TITLE: DETERMINATION OF DEFLECTION FOR BEAM WITH DIFFERENT MATERIALS.

INTRODUCTION:

This experiment is done to allow the students to familiarize themselves with the apparatus’s sensitivity and its accuracy. This experiment uses simple TQ beam apparatus, two vial gauges, two load hangers, beams the weights. In this experiment, it requires the student to measure the length between the two supports of the beam, the thickness and also the width of the beam. The mid span of the beam should be marked as a reference.

APPARATUS:

 

Two vial gauges, two load hangers, beams the weights.

PROCEDURE:

1. First the beam is placed in position above the support with overhang at either end.

2. The dial gauge is placed in position on the upper cross member so that the ball rest on the centre line of the beam. The stem is checked whether it is vertical or not and assurance is made that the bottom O-ring has been moved down the stem.

3. The dial gauge is adjusted to zero then the bezel is locked in position.

4. Two hangers are adjusted equidistant (‘A’ length) from the mid point of the beam (the curser may press lightly against the scale).

5. A load is applied to hangers and the beam deflection value read on the dial gauge is recorded.

6. Then the load is increased and the new dial gauge reading (deflection) is recorded. It is done at least three times.

7. The load then is decreased by the same value as step 6 and the beam deflection at each is recorded.

8. The experiment is repeated by taking three different distances for ‘A’ for each different types of beam.

9. A graph of deflection against load for each beam is plotted. The gradient for each graph is determined.

NOTE: The scale on the dial gauge is 0.1 mm. when the loads are applied to or removed from the hangers in a systematic manner, the beam is tapped very gently and then the reading is taken.

Some data value taken: E_S = 21 x10¹⁰ N/m²

E_B = 10.5×10¹⁰ N/m²

E_A = 7.6×10¹⁰ N/m²

 

 

 

 

 

 

DISCUSSION:

For brass beam:

1. From the experiment, we can see that the value of the deflection from the dial gauge is higher the other beam. But as we calculate using the formula which means its theorical value, it is even higher.

2. Beside that, we also can see that most of the value of while unloading is more than when it is loading. This may be due to the characteristics of brass that have lower tensile strength.

3. The value of deflection also increased as the load increased. This means that it is linear relations between the brass beam and the load. The graph plotted also shows the same conclusion.

For steel beam:

1. From the experiment, the deflection value is lower than the brass deflection value at the equivalent load. This means that steel have more resistance to deflection than brass.

2. Besides that, steel also shows that it has higher tensile strength than brass. It is proven by the smaller value of deflection compared to brass.

3. The value of deflection also increased with load. This shows a linear relationship between the steel beam and the load. The graph plotted also shows the same conclusion.

CALCULATION:

Modulus of elasticity for

1. Brass = 10.5 × 10¹º N/m²

2. Steel = 21 × 10¹° N/m²

3. Aluminium = 7.6 × 10¹° N/m²

1. 1. Brass Deflection

Thickness = 3mm

Width = 19mm

Length = 750mm

I = (0.019 × 0.003³)/12 = 4.275 × 10־¹¹ m

A = 15cm

Load (N)

Loading Deflection (mm)

Unloading Deflection (mm)

Mean of

Theorical value of

1. 1. Steel Deflection

Thickness = 3mm

Width = 19mm

Length = 750mm

I = (0.019 × 0.003³)/12 = 4.275 × 10־¹¹ m

A = 15cm

Load (N)

Loading Deflection (mm)

Unloading Deflection (mm)

Mean of

Theorical value

1. 1. Aluminium Deflection

Thickness = 6mm

Width = 19mm

Length = 750mm

I = (0.019 × 0.006³)/12 = 3.42 × 10־¹º m

A = 15cm

Load (N)

Loading Deflection (mm)

Unloading Deflection (mm)

Mean of

Theorical value

Discussion

1. From the calculated value above, deflection for brass is the greatest when load 10 N is exerted.

2. Deflection for brass is 2.95mm, 1.81mm for steel and 0.52 for aluminium.

3. Here we see that, the deflection is relatively depend on modulus of elasticity (E) where higher value of E for each material, deflection will decreased.

4. External for may be exerted when we tapping the beam and this affect our reading when we undergo the experiment because dial gauge is very sensitive device.

