Strength Analysis in Geomechanics Part 9 ppsx

5.134 At small µ that value must be compared to ultimate load pu which follows from 5.131 atµ → 0 as pu=σyi1− 1/β whereσyiis a yielding point at an axial tension or compression and the s

For the crack and punch we have two other border demands as f(π) = 0,

f(0) = 1 and f(π) = 1, f(0) = 0 respectively It is interesting to notice that

once again the approximate solution gives at m = 1 the rigorous results Now we find f, f(see calculations in Appendix J) and according to (5.92) – stressesσθ,σr, τr θ,τe., strainεθ, displacement ur, integral J and factor K as

Km+1= (κ + 1)πτ2l/2I(m)/f (π)/m where I(m) is the same as in Sect 5.2.1 Conditionτe= constant gives equation for r in form

2r(2τe)m+1GΩ(t)/τ2(κ + 1)πl = Fm+1/I(m)/f (π)/m. (5.100) Diagrams σθ/σyi, σr/σyi, τrθ/σyi are given in Fig 3.11 by the same lines as for m = 1 but with index 0 From the figure we can see that with the growth

of m the distribution of stresses changes very strongly In the same manner the problem of the punch horizontal movement can be considered The curves for the stresses can be received by reflection of the previous ones relatively to axis θ = π/2.

5.3 Axisymmetric Problem

5.3.1 Generalization of Boussinesq’s Solution

As in Sect 3.3.2 we suppose for incompressible material (ν = 0.5) σχ=σθ =

τρχ = 0, and from the first static equation (2.77) as well as from rheological law (1.29) at α = 0 we have for stress and strain following relations

σρ= f(χ)/ρ2,ερ= g(χ)Ω(t)ρ−2m (5.101)

where g = fm Since for this case in (2.79)εχ =εθ we find easily

uχ= U(ρ) sin χ, uρ=ϕ(χ) + g(χ)Ω(t)ρ1−2m/(1 − 2m). (5.102) Using conditionερ=−2εχ we derive

0.5Ω(t)(3 − 2m)g(χ)ρ1−2m /(1 − 2m) + ϕ(χ) + U(ρ) cos χ = 0. (5.103) Putting uρ, uχ from (5.102) into conditionγρχ = 0 we determine

ϕ(χ) + g(χ)Ω(t)ρ1−2m /(1 − 2m) + ρ2(sinχ)∂(U(ρ)/ρ)/∂ρ. (5.104) Excluding ϕ(χ) from (5.103), (5.104) we obtain the expression in which both parts must be equal to the same constant, say n, since each of them depends only on one variable (neglecting t as a parameter) in form

Ω(t)g(χ)/2 sin χ = ρ2mdU(ρ)/dρ = −n

150 5 Ultimate State of Structures at Small Non-Linear Strains

P

Z

Fig 5.17 Computation of constant n

with obvious solutions

f(χ) = Ω−µ(C + 2n cosχ)µ, U( ρ) = D − nρ1−2m /(1 − 2m). (5.105) Since at χ = π/2 we have σρ= 0 we must put in the first (5.105) C = 0 and constant n should be found from condition (Fig 5.17)

P =−2

λ

0

σρρ2sinχ cos χdχ. (5.106)

Putting here σρfrom (5.101) we find after calculations

σρ=−P(µ + 2) cosµχ/2π(1 − cosµ+2λ)ρ2. (5.107) Taking in the second relation (5.105) D = 0 we get the displacement as

uχ=−Ω(t)(P(µ + 2)/2π(1 − cos2+ µλ))mρ1−2m(sinχ)/(1 − 2m). (5.108) The most interesting case takes place at λ = π/2 when we receive from expressions (5.107), (5.108)

σρ=−P(2 + µ)(cosµχ)/2πρ2,

uχ=−Ω(t)(P(2 + µ)/2π)mρ1−2m(sinχ)/(1 − 2m). (5.109)

