• Nie Znaleziono Wyników

Draw probable plots of the shearing forces Q(ξ) and moments M(ξ) along the beam. Compare with the sym- metrical case, where the forces are both equal to +1000kN, r(x) = const.

N/A
N/A
Protected

Academic year: 2021

Share "Draw probable plots of the shearing forces Q(ξ) and moments M(ξ) along the beam. Compare with the sym- metrical case, where the forces are both equal to +1000kN, r(x) = const. "

Copied!
5
0
0

Pełen tekst

(1)

TESTING QUESTIONS AND EXERCISES to Lectures no. 1,2,3,4

Question: Why the use of the Winkler model is recommended only if thickness H of deformable soils under foundation is “small”, say H/B < 1,0÷1,5 ?

Question: Antisymmetric beam of the length 2π with the ends at ±π is loaded by two opposite forces ±1000kN applied at the cross-sections ±π/2. Subsoil reaction r(x) is linear along the beam, r(-x) = r(x), of course r(0) = 0.

Draw probable plots of the shearing forces Q(ξ) and moments M(ξ) along the beam. Compare with the sym- metrical case, where the forces are both equal to +1000kN, r(x) = const.

Question: Is it true that for a finite Winkler beam 6 virtual Bleich forces provide higher accuracy than 4 such forces? Justify.

Question: Is it true that a half-finite Winkler beam can never be solved by making use of only one virtual force?

Justify.

Question: Two silt layers are found under a foundation beam which is B=1,0m wide (B<< L); the upper one 1,5m thick is stronger I

L

= 0,07, the lower one 2,0m thick is much weaker I

L

= 0,39. In GEO bearing capacity calculations it was found that the lower soil layer has a greater margin of safety than the stronger upper one.

Explain how this can happen (there are two reasons).

Question: Plot of bending moments along the beam loaded by a concentrated force has a sharp edge under the force. Explain why.

Question: Draw probable solutions y(x), M(x), Q(x) along a beam on elastic subsoil which has a perfect hinge under the concentrated force P, like in Fig.

Question: Draw mining corrections ∆M, ∆Q, ∆r along a beam of the length L for -L/2 < x < L/2 if R > 0 happens (convex case). Pay special attention to signs of the corrections. Why

//

∆ = 0?

Question: the H.Bleich method for beams and plates on the Winkler subsoil:

Present one basic common feature and one major difference.

Question: Is this true that the relations Q = -EI⋅d

3

y(x)/dx

3

yields from the Euler-Bernoulli hypothesis? Justify.

Question: Underground mining exploitation at a depth H in 2D (long mining front) causes continuous deformations of the surface. Which situation is more dangerous for a CE-infrastructure on the surface:

H=400m or H=800m? There is no other difference neither in type of rocks (tgβ), thickness of the mineral deposit (h), nor mining technology (a) etc.

Question: A certain CE-object is sensitive to mining-caused tilt T. Present two (different) possible measures to be undertaken on the exploitation level which can reduce the potential danger (2D case), so can reduce the tilt.

Question: Present a situation in which the biggest losses are caused by uniform mining subsidences on a certain vast area.

Question: Present a situation in which a danger is caused by compressive mining strains i.e. ε < 0.

Question: How many unknowns operate in the ZEM_SIN algorithm, if the beam is subdivided into 17 calculation segments?

Question: The beam is subdivided into 21 calculation segments in ZEM_SIN. Is this true that y

18

< y

78

? How about y

88

> w

88

? And maybe w

88

= w

22

?

−∞

P

C

y( ξ )

ξ 0

+∞

(2)

Question: The fundamental solution of the first kind y(ξ) for the Winkler beam appears to be an even function (symmetrical). Explain how this yields a conclusion that moments M(ξ) are even, too.

Question: The fundamental solution of the second kind y(ξ) for the Winkler beam loaded by a moment M at ξ=0 appears to be an odd function. Justify.

Question: What’s the use of the tilted object rectification in mining area protection?

