CHAPTER FOUR

CHAPTER FOUR

SCIENCEQUESTION ONE

Three ways by which heat is transferred are:

1. Convection

2. Radiation

3. Conduction

1.CONVECTION: Is the process by which heat is transferred through a fluid medium. Convection has two practical uses in building and that is it removes state air fumes and keeps up the supply of fresh air and it causes water to circulate naturally in hot water supply system.

2.RADIATION: Is the process by which heat is transferred through air/gas. It does not need a solid or liquid medium. This means radiation is the only process which takes place in vacuum.

3.CONDUCTION: Is the process whereby heat is transferred from one material to another through a solid medium such as metals which generally transmit heat by conduction.

(II)

1.BENDING MOMENT: Is the algebraic sum of the moment about the section of all the forces acting at one of the section.

2.BULK DENSIT Y: It is the mass per unit of the apparent volume of a substance.

3. SOULD DENSITY: Is defined at the mass per unit of the absolute volume of a substance

(III) Data:Density = 2000kg/m

Volume = 2m x 1.25m x 5m=1.25m

To find the weight of a block

Density = Mass

Volume

Mass = Density x Volume

2000kg/m x 1.25

2500kgf

9.8 x 2500 = 24500N

24500 1000

= 24.5KN

(IV)

a. Porosity in materials reduces durability.b. porosity in materials reduces specific gravity.

c. It reduces the strength of materials

d. It also increases heat insulation materials.

e. Porosity reduces sound insulation in materials.

(V)

Data

Bulk Density = 2080kg/m

Solid Density = 2660kg/m

To calculate the percentage porosity of stone

Porosity = Solid Density Bulk Density x 100

Solid Density

= 2660kg/m 2080kg/m x 100

2660kg/m

= 580 x 100

2660

= 21.8045

= 21.8%

QUESTION 2

(i) Data

Oven dry weight = 300g

Wet Weight = 3450g

Calculate the percentage by weight

Weight percentage = wet weight Dry weight x 100

Dry weight

= 3450 300 x 100

300

= 3150 x 100

300

= 1050%

(ii) (a)Water absorption is the taking in of water through the pores of materials and heeding the water in the materials because of the voids. Example is when rain falls on dry concrete the first few drops disappear into the concrete.

(b)Permeability is defined as the rate at which water will flow through a saturated material under constant head of water.

(c)Moisture content is the amount of water present in a material.

(d)Specific gravity is defined as the reaction of any volume of a substance to the mass of an equal volume of water.

(iii) Data

Initial Weight = 20N

Time = 24Hrs

Percentage absorption = 21 + 100 = 121

To calculate the final weight of the brick

Final Weight = Percentage of absorption x Initial Weight

100

= 121 x 20N

100

= 24.2N (iv) DataSpecific gravity = 7.8

Density of water = 1000kg/m

To calculate for density of a metal in kg/m and weight per m (kg/m)

Specific Gravity = Density of a substance

Density of water

Density of water = Density of Substance Specific gravity

= 1000kg/m3 7.8

=128.21kg/m

(v) Two examples of how defects can rise due to drying shrinkage are:

(a) Over seasoning of timber.

(b) When the timber is not grownQUESTION THREEDATAWeight of iron boiler = 150kgf

Water= 2000kg

Change in temperature $sc -50 =70c

To find how much heat a joule is required to raise the temperature of both boiler and its content.

SOLUTION

Iron boiler MCKDT (Weight x water x change in temperature)

= 150kgf x4600 j/kg x 70c

= 48.300,000J

Water = MCDT (Mass x water x change in temperature

= 200kg x 4600 kg x 70 = 64.400,000JBoth = Iron boiler +water

= 48.300,000J +64,400,000J

= 112,700,000J(ii)

a. MOMENT OF A FORCE: It is the force about a point which is the product of the force and the perpendicular distance of its line of action form the point.

b. RESULTANT FORCE: Is the multiple of two or more forces that produces a single force.

c. MECHANICAL ADVANTAGE: It is the ratio of load to effort

d. MASS: It is the quantity of matter contained in a substance

e. WEIGHT: It is the gravitational pull exert on any Objects by the earth.

f. EFFORT: Is defined as the force applied a load to overcome a resistance.