It is easy to notice that the highest value ofσρat ρ = constant is on the line

χ = 0 It is not difficult to find that there stress σρat m = 1 is 1.5 times more than at µ = 0 The biggest value of uχ is atχ = π/2 but its dependence on

m is more complex However the second relation (5.109) allows to calculate the displacements in some distance from the structure loaded by forces with

a resultant P

0 0.2 0.4

2

4

z/a

Fig 5.18 Comparison of stress distribution

To appreciate a practical meaning of the results we compare for m = 1 the distribution of stress σz on axis z for the concentrated force P = qa2 and for the circular punch of radius a when we have from (5.109) and (3.122) respectively

/σz//q = 3(a/z)2/2 π, σz/q = 1 − (1 + (a/z)2)−3/2 (5.110) From Fig 5.18 where by solid and broken lines diagrams /σz/(z) are shown

we can see that at z/a > 3 the simplest solution for concentrated force can be

used Since at µ < 1 a distribution of stresses becomes more even we can

expect better coincidence of similar curves with the growth of a non-linearity

It is interesting to notice that according to Figs 5.18, 5.4 vertical stress in axisymmetric problem is approximately twice less than in the plane one This explains higher load-bearing capacity of compact foundations

5.3.2 Flow of Material within Cone

Common Equations

We solve this problem at the same suppositions as that in Sect 4.3.3 From (2.79) at uχ= 0 we compute

ερ=−2U/ρ3,εθ=εχ= U/ρ3,

γρχ ≡ γ = dU/ρ3dχ, γm= g/ρ3 (5.111) where U = U(χ) and

g(χ) =9U2+ U2

Similar to (5.18) and (5.68) we use representations

εχ=−g(cos 2ψ)/3ρ3,γρχ = g(sin 2ψ)/ρ3 (5.112)

152 5 Ultimate State of Structures at Small Non-Linear Strains

putting which into the first law (2.82) we have equations

(g cos 2ψ) + 3g sin 2ψ = 0, dg/gd χ = 2(dψ/dχ − 3/2) tan 2ψ. (5.113)

The latter gives boundary condition dψ/dχ = 3/2 at ψ = π/4 From expres-sions for strains above we can also find

dln/U//dχ = −3 tan 2ψ, g(χ) = −3U/ cos 2ψ, (5.114) From (5.17) and (5.112) we derive representations

τρχ =τ = ω(t)ρ−3µgµsin 2ψ,

σ ρ

σ χ =ω(t)(C + ρ−3µ(K+2

−1x2gµ(cos 2ψ)/3))

(5.115)

where C is a constant and function K(θ) can be found from the first static equation (2.77) as follows

3µK = (gµsin 2ψ)+ cotχ(gµsin 2ψ) + 4(1− µ)gµcos 2ψ. (5.116) Putting (3.115) into (2.78) we derive

(gµsin 2ψ)+ (gµsin 2ψ)cotχ + (9µ(1 − µ) − 1/ sin2χ)gµsin 2ψ

Combining (5.113), (5.117) we have at Ψ according to (5.49) two differential equations

(cot 2ψ)dΘ/d χ − 2(µ − 1 + 2µ/Ψ) + Θ((6µ2

+ 3µ + 4(1 − µ) cos2

2ψ)/Ψ

− cot χ cot 2ψ) − 3Θ2

(3µ2− (µ + µ tan 2ψ cot χ + 1/3 sin2χ) cos2

2ψ)/2Ψ = 0

(5.119) the second of which should be solved at different Θo= Θ Then we integrate (3.118) at border demandχ(0) = 0 The searched function must also satisfy condition χ = λ at ψ = π/4 Now we receive from (5.113), (5.114), (2.65)

U(χ), g(χ) and τe

Putting stressσρfrom (5.115) into integral static equation (4.105) we find

at−q ∗=σρ(a,λ) = σχ(a,λ) expression for max τe as

maxτe= 3µq∗maxgµ χ)/((gµ θ) sin 2ψ)

χ=λ− gµ λ) cot λ − 2J3/ sin2λ) where

J3=

λ

0

gµ θ)(sin 2ψ sin2χ + 2 cos 2ψ sin 2χ)dχ.

