材料力学(双语)强度理论

MECHANICS OF MATERIALS材料力学（双语）Chapter 7

Stress State and Theory of Strength

σ3
σ1 = 44.3 MPa σ2 = 0 σ 3 = −20.3 MPa
ε2 = − ν
E
σ1
(σ 3 + σ 1 )
0.3 6 −6 ( 22.3 44.3) 10 34.3 10 =− − + × = − × 210 ×109
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Example A thin-walled container subjected to inside pressure is shown in Fig.a.
In order to determine the value of the inside pressure the hoop strain(环向应变) tested on the surface of the container with strain foil(应变片) is ε t =350×l06. If the mean diameter of the container is D=500 mm, thickness of its wall is δ=10 mm, E=210GPa，v =0.25. Try to : 1.derive the expressions of stress in the lateral and longitudinal sections of the container . 2.calculate the inside pressure of the container .
y
σt σm
p p L
D
p x
x A
p O B
Fig.a
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Solution: Expressions of the circumferential and longitudinal stress of the container reservoir 1) Longitudinal stresses Cutting the container along a section shown and in Fig.b considering the equilibrium of the right part, we get:
σm
p D x
σ m (πD δ )= p×πD 2 4
pD σ m= 4δ
σm
Fig.b
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y dq z
σ
t
2) Circumferential stress
D Cutting off the container along longitudinal p(l⋅ ⋅dθ ) section, and take the upper pant as the study 2
q
σt
object. The free body diagram is shown in Fig.c
O p D
Fig. c
σ t (l×2δ )= p×Dl
pD σt= 2δ
3) Determine the inside pressure （by stress-strain relation）
σt
External surface
σm
1 pD ε t = [σ t −νσ m ] = [ 2 −ν ] E 4δ E 4δ Eε t p= D(2 −ν )
4 × 210 ×109 × 0.01× 350 ×10−6 = 0.5 × (2 − 0.25) = 3.36MPa
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7.5 Strain Energy Density under Triaxial Stress
σ1
1 1 1 vε = σ 1ε 1 + σ 2ε 2 + σ 3ε 3 2 2 2 1 2 2 2 ⎡ σ σ σ ν (σ1σ 2 + σ3σ 2 + σ1σ3 ) ⎤ = + + 1 2 3 −2 ⎣ ⎦ 2E
σ2
1 σ m = (σ 1 +σ 2 +σ 3 ) 3
σ3
Fig. a
σm
σ 1 -σm
σm σ2 -σm σm
Fig. b
σ3 -σm
Fig. c
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1−2μ Θa = (σ 1 +σ 2 +σ 3 )=Θb E
Θc =0

Strain energy of the element shown in Fig.b is
1 2 2 2 2 2 2 vv = +σm +σm − 2ν (σ m +σm +σm [σ m )] 2E 3(1 − 2ν ) 2 1 − 2ν = σm = (σ 1 + σ 2 + σ 3 ) 2 2E 6E
Where vv is called the strain-energy density corresponding to the volume change(体积改变能密度). σm
σm σm
Fig. b
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Strain energy of the element shown in Fig.c is
1 +ν ⎡ 2 2 2 vd = σ 1 − σ 2 ) + (σ 2 − σ 3 ) + (σ 3 − σ 1 ) ⎤ ( ⎦ 6E ⎣
Where vd is called the strain-energy density corresponding to the distortion(形状改变能密度).
σ 1 -σm
σ2 -σm σ3 -σm
Fig. c
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Example Prove the relations between three elastic constants by the energy method.
n Specific strain energy of the element in pure shear is:
A
τxy
σ3
1 τ2 v ε = τγ = 2 2G
o Expression of the specific strain energy of the element in pure shear by principal stresses is:
1 2 2 2 ⎤ νε = ⎡ σ σ σ + + 1 2 3 − 2μ (σ1σ 2 + σ 3σ 2 + σ1σ 3 ) ⎦ ⎣ 2E
1 2 2 ⎡ = + + − − 2μ ( 0 + 0 + (−τ )τ ) ⎤ τ τ 0 ( ) ⎣ ⎦ 2E
σ1
1+ μ 2 = τ E
E G= 2(1+ μ )
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