Introduction

The particle size analysis of a soil is carried out to determine the weight percentage falling within bands of size distribution represent size of soil. Where the sample contain fine-grained particle, wet sieving procedure is first to carry out to remove these and determine the combined clay/silt fraction percentage. Sub-sample is then immersed in water containing dispersion agent (sodium hexametaphosphates) before being washed through 63 μm mesh sieve. Sedimentation process in carried out when particle size distribution in the fine-grained fraction is not possible by sieving method.

Objective

To determine particle size distribution of fine-grained soil using dispersion of soil test and sedimentation test (pipette method).

Apparatus

1. 10ml Pipette

2. Sedimentation tube

3. Weighing bottles

4. Temperature bath

5. Mechanical shaker

6. British Standard sieves : 2mm, 600μm, 212μm, 63μm and appropriate receiver

7. A thermostatically controlled drying oven

8. A stop clock

9. A desiccator

10. A 650/1000 ml conical beaker and cover glass

11. A centrifuge capable of holding 250ml capacity bottles

12. A 100ml measuring cylinder

13. A glass filter funnel

14. A wash bottle

15. A length of glass rod

Reagent

1. Hydrogen Peroxide

2. Sodium hexametaphosphate (33g of sodium hexametaphosphate and 7g of sodium carbonate dissolved in distilled water to make 1 litre of solution)

 

Procedure

Pretreatment of Soil

1. 30g of soils (clay and sand) added into a 650ml/1000ml conical beaker named beaker A.

2. 50ml of distilled water added into beaker A.

3. Solution in beaker A is heated until its composition decreased to 40ml.

4. Beaker A then let cooled and 75ml of hydrogen peroxide is then added.

5. Closed the lid for one night

6. Beaker A heated until its composition decreased to 50ml.

7. Solution in Beaker A then transferred into a centrifuge bottle where its mass already being weight and distilled water added until the solution is 200ml.

8. The centrifuge bottle is closed and centrifuged 20 minute with 2000 rev/min.

9. Centrifuge bottle kept in an oven for a night until it’s completely dry.

10. Then the centrifuge bottle cooled in desiccators and the left over soil in the centrifuge bottle is weight.

Dispersion of Soil

1. 100ml of distilled water added into the soil in the centrifuged bottle obtain from previous step and then shakes vigorously to bring the soil into suspension.

2. 25ml of sodium hexametaphosphate added into the centrifuged bottle and then shakes on the mechanical device for 20 minutes.

3. The suspension then transferred to the 63μm BS test sieve placed on the receiver and the soil washed on the sieve using distilled water.

4. The material retained on the 2mm, 600μm, 212μm and 63μm then weight. Each mass recorded on these sieve then distributed as mass gravel, coarse, medium and fine sand in the sample (mg, mcs, mms, mfs respectively)

5. Material passing 63μm sieve is then transferred into a sedimentation tube.

Sedimentation

1. 25ml of sodium hexametaphospate added into a sedimentation tube and distilled water is added until the composition in the sedimentation tube is ≤ 500ml.

2. The sedimentation tube is put in a temperature bath at 25˚C (temperature at temperature bath is make sure higher than temperature at the sedimentation tub).

3. Approximately 1 hour, temperature at the temperature bath equal with the temperature at the sedimentation tube. Sedimentation tube is taken out from the temperature bath and shakes. Then the tube is put back into the temperature bath. Time is recorded as the tube inside the temperature bath. 3 sample from the sedimentation tube(10ml) taken using a pipette.

Sample 1, M1 after 4 min 5 sec

Sample 2, M2 after 46 min 0 sec

Sample 2, M2 after 6 hours 54 min

1. Each sample inserted into a different bottle that already been cleaned and weight its mass. The samples then let dry in an oven and cooled after that in a desiccators. The left over is weight as m1, m2 and m3.

 

By: Basyid Hamid HK2005-4056

Awang Azamawy bin Junggal HK2005-4065

RESULTS

A. Pretreatment of Soil

1. Weight of sample : 30g

2. Volume of distilled water : 50ml

3. Volume of hydrogen peroxide : 75ml

4. Weight of centrifuge bottle : 177.19g

5. Weight of centrifuge bottle + dried soil : 207.19g

m= (5)-(4) : 30g

B. Dispersion of Soil

1. Volume of distilled water : 100g

2. Volume of sodium hexametaphosphate solution : 25ml

3. Shaking time : 20min

4. Second wet sieving (‘final weight’ taken after drying the 2^nd wet sieving)

Sieve size

Initial weight

Final weight

Weight retained

C. Sedimentation

1. Volume of sodium hexametaphosphate solution : 25ml

2. Temperature of the bath : 29⁰C

3. Volume of pipette and tap (V_p) : 10ml

4. Dried weight of soils

Time (after shaking)