Then the criteria maxτe =τuand dγm/dt → ∞ must be used as before For

the latter we have

ε∗ = 1/ α, Ω(t ∗) = (αe2 max τe)−m

Some Particular Cases

Atµ = 0 we have the solution of Sect 4.3.3

Ifµ = 1 we compute from (5.113), (5.117) equation

(g sin 2ψ)+ (g sin 2ψ)cotχ + (6 − 1/ sin2χ)g sin 2ψ = 0

with obvious solution

where D is a constant Then from (5.112)

g cos 2ψ = 3D(cos 2χ − cos 2λ). (5.121) From (5.120), (5.121) we receive

tan 2ψ = 2(sin 2χ)/3(cos 2χ − cos 2λ) Diagrams ψ(χ) at different λ according to this relation are drawn in Fig 5.19 Similar to the general case we have ultimate condition as

maxτe= q∗x12/3 tanλ (λ>

< 33.7 ◦ (5.122)

Atµ = 2/3 we calculate from (5.117) equation

(g2/3sin 2ψ)+ (g2/3sin 2ψ)cotχ + (2 − 1/ sin2χ)g2/3sin 2ψ = 0 with obvious solution solution

where H is a constant Putting (5.123) into (5.113) we derive differential equation of the first order

0 15 30

Fig 5.19 Dependence ψ(χ) at µ = 1

154 5 Ultimate State of Structures at Small Non-Linear Strains

1.28 0.58

1

15

30

χ o

ψ o

Fig 5.20 Diagrams ψ(χ) at µ = 2/3 and different Θo

dχ/dψ = 2(2 + 3 cot22ψ)/3(2 + cotχ cot 2ψ) (5.124) that should be integrated at different Θ(0) = Θo Diagramsχ(ψ) at Θo-values

in the curve’s middles andλ at their tops are given in Fig 5.20

Putting σρ from (5.113) into (4.105)we find C and from condition

dτe/dχ = 0 with consideration of (5.124) – equality tan 2ψ = 3 tan χ which gives to maxτe (it increases with a growth ofχ) value

maxτe= q∗sin2λcos2λ + 9 sin2λ/(3 cos λ − cos3λ − 2 − 12J4).

Here as before−q ∗=σρ(a,λ) = σχ(a,λ) and

J4=

λ

0 (sin2χ cos χ)(tan 2ψ)−1dχ

Diagrams J4(λ) and max τe(λ) are shown in Figs 5.21, 5.22 respectively The broken line in the latter picture refers to the case µ = 1 (computations for

µ = 1/3 see in Appendix K-interrupted by points curve in the figure) and

pointed line refers to solution (4.106)

5.3.3 Cone Penetration and Load-Bearing Capacity

of Circular Pile

Here common relations (5.111) (5.119) are valid We put stresses according

to (5.115) into integral static equations (4.107) to detail the constants and according to (2.65) we compute maxτ at a =ρ as

0 0.4 0.8 1.2

J4

Fig 5.21 Diagram J4(λ) for µ = 2/3

0 1 2

Fig 5.22 Diagram maxτe(λ)

maxτe= 3µ(P/π − p ∗(a + 1)2sin2λ)(maxgµ χ))/a(a(2(gµ χ) sin 2ψ)

χ=λsin

2λ + gµ λ)(1 + 3µ) sin 2λ((1 + l/a)2−3µ − 1)/(2 − 3µ)

where p∗ is the strength of soil in a massif at compression and

J5=

λ

0

gµ χ)((1 + 3µ) sin 2ψ sin2χ + 2 cos 2ψ sin 2χ)dχ.

Atλ → π, a → ∞ we find for a circular pile

maxτe= (P/π − p ∗b2)maxgµ χ)/l(bgµ λ) + J5(λ))(2 + l/a) (5.126)

In the same manner we consider the particular cases and consequently for

µ = 0 we receive from (5.117), equation (4.103) and hence the solution of Sect 4.3.4

156 5 Ultimate State of Structures at Small Non-Linear Strains

Atµ = 1 we have

maxτe= 4.5(P/π − p∗(a + 1)2sin2λ)x1

2/3 tanλ/la(2(5 − 6 sin2λ)/(1 + l/a)

− (2 + 3 sin2λ)(2 + l/a)) λ<146 ◦

λ<146 ◦

and for the pile the yielding (the first ultimate) load is

Pyi=πb(2τyil + p∗ b). (5.127)

We can see that this result has obvious structure and coincides with approxi-mate relation (4.110) for ideal plasticity

Similarly we compute forµ = 2/3

maxτe= (P/π − p ∗(a + 1)2sin2λ)

cos2λ + 9 sin2λ/6a(2a cos λ sin2λ ln(1 + l/a)

− l(2 + l/a)(1 − cos λ + 2J4)).

