material teknik universitas gunadarma hand out

Material Teknik(Engineering Materials)

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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

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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

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Material

• Material adalah sesuatu yang disusun/dibuat oleh bahan.

• Material digunakan untuk transportasi hingga makanan.

• Ilmu material/bahan merupakan pengetahuan dasar tentang struktur, sifat-sifat dan pengolahan bahan.

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Jenis Material• Logam

Kuat, ulet, mudah dibentuk dan bersifat penghantar panas dan listrik yang baik

• KeramikKeras, getas dan penghantar panas dan listrik yang buruk

• Polimerkerapatan rendah, penghantar panas dan listrik buruk dan mudah dibentuk

• Kompositmerupakan ganbungan dari dua bahan atau lebih yang masing-masing sifat tetap

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Logam

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Keramik

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Polimer

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Komposit

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Struktur dan Ikatan Atom

Material Teknik

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Pendahuluan• Atom terdiri dari elektron dan inti atom• Inti atom disusun oleh proton dan neutron• Elektron mengelilingi inti atom dalam orbitnya masing-

masing• Massa elektron 9,109 x 10-28 g dan bermuatan –1,602 x 10-

19 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 unsurNo. Element K L M N O P Q

1 2 3 4 5 6 7 s s p s p d s p d f s p d f s p d f s

1 H 12 He 23 Li 2 14 Be 2 25 B 2 2 16 C 2 2 27 N 2 2 38 O 2 2 49 F 2 2 510 Ne 2 2 611 Na 2 2 6 112 Mg 2 2 6 213 Al 2 2 6 2 114 Si 2 2 6 2 215 P 2 2 6 2 316 S 2 2 6 2 417 Cl 2 2 6 2 518 Ar 2 2 6 2 619 K 2 2 6 2 6 - 120 Ca 2 2 6 2 6 - 221 Sc 2 2 6 2 6 1 222 Ti 2 2 6 2 6 2 223 V 2 2 6 2 6 3 224 Cr 2 2 6 2 6 5* 125 Mn 2 2 6 2 6 5 226 Fe 2 2 6 2 6 6 227 Co 2 2 6 2 6 7 228 Ni 2 2 6 2 6 8 229 Cu 2 2 6 2 6 10 1*30 Zn 2 2 6 2 6 10 231 Ga 2 2 6 2 6 10 2 132 Ge 2 2 6 2 6 10 2 233 As 2 2 6 2 6 10 2 334 Se 2 2 6 2 6 10 2 435 Br 2 2 6 2 6 10 2 536 Kr 2 2 6 2 6 10 2 6

* l

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Tabel Periodik

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Elektronegatip dari Unsur

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Ikatan Atom Ionik

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Ikatan Atom Kovalen

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Ikatan Atom Logam

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Ikatan Atom Hidrogen

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Bilangan Koordinasi utk Ikatan Atom

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Struktur dan Cacat Kristal

Material Teknik

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Pendahuluan

• Kristal adalah susunan atom-atom secara teratur dan kontinu pada arah tiga dimensi

• Satuan sel adalah susunan terkecil dari kristal

• Parameter kisi struktur kristal– Panjang sisi a, b, c– Sudut antara sumbu

b

ac

x

y

z

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Sistem Kristal Parameter kisi diklasifikasikan dalam tujuh sistem kristal dan empat belas kisi kristal• Arah kristal

dinyatakan sebagai vektor dalam [uvw]

• uvw merupakan bilangan bulat

• Himpunan arah <111> terdiri dari [111], [111], [111], [111], [111], [111], [111], [111]

[100]

b

ac

x

[111]

[110]

z

y

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Menentukan Indeks Miller Arah Kristal• Prosedur menentukan arah

kristalx y z

Proyeksi a/2 b 0Proyeksi (dlm a, b, c)

½ 1 0Reduksi 1 2 0Penentuan [120]

cy

b

a

x

Proyeksi pd sb y: b

z

Proyeksi pd sb x: a/2

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Bidang Kristal

• Dinyatakan dengan (hkl)

• hkl merupakan bilangan bulat

bac

x

Bid (110) mengacu titik asal O

Bid. (110) ekivalen

z

y

bac

x

Bid (111) mengacu titik asal O

Bid. (111) ekivalen

z

y

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Menentukan Indeks Miller Bidang Kristal• Prosedur menentukan bidang

kristalx y z

Perpotongan ~a -b c/2Perpotongan (dlm a, b dan c)

~ -1 ½Resiprokal 0 -1 2Penentuan (012)

cy

b

a

x

z

bid.(012)

z’

x’

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14 kisi kristal

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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.

