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ENGINEERING SURVEYING (221 BE)

Levelling-TechnicalLevelling-Technical

Sr Dr. Tan Liat Choon

Email: tanliatchoon@gmail.com

Mobile: 016-4975551

LEVELLING

TWO PEG TEST

� Set 2 marks at 30-40 metre apart, also mark centre point in a relatively flat area

� Set level at midpoint and take readings at each end

� Determine difference in readings (difference in elevation)

Move level to one end and setup so that level is just in front of rod on point� Move level to one end and setup so that level is just in front of rod on point

� Read rod by looking backward through scope (x-hair not visible), hold pencil on rod to determine reading

� Read rod at other end in normal manner

� Difference in readings should equal of 3

� If values are not equal, there is error• Most instruments have adjustment screws• Adjust and repeat test as a check

2

Two Peg Test

x

S1

S1’

Line of Collimation

Horizontal Line

3

L / 2 L / 2

L

A

B

S2

S2’

x x

S1

S1’

Line of Collimation

Horizontal Line

Two Peg Test

L / 2 L / 2

L

A

B

= S1’ - S2’

The APPARENT height difference δ hA

The TRUE height difference hTδ

= S1 - S2

S1= S1’ + x and S2

= S2’ + x 4

x

S1

S1’

Line of Collimation

Horizontal Line

S2

S2’

x

Two Peg Test

L / 2 L / 2

L

A

B

δThe TRUE height difference hT = S1’ - S2’

= S1 - S2The APPARENT height difference δ hA

S1= S1’ + x and S2

= S2’ + x hA = (S1’ + x) - (S2’ + x )δ 5

x

S1

S1’

Line of Collimation

Horizontal Line

S2

S2’

x

Two Peg Test

L / 2 L / 2

L

A

B

δThe TRUE height difference hT = S1’ - S2’

= S1 - S2The APPARENT height difference δ hA

S1= S1’ + x and S2

= S2’ + x hA = S1’ - S2’δ = δ hT6

hA δ = δ hT

Therefore :

This is true since the instrument is the same distance from both staff positions and

the errors x are equal and cancel out

Two Peg Test

7

S3’

S3

Now move the instrument outside the “odd numbered” peg

Two Peg Test

A

B

L / 10

8

S3

S3’

S4

S4’

Two Peg Test

A

B

L / 10

= S3 - S4The APPARENT height difference δ hA

δBut the TRUE height difference hT We already know

9

S3

S3’

S4

S4’

Two Peg Test

A

B

L / 10

= S3 - S4

= S1 - S2

then the instrument is OK

If NOT then the error is e =

The APPARENT height difference δ hA

δBut the TRUE height difference hT

δ δTherefore if hA = hT

(S1 - S2) - (S3 - S4) / L mm / m10

Summary

Place two pegs about L = 30m (to 40m) apart

Set up level midway between the two pegs

Read staff on each peg, and calculate true height difference

Move level about L / 10 = 3m (or 4m) beyond one of the pegs

Read staff on each peg again, and calculate height differenceRead staff on each peg again, and calculate height difference

Collimation Error e = difference in the differences

and is expressed as a number of mm per L m

Acceptable errors

Uren and Price 1mm per 20m

Wimpey 4mm per 50m

Test should be carried out regularly say once per week or two11

L / 2 L / 2

L

S1

AB

S2

S3

S4 S3

AB

L / 10Collimation error,

e = (S1 - S2) - (S3 - S4) mm / Lm12

DATUM

Could be our own Datum - Assumed Datum

- Arbitrary Datum

- Site DatumOr

A National Datum - Ordnance Datum

Above Assumed Datum

In the UK we have a national organisation known as

The Ordnance Survey (O.S.)

The O.S. has established a ZERO Datum at Newlyn in

Cornwall.

Based on the Ordnance Datum - points of known height above

or below Zero height have been established around the U.K.