Here J4 is given in Sect 5.3.2 For the circular pile this relation predicts big values of ultimate load and so we can take in the safety side

5.3.4 Fracture of Thick-Walled Elements Due to Damage

Stretched Plate with Hole

We consider plate of thickness h with axes r, θ, z (Fig 5.23) and use the Tresca-Saint-Venant hypotheses Since hereσθ >σr>σz= 0 we haveεr= 0 and from (2.32) atα = 0, σeq= 2τe = σθ

ε ≡ εθ= 3Ω(t)σθm/4

whereε = u/r and radial displacement u depends only on t.

Putting (5.129) into the static equation of this task

integrating it at h = constant as well as at boundary conditions σr(a) = 0,

σr(b) = p and excluding factor (4u/3Ω)µ we receive with the help of (5.129)

σθ= p(1− µ)(b/r)µ/(1 − β µ−1) (5.131) where β = b/a Putting σθ into (2.66) we find for the dangerous (internal) surface

e−αε ε = 3β(1 − µ)mΩ(t)pm(1− β µ−1)m/4. (5.132) Applying to (5.132) criterion dε/dt→ ∞ we have

b

r u

p

Fig 5.23 Stretched plate

ε∗ = 1/α, pmΩ(t∗) = 4(1− β µ−1)m/3 β(1 − µ)mαe. (5.133) When the influence of time is negligible we compute from (5.133) at

Ω = constant critical load

p∗ = (4/3)µ(1− β µ−1 )/(1 − µ)(αβΩe)µ. (5.134)

At small µ that value must be compared to ultimate load pu which follows from (5.131) atµ → 0 as

pu=σyi(1− 1/β)

whereσyiis a yielding point at an axial tension or compression and the small-est value should be taken At m near to unity we must compare p∗ with yielding load which follows from (5.131) at m = 1 in form:

pyi= (σyi/β) ln β and the consequent choice should be made

Sphere

For a sphere under internal q and external p pressures (Fig 3.23) we denote the radial displacement also as u and according to relations (2.80) we compute

εθ = u/ρ, ερ= du/dρ and from the constant volume demand (2.81) we find

u = C/ρ2,ερ=−2C/ρ3,εθ= C/ρ3= u/ρ (5.135) where constant C is to be established from boundary conditions Now from (2.32) at α = 0 and σeq=σθ− σρwe deduce

158 5 Ultimate State of Structures at Small Non-Linear Strains

εθ = Ω(t)(σθ− σρ m/2

or with consideration of (5.135)

Putting (5.136) into static equation (2.80) we get after integration at border demands σρ(b) =−p, σρ(a) =−q and exclusion of constants

σθ− σρ= 3(q− p)µ(b/ρ)3 µ/2(β3 µ− 1). (5.137)

Now we use constitutive law (2.32) which for our structure is

e−αε ε = 0.5Ω(t)(σθ− σρ m (5.138)

where ε = εθ Using hereσθ − σρ from (5.136) and criterion dε/dt→ ∞ we deduce

ε∗ = 1/α, (q − p)mΩ(t) = 2(2m/3)m(1− β −3µ)m/ αe. (5.139)

When the influence of time is not high critical difference of the pressures at

Ω = constant can be got

(q− p) ∗= 21+µm(1− β −3µ )/3(αΩe)µ. (5.140)

At small µ this value should be compared with (q − p)u according to (4.97) and the smaller one must be taken Similar choice have to be fulfilled between (q− p) ∗ and (q− p)yi given by (4.94) atµ near unity

Cylinder

In an analogous way the fracture of a thick-walled tube can be studied From (2.32) at α = 0, εx= 0, εθ ≡ ε and σeq=σθ− σrwe have