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Kristal Kubik Berpusat Bidang

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Kristal Heksagonal Tumpukan Padat

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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 volum31

Cacat Titik

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Dislokasi Garis

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Dislokasi Ulir

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Batas Butir

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Permukaan

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Inklusi

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Sifat MekanikMaterial Teknik

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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

Material

Gaya/beban

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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

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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)

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Konsep tegangan (stress) dan regangan (strain)

• Pembebanan statik:– Tarik– Kompressi– Geser

F

F

F

F

Beban tarikBeban kompressi

F

FBeban geser 42

Uji tarikStandar sampel untuk uji tarik

• Tegangan teknik, = F/Ao (N/m2=Pa)• Regangan teknik, = (li-lo)/lo

• Tegangan geser, = F/Ao

2¼’

2’

¾’ 0,505’

R 3/8’

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Deformasi elastis

• Pada pembebanan rendah dalam uji tarik, hubungan antara tegangan dan regangan linier

Teg.

Reg.

Modulus elastis

Pembebanan

Beban dihilangkan

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Mesin uji tarik (Tensile Test)

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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

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Paduan logam

Modulus elastis (104 MPa)

Modulus geser (104 MPa)

Ratio Poisson

Al 6,9 2,6 0,33Cu-Zn 10,1 3,7 0,35

Cu 11,0 4,6 0,35Mg 4,5 1,7 0,29Ni 20,7 7,6 0,31

Baja 20,7 8,3 0,27Ti 10,7 4,5 0,36W 40 7 16 0 0 28

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• Hubungan tegangan geser dan regangan geser dinyatakan dengan = G

• Dengan = teg.geser = reg.geserG = modulus geser

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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

z

x

y

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= x/y

• Harga selalu positip, karena tanda x dan yberlawanan.

• Hubungan modulus Young dengan modulus geser dinyatalan dengan

E = 2 G (1 + )• Biasanya <0,5 dan utk logam umumnya

G = 0,4 E

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Deformasi plastis

• Utk material logam, umumnya deformso elastis terjadi < 0,005 regangan

• Regangan > 0,005 terjadi deformasi plastis (deformasi permanen)

Teg.

Teg.

Reg.

Reg.

ys

ys

Titik

luluh atas

Titik

Luluh bawah

0,002

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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

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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)

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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%

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Ketangguan (Toughness)

• Perbedaan antara kurva tegangan dan regangan hasil uji tarik utk material yang getas dan ulet

• ABC : ketangguhan material getas

• AB’C’ : ketangguhan material ulet

Teg.

Reg.A

B

B’

C C’

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Logam Kekuatan luluh (MPa)

Kekuatan tarik (MPa)

Keuletan %El.

Au - 130 45Al 28 69 45Cu 69 200 45Fe 130 262 45Ni 138 480 40Ti 240 330 30Mo 565 655 35 56

Tegangan dan regangan sebenarnya• Pada daerah necking,

luas tampang lintang sampel uji material

• Tegangan sebenarnyaT = F/Ai

• Regangan sebenarnyaT = ln li/lo

Teg.

Reg.

teknik

sebenarnya

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Bila volum sampel uji tidak berubah, maka Aili = Aolo

• Hubungan tegangan teknik dengan tegangan sebenarnya

T = (1 + )• Hubungan regangan teknik dengan

regangan sebenarnyaT = ln (1+ )

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Uji Kekerasan (Hardness Test)

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Uji Mulur (Creep Test)

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Uji Kelelahan (Fatique Test)

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Patahan (Failure)

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Diagram FasaMaterial Teknik

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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.

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• 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.

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• 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.

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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

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Diagram Cu-Ni

• Jumlah persentasi cair (Wl) = S/(R+S)x100%

• Jumlah persentasi a (W) = R/(R+S)x100%

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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

B

C

E

F

G

H

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Diagram Fasa Pb-Sn• Reaksi eutektik

Cair (61,9%Sn) (19,2%Sn)+(97,6%Sn)

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Diagram Fasa Cu-Zn

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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

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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

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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 Al-

Si mengandung Si <12,6%

• Paduan eutektik Al-Si mengandung Si sekitar 12,6%

• Paduan hipereutektik Al-Si mengandung Si >12,6%

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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

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• 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

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Anil Sempurna

• Bahan: baja• Tujuan: Pelunakan

sebelum pemesinan• Prosedur: austenisasi

2-30C• Perubahan

strukturmikro: pearlit kasat

+Fe3C

700

800

900

C0,77%C

anilnormalisasi

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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

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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.