These points around the country are known as Bench Marks

Above Ordnance Datum

13

Levelling

Measured and CalculatedReducedRL A (known)ReducedLevel of BRL B

(unknown)

the Plane of CollimationHeight of (HPC)S1

Levelling Staff

S2

AB

Measured and CalculatedLevel of A ReducedLevel of A

RL A (known) Level of BRL B

DATUM

HPC = RL A + S1HPC = RL A + S1

RL B = HPC - S2

14

AB

C

Some Terminology

RL A RL BRL C

S1

Level staff on A Back Sight (BS) reading is first reading

BS

Levelling

Level staff on A Back Sight (BS) reading is first reading

15

RL A RL B

AB

C

RL C

Level staff on A Back Sight (BS) reading is first reading

S2FS

Levelling

Level staff on A Back Sight (BS) reading is first reading

Level staff on B Fore Sight (FS) reading is last reading

Move instrument to new position

16

RL A RL BRL C

AB

C

CP

S3BS Levelling

Move instrument to new position Level staff stays on B

The instrument has changed its position about point B

Point B is known as a Change Point (CP)

CP

2nd instrument position starts with BS to B

17

FSS4S3

BS

RL A RL BRL C

AB

C

Levelling

and finishes with FS to C

18

RL A RL B

AB

C

RL C

BS FS

BS FS

RL A is known

HPC

(CP)

HPCLevelling

RL A is known

HPC = RL A + BS RL B = HPC - FS

Now the RL B is known So we can repeat the process

HPC = RL B + BS RL C = HPC - FS

Generally : HPC = Known RL + Back Sight

Unknown RL = HPC - Fore Sight 19

RL A RL B

AB

C

RL C

BS FS

BS FSHPC

HPC

(CP)

Levelling

RL A is known

HPC = RL A + BS RL B = HPC - FS

RL B + BS RL C = HPC - FS HPC =

Generally : HPC = Known RL + Back Sight

Unknown RL = HPC - Fore Sight

Now the RL B is known So we can repeat the process

20

PLANE AND COLLIMATION METHOD

� This method is simple and easy

� Reduction of levels is easy

� Visualization is not necessary regarding the nature of the ground

� There is no check for intermediate sight readings

� This method is generally used where more number of readings can be taken

with less number of change points for constructional work and profile

levelling

� To check:

∑ BS - ∑ FS = Last RL – First RL

21

PLANE AND COLLIMATION METHOD

Determine the RLs of various points if the reduced level (RL) of a point on which the first reading was taken is

136.440 gives the Height of Plane and Collimation method and applies the check

Station BS IS FS HLC RL Remarks

1 0.585 137.025 136.440 BM A RL=136.440

HLC = RL + BS

= 136.440 + 0.585 = 137.025

RL = HL – BS

Check(Summation of BS)-(Summation of FS) = Last RL – First RL

2.670 – 9.275 = 129.835 – 136.440

-6.605 = -6.605

22

2

3

4

5 0.350

1.010

1.735

3.295

3.775 133.600

136.015

135.290

133.730

133.250 CPI

6

7

8

9

10 1.735

1.300

1.795

2.575

3.375

3.895 131.440

132.300

131.805

131.025

130.225

129.705 CP 2

11

12

0.635

1.605

130.805

129.835 BM B RL=129.835

Sum of BS=2.670 Sum of FS = 9.275

2.690-9.275 = -6.605 129.835-136.440= -6.605

How Levelling is Conduct

23

How Levelling is Conduct

24

Calculation checks

∑ FS - ∑ BS = 1st RL - Last RL

∑ IS + ∑ FS + ∑ (RLs except first)

= ∑ (each HPC x number of applications)

Simple check

Full check

Check Misclosure

Allowable Misclosure = 5 √N mm. ("Rule of Thumb")

When calculations are checked and

if the misclosure is allowable

Distribute the misclosure

25

RISE AND FALL METHOD

� This method is complicated and is not easy to carry out

� Reduction of levels takes more time

� Visualization is necessary regarding the nature of the ground

� Complete check is there for all readings

� This method is preferable for check levelling where number of

change points is more

� To check:

∑ BS - ∑ FS = ∑ Rise - ∑ Fall = Last RL – First RL

26

RISE AND FALL METHODDetermine the RLs of various points if the reduced level (RL) of a point on which the first reading was taken is