ε = (3/4)Ω(t)(σθ− σr)m (5.141)

and providing the procedure above for the disk and the sphere we find /17, 27/

σθ− σr= 2µ(q − p)(b/r)2µ/(β2µ− 1). (5.142) Equation (2.32) for this structure is

e−αεε = 3Ω(t)(σθ− σr)m/4. (5.143)

Using here expression (5.142) at r = a and the criterion dε/dt → ∞ we derive

ε∗ = 1/α, (q − p)mΩ(t∗ ) = 4(m/2)m(1− β −2µ)m/3 αe. (5.144) When influence of time is negligible we can find as before critical difference

of pressures as

(q− p) ∗ = (4/3)µm(1− β −2µ )/2(αΩe)µ and again for µ near to zero this value must be compared with (q − p)u according to (4.99) and smaller one have to be taken The similar choice should be made between (q− p)∗ and (q − p)yi from relation (4.98) at m near unity

From Fig 5.24 where atα = 1, m = 1 and m = 2 by solid and broken lines

1, 2, 3 for plate, sphere and cylinder curves t∗ β) are represented respectively

we can see that the critical time for a tube is less than the consequent one for the sphere and higher that of the plate

Cone

We consider this task at the same suppositions as in Sect 3.3.1 and Sect 4.3.1 Using the scheme above, relations for strains (3.117) and stresses (3.116) as well as law (5.141) atσr→ σχ we find

σθ− σχ= (q− p) sinµχ/J6sin2µχ (5.145) where

J6=

λ

ψ (cos1+µχ/ sin2µ+1χ)dχ. (5.146)

1 0 0.5

Ω* (q − p) m

4

3 1 2 2

β

Fig 5.24 Dependence of t∗onβ and m

160 5 Ultimate State of Structures at Small Non-Linear Strains

The computations for m = 1 when J6 = A/2 from Sect 3.3.1 at λ = π/3,

ψ =π/6 show that integral J6can be easily calculated For example at m = 2 and the shown meanings ofλ, ψ its value is 0.9

In order to appreciate the moment of fracture we put (5.145) into (5.143) and use criterion dε/dt → ∞ when we have for a dangerous (internal) surface

ε∗ = 1/α, Ω(t ∗)(q− p)m= 4(J6)m(sin2ψ)/3eα cos ψ. (5.147)

If the influence of time is negligible we derive from (5.147) at Ω = constant

(q− p) ∗ = (4/3)µJ6(sin2ψ/αeΩ cos ψ)µ. (5.148) Once more for small µ this value must be compared with (q − p)u accord-ing to (4.100) and smaller one should be taken At µ near to unity (q − p) ∗

must be compared to (q− p)yi from Sect 4.3.1 and similar choice should

be made

Conclusion

The results of the solutions of this paragraph can be used for a prediction of

a failure not only of similar structures but also of the voids of different form and dimension in soil massifs

Ultimate State of Structures at Finite Strains

6.1 Use of Hoff ’s Method

6.1.1 Tension of Elements Under Hydrostatic Pressure

This approach takes as the moment of a fracture time t∞when the structures’ dimensions become infinite We consider as the first example a plate in tension

by stresses p under hydrostatic pressure q (Fig 6.1) Since here σ1=σ2= p,

σ3=−q we have from (2.31)

dε/dt = 0.5B(po)m(eε+κo)m (6.1) whereε = ε1,κo= q/po and according to (1.42) p = poeεThe integration of (6.1) in limits 0≤ ε ≤ ∞, 0 ≤ t ≤ t ∞gives

B(po)mt∞= 2(ln(1 +κo) + (m−1)!

m−1

i=1 (−1)i (1−i/(1+κo)i)/i!i(m −1−i)!)/(κo)m.

From Fig 6.2 where for someκocurves tu(µ) according to the latter expression are given by broken lines we can see that tu≡ t ∞ diminishes with an increase

of hydrostatic component

In a similar way the fracture time can be found for a bar in ten-sion by stresses q under hydrostatic pressure p (Fig 6.3) In this case /17/

σ1= q,σ2=σ3=−p and hence in (2.31)

S1= 2(p + q)/3,σeq= p + q.

Comparing this data to the previous ones we can see that rate dε/dt in the

latter problem is twice of that for the plate Hence t∞ for the bar is one half

of that in the plate case