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Normalisasiterdiri 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-

praeutektoid82

Anil

83

Recovery, Rekristalisasi, Pertumbuhan Butir

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Proses Presipitasi

• Pengerasan presipitasi dilakukan dengan memanaskan logam hingga unsur pemadu larut, kemudian celup cepat, dan dipanaskan kembali pada suhu relatip rendah

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Diagram Transformasi-Isotermal

86

Diagram Transformasi-Isotermal untuk Baja Eutektoid

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Logam BesiMaterial Teknik

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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%<C<0,6%, utk roda gigi, axle, batang penghubung, crankshaft, rel, komponen utk mesin pengerjaan logam

– Baja karbon tinggi: 0,6%<C<1,0%, utk mata pahat, kabel, kawat musik, pegas

90

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 tinggiKekuatan luluh Komposis kimia Deoksidasi

103 Psi MPa

35 240

S = kualitas struktur

X = paduan rendah

W = weathering

D = fasa ganda

F = kill + kontrol S

K = kill

O = bukan kill

40 275

45 310

50 350

60 415

70 485

80 550

100 690

120 830

140 970

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


• 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


• 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 BesiMaterial 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

AlimuniumProduk Wrough

1xxx Al murni: 99,00%2xxx Al+Cu3xxx Al+Mn4xxx Al+Si5xxx Al+Mg6xxx Al+Mg+Si7xxx Al+Zn8xxx Al+unsur lain

107

AlimuniumProduk Cor

1xx.x Al murni: 99,00%2xx.x Al+Cu3xx.x Al+Si, Cu, Mg4xx.x Al+Si5xx.x Al+Mg6xx.x Tidak digunakan7xx.x Al+Zn8xx.x Al+Pb

108

Perlakuan utk produk aluminium wrough dan corF 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)

T Perlakuan panas

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 corPaduan Komposisi (%) Kondisi Pembentukk

anAl Zn Mn Zr

AZ31B 3,0 1,0 0,2 F H24 Ekstrusi lembaran & pelat

AZ80A 8,5 0,5 0,2 T5 Ekstrusi & tempa

HK31A 0,7 H24 Lembaran & pelat

ZK60A 5,7 0,55 T5 Ekstrusi & tempa 111

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-T6A AlZ Zn9 9%Al1 1%ZnC Standar CT6 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

KeramikMaterial 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-anionBilangan koordinasi

Perbandingan jari-jari kation-anion

Geometri koordinasi

2 <0,155

3 0,115-0,225

4 0,225-0,414

6 0,414-0,732

8 0,723-1,0 120

Jari-jari kation dan anionKation 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 AXCth.; 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

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 Structures

• Polymer = many mers

C C C C C CHHHHHH

HHHHHH

Polyethylene (PE)

mer

ClCl Cl

C C C C C CHHH

HHHHHH

Polyvinyl chloride (PVC)

mer

Polypropylene (PP)

CH3

C C C C C CHHH

HHHHHH

CH3 CH3

mer

Adapted from Fig. 14.2, Callister 6e.

Polymer Microstructure

Polyethylene perspective of molecule

A zig-zag backbone structure with covalent bonds

• Covalent chain configurations and strength:

Direction of increasing strengthAdapted from Fig. 14.7, Callister 6e.

Branched Cross-Linked NetworkLinear

secondarybonding


Van der Waals, H More rigid

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…

• 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 Classification

Common Chemical Functional Groups

Saturated hydrocarbons(loose H to add atoms)

C CH

H H

H

Ethylene(ethene)

C CH

H C

H

HH

H

Propylene(propene)

=

1-butene

2-butenetrans cis

Acetylene(ethyne)

C CH H

Unsaturated hydrocarbons(double and triple bonds)

Alcohols Methyl alcohols

Ethers Dimethyl Ether

Acids Acetic acid

Aldehydes Formaldehyde

Aromatic hydrocarbons

Phenol

Common Hydrocarbon Monomers

Some Common Polymers

C CCN

H

H H

Polyacrylonitrile (PAN)

C C

H

H H

XC C

H

H X

H

Vinyl polymers (one or more H’s of ethylene can be substituted)

Common backbone with substitutions

Monomer-based naming:poly________

e.g. ethylene -> polyethylene

if monomer name contains more than one word:poly(_____ ____)

e.g. acrylic acid -> poly(acrylic acid)

Monomer name goes here

Monomer name in parentheses

Note: this may lead to polymers with different names but same structure.

