UNIVERSITAS GADJAH MADA FAKULTAS MIPA/JURUSAN FISIKA/PRODI GEOFISIKA Sekip Utara, Po. Box. 21 Yogyakarta 55281, Indonesia
Buku 2: RKPM (Rencana Kegiatan Pembelajaran Mingguan) Modul Pembelajaran Pertemuan ke VIII
GEODINAMIKA Semester 5 /3 sks/ MFG 3919 Oleh Muhammad Darwis Umar, SSi, Msi Dr.-Ing. Ari Setiawan, MSi
Didanai dengan dana BOPTN P3-UGM Tahun Anggaran 2013 Desember 2013
BAB VIII. STRESS DAN STRAIN PENDAHULUAN
Dalam pokok bahasan mengenai stress dan strain mahasiswa dapat menjelaskan: Konsep dasar tentang stress di dalam bumi, Body force yang melalui volume sebuah solid, Surface force beraksi pada permukaan elemen volum, Penyeimbang gaya pada continental block, Menentuka Fc, Normal dan tangensial gaya permukaan pada elemen area bidang patahan strike-slip-fault PENYAJIAN
Stress dan Strain dalam dalam solid Plate tectonic – konsekuensi dari gaya gravitasi pada mantel dan crust Tujuan utama chapter ini – memperkenalkan konsep dasar yang diperlukan untuk memahami secara kuantitatif dari stresses di dalam Bumi Stress adalah gaya perunit area Normal stress ditransmisikan tegaklurus permukaan Shear stress ditransmisikan parallel permukaan Harga rata-rata normal stresses dinamakan tekanan – pressure Stress di dalam elastic solid menghasilkan strain atau deformation Normal strain – ratio perubahan panjang dasri solid terhadap panjang aslinya Shear strain – setengah penurunan sudut dalam right angle pada balok jika terdeformasi
Stress-strain relation: Elastic domain • Stress-strain relation is linear • Hooke’s law applies
Beyond elastic domain • Initial shape not recovered when stress is removed • Plastic deformation • Eventually stress > strength of material => failure
Failure can occur within the elastic domain = brittle behavior Strain as a function of time under stress • Elastic = no permanent strain • Plastic = permanent strain
Body force dan surface force Body force Melalui volume sebuah solid Besarnya body force pada elemen proporsional terhadap volumenya atau massanya Contoh gaya gravitasi Elemen berat = elemen massa x g Elemen F = g Velemen Densitas beberapa batuan secara umum dapat dilihat pada apendiks 2E (Turcot) Densitas batuan tergantung pada tekanan Dibawah kedalaman mantel tekananya tinggi Batuan lebih besar densitasnya 50% dibanding tanpa tekanan Surface force beraksi pada permukaan yang membatasi elemen volume Muncul dari gaya interatomik yang didesak oleh material pada salah satu sisi permukaan terhadap material pada sisi yang berlawanan
Besarnya proporsional pada luas permukaan dimana surface beraksi
Contoh:
surface Y=0 A
Y gyA
yy A yy = gy Stress = yy = gaya permukaan perunit area yang bekerja tegak lurus permukaan horizontal Normal force perunit area pada bidang horizontal naik linear terhadap kedalaman Normal stress dari batuan yang membebani (overburden) dinamakan lithostatic stress (pressure) Selain gaya normal permukaan perunit area pada bidang horizontal juga ada gaya normal permukaan perunit area pada bidang vertical Komponen horizontal normal stress xx dan zz dapat mencakup skala yang besar dari gaya tektonik dimana xx ≠ yy ≠ zz Pada sisi lain banyak contoh batuan yang terpanaskan sehingga temperaturnya cukup tinggi atau cukup cair sehingga ketiga stress xx, yy dan zz adalah sama terhadap besar overburden PL = xx = yy = zz = gy Keseimbangan antara tekanan dan berat overburden dinamakan “lithostatic state” dari stress
z z
x
A
xx A
A zz A y Penyeimbang gaya pada continental block
Continental cC
b
h Fc
Fm
Mantel mm Gaya horizontal bekerja pada sisi block F Distribusi vertical dari tekanan dapat dilihat pada gambar berikut
xx atauxx = gy
y Figure 2.9 The area under the stress versus depth profile is proportional to the total horizontal force on a vertical plane.
