ohhhh enaknya pagi-pagi minum milo campur quekeroat hmm...skrg w iseng-iseng baca-baca tentang material
Untuk material pasti ada di setiap strandard untuk ASME yang w tau ada di ASME II partd D 2007 kalau ada yang mau silahkan sedot disini ..
kalau di ASME IX ada di qw-420 ttg grouping material silahkan baca yahh atau sedot disini untuk ASME IX
w ada baca juga tentang mechanical behavior of material, disini banyak ngulas tentang sifat dari material yaitu struktur mikro,stress, kekerasan dll kalau ada yang mau silahkan email w d bahtiarparulian@gmail.com penting nih e book diforumnya aja smua pada bilang big thanks...
contoh yg di ebook
Principal stresses. It is always possible to find a set of axes (1, 2, 3) along which the
shear stress components vanish. In this case the normal stresses, σ 1, σ 2, and σ 3, are called
principal stresses and the 1, 2, and 3 axes are the principal stress axes. The magnitudes
of the principal stresses, σ p, are the three roots of
σp
3
− I1σp2 − I2σp − I3 = 0, (1.11)
where
I1 = σxx + σyy + σzz,
I2
= σ
yz
2
+ σzx2 + σxy2 − σyyσzz − σzzσxx − σxxσyy, (1.12)
and
I3
= σ
xxσyyσzz + 2σyzσzxσxy − σxxσyz2 − σyyσzx2 − σzzσxy2.
The first invariant I1 = −p/3, where p is the pressure. I1, I2, and I3 are independent of the
orientation of the axes and are therefore called stress invariants. In terms of the principal
stresses, the invariants are
I1 = σ1 + σ2 + σ3,
I2
= −σ22σ33 − σ33σ11 − σ11σ22, (1.13)
and
I3
= σ11σ22σ33.
Example problem 1.2: Find the principal stresses in a body under the stress state σ x
= 10, σ y = 8, σ z = −5, τ yz = τ zy = 5, τ zx = τ xz = −4, and τ xy = τ yx = −8, where
all stresses are in MPa.
SOLUTION: Using Equation (1.13), I1 = 10 + 8 − 5 = 13, I2 = 52 + (−4)2 + (