QUESTION FOUR

DATA

Room dimension =35mm x 40mm

Scale=1:100

The actual dimension of the room

SOLUTION

IF 1= 100

IF 1=100

35 =?

40=?

35 X100

40 X 100

1

1

= 3500mm

=4000mm(B)DATA

Length =350mm 1000=0.35m

Width = 150mm -1000 =0.150mm

Thickness =20mm 1000 =0.020m

Density of metal

SOLUTION

Density =mass

Volume (length x width x thickness

10.00150m3

= 150,000 kg/mC DATA

Length =5m

Square section =20mm -1000 = 0.020m

Density = 2.65 x 103kg/m3

The mass of a square section aluminum rod

SOLUTION

Density = Mass

Volume

Volume = density x (Length x square section)

Volume = 5m x 0.020m x 0.020m = = 0.2m3

Mass= 2650kg/;m3 x 0.002m3 = 5.3kg

QUESTION FIVE

BEARING CAPCITY: Is the maximum stress of the maximum load per unit area that the soil can carry without unequal settlement.

BEARING PRESSURES: It is the pressure produced on the ground due to load acting on it.

FACTOR OF SAFETY : It sit he ratio of the ultimate stress to the working stress

SLENDERNESS RATIO: It is the ratio of the actual length of strut to its least lateral dimension.

HYDROSTATIC PRESSURE: Is defining as the pressure exerted in liquids.

MADE UP GROUND: It is the radiation of portion of the land by filling to an appreciable level either below of above datum.

Data given

Load = 20kg

Distance = 3m

1kgf =9.84

Required: to calculate the % efficiency of the machine

Solution

Percentage efficiency = mechanical advantage x 100

Velocity ration

= 20kg x 9.84

3

= 65.6

=66

QUESTION SEVEN

Efflorescence would be treated in affected block wall by removing the crystal deposit and brushing the surface dry with metal brush.

Washing down merely shift the crystal from place to place unless continuously does for a long time

(B)

Data given

Ratio = 2:1

Bearing capacity = 240kn/m2Load = 70 KN

Factor of safety = 1.5

Required: To calculate the size of the base according to the ratio

Solution

Working stress = Ultimate stress

Factor of safety

=240KN/m2

1.5

= 160 KN/M2Area of the base = Load

=70KN

Stress =0.438m2Size of base =2a x a=0.438

Therefore

a2 =0.219

A= (0.219) =0.46m

2a = 0.936m

QUESTION SEVEN (C) Data given

Bearing capacity = 200KN/m

Load = 65KN

Factor of safety = 2.5

Required: To calculate the recommended

Width for the strip foundation

Solution

Working stressUltimate Stress

Factor is safety

= 200KN/M

25

= 80KN/mTherefore width = Load

Working stress

=65KN

80KN

= 0.88125.5

=812.5

= 813mm

(D)

Data given

Mass =50kg

Area = 50mm2

Diameter = 8mm

Required: to calculate for

a. The force on the surface on which she stands.

b. The pressure on the surface on which she stands.

c. The force on the surface.

SOLUTION

a. Force = Mass x acceleration due to gravity

= 50kg x 9.81

= 490.5 N

i. Pressure Force Area

= 490.5N

50mm2

= 9.81 N/MM2Force on the surface = Area = X d2

4

3.142x 8mm24

=201.088mm2

4

= 50.272mm2Pressure = Force

Force = Pressure x Area

Area

=9.81 N/mm2 x 50. 272 m2QUESTION EIGHT

SOLUTION

Taking moment about RA Sum of clockwise = Sum of anti Clockwise

(60 KN x 1.2m) + (40KN x 3.2m) =RB x 4m

72KNm + 12% KNm= RB x 4m

RB = 200 KNm

4m

RB = 50KN

In equilibrium

Sum of upward force = Sum of downward force

= RA + RB =100KN

= RA + 50KN =100KN

= RA = 50 KN.