Weighing bottle, g

Weighing bottle + dried sample, g

Dried sample, g

By: Ching Ghin Wei

CALCULATION

The mass of pretreated soil (m) shall be used to calculate the percentages below:

1. the percentage of gravel in the original sample shall be calculated from the following equation;

Percentage gravel (over 2.0mm) =

=

=2.13%

1. the percentage of coarse sand in the original sample shall be calculated from the following equation;

Percentage gravel (2.0mm to 0.6mm) =

=

=14.63%

1. the percentage of gravel in the original sample shall be calculated from the following equation;

Percentage gravel (0.6mm to 0.2 mm) =

=

=30.57 %

1. the percentage of gravel in the original sample shall be calculated from the following equation;

Percentage gravel (0.2mm to 0.06mm) =

=

=19.33%

1. The mass of solid material in 500 ml of suspension for each respective sampling time shall be calculated from the equation:

W₁=

=

=10g

W₂=

=

=8.5g

W₁=

=

=7.5g

W₁=

=

=7.0g

W₁ - is the material in 500 ml from the first sampling (g)

W₂ - is the material in 500 ml from the second sampling (g)

W₃ - is the material in 500 ml from the third sampling (g)

W₄ - is the mass of sodium hexametaphosphate in 500 ml (g)

Vp is the calibrated volume of the pipette (ml)

2. The percentage of medium silt in the original sample shall be calculated from the following equation :

Percentage medium silt (0.02mm to 0.006 mm) = %

= %

= 5%

3. The percentage of fine silt in the original sample shall be calculated from the following equation:

Percentage fine silt (0.006mm to 0.002 mm) = %

= %

= 3.33%

4. The percentage of clay in the original sample shall be calculated from the following equation:

Percentage clay (less than 0.002 mm) = %

= %

= 1.67%

5. The percentage of medium silt in the original sample shall be calculated from the following equation:

Percentage coarse silt (0.06 mm to 0. 02 mm)

=x 100%

=x 100%

=23.33%

Ho King Foo HK2005-

Discussion:

1. Every step is to be taken with great accuracy, especially when it comes to using the pipette because the capillary action of water and the surface of the glass will stick, and there will be meniscus, so, have to look where the curve touch

2. The condition of the water bath cannot be kept at 25 C because it is lower than the room temperature, which is 29 C, therefore, we have to calibrate the temperature so that it won’t change entire course of the experiment

3. When the 10ml of the solution is extracted, the solution is made sure not to be disturbed, so that we won’t have to restart our stopwatch every time we take a reading, so, the time would be cumulative

4. The use of sodium hexametaphospate is to act as dispersing agent and prevent the clay size particles from forming flocs during the hydrometer test

5. The method provides an inexpensive and reliable estimate of soil texture, useful in soil-quality assessment.

6. The sample was pretreated with hydrogen peroxide in order to keep the organic substance away. This is because the existence of the organic substance will give the affect to the result.

7. There is disadvantage of the pipette method. That is the time consuming in this type of experiment is too long.

8. The advantage of the pipette method is that it can separate further sub-division of particle-size distribution in the fine-grained fraction compare to the sieving method which can only determine the particle size until 63µm. In pipette method, it can be used to determine the soil sample and the percentage of particle-size obtained is more reliable.

9. Precautions during the pipette method experiment,

• Student’s eyes sight must be perpendicular as there is a parallax error if more or less than 90 degrees.

• All the container such as glass sedimentation tubes, conical beaker, centrifuge and sieves should be clean and dried in oven before experiment.

• All the samples were making sure fully dry in the oven which the samples should be dried in the oven for at least 24 hours.

• The soil sample should handle with care and fan must be off when filling soils sample into the container to avoid some soils sample might be flying away by wind.

• Beside that, when we taking the reading of the mass of the soils sample with electronic balance, the ceiling fan should be turned off because the electronic balance is very sensitive.

• Careful should be taken during shaking process to avoid some of the sample flew away.

1. 1. Errors encountered during the experiment;

 

1. The soils sample prepared might not 100% in dry condition.

2. The scale affects the result because miss taken the reading or the it cannot give the accurate reading.

3. Parallax error may be encountered during experiment, due to the incorrect eye sight position.

4. Besides, the ambient temperature also makes the result not accurate, because it causes the rate of the evaporation of water getting higher due to the high temperature and high wind blow.

