英国名校结构力学第3讲

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4
Eqns 3+4 one way to find σθ and τθ, alternative, use Mohr’s Circle
3.17
Definition of “principal planes” Planes on which τθ = 0.
τθ = ( σ x − σ y ) sin2θ − τ xy cos2 θ = 0 2
In general, structural components are under several states of stress, i.e. forces acting in all different directions. So far, we have dealt with forces acting on components in only one direction at any one time. Now, consider bi-axial (i.e. two-axes) stresses, i.e. forces on a 2D block of material.
σ1 σ1
pd σ1 = 4t
σ1
σ1 σ1
σ1 σ1
2D loading σ/E − νσ/E σ1
strain = σ1 (1 − v ) E pd (1 − v ) strain = 4tE
3.6
3
Cylindrical vessel, σ1
σ1 σ2 σ1 σ1 = circumferential (hoop) stress For σ1, cut cylinder horizontally σ2 = longitudinal stress p fluid force = vessel force pressure × area = stress × area σ1 σ1 L d t σ2 • ignore ends, “long” cylinder • 2 stresses due to axi-symmetry axis • diameter d, thickness t (length L not important)
ε=σ/E
lateral
3.3
longitudinal
Strain in 2D loading (ii)
σ
Now consider a σ in the other direction Block elongates by ε in direction of stress
ε=σ/E
Block contracts by νε in lateral direction (Poisson ratio effect)
A single constant σ1 in all directions due to symmetry in eqm with internal fluid pressure p σ1
Cut sphere into halves, examine eqm fluid force = vessel force
3.16
8
Resolve perpendicular to plane
A
σx τyx=τxy
θ
σθ All forces except τθ have component
θ
θ C
τθ
σ θ (AB ⋅ 1) = σ x cos θ (AC ⋅ 1) + τ xy cos θ (BC ⋅ 1)
+ τ xy sin θ (AC ⋅ 1) + σ y sin θ (BC ⋅ 1) = σ x cos θ (ABcos θ ) + τ xy cos θ (ABsin θ ) + τ xy sin θ (ABcos θ ) + σ y sin θ (ABsin θ )
p (d L ) = 2 (σ1 × L t ) σ1 =
3.7
pd 2t
Cylindrical vessel, σ2
σ1 σ2 σ1 σ1 = circumferential (hoop) stress σ2 = longitudinal stress For σ2, cut cylinder vertically fluid force = vessel force pressure × area = stress × area
θ
τyx σx
Consider unit width element loaded with σx, σx and τxy (=τyx) Find σθ and τθ on inclined plane Resolve parallel & perpendicular to inclined plane.
σ o = σ θ tan 2 θ + σ θ = σ θ sec 2 θ
⇔
∴
3.14
σ θ = σ o cos 2 θ τ θ = σ o sin θ cos θ
7
Element under biaxial & shear stress
consider a plane A
σy
θ
τxy σθ
θ
σx
p
pressure × area = stress × area
π d2 p 4 = σ1 (π d t ) pd σ1 = 4t
σ1
3.2
1
Strain in 2D loading (i)
Remember: E = σ / ε Original square block
EN1030 Structural Mechanics
Week 3: Pressure Vessel & Biaxial Stresses 12 February 2007
3.1
Pressure vessels
• 3 types (sphere, cylinder, combined) • “thin” shells, t < 1/10 radius • constant through-thickness stress σ1 σ1 σ1 σ1
τyx
τθ
B
C
τxy
σy
3.15
Resolve parallel to plane
A
σx τyx=τxy
θ
σθ All forces except σθ have component
θLeabharlann θ Cτθτ θ (AB ⋅ 1) = σ x sin θ (AC ⋅ 1) + τ xy sin θ (BC ⋅ 1) − τ xy cos θ (AC ⋅ 1) − σ y cos θ (BC ⋅ 1) = σ x sin θ (ABcos θ ) + τ xy sin θ (ABsin θ ) − τ xy cos θ (ABcos θ ) − σ y cos θ (ABsin θ )
This ratio of t2/t1 causes no distortion at the junction ALL other values of t2/t1 will cause distortion (bending)
5
Bi-axial Stress
3.11
Bi-axial stress
3.12
6
Bar in tension
width out of page = 1
Consider a unit width bar in tension Find σ, τ on inclined plane σo σo
θ
b
A
θ
σo
θ
σθ
θ
τθ
B
AB =
b cos θ
C
3.13
Resolving
eqm = balance of force, not stress ←
τxy σB y
τ θ = σ x cosθ sin θ + τ xy sin 2 θ -τ xy cos 2 θ − σ y cos θ sin θ sin2 θ sin2 θ − τ xy ( cos 2 θ -sin 2 θ ) − σ y 2 2 sin2 θ = (σx − σ y ) − τ xy cos2 θ 3 2 = σx
t1
t2
p
For compatibility, equal radial strains at junction, ε1 = ε
pd (2 − v ) = p d (1 − v ) 4 t1 E 4 t2 E ⇒ t 2 1− v = 0 .412 for v = 0 .3 = t1 2 − v
3.10
σ2
p
σ2
3.8
4
Strain in cylindrical vessel
From before
σ1
σ σ
ε − νε = σ/E−νσ/E
σ2 σ
σ2
ε − νε σ
σ2/E − νσ1/E σ1
σ2 σ 1 pd pd pd (1 − 2v ) -v 1 = −v = E E E 4t 2t 4tE σ 1 pd pd pd σ (2 − v ) ε1 = 1 - v 2 = −v = E E E 2 t 4 t 4 t E 3.9 ε2 =
A
σo
θ
σθ
θ
θ
τθ
C B
⇔
b b σ o (b ⋅ 1) = τ θ sin θ ⋅ 1 + σ θ cos θ ⋅1 cos θ cos θ σ o = τ θ tan θ + σ θ 1 b b 0 = τ θ cos θ ⋅ 1 − σ θ sin θ ⋅1 cos θ cos θ τ θ = σ θ tan θ 2