Bending moment = Reaction moment - load moment

BM max (A) BM max (D)

= OKN

BM max (B)

= Rational moment Load moment

= (50KN x 1.2m) (60KN x om)

= 60 KNm

= 60 KN/m

BM max (c)

= Relational moment Load moment

= (50KN x 3.2m) (60KN x 2m)

(40KN x 0)

= 160KNm 120KNm -o

= 40KNm

Shear force max = Rational Load

SF max (A) =500KN O

50KN

SF max (B) = Reactional - load

= 50KN O -60KN

=-10KN

SF max (c) = Reaction _Loan

= 50KN-O-60 KN 40KN

=-50 KN

SF max (D) = Reaction Load

=50Kn -0-600KN

= 40KN +50KN

=OKN

BENDING MOMENT DIAGRAM (attached)QUESTION NINE

(a) INDUCED IS SYPHONAGE: Is the action that takes place when a column of water in a soil stack passing causes other appliances water seal to get broken,

(b) SELF SYPHONAGE: Is defined as the action that takes place when appliance causes its own water seal to get broken.

(c) BACKPRESSURE: It is the pressure in pipes which exceed the atmospheric level and cause the admission of drawn air into the room.

QUESTION TEN

Data given

Bearing Capacity = 250N/m2Square size = 1.8m = 1.8m x 1.8m = 3.24m2Factor of safety = 2

Required to calculate the maximum load to be carried by the column.

Solution

Working stress = ultimate stress

Factor of safety

= 250 KKN/m2

2

= 125 KN/m2

Maximum load = working stress x square side

= 125KN/m2 x 3.24m2

= 405 KN

(B)

Data given

Width = 6.6m

Thickness = 0.225m

Height = 3.75m

Density = 210kg/m

Required: To calculate the mass of brick

Work of the gable and of a wall.

Solution

Volume of gable = 1.2 x 6. 6m x 3.75m x 0.225m

= 2. 784m3

Mass = Density x volume

= 210kg.m3 x 2.784m3

= 584. 64kg(C)( C) Reactions of the support

BA = RB= 5KN + 20KN + 5KN

RA =RB = 30 KN

2

RA = RB = 15KN

Since the load is symmetric, the reactions are equal

RA = 15N

RB = 15KN

MEMBERMANINITUDESENESSDISTANCE

A F

B G

C H

D J

E J

E F

E F

G H

H J55 KN

41 KN

38 KN

55 KN

33 KN

47 KN

20 KN

17 KN

9 KN

Compressive

Compressive

Compressive

Compressive

Tensile

Tensile

Compressive

Tensile

Compressive

QUESTION ELEVEN (A)

Data given

Reduced level of the lower ground = 33.6m

Height of traveler = 6.7m

Inverted level (lower) = 29.0m

Distance between sight rails (AB) =60m

Required: To calculate

(i) Gradient of the sewer trench

(ii) The right of size of A

(iii) The invert depth at A

(iv) The invert depth at B

(v) Reduced level on the ground at B

(vi) Height of sight rails at B

(vii) Adjustment at A to obtain HI

(viii) Adjustment at is to obtain HPC

Solution

(a) Gradient of the sewer trench

Vertical distance

Horizontal distance

But vertical distance = inverted level (upper)

= inverted level (lower)

= 29.5m 29.OM

= 0.5m

Gradient = 0.5m = 5 = 1:120

60m 600

(ii) The height of sight rail at A

= (inverted level (lower) + (height of traveler) Reduced level on the lower ground

