crystalline and part amorphous. • The more “lumpy” and branched the polymer, the less dense and less crystalline. • The more crosslinking the stiffer the polymer. And, networked polymers are like heavily crosslinked ones. • Many long-chained polymers crystallize with a Spherulite microstructure - radial crystallites separated by amorphous regions. • Optical properties: crystalline -> scatter light (Bragg) amorphous -> transparent. Most covalent molecules absorb light outside visible spectrum, e.g. PMMA (lucite) is a high clarity tranparent materials.
Komposit (Composites) Material Teknik
181
Many engineering components are composites
182
ISSUES TO ADDRESS... • What are the classes and types of composites? • Why are composites used instead of metals, ceramics, or polymers? • How do we estimate composite stiffness & strength? • What are some typical applications?
183
Classification of Composites • Composites: - Multiphase material w/significant proportions of ea. phase.
woven fibers
• Matrix: - The continuous phase - Purpose is to:
0.5mm
transfer stress to other phases protect phases from environment
- Classification: MMC, CMC, PMC metal
ceramic
cross section view
polymer
• Dispersed phase: -Purpose: enhance matrix properties. MMC: increase sy, TS, creep resist. CMC: increase Kc PMC: increase E, sy, TS, creep resist.
0.5mm From D. Hull and T.W. Clyne, An Intro to Composite Materials, 2nd ed., Cambridge University Press, New York, 1996, Fig. 3.6, p. 47.
-Classification: Particle, fiber, structural 184
COMPOSITE SURVEY: Particle-I Particle-reinforced Fiber-reinforced • Examples: -Spheroidite matrix: ferrite () steel (ductile)
60m
-WC/Co cemented carbide
matrix: cobalt (ductile) Vm: 10-15vol%!
Structural particles: cementite (Fe3C) (brittle)
Adapted from Fig. 10.10, Callister 6e.
particles: WC Adapted from Fig. (brittle, 16.4, Callister 6e. hard) 600m
-Automobile matrix: rubber tires
particles: Adapted from Fig. C 16.5, Callister 6e. (stiffer)
(compliant) 0.75m
185
COMPOSITE SURVEY: Particle-II Particle-reinforced Fiber-reinforced
Structural
• Elastic modulus, Ec, of composites: -- two approaches. E(GPa) 350 Data: Cu matrix 300 w/tungsten 250 particles 200 150
upper limit: “rule of mixtures” Ec VmEm VpEp
0
(Cu)
lower limit: 1 Vm Vp Ec Em Ep Fig. 15.3
20 40 60 80 100 vol% tungsten
(W)
• Application to other properties: -- Electrical conductivity, se: Replace E by se. -- Thermal conductivity, k: Replace E by k. 186
COMPOSITE SURVEY: Fiber-I Particle-reinforced Fiber-reinforced Structural • Aligned Continuous fibers • Ex: --Metal: g'(Ni3Al)-a(Mo) --Glass w/SiC fibers by eutectic solidification.
matrix: (Mo) (ductile)
formed by glass slurry Eglass = 76GPa; ESiC = 400GPa.
(a)
2m
fibers:’ (Ni3Al) (brittle) From W. Funk and E. Blank, “Creep deformation of Ni3Al-Mo in-situ composites", Metall. Trans. A Vol. 19(4), pp. 987-998, 1988.
(b)
fracture surface From F.L. Matthews and R.L. Rawlings, Composite Materials; Engineering and Science, Reprint ed., CRC Press, Boca Raton, FL, 2000. (a) Fig. 4.22, p. 145 (photo by J. Davies); (b) Fig. 11.20, p. 349 (micrograph by H.S. Kim, P.S. Rodgers, and R.D. Rawlings).
187
COMPOSITE SURVEY: Fiber-II Particle-reinforced Fiber-reinforced • Discontinuous, random 2D fibers • Example: Carbon-Carbon --process: fiber/pitch, then burn out at up to 2500C. --uses: disk brakes, gas turbine exhaust flaps, nose cones.
(b)
Structural
C fibers: very stiff very strong C matrix: less stiff view onto plane less strong fibers lie in plane
(a)
• Other variations: --Discontinuous, random 3D --Discontinuous, 1D
188
Elasticity of Composites Stress-strain response depends on properties of • reinforcing and matrix materials (carbon, polymer, metal, ceramic) • volume fractions of reinforcing and matrix materials • orientation of fibre reinforcement (golf club, kevlar jacket) • size and dispersion of particle reinforcement (concrete) • absolute length of fibres, etc.
concentration
size
distribution
shape
orientation
189
Families of Composites: particle, fibre, structural reinforcements
ceramics Twisting, Bending Orientation dependence
190
Two simplest cases: Iso-load and Iso-strain Isostrain:
V V VTot
Load & Reinforcements Aligned F Strain or elongation of matrix and fibers are the same!
Ec 1, N E V%
Volume fraction
F Isoload:Load & Reinforcements Perpendicular (Isostress below) F Load (Stress) across matrix and fibers is the same!