Gaya horizontal Fm didapat dengan mengintegralkan tekanan lithostatic
Gaya ini adalah perunit lebar dari block sehingga dimensinya gaya perunit panjang Berikutnya menentukan gaya horizontal perunit lebar pada penampang melintang continental block Fc Asumsi: Horizontal normal stress pada kontinu xx terdiri dari dua bagian: kontribusi lithostatic gy dan constant tectonic xx xx= gy + xx tectonic kontribusi dinamakan “deviatoric stress” Fc didapat dengan mengintegralkan horisontal normal stress
This force is per unit width of the block so that it has dimensions of force per unit length. The total force per unit width is proportional to the area under the stress distribution given in Figure 2–9. We next determine the horizontal force per unit width acting at a typical cross section in the continental block Fc. We assume that the horizontal normal stress acting in the continent σxx is made up of two parts, the lithostatic contribution ρcgy and a constant tectonic contribution _σxx, σxx = ρcgy + _σxx. (2.15) and that the pressures in the water, in the oceanic crust, and in the mantle beneath the oceanic crust are p0 = ρwgy 0 ≤ y ≤ hw = ρwghw + ρocg(y − hw) hw ≤ y ≤ hw + hoc = ρwghw + ρocghoc + ρmg(y − hw − hoc) hw + hoc ≤ y ≤ hcc.
Figure 2.10 Normal and tangential surface forces on an area element in the fault plane of a strike–slip fault. Find the net difference in the hydrostatic pressure force between the continental and the oceanic crusts F by integrating the pressures over a depth equal to the thickness of the continental crust. The result is
(2.20) Calculate F for hw = 5 km, ρw = 1000 kg m−3, hoc = 7 km, ρoc = 2900 kg m−3, ρcc = 2800 kg m−3, and ρm = 3300 kg m−3. Find hcc from Equation (2– 5). If the elastic stresses required to balance this force are distributed over a depth equal to hcc, determine the stress. If the stresses are exerted in the continental crust, are they tensional or compressional? If they act in the oceanic lithosphere, are they tensional or compressional? Surface forces can act parallel as well as perpendi-cular to a surface. An example is provided by the forces acting on the area element δA lying in the plane of a strike–slip fault, as illustrated in Figure 2–10. The normal compressive force σxxδA acting on the fault face is the consequence of the weight of the overburden and the tectonic forces tending to press the two sides of the fault together. The tangential or shear force on the element σxzδA opposes the tectonic forces driving the left-lateral motion on the fault. This shear force is the result of the frictional resistance to motion on the fault. The quantity σxz is the tangential surface force per unit area or the shear stress. The first subscript refers to the direction normal to the surface element and the second subscript to the direction of the shear force. Another example of the resistive force due to a shear stress is the em138 Stress and Strain in Solids
Figure 2.11 Normal and tangential forces acting on a rock mass displacedhorizontally to the right in a low-angle overthrust fault. placement of a thrust sheet. In zones of continental collision a thin sheet of crystalline rock is often overthrust upon adjacent continental rocks on a low-angle thrust fault. This process is illustrated in Figure 2–11, where the thrust sheet has been emplaced from the left as a consequence of horizontal tectonic forces. Neglecting the influence of gravity, which is considered in Section 8–4, we can write the total horizontal tectonic force FT due to a horizontal tectonic stress _σxx as FT = _σxxh, (2.21) where h is the thickness of the thrust sheet and FT is a force per unit width of the sheet. This tectonic driving force is resisted by the shear stress σyx acting on the base of the thrust sheet. The total resisting shear force per unit width FR is FR = σyxL, (2.22) where L is the length of the thrust sheet. In many cases it is appropriate to relate the shear stress resisting the sliding of one surface over another to the normal force pressing the surfaces together. Empirically we often observe that these stresses are proportional to one another so that σyx = fσyy, (2.23) where σyy is the vertical normal stress acting on the base of the thrust sheet and f, the constant of proportionality, is known as the coefficient of friction. Assuming that σyy has the lithostatic value σyy = ρcgh, (2.24) and equating the driving tectonic force FT to the resisting shear force, we
Figure 2.12 Gravitational sliding of a rock mass. find that σxx = fρcgL. (2.25) This is the tectonic stress required to emplace a thrust sheet of length L. Taking a typical value for the tectonic stress to be _σxx = 100 MPa and assuming a thrust sheet length L = 100 km and ρc = 2750 kg m−3, we find that the required coefficient of friction is f = 0.036. The existence of long thrust sheets implies low values for the coefficient of friction.
PENUTUP Diketahui, Gaya horizontal per unit lebar pada tepi blok:
Fm
1 m gb 2 2
Gaya horisontal per unit lebar pada penampang melintang tipis pada continental blok : 1 Fc c gh 2 xx h 2 Untuk memenuhi kesetimbangan statik, kedua gaya tersebut harus sama, buktikan bahwa:
1 xx c gh1 c 2 m