C C C CH

H

H

H

H

H

H

H

…… C C C CH

H

H

H

H

H

H

H

……

polyethylene polymethylene

Nomenclature

Polymerization Methods

H H

A. Free Radical Polymerization

1. Initiation

Free radical initiator (unpaired electron)

C CH

H H

H

monomer

C C

H

H HR

H

R Radical transferred

CC

HH

bonds bond

R

H

HC

HH

C

R

sp2 carbons

sp3 carbon

Polymerization MethodsA. Free Radical Polymerization

2. Propagation

C CH

H H

HC C

H

H HR

H

C C

H

H H

CH

H

H

C

H

H

R

C CH

H H

H

C C

H

H HC

H

H

H

C

H

H

C

H

H

C

H

H

R

H

HC

HH

C

R

H HCC

HH

Both carbon atoms will change from sp2 to sp3.


A. Free Radical Polymerization

3. Termination

C C

H

H HR

H

C C

H

H HR

H

R + C C

H

H H

RH

R

C C

H

H HR

H

+ C

H

H

C

H

H

C

H

H

C

H

H

R R

Intentional or unintentional molecules/impurities can also terminate.


B. Stepwise polymerization

RC

OH

ONH2

+R

COH

ONH2 R

CNH

ONH2 R

COH

O

C. Other methods

Anionic polymerization, cationic polymerization, coordination polymerization…

RCO

NH

n

HO

H+

HO

H+ (n-1)

Loses water(condensation)

Proteins (polypeptides have similar composition)

CH C

ONH

R nVarious R groups…

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).

20 mers 16 mers

10 mers

Average molecular weight = monomermonomer MM 3.153

101620

This is what is called number average molecular weight.

Number average molecular weight:

Mn

N jM jj

N jj

mo N j jj

N jj

N jM jj Note: Total weight

N jj Total # of polymer chains

Weight average molecular weight:

Mw

W jM jj

W jj

N jM j2

j

N j M jj Wj N jM j

In general:

M

N jM j1

j

N j M j

j

MnIf = 0 then If = 1 then Mw

Nj = # of polymer chains with length j

Mj = jmo mass of polymer chain with length j(mo = monomer molecular weight).

Molecular Weight: Different Notations

Mn

N jM jj

N jj

Mn xiMi

i

xi NiN j

j

Mw

N jM j2

j

N j M jj

Mw wiMi

i

wi NiMiN j M j

j

In Lecture Notes In Callister Textbook

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.

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

Polydispersity and Degree of Polymerization

Polydispersity:

M wM n

1

When polydispersity = 1, system is monodisperse.

Degree of Polymerization:

nn

M nmo

Number avg degree of polymerization

nw

M wmo

Weight avg degree of polymerization

Compute the number-average degree of polymerization for polypropylene, given that the number-average molecular weight is 1,000,000 g/mol.

What is “mer” of PP?

Mer molecular weight of PP is

Example 1

C3H6

mo=3AC+6AH=3(12.01 g/mol)+6(1.008 g/mol)= 42.08 g/mol

Number avg degree of polymerization

nn

M nmo

106g /mol

42.08g /mol 23,700

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 Relative abundance10 5100 25500 501000 305000 1050,000 5

nn Mn

mo

m0

m0

jN jjN jj

jN jjN jj

5 *10 25 *100 50 * 500 30 *1000 10 * 5000 5 * 50000

5 25 50 30 10 5 2860.4

nw Mw

mo

1

mo

( jmo )2 N jjN j ( jmo )

j

j2N jjjN jj

5 *102 25 *1002 50 * 5002 30 *10002 10 * 50002 5 * 500002

5 *10 25 *100 50 * 500 30 *1000 10 * 5000 5 * 50000 35,800

Note: m0 cancels in all these!

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

M n nnmo 2860.4(100g /mol) 286,040g /mol

M w nwmo 35,800(100g /mol) 3,580,000g /mol

M wM n

3,580,000286,040

~ 12.52Polydispersity:

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?

Add 5 more monomers of length 10 …. nn =Mn

mo

=jNjjN jj

=10 *10 + 25 * 100 + 50 * 500 + 30 *1000 + 10 * 5000 + 5 * 50000

10 + 25 + 50 + 30 + 10 + 5= 2750

Mw

Mn

=3,580,000

275000~ 13Polydispersity:

nw M wmo

j2N jjjN jj

35,800

Note: significant change in number average (3.8 %) but no change in weight average!