= (29m + 6.7m) -33.6m

= 35.7m 33.6m

= 2.1m

(iii) The invert depth at A

= Height of traveler height of sight rails of A

= 6.7m -2.1m

= 4.6m

(iv) The invert depth at B

= invert depth at A + vertical distance

= 4.6m + 0.5m

= 5.1m

(v) Reduced level on the found at B

=inverted level (upper) + invert depth at B

= 29.5 + 5.1m

=34.6m

(vi) Height of sight rail at

= (inverted lower ground + height of traveler)

Reduced level at B

= (29m + 6.7m) -34.6m

= 1.1m

(vii) Adjustment at A to obtain Height of instrument

There is no adjustment at A to obtain Height of instrument.

(viii) Adjustment t B to obtain HPC

Height of sight rail at A- Height of sight rail at B

= 2.1m lm

= 1m

(B)

Data givenUltimate tensile stress = 450N/mm2Factor of safety = 3

Required: To calculate the working stress of the materials

Solution Working stress = ultimate stress

Factor of safety

= 450N./mm2

3

= 150N/mm2

QUESTION TWELVESolution

Taking moment about RA

Clockwise moment =Anti- clockwise moment

(3KN x5m) + (5KN x 16m) + (2KN x 25m) = RB x 3cm

15KN + 80KNm + 50Knm = RB x 30

145 KNm =RB x 30m

RB =145KNm

30m

=4.83KN

In equilibrium

Sum of upward forces = Sum of downward forces

RA + RB = 3KN + 5KN +2KN

RA+RB = 10KN

RA + RB +4.83KN = 10KN RA = 10KN 4.83KN

RA = 5.17KN

Bending moment values

= Reaction moment load moment

= BM max A = BM max E = OKN

Bmmax B = (5.17KN x 5m) (5KN x 0) = 25.85KN

Bmmax C = (5.17KN x 16m) (3KN x 5) (5KN x 0)

= 67.72KN

BMmax D = (5.17KN x 25m) (3KN x 16m) (5KN x 11m) (2KN x 0)

= 26.25KN

Shear force moment

Shear force max (A) = Reaction load

= 5.17KN 0

= 5.17KN

Shear force max (B) = Reaction load

= 5.17KN 0 3KN

= 2.17KN

Shear force max (C) = 5.17KN 0 3KN 5KN

= -2.83KN

Shear force max (D) = 5.17KN 0 3KN 5KN

= - 4.83KN

Shear force max (E) = 5.17KN 0 3KN 5KN - 2KN + 4.83KN

= 0KN

(B)

DATA GIVEN

Ultimate stress = 50N/mm2Factor of safety = 10

Required; to calculate the working stress of the material

SOLUTION

Working Stress = Ultimate stress

Factor of safety

= 50N/mm2

10

= 5N/mm2 (C)

Data Given

Total load = 30 tones / force

= 30 x 1000 x 9.81

= 294,300N

Ultimate stress = 28N/mm2 Factor of safety = 4

Required; to calculate the minimum size of a short square column SOLUTION

Working stress = Ultimate stress

Factor of safety

= 28N/mm2 4

= 7N/mm2 Area = Load Working stress

= 294,300N

7N/mm2 = 42,042.9mm2 Size of a short column = Area

= 4242.9mm

= 205mm

=205mm

(D)

Data given

Force = 100N

Area = 5m2Required To calculate the pressure on the surface

Pressure = Force

Area

= 1000N

5m2= 200N/m2QUESTION THIRTEEN Data given

Impose lead = 34N./m

Factor of safety = 4

Bearing capacity = 200KN/m2Required To calculate the width f the concrete stripSolution

Bearing pressure = bearing capacity

Factor of safety

= 500KN/m2Area = Load

Bearing pressure

= 34/Nm

1000

= 0.034KN/m2

50KN/m2 Width =0.00068m

(B)

Data given

Total load = 284.45 K/n

Safety bearing load = 159.33KN

Required To calculate the size of the foundation required for a squire brick.