5. Small amount of the soils sample might fly away by wind when filling soils sample into the container. This can cause a reduction of the weight of soils sample and then affect the reading and results.

6. There are sometimes the airs bubbles trapped in the pynometer which will affect the total weight of water.

7. Time taken to shake the sample is not the same, and this will lead the samples are not in the same condition during the experiment.

8. It is hard to record the surrounding temperature, so temperature variation during testing was not recorded.

Conclusion:

As

Attachment:

Andreasen Pipette and Stand with Constant Temperature Bath

Mechanical Analysis Stirrer

Reference

Roy Whitlow, Basic Soil Mechanics Prentice Hall, 2001

http://www.ele.co.uk/pdfs/18-19.pdf

ASTM D422-63(2002)e1

Clement Wong;  7-10-2007, Sunday, 19:42:56

Terrace house

 BEAM MARK: gb7

 DETAILED CALCULATION CORRESPONDING TO BRIEF BEAM OUTPUT FILE *.OO1

 CodeOfPractice    fcu      fys      fyv      cover  span  cantilever
    BS8110:1997     25      460      250       35      1         Nil

Span No    Span-m   Width-mm Depth-mm F-width F-depth

    1   3.35      150      450      150        0

 Maximum Concrete strain, Ecc = 0.0035
 Average Concrete Stress Above Neutral Axis, k1 = 10.20 N/mm^2
 Concrete Lever Arm Factor, k2 = 0.4557
 Limiting Effective Depth Factor, cb = 0.50
 Limiting Concrete Moment Capacity Factor, kk1
          = cb*k1*(1-cb*k2) = 0.50*10.20(1-0.50*0.4557) = 3.9379 N/mm^2
 k2/k1 Factor, kkk = 0.0447

 Span No = 1 ; Location =  1/4 Span ; Bending Moment, M: 0.0 kNm

 kk1 = cb*k1*(f1-cb*k2) = 0.50*10.200*(1-0.50*0.456) = 3.938

 Mu/bd^2 = 0.0*1000000/(150*403^2) =  0.000

 For Singly Reinforced Design, limit Mu/bd^2 < kk1 ;  Mu/bd^2 = 0.000 ; kk1 = 3.938
 Design as Singly Reinforced Rectangular Beam : b = 150 mm;  d = 403 mm

 Concrete Neutral Axis, x = 0.0 mm

 Concrete Compression Force, Fc = k1.b.x/1000 = 10.20*150*0.0 = 0.00 kN

 Steel area required, As = Fc*1000/(fyy*fy) = 0.00/(0.95*460) = 0 mm^2

 Moment capacity = Fc(d-k2.x) = 0.00(403-0.4557*0.0)/1000 = 0.0 kNm

 Tension Steel area required, As = 0 mm^2
 Compression Steel area required, As = 0 mm^2

 Tension Steel area provided, As = 101 mm^2
 Compression Steel area provided, As = 88 mm^2

 Span No = 1 ; Location = Span ; Bending Moment, M: 28.0 kNm

 kk1 = cb*k1*(f1-cb*k2) = 0.50*10.200*(1-0.50*0.456) = 3.938

 Mu/bd^2 = 28.0*1000000/(150*409^2) =  1.114

 For Singly Reinforced Design, limit Mu/bd^2 < kk1 ;  Mu/bd^2 = 1.114 ; kk1 = 3.938
 Design as Singly Reinforced Rectangular Beam : b = 150 mm;  d = 409 mm

 Concrete Neutral Axis, x = 47.1 mm

 Concrete Compression Force, Fc = k1.b.x/1000 = 10.20*150*47.1 = 72.13 kN

 Steel area required, As = Fc*1000/(fyy*fy) = 72.13/(0.95*460) = 165 mm^2

 Moment capacity = Fc(d-k2.x) = 72.13(409-0.4557*47.1)/1000 = 28.0 kNm

 Tension Steel area required, As = 165 mm^2
 Compression Steel area required, As = 0 mm^2

 Tension Steel area provided, As = 165 mm^2
 Compression Steel area provided, As = 88 mm^2

 Span No = 1 ; Location = Left Support ; Bending Moment, M: 0.0 kNm

 kk1 = cb*k1*(f1-cb*k2) = 0.50*10.200*(1-0.50*0.456) = 3.938

 Mu/bd^2 = 0.0*1000000/(150*403^2) =  0.000

 For Singly Reinforced Design, limit Mu/bd^2 < kk1 ;  Mu/bd^2 = 0.000 ; kk1 = 3.938
 Design as Singly Reinforced Rectangular Beam : b = 150 mm;  d = 403 mm