For an asymmetric monomer

T H T H+

T H T H

T H H T

H T T H

C

H

F

C

H

H

C

F

H

C

H

H

C

H

F

C

H

H

C

H

H

C

F

H

e.g. poly(vinyl fluoride):

H to T T to TH to H

Random arrangement

e.g. PMMA

C

C

CH3

C

H

H

C

C

CH3

C

H

H

O OCH3 O

CH3O

C

C

CH3

C

H

H

C

C

CH3

C

H

H

OO

CH3OCH3

O

H to T H to TH to T

Exclusive H to T arrangement (Why?)

Sequence isomerism

• Regularity and symmetry of side groups affect properties

• Stereoisomerism: (can add geometric isomerism too)

Polymer Molecular Configurations

IsotacticOn one side

SyndiotacticAlternating sides

AtacticRandomly placed

- Conversion from one stereoisomerism to another is not possible by simple rotation about single chain bond; bonds must be severed first, then reformed!

PolymerizeCan it crystallize?Melting T?

• Regularity and symmetry of side groups affect properties

Polymer Geometrical Isomerism

cis-structure trans-structure

with R= CH3 to form rubberCis-polyisoprene trans-polyisoprene

HH

-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 IsomerismSome polymers contain monomers with more than 1 reactive site

e.g. isoprene

CH2

CCH

CH2

CH3

trans-isoprene

trans-1,4-polyisoprene

CH2

CCH

CH2

CH3

1

42

trans-1,2-polyisoprene

n

CH2

C

CHCH2

CH3

n

3

3,4-polyisoprene

CH2

CH

CCH2

CH3

n

Note: there are also cis-1,4- and cis-1,2-polyisoprene

• Covalent chain configurations and strength:

Direction of increasing strengthAdapted from Fig. 14.7, Callister 6e.

Branched Cross-Linked NetworkLinear

secondarybonding


Van der Waals, H More rigid

Short branching

Long branchingStar branching Dendrimers

• Random, Alternating, Blocked, and Grafted

CoPolymers

• 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 benzene ratio

+… …

styrene

divinyl benzene crosslinked polystyrene

Monomers with trifunctional groups lead to network polymers.

Molecular Structure Branching in polyethylene (back-biting)

CH2 CH2 RCH2

CH2

CH2

CH2

CH2

C

H

H

RC

CH2

CH2

CH2

C

H

HH

H

Same as

Radical moves to a different carbon

(H transfer) RC

CH2

CH2

CH2

C

HH

H

H

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 3Nitrile rubber copolymer, co-poly(acrylonitrile-butadiene), has

Calculate the ratio of (# of acrylonitrile) to (# of butadiene).

Mn 106,740g /mol nn 2000

3 C = 3 x 12.01 g/mol3 H = 3 x 1.008 g/mol1 N = 1 x 14.007 g/mol

m0= 53.06 g/mol

4 C = 4 x 12.01 g/mol6 H = 6 x 1.008 g/mol

m0= 54.09 g/mol

1,4-addition product

mo

Mnnn

106,740

2000 53.57g /molWe need to use an

avg. monomer MW:

mo f1m1 f2m2 f1(m1m2)m2

f1

m0 m2m1m2

53.37 54.0953.0654.09

0.7 f2 1 f1 0.3

f2f1

0.70.3

7 : 3

• Crosslinking in elastomers is called vulcanization, and is achieved by irreversible chemical reaction, usually requiring high temperatures.

Vulcanization

• 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 Stress-strain curves

+ (m+n) S (S)n

(S)m

Single bonds

Double bonds

See also sect. in Chpt. 8

• 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


Molecular Weight and Crystallinity

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 )s (c a )

100%

s = density of specimen in questiona = density of totally amorphous polymerc = density of totally crystalline polymer

%crystallinity

McrystallineMtotal

100% cVcsVs

100% cs

fc 100%

Volume fraction of crystalline component.

MtotalMcrystalline Mamophous

Ms Mc MasVs cVc aVa

s cVcVs

aVaVs

cfc afa cfc a(1 fc ) fc (c a ) a

Using definition of volume fractions:

fc

VcVs

fa VaVs

fc

s ac a

Substituting in fc into the original definition: %crystallinity c (s a )s (c a )

100%

Polymer CrystallinityDegree 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

Smaller/simper side groups

Linear

Isotactic or syndiotactic

Less Crystalline

Larger/complex side groups

Highly branched

Crosslinked, network

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.

fc

s ac 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 PolymersSpherulites: 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:

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.F. Bates, Science 1991.