Solution

Pressure = Force

Area

Area = Force

Pressure

= 284.45KN

169.33KN

= 1.6799

Size of foundation = size Area

169. 33m

130.0

Size of foundation = 13/.)m x13.0m

(C)

Data given

Square section =100 x100mm

Working stress= 7N/mm2

1 kg/f = 9.8N

Required: To find the maximum local which may be applied to the timber struct?Solution

Stress=load/Force

Area

Load =stress x Area

= 7N/mm2 x 100 x 100mm

= 7000N

=7000N 98N

= 7142.857 kg

(D)

Data given

Bearing capacity = 200KN/m2

Load = 65Kn

Factor of safety =2.5

Required: To calculate the recommended

Width for the strip foundation.

Solution

Working stress= ultimate stress

Factor of safety

= 200KN/m2

L2.5

= 80KN/m2Width of foundation = load

Working stress

= 65KN

80KN/m2

= 813m

Area = 2

0.813m2=9

0.813

=9

0.90166m

Area = 2

0-813m 2=9

0.813

=9

= 0.90166m

Solution

Uniformly distributed load =UDL

=43N/m x 4

=16 KN

Taking moment about RA

Sum of clockwise moment = sum of Anti-clockwise moment

(8KN x 3m)+(16KN x 5m) +(12Kn x7m) =RB x 10m

24KNm + 80KNm = 84KNm = RB x10m

188KNm = Rb x10m

RB =188KNm

10m

RB =18.8KN

In equilibrium

Sum of upward forces = sum of down wared forces

RA +RA =8KN +16KN +12KN

RA +18.8KN =36KN

RA=36KN 18.8KN

RA =17.2KN

Bending moment values

Rational moment Load moment

BM max (A) =BM max E

=OKN

BM max (B) = RM Lm

=(17.2Kn x3m) (8KN x0)

= 51.6KNm OKN

=51.6KNm

BM ma=bm-Lm

= (17.2Kn x5m)-(8Kn x3m)-(16Kn x0)

=86KNm -24kNm

= 62Knm

BM max (D)= RM Lm

=(17. 2KN x 5m) ( 88Kn x 5m) (16kn x2m) (12KN x0)

= 120.4knm 40knm -32kN

= 48.4KNm

Shear force = R-l

SF max (A) =17.2KN O

=17.2KN

SF max (B)=R-L

= 17.2KN O-8KN

=9.2KN

SF max (c) =R-L

= 17.2KN O -8KN -16KN -12KN

=-18.81KN

SF max (E)=R-L

=17.2KN-0-8KN -16KN-12KN+18.8KN

(B)

a. Dew point: is defined as the temperature at which water vapor in the atmospheric is just sufficient saturates it.

b. Atmospheric Pressure: is defined as the air having weight and awing to its weight the atmospheric exerts a pressure on all surface to the earth

c. Ambient temperature: is defined as the temperature of the surrounding air.

d. Vebe/ consitometre test

e. Slump test: is defined as the test conducted on site where quick checks on control are necessary to provide a comparisim of workability between similar. Mixes.

(c)AREA Q/5 =Tan 30

X =5x Tan 3-

X = 5 x0.577350269m

= 2.886751346m

Area R = L x B

=10m2

30m

Total Area Q +R

=2.88675136m +30m

= 32.58675135135m2Data given

Joint =470mm x 245mm x 150mm

Allowance = 20% for waste

Required : To calculate the quantity of blokes required for the wall

Solution

Area of 1 block with mortar

=947mm-1000) x (245mm -1000)

= 0.470m x0.245m

=0.1151m

Number of blocks required

= Total area.