 Concrete Neutral Axis, x = 0.0 mm

 Concrete Compression Force, Fc = k1.b.x/1000 = 10.20*150*0.0 = 0.00 kN

 Steel area required, As = Fc*1000/(fyy*fy) = 0.00/(0.95*460) = 0 mm^2

 Moment capacity = Fc(d-k2.x) = 0.00(403-0.4557*0.0)/1000 = 0.0 kNm

 Tension Steel area required, As = 0 mm^2
 Compression Steel area required, As = 0 mm^2

 Tension Steel area provided, As = 101 mm^2
 Compression Steel area provided, As = 88 mm^2

 Span No = 1 ; Location = Right Support ; Bending Moment, M: 0.0 kNm

 kk1 = cb*k1*(f1-cb*k2) = 0.50*10.200*(1-0.50*0.456) = 3.938

 Mu/bd^2 = 0.0*1000000/(150*403^2) =  0.000

 For Singly Reinforced Design, limit Mu/bd^2 < kk1 ;  Mu/bd^2 = 0.000 ; kk1 = 3.938
 Design as Singly Reinforced Rectangular Beam : b = 150 mm;  d = 403 mm

 Concrete Neutral Axis, x = 0.0 mm

 Concrete Compression Force, Fc = k1.b.x/1000 = 10.20*150*0.0 = 0.00 kN

 Steel area required, As = Fc*1000/(fyy*fy) = 0.00/(0.95*460) = 0 mm^2

 Moment capacity = Fc(d-k2.x) = 0.00(403-0.4557*0.0)/1000 = 0.0 kNm

 Tension Steel area required, As = 0 mm^2
 Compression Steel area required, As = 0 mm^2

 Tension Steel area provided, As = 101 mm^2
 Compression Steel area provided, As = 88 mm^2

 DEFLECTION CHECK FOR SPAN NO. 1:
 Refer to BS8110:Part 1: Table 3.10, Table 3.11 & Table 3.12

 Eqn. 8, fs = 5fy*As,reqd/(8As,prov) = 5*460*165/(8*226) = 223.8 N/mm^2
 Eqn. 7, Tension Modification Factor, TMF = 0.55 + (477-fs)/(120*(0.9+M/bd^2))
      = 0.55 + (477-223.8)/(120*(0.9+27950674/(150*409.0^2))) = 1.60

 Actual Beam span/depth ratio = 3350/409.0 = 8.2

 Eqn. 9, Compression Modification Factor, MF1 = 1+As/(3+As/)  = 1+0.37/(3+0.37) = 1.11
 Allowable span/depth ratio =  TMF*MF1*BasicRatio  = 1.60*1.11*20 = 35.5

 Modification fac = 1.77; Deflection ratio = 4.33; Steel = 0.37 percent

 Actual Beam span/depth ratio < Allowable span/depth ratio, i.e. 8.2 < 35.5 –> Deflection O.K.

 SHEAR CHECK:
 Span No 1 at Left Support ; Shear, V = 33.4 kN

 Shear Stress, v = V/bd = 33.4*1000/(150*403) = 0.553 N/mm^2

 Refer to BS8110:Part 1:1997 Table 3.9
 Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
 Effective depth ratio = max(1,400/d) = max(1,400/403) = 1.000
 Concrete Grade ratio  = min(40,fcu)/25 = min(40,25)/25 = 1.000
 Steel Percentage, 100As/(bd) = min(3,0.37) = 0.37
 vc = ( 0.79*(0.37)^1/3*(1.000)^1/4*(1.000)^1/3 )/1.25 = 0.456 N/mm^2

 Shear Stress - Shear Capacity = v - vc = vd = 0.553 - 0.456 = 0.097 N/mm^2
 vd < 0.40 N/mm^2 –> Design for vd = 0.40 N/mm^2

 Steel area provided by Link size 10 = 2*pie*dia*dia/4 = 2*3.1416*10*10/4 = 157.1 mm^2
 Link spacing required for dia. 10 = 622
 Shear Capacity provided by Link = 0.95*250*157.1/(622*150) = 0.400 N/mm^2

