GROUTING OF ANCHORS TO RESIST HYDROSTATIC UPLIFT AT ...

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1GROUTING OF ANCHORS TO RESIST HYDROSTATIC UPLIFT AT BURNLEY TUNNEL, MELBOURNE, AUSTRALIA Devon Mothersille1 and Stuart Littlejohn2 1Managing Director, Geoserve Global Ltd & Single Bore Multiple Anchor Ltd, 51 Inchmery Road, London, England , SE62NA; 2Emeritus Professor of Civil Engineering, University of Bradford, Bradford, England BD71DP; ABSTRACT: Failure of the unreinforced concrete invert of the Burnley Tunnel during a field pressure test to 600kPa required the installation of some 5200 permanent ground anchors for structural remediation.

• tendon
• austress freyssinet austress freyssinet
• bond stresses of 3.83mpa
• anchors
• anchor lengths
• grout
• test bond stress of 4.6mpa at the grout
• bond length
• tunnel

Sujets

• University of Bradford
• Múltiple
• Anchor
• Burnley
• Tunnel
• Tendon

Informations

• Publié par: osso0
• Nombre de lectures: 40
• Langue: English

Extrait

MATH 6210:SOLUTIONS TO PROBLEM SET #2

Rudin, Chapter 2, Problem #5.Show that the Cantor set has measure 0 but is uncountable.

LetE k denote thekth step in the construction ofE. SoE 0 = [0,1],E 1 = T [0,1/3]∪[2/3,1], and so on; andE=E i . Addup size of intervals removed to i formE i to get 2 23i−1i (1/3) + 2(1/2 (13) +/3) +∙ ∙ ∙(1+ 2/3). SoEis the complement of a set of size 2 23 (1/3) + 2(1/3) +2 (1/3) +∙ ∙ ∙= 1, and hence has measure zero. Finally, the elements ofEare in 11 correspondence with infinite base three decimals, soEis uncountable.˜

Rudin, Chapter 4, Problem #7.Construct a totally disconnected compact set K⊂Rsuch thatµ(K)>0.

Repeat the constuction of the Cantor set from the interval [0,1] recalled in the previous problem, but defineK i inductively by removing (open) middle segments i fromK i=1 of lengthε(forε <a simple argument (along the lines in the1). Then previous argument) shows that the resulting intersectionKhas measure 1−3ε . 1−2ε Kis totally disconnected and compact for the same reasons that the Cantor set is. ˜

′ Rudin, Chapter 4, Problem #6.Fix 0<< εan open dense subset of1. Find ′ [0,1] of measureε.

Consider the complement of the set constructed in Problem #7.It is dense and ′ can be made to have measureε.˜

Rudin, Chapter 4, Problem #8.Construct a Borel setE⊂Rsuch that 0< µ(I∩E)< µ(I) for every nonempty intervalI. Suppose we can findE⊂[0,1] satisfying the requirement for eachI⊂[0,1]. Then repeatingEalong the real line gives a solution to the problem. To find suchE⊂[0,1], start with the generalized Cantor set, sayE 1 , of measure εnumber of “holes” inThen in each of the (countable!)constructed in Problem #7. E 1 with appropriately scaled generalized Cantor sets so that the result has meaure 1

2 ε+ε/the result2. CallE 2 fill the countable number of holes in. NextE 2 with scaled generalized Cantor sets so that the result now has measureε+ε/2 +ε/4. Continue. The union of allE i will have the required properties and will have measure 2ε. As we noted, if we repeatEalong the real line, it has the required properties of the problem.(By shrinking its measure as we repeat, we can insure the result has finitetotal measure!) ˜

Rudin, Chapter 4, Problem #9.Construct a sequence of continuous functions f n on [0,1] such that Z 1 limf n (x)dx= 0, n→∞ 0 but so thatf n (x) converges for nox∈[0,1].

Recall a tent function centered atcof heightaand widthbis the piecewise linear function ( 0 ifx /∈[c−b/2, c+b/2] f(x) = a(1− |2(x−c)/b|) ifx∈[−b/2, b/2]. Letg n + 1be the tent function centered at 1/2 +∙ ∙ ∙+ 1/n(taken modulo 1) with height 1 and width 1/ n, and letf n =χ [0,1] g n . Thissequence is a “pulse” of overlapping tent functions which wrap around the unit interval.The width of each tent keeps shrinking (so the integral tends to zero), but if we fixx∈[0,1],xtakes values 0 and arbitrarily close to 1 infinitely often. ˜

Rudin, Chapter 3, Problem #3.Ifφis a continuous function on (a, b) such that   x+y φ(x)φ(y) φ≤+, 2 22 thenφis convex. The best proof (like that of Theorem 3.2) is by drawing a picture.I omit the details. As Rudin pointsout, you need to be a little careful since the result fails if φis not assumed to be continuous (for example, by takinga= 0, b= 2,φ(x) = 0 forx <1, andφ(x) = 3x+ 1forx≥1).˜