V% 1 1, N E Ec F
191
Isostrain Case in Ideal Composites F Isostrain Case:
strain
c m r
forces
Fc Fm Fr F
Load is distributed over matrix and fibers, so cAc = mAm + fAf.
c m (Am / A ) f (A f / A ) c
c
% or c mV% m f Vf
*if the fibers are continuous, then volume fraction is easy.
% % % For Elastic case: c c Ec m EmV% m f E f V f c (E mVm E f V f ) Composite Property:
Pc 1, N P V%
* like law of mixtures
Properties include: elastic moduli, density, heat capacity, thermal expansion, specific heat, ... 192
Iso-Load Case for Ideal Composites Isoload Case:
strain
c m r
forces
Fc Fm Fr
Without de-bonding, loads are equal, therefore, strains must add, so % % *if the fibers are continuous or % % elastic case c mVm f V f Vm V f planar, then area of applied Em Ef stress is the same.
Composite Property:
V% 1 1, N Pc P
* like resistors in parallel.
Properties include: elastic moduli, density, heat capacity, thermal expansion, specific heat, ... 193
ISOSTRAIN Example Suppose a polymer matrix (E= 2.5 GPa) has 33% fibre reinforcements of glass (E = 76 GPa). What is Elastic Modulus?
˜ E (1 V ˜ )Em V ˜ E ˜ mE m V ˜ E V Ec V f f f f f f f = 26.7 GPA
~ 25 GPA
* Stiffness of composite under isostrain is dominated by fibers.
194
ISOLOAD Example Suppose a polymer matrix (E= 2.5 GPa) has 33% fibre reinforcements of glass (E = 76 GPa).
˜ ˜ V V 1 What is Elastic Modulus? m f Ec Em Ef Rearrange:
E C
EmEf ˜ E (1 V ˜ )E V f m f f
Em ˜ ) (1 V f
= 3.8 GPA
* Elastic modulus of composite under isoload condition strongly depends on stiffness of matrix, unlike isostrain case where stiffness dominates from fibers. 195
Modulus of Elasticity in Tungsten Particle Reinforced Copper
isostrain
isoload
•Particle reinforcements usually fall in between two extremes. 196
Simplified Examples of Composites
Are these isostrain or isoload? What are some real life examples?
197
Simplified Examples of Composites A
B
||
Load, F • Material A and B are different • e.g., walkway, trapeze bar,…
• Fiber reinforced epoxy cylinder • e.g, pressure cylinder
unloading
3" 4" 6"
200,000
B A
F
F
60,000 permanent strain.
0
• welded tubular composite 198
Self-Assessment Example: isostrain A
B
Load, F
A platform is suspended by two parallel rods (A and B). Yielding of either rod of this “composite” constitutes failure, such as the falling (and possible death) of the trapeze artist, the people using the walkway, etc. Each rod is 1.28 cm in diameter. Rod A is 4340 steel, with E= 210 GPa, ys= 855 MPa. Rod B is 7075-T6 Al alloy, with E= 70 GPa, ys= 505 MPa.
(a) What uniform load can be applied to the platform before yielding will occur? If not elastic, then composite fails, due to permanent deformation! Hence A ys
B 855MPa 505MPa 3 A 4.07x 10 B ys 7.21x 103 E A 210GPa EB 70GPa
C A B
F FA FB Arod ( A B ) Arod (E AA EB B ) Arod rod (E A EB ) F
(1.28x 102 m )2 4
(4.07x 103 )(210 70)GPa 146.6kN
(b) Which rod will be first to yield? Justify and explain your answer.
C A B
Steel yieldsfirst! To justify, consider how much load is carried FA/FB relative to that expected from the YS.
199
COMPOSITE SURVEY: Fiber-III Particle-reinforced Fiber-reinforced Structural • Critical fiber length for effective stiffening & strengthening: fiber strength in tension
f d fiber length 15 c
fiber diameter shear strength of fiber-matrix interface
• Ex: For fiberglass, fiber length > 15mm needed • Why? Longer fibers carry stress more efficiently! Shorter, thicker fiber: d fiber length 15 f (x)
c
Longer, thinner fiber: d fiber length 15 f (x)
c
Adapted from Fig.15.7
Poorer fiber efficiency
Better fiber efficiency 200
COMPOSITE SURVEY: Fiber-IV Particle-reinforced
Fiber-reinforced
Structural
• Estimate of Ec and TS:
d --valid when fiber length 15 f c
-- Elastic modulus in fiber direction:
Ec EmVm KEfVf efficiency factor: --aligned 1D: K = 1 (anisotropic) --random 2D: K = 3/8 (2D isotropy) --random 3D: K = 1/5 (3D isotropy)
Values from Table 15.3
--TS in fiber direction:
(TS)c (TS)mVm (TS)f Vf
(aligned 1D)
201
COMPOSITE SURVEY: Structural Particle-reinforced
Fiber-reinforced
Structural
• Stacked and bonded fiber-reinforced sheets -- stacking sequence: e.g., 0/90 -- benefit: balanced, in-plane stiffness Fig. 15.16
• Sandwich panels -- low density, honeycomb core -- benefit: small weight, large bending stiffness face sheet adhesive layer honeycomb
Fig. 15.17
202
Composite Benefits • CMCs: Increased toughness Force
103 E(GPa) PMCs 102 10
particle-reinf
fiber-reinf
ceramics
1
un-reinf
Bend displacement 10-4 6061 Al ss (s-1) • MMCs: 10-6
Increased creep resistance
• PMCs: Increased E/r
10-8
metal/ metal alloys
.1 G=3E/8 polymers .01 K=E .1 .3 1 3 10 30 Density, [Mg/m3]
6061 Al w/SiC whiskers
Adapted from T.G. Nieh, "Creep rupture of a silicon-carbide reinforced aluminum composite", Metall. Trans. A Vol. 15(1), pp. 139-146, 1984.