122.156 gives the Height of Rise and Fall method and applies the check

Station BS IS FS Rise Fall RL Remarks

1

2

1.536

0.974 2.072 0.536

122.156

121.620

BM A RL=122.156m

CP1

R/F = BS - FS= 1.536 – 2.072 = 0.536

RL = HL – IS

Check(Summation of BS)-(Summation of FS) = Last RL – First RL

12.172 – 12.725 = 121.605 – 122.156

-0.551 = -0.551

27

2

3

4

0.974

1.124

1.768

2.072

2.700

0.536

0.794

0.932

121.620

120.826

119.894

CP1

CP2

5

6

7

8

9

2.236

1.413

1.994

1.639

1.256

2.362

1.302

0.874

0.825

1.120

0.934

0.539

1.169

0.519

1.238 118.656

119.590

120.129

121.298

121.817

CP 3

CP 4

CP 5

CP 6

CP 7

10 1.468 0.212 121.605 BM B RL=121.605

Sum of BS=12.172 Sum of FS =

12.723

3.161 3.712

12.723-12.172=0.551 3.712-3.161=0.551 122.156-

121.605=0.551

How Levelling is Conduct

28

COMPARISON

� Plan and Collimation Method

• Quicker

• Good for a lot of IFSs

� Rise and Hall Method

• More accurate

• More calculation

• Intermediate RLs are known

29

ACCURACY IN LEVELLING

For normal engineering works and site surveys

Allowance misclosure = ± 5 √ N mm

Where N = Number of instrument positions

OR

Allowance misclosure = ± 12 √ K mm

Where K = length of levelling circuit in KM

If actual misclosure > allowance misclosure, levelling should be repeated

If actual misclosure < allowance misclosure, misclosure should be equally

distributed between the instrument positions

30

CORRECTION IN LEVELLING

Correction = (Misclosure / No. of Station) x n, n+1, n+2 and ………

For example:

Loop 1 = (0.117 / 3) x 1 = 0.039m

Loop 2 = (0.117 / 3) x 2 = 0.078m

Loop 3 = (0.117 / 3) x 3 = 0.117mLoop 3 = (0.117 / 3) x 3 = 0.117m

31

Example

32

Example

33

Example

34

Example

35

Example

36

Example

37

Example

38

Summary of work:

Check tripod is on stable ground or dig feet well in

Use pond bubble to set approximately vertical standing axis

LEVELLING WORK

Eliminate PARALLAX every time we sight the staff

check that the compensators are functioning every

time we sight the staff.

and

39

At every instrument set up - always start with a BS to a

point of known RL.

At every instrument set up - always finish with a FS.

Either the instrument moves or the staff moves

LEVELLING WORK

Either the instrument moves or the staff moves

NEVER BOTH

ALWAYS CLOSE levelling to a point of KNOWN RL

40

TBM

9.09 m A.A.D.

Main Gate

Burnaby Building

Approximate North

TBM

10.00 m A.A.D.

Start at a TBM outside the main entrance of Burnaby Building and obtain the

RL values of three points before closing onto another TBM near the main gate.

Point 1

Ground level at entrance to structures laboratory

Top of door level at entrance to structures laboratory

41

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD

It is important to complete details at the top of booking forms or

on every page of field books.

42

TBM Level Posn. BS

Key

43

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group

Good Group

TBM 10.00m AAD10.0001.546 11.546

HPC = RL + BS HPC = 10.000 + 1.546 = 11.546

We now signal to the staff person to move to the next point.

As the next required point is too far away (it is also round a corner)

we will eventually need to move the instrument.

So, we must move the staff to a change point (CP), to allow us to move the

instrument to a better position later on.

44

TBM CP Level Posn. BS FS

Key

45

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD1.546 10.00011.546

C.P.

New staff

position

therefore

a new row.

Each rowrepresents

1.562 9.984

represents

a staff

position.

RL = HPC - FS RL = 11.546 - 1.562 = 9.984

After we make a FS and we have calculated the new RL we are finished

with that instrument position.