Area of 1 block with mortar

=55m2

0.11515m

=0.,111515m

=478 blocks

Allowing 20%

BUILDING CALCULATION

QUESTION ONE (A)

(b) Solution

Area of vegetable to the stripped

Foundation spread = width if foundation -wall thickness

2

= 450m -150m

2

=150m -0.15m

Stripped length = length + 2(spread)

14m +2m (0.15m)

=14.3m

Stripped length =6+2 (spread)

=10m +2 (0.15m)

=10.3m

Area of vegetables to the stripped

=14.3m x 10.3m

147.29m2(ii) Solution

Death of excavation

= (1021.5 -1019 .5) +0.225

=2.0 +0.225

= 2.225m

Actual depth of excavation

=2.225 -0.15

= 2.075m

MG =2(L + B) 4-pertition

=2(14 +10) -4(0.15) + (10-0.3)

=48 -06 +9.7

= 57.1m

Volume of trench excavation

=GM x Actual foundation depth x foundation width

= 57.1m x 2.0 75m x 0.45m

= 53.32m

20%bulking

=120 x 53m2

100

=63.984

= 63.98m2(iii) Volume at foundation concrete

= MG X foundation width x foundation thickness

= 57 1m x 0.225m x 0.45m

=5.78m

3% residual waste = 103 x 5.78m

100

=5.95m

(c) Volume of site concrete

Length = 14-210.15 =13.7m

Width= 10-2(0.15) =9.7m

Depth = 0.1

Volume = 13.7m x 9.7m x 0.1m

= 13.289m3

Allow 4% residual and transit waste

104 x 13.289 100

= 13.8206m3

(V) Volume of hard-core filling

Length = 13.7m

Width = 9.7m

Depth = 0.3m

Volume = 13.7m x 9.7m x 0.3m

= 39.867m3

Allowing 20% compaction

120 x 39.867 100

= 47.8404m3

(IV) Number of sand Crete blocks

= 450mm x 225mm

= 0.45m x 0.225m

= 0.10125m2

Number of sand Crete blocks

= Area of wall Area of block

= 0.9m

0.10125m

= 8.888888889 or 9 blocks

Allowing 4% cutting waste

= 104 x 9

100

= 9.36 blocks

(VII) Volume of mortar required

= 0.9m x 14m x 10m

= 126m3

Allowing 3% residual waste

= 103 x 126m3 100

= 129.78m2QUESTION ONE (B)

(I) 7m of 225mm copper pipe at 400.00/m

= 225mm = 0.225m

= 0.225m x 400/m

7

= 12.86m2

(II) 5m of 15mm copper pipe at 300.00/m

= 15mm = 0.015m

= 0.015m x 300m

5

= 0.9m2

(III) 3 No. 22mm copper capillary

Fittings at 700.00each

=22mm = 0.022m

= 0.022m x 700

3

= 5.1333

(iv) No.15mm copper capillary fittings at 700.00each

= 15mm =0.015m

=0.0.15 x 700

4

=2.625

(iv) plumber + hrs. at 500.00 per hour

= 1 hour =500.00

=4 hour

4x 500

=2000.00

(vi) Plumbers assistant 4 hirs. At 300.00 per hour

=1 hour 300.00

= 4 hour

300+4 =1200.00

12% over head profit

= 112 x 1200

100

=1,344.00

QUESTION ONE (c)

(i) Data given

Length = 10m

Width = 6

Height= 4m

Opening =10%

Required To calculate the quantify of paint required in

(a) Litres(b) Gallons

Taking

1st cost = 1litre covers 5.5m

2nd coat =1litre covers 6.5m2

3rd cost =1litre cover 8.5m

1 Gallons =5 litres

Solution

Surface Area = 2(L +B) H

=2(10m + 6m) 4m

= 2(16m) 4m

=32m x 4m

= 128m

Area of ceiling

=L XB

=10m x 6m

60m

Area of wall

=10 x 128

100

=13m

Total area of wall

1238 -13

= 115m2

Total area if room to be painted

= Total area of wall + Area of ceiling= 115m +60m

=175m

1st coat

5.5m = 1 litre

175m = ?