Question: a rigid smooth foundation beam is loaded by a vertical force P [kN/m]

acting on a small eccentricity e

B

.Draw probable plots of subsoil reactions along the width B for:

a) the Winkler subsoil model

b) the Pasternak subsoil model

c) the elastic half-space.

(3)

Calculation job:

Hints:

1) there are 3x5 identical square segments 1m x 1m distinguished on the founding level – 6 real footings in this number; F

ki

can be used to model interactions between the segments if the superposition law is assumed;

action of „b” on „a” or of „C” on „B” etc. uses F

02

= 0,511 action of „c” on „A” or of „a” on „C” uses F

24

= 0,225

action of „B” on „B” or of „c” on „c” etc. uses F

00

= 2,974 and so on.

(some values are not used here, like F

01

, F

03

, F

14

, F

23

…, because there is no such pair of footings).

2) of course, focus on the columns’ heads level (2.5m above the footings), where the head settlements are equal.

Extra Question:

What will change if 4 outer footings subside of ∆

o

(mining effects) and 2 central ones of ∆

c

?

A rectangular tank is placed on a 2m x 4m rigid plate, under which there are 2x3 identical vertical steel columns of the length (height) H = 2.5m and stiffness EA = 500 MN.

Each column rests on an identical 1m x 1m prefabricated square footing. There are two symmetry axes,

and the load is constant q = 0.150 MPa.

Calculate the axial forces in the columns [MN].

It is the elastic half-space, ν = 0,3 and E

o

= 30 MPa.

Average settlement w

ki

of the footing k, which is d

k

x b

k

= 1m x 1m, caused by the footing i which is d

i

x b

i

= 1m x 1m and is loaded by a force R

i

[MN] can be expressed as follows:

= ∙ ∙ F ki

Where the (symmetrical) influence coefficients F

ki

= F

ik

for settlements are presented in Table below; note that for these dimensions and distances, numbers k and i mean simply distances between centers of footings along x,y, expressed in meters.

For simplification assume

∙ ∙ = 0.01 m/MN.

q=150kPa

„A”

„a”

„B”

„b”

„C”

„c”

2m 2m

2m

Table. Influence coefficients F

ki

for identical square segments

Use

asymptotic values as …

(4)

Calculation job:

Solve a half-infinite beam loaded by a force P at the distance of π/2 from a free end of the beam, i.e. find both T

1

= ? and T

2

= ? Fundamental solution to be used is presented on the right.

Calculation job :

Calculation job:

A foundation beam has a trapezoidal shape: length L=16m, width B

1

= 1,0m at the left end, B

2

= 2,0m at the right one. Find maximal mining force Z

max

[kN] caused by tensile mining ε and draw probable shape of Z(x).

Ignore Z

b

on both sides. Take pre-ultimate distance x

θ

= 0 and θ= 30 kPa.

Calculation job:

Draw plot of a mining tensile force Z [kN] along the beam of L=16m and next calculate Z

max

. Take x

θ

=2.0m, θ = 30kPa. The beam is symmetrical but is not prismatic. It consists of 3 intervals:

B=2.0m in the central part 0±4.0m and B=1.5m near the ends, 4.0m long each.

Assume no Z

b

forces on both sides of the beam.

ξ =

2"#$ % ∙ & ξ ∙ cos ξ + sin ξ , ξ =

2"#$ % ∙ & ξ ∙ −2 sin ξ . ξ = $ %

4 ∙ & ξ ∙ cos ξ − sin ξ 0 ξ = −

2 ∙ & ξ ∙ cos ξ P

+∞

π / 2

π / 4

π/ 4 T 2

T 1

P

y(ξ)

ξ 0

- ∞

π/2

+ ∞

-π/2

P

The Winkler beam, continuous and doubly-infinite, is loaded by two forces P at ξ

1

= 0 and ξ

2

= +π/2.

At the cross-section ξ = -π/2 there is also a perfect support, so no settlement happens there. Solution for the standard unsupported beam (fundamental solution of the first kind) is presented above for 0 < ξ < ∞ in the local system of coordinates (odd/even curves).

