8/17/2019 Tugas102 Mekanika Material Lanjut

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SOAL No 2 September 2011

TUGAS :MEKANIKA MATERIAL LANJUT

Nama : Toni Hartono BagioNIM : 3111301009

axes Oxyz, is given by :

σ xx τxy τxz 200 100 0σ ij = τyx σ yy τ yz = 100 0 0

τ zx τ zy σ zz 0 0 500unit in Mpa

If new set of axes Ox'y'z' is formed by rotating Oxyz 60⁰

find the matrix of the axis Ox'y'z' through the same point.

JAWABσ x = 200 Mpa τxy = 100 Mpaσ y = 0 Mpa τxz = 0 Mpaσ z = 500 Mpa τyz = 0 Mpa

x z'

x' θ

θ z

y'

ℓ 1 = cos < (x',x) = cos 60⁰ = 0.5

ℓ 2 = cos < (y',x) = cos 90⁰ = 0

ℓ 3 = cos < (z',x) = cos (90⁰ - 60⁰ ) = 0.86603

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8/17/2019 Tugas102 Mekanika Material Lanjut

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SOAL No 2 September 2011

TUGAS :MEKANIKA MATERIAL LANJUT

Nama : Toni Hartono BagioNIM : 3111301009

= < ' = ⁰ =

1 ,

m 2 = cos < (y',y) = cos 0⁰ = 1

m 3 = cos < (z',y) = cos 90⁰ = 0

n1 = cos < (x',z) = cos (90⁰ + 60 ⁰ ) = -0.866

n2 = cos < (y',z) = cos 90⁰ = 0

n3 = cos < (z',z) = cos 60⁰ = 0.5

ℓ m n

1 0.5 0 -0.866 200 100 02 0 1 0 100 0 0. .

σ x' = σ x.ℓ 12 + σ y.m 12 + σ z.n12

+ 2τ xy. (ℓ 1.m1) + 2 τ yz.(m 1.n1) + 2 τ xz.(ℓ 1.n1)200*0.5^2 + 0*0^2 + 500*-0.866^2 +2*100*(0.5*0) + 2*0*(0*-0.866) + 2*0*(0.5*-0.866)

425 MPa

σ y' = σ x.ℓ 22 + σ y.m 22 + σ z.n22

+ 2τ xy. (ℓ 2.m2) + 2 τ yz.(m 2.n2) + 2 τ xz.(ℓ 2.n2)200*0^2 + 0*1^2 + 500*0^2 +2*100*(0*1) + 2*0*(1*0) + 2*0*(0*0)

0

σ z' = σ x.ℓ 32 + σ y.m 32 + σ z.n32

+ 2τ xy. (ℓ 3.m3) + 2 τ yz.(m 3.n3) + 2 τ xz.(ℓ 3.n3)200*0.866^2 + 0*0^2 + 500*0.5^2 +2*100*(0.866*0) + 2*0*(0*0.5) + 2*0*(0.866*0.5)

275 MPa

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SOAL No 2 September 2011

TUGAS :MEKANIKA MATERIAL LANJUT

Nama : Toni Hartono BagioNIM : 3111301009

τ x'y' = σ x. 1. 2 + σ y.m 1.m 2 + σ z.n1.n2 +τ xy. (ℓ 1.m 2 + ℓ 2.m1) + τ yz .(m 1.n2 + m 2.n1) + τ xz.(ℓ 1.n2 + ℓ 2.n1)200*0.5*0 + 0*0*1 + 500*-0.866*0 +100*(0.5*1 + 0*0) + 0*(0*0 + -0.866*1) +0*(0.5*0 + 0*-0.866)

a

τ x'z' = σ x.ℓ 1.ℓ 3 + σ y.m 1.m 3 + σ z.n1.n3 +

τ xy. (ℓ 1.m 3 + ℓ 3.m1) + τ yz .(m 1.n3 + m 3.n1) + τ xz.(ℓ 1.n3 + ℓ 3.n1)200*0.5*0.866 + 0*0*0 + 500*-0.866*0.5 +100*(0.5*0 + 0.866*0) + 0*(0*0.5 + -0.866*0) +0*(0.5*0.5 + 0.866*-0.866)

-129.9 MPaτ y'z' = σ x.ℓ 2.ℓ 3 + σ y.m 2.m 3 + σ z.n2.n3 +

τ xy. (ℓ 2.m 3 + ℓ 3.m2) + τ yz .(m 2.n3 + m 3.n2) + τ xz.( ℓ 2.n3 + ℓ 3.n2)200*0*0.866 + 0*1*0 + 500*0*0.5 +100*(0*0 + 0.866*1) + 0*(1*0.5 + 0*0) +0*(0*0.5 + 0.866*0)86.6025 MPa

σ x' τx'y' τx'z'σ ij = τy'x' σ y' τ y'z'

τ z'x' τ z'y' σ z'

425 50 -129.9

σ ij = 50 0 86.6025

-129.9 86.6025 275

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