Tugas104

SOAL No 4 September 2011Kondisi tegangan tiga demensi disuatu titik dalam kordinat/system sumbu Oxyz, diketahui sebagai berikut :

σxx τxy τxz 110 60 -75

σij = τyx σyy τyz = 60 -86 0 Mpa

τzx τzy σzz -75 0 55

Tentukana) Tegangan Prinsipal dan arahnya masing-masingb) Tegangan Geser Prinsipal dan arahnya masing-masing

JAWABσx = 110 Mpa τxy = 60 Mpa

σy = -86 Mpa τxz = -75 Mpa

σz = 55 Mpa τyz = 0 Mpa

Tegangan Prinsipal

σ3 - I1 σ2 + I2 σ - I3 = 0

dimana :I1 = σx + σy + σz

I2 = σx.σy + σx.σz + σy.σz - (τxy)2 - (τxz)2 - (τyz)2

I3 = σx.σy.σz + 2τxy.τxz.τyz - σx.(τyz)2 - σy.(τxz)2 - σz.(τxz)2

I1 =I2 =I3 =

Cubic EquationAx3 + Bx2 + Cx + D = 0

-234,550

TUGAS :MEKANIKA MATERIAL LANJUT

Nama : Toni Hartono BagioNIM : 3111301009

79-17,365

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SOAL No 4 September 2011



dimana A =B =C =D =

Cara mencari Cubic EquationW = 3 W = 3AE = -26.333 E = B/WV = -58336 V = WC - B^2P = -6481.8 P = V / (W^2)X = -7E+06 X = 2*B^3 - 3*W*B*C + 3*D*W^2Q = -259250 Q = X / W^3

M = (X^2+4*V^3)/W^6M < 0

F+0⁰ = 0.6073 rad F1 = (ACS(Q/2/sqr(-(P^3)))) / 3F+120⁰ 2.7017 rad F2 = (ACS(Q/2/sqr(-(P^3)))) / 3 + 2/3*πF+240⁰ 4.7961 rad F3 = (ACS(Q/2/sqr(-(P^3)))) / 3 + 4/3*πR = -161.02 R = -2*SQR(-P)S1 = -105.9 S1 = R*COS(F1) - ES2 = 172.02 S2 = R*COS(F2) - ES3 = 12.876 S3 = R*COS(F3) - E

PRINCIPAL STRESSσ1 = 172.02 MPa

σ2 = 12.876 MPa

σ3 = -105.9 MPa

KONTROLI1 = σ1 + σ2 + σ3

I2 = σ1.σ2 + σ1.σ3 + σ2.σ3

I3 = σ1 . σ2 . σ3

79

-17,365

-234,550

1-79

-17,365234,550

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SOAL No 4 September 2011



Persamaanσ1 = 172.02 MPa

62.021 -60 75 0-60 258.02 0 = 075 0 117.02 0

Arah ℓ = 0.8262m = 0.1921n = -0.5295

σ2 = 12.876 MPa -97.124 -60 75 0

-60 98.876 0 = 075 0 -42.124 0

Arah ℓ = -0.4694m = -0.2849n = -0.8358

σ3 = -105.9 MPa -215.9 -60 75 0

-60 -19.897 0 = 075 0 -160.9 0

Arah ℓ = -0.3114m = 0.9391n = -0.1452

PRINCIPAL SHEAR STRESS τ1 = (σ2 - σ3 ) / 2 = 59.386 MPa

Arah 45⁰ thd σ2 dan σ3

τ2 = (σ1 - σ3 ) / 2 = 138.96 MPa Arah 45⁰ thd σ1 dan σ3

τ3 = (σ1 - σ2 ) / 2 = 79.572 MPa Arah 45⁰ thd σ1 dan σ2

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Tugas104