(MPa) 10-10 20 30 50 100 200 203
Laminate Composite (Ideal) Example Gluing together these composite layers composed of epoxy matrix (Em= 5 GPa) with graphite fibres (Ef= 490 GPa and Vf = 0.3). Central layer is oriented 900 from other two layers. Case I - Load is applied parallel to fibres in outer two sheets. Case II - Load is applied parallel to fibres of central sheet.
What are effective elastic moduli in the two case? • First need to know how individual sheets respond, then average. 0.3 0.7 1 E 7.1 GPa E 490 GPa 5 GPa E|| 0.3(490 GPa) 0.7(5 GPa) E|| 150.5 GPa
For isoload case. For isotrain case.
Case I: Elam=(2/3)(150.5 GPa) + (1/3)(7.1 GPa) = 102.7 GPa Case II: Elam=(1/3)(150.5 GPa) + (2/3)(7.1 GPa) = 54.9 GPa 204
Mechanical Response of Laminate: Complex, NOT Ideal 3 Conditions required: consider top and bottom before laminated • strain compatibility- top and bottom must have same strain when glued. • stress-strain relations - need Hooke’s Law and Poisson effect. • equilibrium - forces and torques, or twisting and bending.
top
Isostrain for load along x-dir:
x
Poisson Effect and Displacements in D:
E
top
bott
x E bott E top top bott y y E bott
• When glued together displacements have to be same! • Unequal displacements not allowed! So, top gets wider (ytop > 0) and bottom gets narrower (ybott < 0). Equilibrium: Fy = 0 = (ybot tbot + ytop ttop)L.
(t = thickness) 205
COMPATIBILITY: When glued, displacements have to be same!
As stress is applied, compatibility can be maintained, depending on the laminate, only if materials twists.
206
Symmetry of laminate composite dictates properties
Elastic constants are different for different symmetry laminates. 207
Orientation of layers dictates response to stresses
Want compressive stresses at end of laminate so there are no tensile stresses to cause delamination - failure!
208
NO delamination - failure!
Apply in-pane Tensile Stress A B +90 +45 +45 –45 –45 +90 –45 +90 +45 –45 +90 +45 Tensile -> delaminate Compressive 209
Why Laminate Composite is NOT Ideal • Depending on placement of load and the orientation of fibers internal to sheet and the orientation of sheets relative to one another, the response is then very different. • Examples of orientations of laminated sheets that provided compressive stresses at edges of composite and also tensile stresses there. >>>> Tensile stresses lead to delamination! • The stacking of composite sheets and their angular orientation can be used to prevent “twisting” moments but allow “bending” moments. This is very useful for airplane wings, golf club shafts (to prevent slices or hooks), tennis rackets, etc., where power or lift comes or is not reduced from bending.
210
Thermal Stresses in Composites • Not just due to fabrication, rather also due to thermal expansion differences between matrix and reinforcements Tm and Tr. • Thermal coatings, e.g. | m – r | TE T TEc
T
At T1
T
At T2
If forced to be compatible, composite will bend and rotate
• Material with most contraction (least) has positive (negative) residual stress. (For non-ceramics, you should consider plastic strain too.) • Ceramic-oxide thermal layers, e.g. on gas turbine engines: • ceramic coating ZrO2-based (lower Tr) • metal blade (NixCo1-x)CrAlY (higher Tm) • Failure by delamination without a good design of composite, i.e. compatibility maintained. 211
Summary • Composites are classified according to: -- the matrix material (CMC, MMC, PMC) -- the reinforcement geometry (particles, fibers, layers).
• Composites enhance matrix properties: -- MMC: enhance sy, TS, creep performance -- CMC: enhance Kc -- PMC: enhance E, sy, TS, creep performance • Particulate-reinforced: -- Elastic modulus can be estimated. -- Properties are isotropic. • Fiber-reinforced: -- Elastic modulus and TS can be estimated along fiber dir. -- Properties can be isotropic or anisotropic. • Structural: -- Based on build-up of sandwiches in layered form. 212