Rudin, Chapter 3, Problem #7.Find necessary and sufficient conditions for p q the inclusion ofL⊂Lto hold. p q The solution I give is based on “Another note on the inclusion ofL(µ)⊂L(µ)” by A. Villani,The American Mathematical Monthly, Vol. 92 (1985), No. 7, 485–487. Proposition 1Letµbe a positive measure inM. LetM 0 denote the sets of nonzero measure.The the following are equivalent p q (a)L(µ)⊂L(µ) for somep < qin (0,∞]; (b) inf E∈M 0 µ(E)>0; and

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p q (c)L(µ)⊂L(µ) for allp < qin (0,∞]; p qpt qt Proof.If(a) implies (b).L(µ)⊂L(µ), then clearlyL⊂Lfor allt >0. So p q assume we can assumep≥1, soLandLare normed.The key isthe hypothesized p q settheoretic inclusionL⊂Lis a continuous linear map.The reason is that, as q p we saw in the proof of the completeness ofL, any convergent sequence inL q converges (inL) to the pointwise limit of some subsequence (almost everywhere). p qq So the image ofLinsideLis closed in theLnorm. Sothe Closed Graph Theorem then implies the inclusion is continuous.Hence there is a constantksuch that for p everyf∈L, ||f|| p ≤k||f|| q . Takef=χ E for a measurable setEto get 1/p1/q µ(E)≤kµ(E), and so −1 µ(E)≥k . 1/q−1/p So the infinium in (b) must be strictly positive. p (b) implies (c).Givenf∈L, letE n denote the set ofxsuch that|f(x)< n. p Sincef∈L,µ(E n ) must tend to zero asnBy (b), this meansbecomes large. ∞q there is someNsuch thatµ(E n ) = 0.Sofis actually inL. Sofis inLtoo. (c) trivially implies (a).˜ There is also a dual assertion (whose proof I leave to you): Proposition 2Letµbe a positive measure inM. LetM ∞ denote the sets of finite measure. The the following are equivalent p q (a)L(µ)⊂L(µ) for somep > qin (0,∞); (b) supµ(E)<∞; and E∈M ∞ p q (c)L(µ)⊂L(µ) for allp > qin (0,∞); Note the ifµ(XIf) is finite, Proposition 2 applies.µis discrete, Proposition 1 applies. Neither propositionapplies to Lebesque measure onR.˜

Rudin, Chapter 3, Problem #11.Supposeµ(Ω) = 1, and supposefandgare positive measurable functions on Ω such thatf g≥1. Then Z Z f dµ∙f dµ≥1 Ω Ω √ √ Sincefandgare positive, we can consider their square rootsfandg. By the Holder inequality, we have ||f|| 2 ||g|| 2 ≥ ||f g|| 1 . The lefthand side is of course Z Z f dµ∙gdµ. Ω Ω By hypothesisf g≥1, and so the righthand side is at least 1∙µ(Ω) = 1. Squaring both sides gives the desired result.

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1iθ Circle Problem. LetS={e|0≤θ <2π}. Foreachn∈Z, define a map 1× χ n :S→Cvia iθ inθ χ n (e) =e . (0) Provethatχ n is a continuous homorphism from the multiplicative group 1× Sto the multiplicative groupC. 1 (1) Supposeχis any continuous homorphism from the multiplicative groupS × to the multiplicative groupCthat there exists an. Provensuch that χ=χ n . 1 (2) Supposeχis a continuous homomorphism fromSto GL(N,C) so thatχ admits no invariant subspaces in the following sense:ifVis a subspace of N Csuch that 1 [χ(x)](v)∈Vfor allx∈Sandv∈V , N thenV={0}orV=C. ProvethatN= 1. 1 Hence the mapsχ n are precisely the set of continuous homomorphism fromSto GL(N,C) so thatχadmits no invariant subspaces.

1 Solution.(0) is trivial.For (1), letH m denote the subgroup ofSconsisting of mth roots of unity.Sinceχis a homomorphism, it is clear that for eachmthere n m exists and integer 0≤n m ≤m−1 (possibly depending onm) so thatχ(z) =z for allz∈H m . Letz m denote a choice of generator forH m . Forprimesp6=q, we know we can takez pq =z p z q . Sowe compute in the groupH pq , n pq n pq n pq n p n q =z q ) =χ(z p z q ) =χ(z p ) z z q pq =χ(z p χ(z q ) =z z . p pq In other words n pq −n p n q −n pq z=z . p q SinceH p andH q intersect only in{1}, n pq =n p modulop and n pq =n q moduloq. Thusn p =n q modp(as well as modq). Soindeed for all primesp,n p is a constant 1 n(independent ofp). Sinceall prime roots of unity are dense inS, the claim follows. iθ For (2), writeA θ forχ(e); soA θ is anN×Nmatrix. SinceA θ cannot be the zero matrix, it has a nonzero eigenvalue.Pick one and call itλ θ . Considerthe 1 matrixB θ =A θ −λ θ Id. Sinceχis a homomorphism andSis abelian, it’s clear thatB θ commutes with allA φ the kernel. HenceKofBis an invariant subspace of N N C. Byhypothesis, this means thatKis either{0}or all orC. Theformer case is impossible since any eigenvector corresponding toλ θ Thusis in the kernel.K N must be all ofC. SoB θ In other words= 0.A θ is the (nonzero) constant multiple N λ θ of the identity.Thusanysubspace ofCis invariant.This is a contradiction unlessN= 1.˜