• Thermoplastics:--little cross linking--ductile--soften w/heating--polyethylene (#2)

polypropylene (#5)polycarbonatepolystyrene (#6)

• Thermosets:--large cross linking (10 to 50% of mers)--hard and brittle--do NOT soften w/heating--vulcanized rubber, epoxies,

polyester resin, phenolic resin

Callister, Fig. 16.9

T

Molecular weight

Tg

Tmmobile liquid

viscous liquid

rubber

tough plastic

partially crystalline solid

crystalline solid


Thermoplastics vs Thermosets

Tm: melting over wide range of Tdepends upon history of sampleconsequence of lamellar structurethicker lamellae, higher Tm.

Tg: from rubbery to rigid as T lowers

• 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.

Packing of Polymers

• Would you expect melting of nylon 6,6 to be lower than PE?

What Are Expected Properties?

N|

H

H|C|

H

6

N|

H

O||C

H|C|

H

4

N|

H

O||C

N|

H

H|C|

H

6

N|

H

O||C

H|C|

H

4

N|

H

O||C

+++

+++

HCH

HCH

HCH

HCH

+++

+++

nylon 6,6 polyethylene

a) What is the source of intermolecular cohesion in Nylon vs PE?b) How does the source of linking affect temperature?

Hydrogen bondsVan der Waals bonds

With H-bonds vs vdW bonds, nylon is expected to have (and does) higher melting T.

Which polymer more likely to crystallize? Can it be decided?


Linear syndiotactic polyvinyl chloride Linear isotactic polystyrene

• 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.

• 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 highly crosslinkcis-isoprene

• Not possible to decide which might crystallize. Both not likely to do so.

• Networked and highly crosslinked structures are near impossible to reorient to favorable alignment.

H++ H20

NetworkedPhenol-Formaldehyde

(Bakelite)



alternatingPoly(Polystyrene-Ethylene)

Copolymer

randompoly(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.

• Soap is a detergent based on animal or vegetable product, some contain petrochemicals

Detergents

grease

water

detergent

• 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)

greasee.g., oxy-clean®

Simple polymer: Elmers glue + Borax SLIME!Chemistry Elmer’s glue is similar to “poly (vinyl alcohol)” with formula:

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

OH OH OH OH OH OH OH OH OH OH OH OH OH OH OH

this is a SHORT, n=15 chain of poly(vinyl alcohol)

Hydrolyzed molecule acts in a condensation reaction with PVA, crosslinking it.

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

Crosslinking ties chains via weak non-covalent (hydrogen) bonds, so it flows slowly.

Crosslinked

Range of Bonding and Elastic Properties

Is “slime” a thermoset or thermoplastic, or neither?

Thermoset bonding

Thermoplastic bonding

• Induced dipolar bonds form crosslinks

Slime?

Stiffness increases

Where is nylon?

• Covalent bonds form crosslinks • H-bonds form

crosslinks

• 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.

Summary

Komposit (Composites)

Material Teknik

181

182

Many engineering components are composites

183

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?

184

Classification of Composites• Composites:

- Multiphase material w/significantproportions of ea. phase.

• Matrix:- The continuous phase- Purpose is to:

transfer stress to other phasesprotect phases from environment

- Classification: MMC, CMC, PMC

• Dispersed phase:-Purpose: enhance matrix properties.

MMC: increase sy, TS, creep resist.CMC: increase KcPMC: increase E, sy, TS, creep resist.

-Classification: Particle, fiber, structural

metal ceramic polymer

woven fibers

cross section view

0.5mm

0.5mmFrom D. Hull and T.W. Clyne, An Intro to Composite Materials, 2nd ed., Cambridge University Press, New York, 1996, Fig. 3.6, p. 47.

185

Particle-reinforced Fiber-reinforced Structural

• Examples:-Spheroidite steel

matrix: ferrite () (ductile)

particles: cementite (Fe3C)

(brittle)

-WC/Co cemented carbide

matrix: cobalt (ductile)

particles: WC (brittle, hard)

-Automobile tires

matrix: rubber (compliant)

particles: C (stiffer)

60m

Vm:

10-15vol%! 600m

0.75m




COMPOSITE SURVEY: Particle-I

186

• Elastic modulus, Ec, of composites:-- two approaches.

• Application to other properties:-- Electrical conductivity, se: Replace E by se.-- Thermal conductivity, k: Replace E by k.