=175 x 1 litre

5.5

=31.8181

=32 litres

2nd coat

6.5m = 1itre

=26.9231

= 27litres

3rd coat

8.5m2 = 1 litre

175m =?

175 x 1 litre

8

=20.5882

21.5882

Total quantity of paint

= 32 + 27+ 21

+ 80litres

1 gallon

5 litres = 1 gallon

80 litres =?

80 x 1 gallon

5

=16gallons

QUESTION (D)

Solution

(i) Area of total trapezium

=1/2 (a +b) x h

A = 85 + 70 =155m

H = 50 + 70 = 120m

= (105 +155) x 120

= 0.5 x 260 x 120

= 15, 600m

Area a = tir

4

= 3. 142 x 70

4

= 3848.95m

Area of plot = Area of trapezium area o quadrant

=15,600-3848 -95

= 11,751.05m2

Volume = Area x thickness

11, 751. 05 x 0.15

= 1,762, 6575m3

AC2=AB2+BC2

AC2=120+50

AC2=14,400+2500

AC=16,900

AC=130m

X=2

4

X=2 x70

4

=1/2 x 3.142 x70

4

=109.97m

(ii) Perimeter of the plot

=105+50+109.97+85+130

=479.97m

QUESTION ONE (E)

Solution

Span=2.5m

Density=2250kg/m3

Size of brick =215mm x 102.5mm x 65mm

Volume=?

Volume=0.215m x 0.1025m

=0.0220375m2

=2.5m x 0.0220375m2

=0.05509375m2Mass=volume x Density

=0.05509375m3 x 2250kg/m3

=123,9609375kg

QUESTION TWO (A)

Data given

Rise =215mm

Going =240mm

Total rise =3010mm

Required: To calculate

(a) Number of risers in the stair.

(ii) Going of the flight.

(iii) Pitch of the stair.

Solution

(b) Number of risers in the stair

=Total rise

Number of step

=3010mm=3.01m

215mm=0.215m

=14m

(ii) Going of the flight

2r+g=240mm

2(0.15)+g=0.24m

0.3+g=0.24m

g =0.24m-0.3

Going=0.3-0.24m

=0.06m

(iii) Pitch of the stair

=Total rise

Total going

=3.01

0.24

=12.54166667

=13

QUESTION TWO (B)

Data given

Length =6m

Width =5m

Height =4.5m

Required :To calculate the number of rolls of wall papers

Solution

Surface Area =2(L+B) x H

=2(6m+5m) x 4.5m

=2(11m) x 4.5m

=22m x 4.5m

=99m2Area of room =99m26m2 =1 roll of paper

Number of rolls =Area of room

1 roll of paper

=99m2 x 1 roll

6m2 =16.5 rolls

15% cutting waste

=115 x 16.5

100

=18.975 rolls

QUESTION TWO ( C)

Data given

Floor =150mm =0.15m

Thickness =40m x 20m

Ratio =1:2:4

Total ratio =7

Solution

Area of floor

=0.15m x 40m x 20m

=120m3

Cost of cement =1 x 120m2 7

=17.14285714m2 =5000.00/m3

17.14285714m2 =291.6666667

Cost of sand =2 x 120m3 7

=34,28571429m2 =6000.00/m2 34,28571429m2 =175.00

Cost of stone =4 x 120m2 7

=68,57142857m2 =10,000.00/m2 68,57142857m2 =145.8333333

Total cost of cement, sand and stone

=145,8333333+175+291,6666667

=612.5

2 masons =8 hours at 6000.00 per hour

=2 x 8hrs

=16hrs

= 16 x 6000

=96,000.000 per hr

12.5% overheads

=112.5 x 28,800

100

Cost of cement

= 1 x 120 m

7

=17.1428514m

= 5000.00/m2

17.14285714m

296666667

Cost of sand

=2x120m

7

=34,28571429m

= 6000.00/m3

34.28571429m

=175.00

Cost of stone

= 4 x 120m

7

= 68,5712857m2

=10,000.000/m

68,57142857m

145.83333333

Total cost of cement, sand and stone

= 145,83333333 + 175 + 291,6666667

= 612.5

2 masons = 8 hours at 6000,00 per hour

= 2 x 8 hrs

= 16 hrs

= 16 x 6000

= 96,000.000 per hr

8 laborers = 8 hours at 450.000 per hour

= 8x 8hour

= 64 hours

=64 x 450.00

=28,8000.00 per hour

12.5% overheads

=112.5 x 28,800

100

=32,400.00 per hour

Gang output = 3 cubic meters/hr.