 Link provided = 1R-10-250

 Span No 1 at  1/4 Span ; Shear, V = 16.7 kN

 Shear Stress, v = V/bd = 16.7*1000/(150*409) = 0.272 N/mm^2

 Refer to BS8110:Part 1:1997 Table 3.9
 Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
 Effective depth ratio = max(1,400/d) = max(1,400/409) = 1.000
 Concrete Grade ratio  = min(40,fcu)/25 = min(40,25)/25 = 1.000
 Steel Percentage, 100As/(bd) = min(3,0.37) = 0.37
 vc = ( 0.79*(0.37)^1/3*(1.000)^1/4*(1.000)^1/3 )/1.25 = 0.453 N/mm^2

 Shear Stress - Shear Capacity = v - vc = vd = 0.272 - 0.453 = -0.181 N/mm^2
 vd < 0.40 N/mm^2 –> Design for vd = 0.40 N/mm^2

 Steel area provided by Link size 10 = 2*pie*dia*dia/4 = 2*3.1416*10*10/4 = 157.1 mm^2
 Link spacing required for dia. 10 = 622
 Shear Capacity provided by Link = 0.95*250*157.1/(622*150) = 0.400 N/mm^2

 Link provided = 1R-10-250

 Span No 1 at Right Support ; Shear, V = 32.9 kN

 Shear Stress, v = V/bd = 32.9*1000/(150*403) = 0.545 N/mm^2

 Refer to BS8110:Part 1:1997 Table 3.9
 Shear Capacity,vc = 0.79*((100As/(bd))^1/3)*(400/d)^1/4)*((fcu/25)^1/3)/1.25
 Effective depth ratio = max(1,400/d) = max(1,400/403) = 1.000
 Concrete Grade ratio  = min(40,fcu)/25 = min(40,25)/25 = 1.000
 Steel Percentage, 100As/(bd) = min(3,0.37) = 0.37
 vc = ( 0.79*(0.37)^1/3*(1.000)^1/4*(1.000)^1/3 )/1.25 = 0.456 N/mm^2

 Shear Stress - Shear Capacity = v - vc = vd = 0.545 - 0.456 = 0.089 N/mm^2
 vd < 0.40 N/mm^2 –> Design for vd = 0.40 N/mm^2

 Steel area provided by Link size 10 = 2*pie*dia*dia/4 = 2*3.1416*10*10/4 = 157.1 mm^2
 Link spacing required for dia. 10 = 622
 Shear Capacity provided by Link = 0.95*250*157.1/(622*150) = 0.400 N/mm^2

 Link provided = 1R-10-250

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One Liner

1. Regular naps prevent old age… especially if you take them while driving.

2. Having one child makes you a parent; having two makes you a referee.

3. Marriage is a relationship in which one person is always right and the other is the husband!

4. They said we should all pay our tax with a smile. I tried- but they wanted cash.

5. A child’s greatest period of growth is the month after you’ve purchased new school uniforms.

6. Don’t feel bad. A lot of people have no talent.

7. Don’t marry the person you want to live with, marry the one you cannot live without… but whatever you do, you’ll regret it later.

8. You can’t buy love. . But you pay heavily for it.

9. True friends stab you in the front.

10. Forgiveness is giving up my right to hate you for hurting me.

11. Bad officials are elected by good citizens who do not vote.

12. Laziness is nothing more than the habit of resting before you get tired.

13. My wife and I always compromise. I admit I’m wrong and she agrees with me.

14. Those who can’t laugh at themselves leave the job to others.

15. Ladies first. Pretty ladies sooner.

16. It doesn’t matter how often a married man changes his job, he still ends up with the same boss.

17. They call our language the mother tongue because the father seldom gets to speak.

18. Saving is the best thing. Especially when your parents have done it for you.

19. Wise men talk because they have something to say; fools talk because they have to say something.

20. Real friends are the ones who survive transitions between address books

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The new wife was being welcomed at the husband’s home in a traditional manner.

As expected she gave a speech;

“My dear family, I thank you for welcoming me in my new home and family,

firstly, my being here does not mean that I would want to change your way of life, your routine.

“No, I will never do that, never in a million years.”

“What do you mean my child?” asked the father-in-law.

“What I mean dad is (looking at her in-laws);

Those who used to wash the dishes must carry on washing them.

Those who used to do the laundry must carry on doing it.

Those who cooked should not stop at my account, AND

Those who used to clean should continue cleaning!!!

“And what are you here for?” enquired the mother-in-law.

SCROLL……………………………. FOR ANSWER

“AS FOR ME, I’M HERE JUST TO ENTERTAIN YOUR SON!!!!!”

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this blog have moved, sorry to say, but I have found an easier platform to blog at http://clement.i.ph

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