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SOAL No 3 September 2011Kondisi tegangan tiga demensi disuatu titik dalam kordinat/system sumbu Oxyz, diketahui sebagai berikut :
σxx τxy τxz 120 -55 -75
σij = τyx σyy τyz = -55 55 33
τzx τzy σzz -75 33 -85all unit in MPa
Tentukana) Tegangan Prinsipal dan arahnya masing-masingb) Tegangan Geser Prinsipal dan arahnya masing-masing
JAWABσx = 120 Mpa τxy = -55 Mpa
σy = 55 Mpa τxz = -75 Mpa
σz = -85 Mpa τyz = 33 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
dimana
-471,680
TUGAS :MEKANIKA MATERIAL LANJUT
Nama : Toni Hartono BagioNIM : 3111301009
90-18,014
6
SOAL No 3 September 2011
TUGAS :MEKANIKA MATERIAL LANJUT
Nama : Toni Hartono BagioNIM : 3111301009
A =B =C =D =
Cara mencari Cubic EquationW = 3 W = 3AE = -30 E = B/WV = -62142 V = WC - B^2P = -6904.7 P = V / (W^2)X = -3E+06 X = 2*B^3 - 3*W*B*C + 3*D*W^2Q = -122740 Q = X / W^3
M = (X^2+4*V^3)/W^6M < 0
F+0⁰ = 0.55932 rad F1 = (ACS(Q/2/sqr(-(P^3)))) / 3F+120⁰ 2.65372 rad F2 = (ACS(Q/2/sqr(-(P^3)))) / 3 + 2/3*πF+240⁰ 4.74811 rad F3 = (ACS(Q/2/sqr(-(P^3)))) / 3 + 4/3*πR = -166.19 R = -2*SQR(-P)
S1 = -110.86 S1 = R*COS(F1) - ES2 = 176.8 S2 = R*COS(F2) - ES3 = 24.0644 S3 = R*COS(F3) - E
PRINCIPAL STRESSσ1 = 176.8 MPa << Max
σ2 = 24.0644 MPa
σ3 = -110.86 MPa << Min
KONTROL
I1 = σ1 + σ2 + σ3
I2 = σ1.σ2 + σ1.σ3 + σ2.σ3
I3 = σ1 . σ2 . σ3
90
-471,680
-18,014
1-90
-18,014471,680
7
SOAL No 3 September 2011
TUGAS :MEKANIKA MATERIAL LANJUT
Nama : Toni Hartono BagioNIM : 3111301009
σ1 = 176.8 MPa 56.7995 55 75 0
55 121.8 -33 = 075 -33 261.8 0
Arah ℓ = 0.83724m = -0.4587n = -0.2977
σ2 = 24.0644 MPa -95.936 55 75 0
55 -30.936 -33 = 075 -33 109.064 0
Arah ℓ = 0.46536m = 0.88355n = -0.0527
σ3 = -110.86 MPa -230.86 55 75 0
55 -165.86 -33 = 075 -33 -25.864 0
Arah ℓ =m =n =
PRINCIPAL SHEAR STRESS τ1 = (σ2 - σ3 ) / 2 = 67.4642 MPa Arah 45⁰ thd σ2 dan σ3
τ2 = (σ1 - σ3 ) / 2 = 143.832 MPa Arah 45⁰ thd σ1 dan σ3
τ3 = (σ1 - σ2 ) / 2 = 76.3676 MPa Arah 45⁰ thd σ1 dan σ2
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