Move the Instrument (about the CP) to a new position where we can see the CP

and also the next point we want the RL value of. 46

TBM CP Level Posn. BS FS ISKey

47

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418

Same staff

position as

last reading

therefore

the same row

11.402

HPC = RL + BS HPC = 9.984 + 1.418 = 11.402

48

TBM CP Level Posn. BS FS ISKey

This reading is not the first so it is not a BSIt is not the last from this position (we can see the next points) so it is not a FS

So it is known as an INTERMEDIATE SIGHT (IS)49

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

New staff

position

therefore

a new row

RL = HPC - IS RL = 11.402 - 1.390 = 10.012

50

TBM CP Level Posn. BS FS ISKey

51

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door

New staff

position

therefore 1.281 10.121

therefore

a new row

RL = HPC - IS RL = 11.402 - 1.281 = 10.121

52

TBM CP Level Posn. BS FS ISKey

53

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

BS IS FS HPC RL Corr Corr RL Remarks

Building A 2

27/09/11 Group 1

Good Group 1

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121 New staff

position Top Struct. Lab Door

position

therefore

a new row

Requires an inverted staff i.e turn the staff upside down

Read and then book the staff with a sign

-2.420

The negative sign will keep all the calculations correct

RL = HPC - IS RL = 11.402 - (-2.420) = 11.402 + 2.420 = 13.822

13.822

54

TBM CP Level Posn. BS FS ISKey

The last point required is the TBM. However it is too long a sight

So we need a CP. This will be the last sighting from this position

Therefore we need a FS

55

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

Building A 2

27/09/11 Group 1

Good Group 1

BS IS FS HPC RL Corr Corr RL Remarks

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121

Top Struct. Lab Door-2.420 13.822

GL Struct. Lab Door1.281 10.121

CP New staff

position

therefore

a new

row

1.321

RL = HPC - FS RL = 11.402 - 1.321 = 10.081

10.081

Last Reading -- FS -- Move the instrument56

TBM CP Level Posn. BS FS ISKey

57

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

Building A 2

27/09/11 Group 1

Good Group 1

BS IS FS HPC RL Corr Corr RL Remarks

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121

Top Struct. Lab Door-2.420 13.822

GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 Same staff

position as

last reading

therefore

the same row

HPC = RL + BS HPC = 10.081 + 1.011 = 11.092

11.092

58

TBM CP Level Posn. BS FS ISKey

59

Site: …………………………………. Instrument: ………………………………….

Date: …………………………………. Observer: ………………………………….

Weather: …………………………………. Booker: ………………………………….

Building A 2

27/09/11 Group 1

Good Group 1

BS IS FS HPC RL Corr Corr RL Remarks

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121

Top Struct. Lab Door-2.420 13.822

GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD

New staff

position

therefore

a new row

2.009

RL = HPC - FS RL = 11.092 - 2.009 = 9.083

9.083

60

Before we look more fully at the results we will complete the

second half of the levelling exercise

TBM CP Level Posn. BS FS ISKey

61

Top of door level at entrance to structures laboratory

Ground level at entrance to structures laboratory

Point 2

TBM CP Level Posn. BS FS ISKey

62

TBM CP Level Posn. BS FS ISKey

63

TBM CP Level Posn. BS FS ISKey

64

TBM CP Level Posn. BS FS ISKey

65

TBM CP Level Posn. BS FS ISKey

66

TBM CP Level Posn. BS FS ISKey

67

TBM CP Level Posn. BS FS ISKey

68

TBM CP Level Posn. BS FS ISKey

69

TBM CP Level Posn. BS FS ISKey

70

TBM CP Level Posn. BS FS ISKey

71

Summary of Levelling Fieldwork

� For levelling fieldwork, the following practice should be adhered to in order to

improve the accuracy of the levelling works

� Levelling should always start and finish at points of known RL so that misclosure can

be detected

� Where possible, all sight lengths should be below 60 metres� Where possible, all sight lengths should be below 60 metres

� The staff must be held vertical by suitable use of a bracket bubble

� BS lengths =/~ FS lengths

� Reading should be booked immediately after they are observed. Important readings,

particularly readings at change points, should be checked

� The rise and fall method of reduction should be used if possible, especially for control

works. This HPC is only a sample

72

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

Levelling Booking & Calculation

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 9.083

The RL value of 9.083m is our measured and calculated value.

It should be 9.09m.

This gives an actual misclosure of 9.083 - 9.09 = -0.007m

This actual misclosure may be because of calculation errors or field errors 73

BS IS FS HPC RL Corr Corr RL Remarks

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121

If it is due to calculation errors we MUST NOT continue.