Fiber-reinforced StructuralParticle-reinforced

Data: Cu matrix w/tungsten particles

0 20 40 60 80 100

150200250300350

vol% tungsten

E(GPa)

lower limit:1

Ec Vm

Em

Vp

Ep

c m m

upper limit:E V E VpEp

(Cu) (W)

“rule of mixtures”

Fig. 15.3

COMPOSITE SURVEY: Particle-II

187

• Aligned Continuous fibersFiber-reinforcedParticle-reinforced Structural

• Ex:

From W. Funk and E. Blank, “Creep deformation of Ni3Al-Mo in-situ composites", Metall. Trans. AVol. 19(4), pp. 987-998, 1988.

fracture surface

matrix: (Mo) (ductile)

fibers:’ (Ni3Al) (brittle)

2m

--Metal: g'(Ni3Al)-a(Mo)by eutectic solidification.

--Glass w/SiC fibersformed by glass slurryEglass = 76GPa; ESiC = 400GPa.

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).

(a)

(b)

COMPOSITE SURVEY: Fiber-I

188

• Discontinuous, random 2D fibersFiber-reinforcedParticle-reinforced Structural

• Example: Carbon-Carbon--process: fiber/pitch, then

burn out at up to 2500C.--uses: disk brakes, gas

turbine exhaust flaps, nosecones.

• Other variations:--Discontinuous, random 3D--Discontinuous, 1D

fibers lie in plane

view onto plane

C fibers: very stiff very strongC matrix: less stiff less strong

(b)

(a)

COMPOSITE SURVEY: Fiber-II

189

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 shape

distribution orientation

190

Families of Composites: particle, fibre, structural reinforcements

Orientation dependence

Twisting,Bending

ceramics

191

Isostrain: Load & Reinforcements Aligned

Isoload:Load & Reinforcements Perpendicular(Isostress below)

Two simplest cases: Iso-load and Iso-strain

Strain or elongation of matrix and fibers are the same!

Load (Stress) across matrix and fibers is the same!

F

F

F

F

Ec E 1,N %V

1Ec

1,N%V

E

V V

VTot

Volume fraction

192

Isostrain Case in Ideal Composites

Isostrain Case:

Fc Fm Fr

c m rstrain

forces

F

F

Composite Property: Pc P 1,N %V

Properties include: elastic moduli, density, heat capacity,thermal expansion, specific heat, ...

* like law of mixtures

Load is distributed over matrix and fibers, so cAc = mAm + fAf.

c m (Am / Ac ) f (Af / Ac )

or c m%Vm f

%Vf

*if the fibers are continuous, then volume fraction is easy.

For Elastic case: c cEc mEm%Vm f E f

%Vf c (Em%Vm Ef

%Vf )

193

Iso-Load Case for Ideal Composites

Isoload Case: strain

forces

Composite Property:1Pc

1,N%V

P

Properties include: elastic moduli, density, heat capacity,thermal expansion, specific heat, ...

* like resistors in parallel.

Without de-bonding, loads are equal, therefore, strains must add, so

c m%Vm f

%Vf Em

%Vm Ef

%Vf

*if the fibers are continuous or planar, then area of applied stress is the same.

elastic case

Fc Fm Fr

c m r

194

ISOSTRAIN Example

Suppose a polymer matrix (E= 2.5 GPa) has 33% fibrereinforcements of glass (E = 76 GPa).

What is Elastic Modulus?

Ec ˜ V mEm ˜ V fEf (1 ˜ V f)Em ˜ V fEf ˜ V fEf

= 26.7 GPA

* Stiffness of composite under isostrain is dominated by fibers.

~ 25 GPA

195

ISOLOAD Example

Suppose a polymer matrix (E= 2.5 GPa) has 33% fibrereinforcements of glass (E = 76 GPa).

What is Elastic Modulus? 1Ec

˜ V mEm

˜ V

f

Ef

ECEmE

f˜ V

fEm (1 ˜ V

f)E

f

Em

(1 ˜ V f)

Rearrange:= 3.8 GPA

* Elastic modulus of composite under isoload condition strongly depends on stiffness of matrix, unlike isostrain case where stiffness dominates from fibers.

196

isostrain

isoload

•Particle reinforcements usually fall in between two extremes.

Modulus of Elasticity in Tungsten Particle Reinforced Copper

197

Simplified Examples of Composites

Are these isostrain or isoload?

What are some real life examples?

198

Simplified Examples of Composites

A B

• Material A and B are different• e.g., walkway, trapeze bar,…

Load, F

||

• Fiber reinforced epoxy cylinder• e.g, pressure cylinder

60,000

200,000

permanent strain. 0

F

B

A

3" 4" 6"

F

unloading

• welded tubular composite

199

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.

Self-Assessment Example: isostrain

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?

(b) Which rod will be first to yield? Justify and explain your answer.