By making use of these functions, calculate the bending moment M at this supported cross-section, so M(-π/2) = ? Next plot probable the solutions y, M, Q.

Hint:

at -π/2 apply a vertical force R in place of the support and find

its value which is required by the context; useful functions are

given in the previous job.

(5)

Calculation job: The subsoil consists of 2 sublayers of the thicknesses H

i

and moduli E

si

.

Using the method of equivalent settlements, estimate the elasticity effective modulus E

s*

for a homogenised strata of thickness 12.0m instead of the real 2m+4m = 6m thick subsoil. Assume B = 2.0m.

Read the required coefficients from the table.

BEDROCK Thickness H2 = 4m, Es2 = 100MPa Thickness = 2m, Es1 = 25MPa

B q

Zo=0

Calculation job:

It was estimated that the ceiling of a tunnel will be deformed (downward deformation, settlement) of ∆

o

= const(x) = 3cm on the interval AB=8m at the depth h

t

= 19m under the ground level, so between x

1

= 0 and x

2

= 8. This way, nonuniform settlements of the existing foundation beams C, D will happen on the founding level h

f

= 1m.

Calculate such additional “mining” settlements of both foundation beams C, D, w

C

= ?cm and w

D

= ?cm.

For soil above the tunnel assume the angle of subsidence propagation β = π/4 + ϕ/2 = 62

o

, tgβ = 1,8

And the radius of mining influences r = H/tgβ, dispersion parameter of the subsidences basin σ = r/√(2π)0,4·r.

Porosity of the soil mass above the tunnel will be constant, more or less, so a ≈ 1.

The complete subsidence curve takes the form w

c

(x) = ∆

o

·Φ(-(x-x

o

)/σ), if initial subsidences ∆

o

occurs along (-∞;x

o

) at the depth H under a protected level (but here they will occur only on AB = (x

1

,x

2

) in this case).

The Laplace integral Φ(s) is presented in the table below.

The points marked in Fig. have the following coordinates (x;y): A(0;19), B(8;19), C(8;1), D(12;1).

s = -∞ -4 -3 -2 -1 -½ 0 ½ 1 2 3 4 +∞

Φ(s) = 0 ≈0 0,001 0,023 0,159 0,309 ½ 0,691 0,841 0,977 0,999 ≈1 1

Hint:

(x

1

;x

2

) = (-∞;x

2

) − (-∞;x

1

)

thus superpose 2 functions of the type w

c

(x) to get an equation for the profile of „mining subsidences” w(x)

which is concentrated only on finite (x

1

;x

2

);

this is the 2D situation of course.

A B

(0;0)

C D

8m

1m

h

t

19m

18m w(x)

3cm

x

4m

z/B= 0 1 2 3 4 6 8 ∞

ω = 0 0,50 0,85 1,10 1,30 1,48 1,65 2,15 Average settlement of the beam of the width B resting on n ≥ 1 elastic layers can be calculated as:

where z

o

= 0

Cytaty

Powiązane dokumenty

Using the definition of the improper integral of the first kind, check if the integrals given below converge, diverge to ∞, −∞, or diverge.. If

The main difference is that in the case of improper integrals of the first kind we have an infinite range of integration, whereas in the case of improper integrals of the second kind

Also the Skorohod type number integral of [10] is a special case of such double integrator conservation integrals.. We think this short note may provide some example on the way to

Find

Instead of a finite beam with two ends A and A', consider infinite beam and additionally apply two virtual forces T1,T2 on the left of A, and another two virtual forces T3,T4 on

In this paper, we survey sufficient conditions for the existence of kernels in the closure of edge coloured digraphs, also we prove that if D is obtained from an edge

The kinematic boundary condition assumes that fluid particles initially forming the in- terface remain on it.. Neglect surface tension at

Find the second component of the substitute nodal force vector for the truss element loaded as shown in the figure.. Solve the truss