Therefore the first thing we always do after reducing our field booking is:

Carry out Calculation Checks

∑ FS - ∑ BS = 1st RL - Last RLSimple Calculation Check:

Top Struct. Lab Door-2.420 13.822

GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 9.083

∑ FS - ∑ BS = 1st RL - Last RL

∑ ∑3.975 4.892

LHS = 4.892 - 3.975 = 0.917

RHS = 10.000 - 9.083 = 0.917

Therefore LHS = RHS Therefore Calculations are OK 74

BS IS FS HPC RL Corr Corr RL Remarks

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121 NOT

CHECKED

NOT

CHECKED

Full Calculation Check:

∑ IS + ∑ FS + ∑ (RLs except first)

= ∑ (each HPC x number of applications)

Top Struct. Lab Door-2.420 13.822

GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 9.083

This Simple Check does not check the calculations for RL values calculated from IS

CHECKED

NOT

CHECKED

∑ FS - ∑ BS = 1st RL - Last RL

∑ ∑3.975 4.892

75

1.418

1.546

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD10.00011.546

C.P.1.562 9.98411.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 9.083

LHS = ∑ IS + ∑ FS + ∑ (RLs except first)

∑ 0.251 ∑ 4.892 ∑ 63.103

= 0.251 + 4.892 + 63.103 = 68.246

RHS = ∑ (each HPC x number of applications)

= (11.546x1+ 11.402x4 + 11.092x1) = (11.546 + 45.608 + 11.092) = 68.246

Therefore LHS = RHS

Therefore the calculations for all the RL values

are correct. 76

BS IS FS HPC RL Corr Corr RL Remarks

Top Struct. Lab Door-2.420 13.822

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

Now we can look at the magnitude of the misclosure

TBM 9.09m AAD2.009 9.083

We have already seen that the

Actual misclosure = 9.083 - 9.09 = -0.007mIs this acceptable ?

Rule of Thumb:

Allowable misclosure = ± 5 √N mmWhere N is the Number of Instrument Positions

which is the same as Number of BS readings

Therefore our Allowable misclosure = ± 5 √3 mm = ± 8.66 say ± 9mm

Therefore Actual < Allowable Therefore our Fieldwork is OK 77

We have carried out the calculation checks and have an acceptable misclosure

The final stage is to apply a correction procedure to distribute the actual misclosure

We assume that we made a similar error every time we set up the instrument

There are 3 backsights, so we set up the instrument 3 times

The actual misclosure was -7mm, so we need to add 7mm in order to correct it

We can add these 7mm to our Reduced Levels in any way, but it is best to assume

that the 7mm error occurred gradually as a set of small errors,

rather than all in one go.

We could divide 7 between 3 like this: 3 2 2

Or like this: 2 3 2

Let use choose the middle method. We will give 2mm to the 1st instrument position,

an extra 3mm to the 2nd position, and an extra 2mm to the 3rd position

We cannot divide our 7mm misclosure evenly between 3 positions,

but we can do our best (we do not use fractions of a millimetre)

Or like this: 2 2 3

We must not correct the initial Reduced Level

We apply the same correction to all readings up to and including each foresight78

BS IS FS HPC RL Corr Corr RL Remarks

TBM 10.00m AAD1.546 10.00011.546

C.P.1.562 9.9841.418 11.402

Point 11.390 10.012

GL Struct. Lab Door1.281 10.121

x

We cannot correct the given TBM value

10.000

2

5

5

9.986

10.017

10.126

Top Struct. Lab Door-2.420 13.822

GL Struct. Lab Door1.281 10.121

CP1.321 10.0811.011 11.092

TBM 9.09m AAD2.009 9.083

5

5

5

7

Corrections are applied with a +ve or -ve sign depending on the sign of the misclosure

10.126

13.827

10.086

9.090

We MUST end up with the correct

final reduced level.

79

T H A N K YO UT H A N K YO U

Q u e s t i o n & A n s w e rQ u e s t i o n & A n s w e rQ u e s t i o n & A n s w e rQ u e s t i o n & A n s w e r

80

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