If not elastic, then composite fails, due to permanent deformation! Hence

A

ysA

EA

855MPa210GPa

4.07x103 B ys

B

EB

505MPa70GPa

7.21x103 C A B

F FA FB Arod (A B ) Arod (EAA EBB ) Arodrod (EA EB )

F (1.28x102m)2

4(4.07x103)(21070)GPa 146.6kN

C A B Steel yieldsfirst! To justify, consider how much load is carried FA/FBrelative to that expected from the YS.

200

(x)

• Critical fiber length for effective stiffening & strengthening:Fiber-reinforcedParticle-reinforced Structural

fiber length 15

f dc

fiber diameter

shear strength offiber-matrix interface

fiber strength in tension

• Ex: For fiberglass, fiber length > 15mm needed• Why? Longer fibers carry stress more efficiently!

fiber length 15 f d

c

Shorter, thicker fiber:

fiber length 15 f d

c(x)

Longer, thinner fiber:

Poorer fiber efficiency Better fiber efficiency

Adapted from Fig.15.7

COMPOSITE SURVEY: Fiber-III

201

• Estimate of Ec and TS:--valid when

-- Elastic modulus in fiber direction:

--TS in fiber direction:

efficiency factor:--aligned 1D: K = 1 (anisotropic)--random 2D: K = 3/8 (2D isotropy)--random 3D: K = 1/5 (3D isotropy)

Fiber-reinforcedParticle-reinforced Structural

fiber length 15 f d

c

Ec EmVm KEfVf

(TS)c (TS)mVm (TS)fVf (aligned 1D)

Values from Table 15.3

COMPOSITE SURVEY: Fiber-IV

202

Fiber-reinforcedParticle-reinforced Structural

• Stacked and bonded fiber-reinforced sheets-- stacking sequence: e.g., 0/90-- benefit: balanced, in-plane stiffness

• Sandwich panels-- low density, honeycomb core-- benefit: small weight, large bending stiffness

Fig. 15.16

Fig. 15.17

honeycombadhesive layer

face sheet

COMPOSITE SURVEY: Structural

203

• CMCs: Increased toughness • PMCs: Increased E/r

• MMCs:Increasedcreepresistance

20 30 50 100 20010-10

10-8

10-6

10-46061 Al

6061 Al w/SiC whiskers (MPa)

ss (s-1)

E(GPa)

G=3E/8K=E

Density, [Mg/m3].1 .3 1 3 10 30

.01.1

1

10102

103

metal/ metal alloys

polymers

PMCs

ceramics

Adapted from T.G. Nieh, "Creep rupture of a silicon-carbide reinforced aluminum composite", Metall. Trans. AVol. 15(1), pp. 139-146, 1984.

fiber-reinf

un-reinf

particle-reinfForce

Bend displacement

Composite Benefits

204

Laminate Composite (Ideal) Example

Gluing together these composite layerscomposed of epoxy matrix (Em= 5 GPa)with graphite fibres (Ef= 490 GPa andVf = 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.

1E

0.3

490 GPa

0.75 GPa

E 7.1 GPa For isoload case.

E|| 0.3(490 GPa) 0.7(5 GPa) E|| 150.5 GPa 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

205

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.

Isostrain for load along x-dir:

Poisson Effect and Displacements in D:

• When glued together displacements have to be same!• Unequal displacements not allowed!So, top gets wider (y

top > 0) and bottom gets narrower (ybott < 0).

Equilibrium: Fy = 0 = (ybot tbot + y

top ttop)L. (t = thickness)

xtop

Etop

Ebott

xbott

ytop

Etop

Ebott

ybott

206

COMPATIBILITY: When glued, displacements have to be same!

As stress is applied, compatibility can be maintained, depending on the laminate, only if materials twists.

207

Symmetry of laminate composite dictates properties

Elastic constants are different for different symmetry laminates.

208

Orientation of layers dictates response to stresses

Want compressive stresses at end of laminate so there are no tensile stresses to cause delamination - failure!

209

NO delamination - failure!

Apply in-pane Tensile StressA B+90 +45+45 –45–45 +90–45 +90+45 –45+90 +45

Tensile -> delaminateCompressive

210

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.

211

Thermal Stresses in Composites• Not just due to fabrication, rather also due to thermal expansion

differences between matrix and reinforcements Tm and T

r.

• Thermal coatings, e.g.

• 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 T

r)• metal blade (NixCo1-x)CrAlY (higher T

m)

• Failure by delamination without a good design of composite, i.e. compatibility maintained.

|Tm –T

r |TE TTEc

At T1 At T2

If forced to be compatible, composite will bend and rotate

212

• 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.

Summary

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