=1.2m x 3.6m

= 4. 32m

QUESTION TWO (d)

VOLUME OF EXCAVATION

= 2(l +b) -4 x thickness of wall

= 2( 7.45 +3.35 ) -4 x 0.25

= 2(10.8) -4 x 0.25

= 21.6 -1m

=20.6m

Volume

Length =20.6m

Weight =0.25m

Height = 7m

=20.6m x0.25m x 7m

= 36. 05m

20% Bulking

=125 x 36.05m

100

=45,0625m

(iii) Volume of concrete used

Length = 20.6m

Weight = 0.1m

Height =0.25m

Volume

=20.6m x 0. 1m x 0.25

=0.515m

3% Residual waste

=103 x 0.515m3

100

=0.53045m3QUESTION THREE (A)

(a) Area A

=length x breath

= 10m x 8m

= 80m

Area B

= r2

2

=3.142 x25

2

=78.55

2

= 39.275m

Total Area of the wall

= Area A + Area B

= 80m2 +39.2 +5m

=119.275m

(ii) Quantity of blocks required

= Area of wall

Area of block

Where : Area of wall = 119.275m

`Area of block =?

Area of a block

=0.470mm x 245mm

=0.47m x 0.245m

= 0.1151m

Quantity of blocks required

= Area of wall

= 119.275m

Area of block

0.11515m

=1,035,82284 blocks

20%cutting waste

=120 x 1,035,82284

100

= 1,242,987408

= 1,243 blocks

QUESTION THREE (B)

Solution

Area = 100m

Size of tiles

= o.26 x 0.16m

0.0416m

End lap of tile 75mm

Side lap of tile =300m

Gauge

=length -end lap of tile

= 0.26 -0.075

= 0.185m

Net width

= width -side lap

= 0.16 0.03m

=0.13m

Area of tile

Gauge x Net width

= 0.185m x0.13m

= 0.02405m

Cost of tile

=800 x 0.02405

100

= 0.1924.00 Cost of tile nails per m2 = 200.00

= 200.00 x100m

M2

=20,000.00

Cost of batten per m =250.00

250.00 x 100m

m2

= 25,000.00

QUESTION THREE (C)

Solution

Area A is a trapezium

=1.2 (a +b) x h

=0.5 (3m +8m) x 8m

= 0.5 x 11m x 8m

= 44m2

Area B = Rectangle

= L x B

= 15m x 8m

= 120m

Area c = trapezium

= (a +b) x h

= 0.5 (3m + 8m) x 8m

= 0.5 x 11m x 8m

=44n2

Area D = Rectangle

= L x B

= 15m x 20m

= 300m

(b) Total area on plan = Area A+B + C + D

=44m + 120m + 44m + 300m

= 508m

(ii) Perimeter

Total perimeter

= 20 + 9.4 + 3 + 8 + 15 + 8 + 3 + 9 .4 + 20 + 15

= 110.8m

(iii) Volume of concrete

= Total area x thickness

= 508m x 0.15m

=76.2m

Question three (D)

Solution

Percentage of moisture content

= Wet weight - Dry weight x 100

Dry weight

= 39.75 - 35. 5 x 100

35.5

=4. 25 x 100

35.5

= 11. 97183099

= 12%

QUESTION THREE ( E)

Data given

Rise =215mm

Going = 240mm

Total rise = 3010

Required : To calculate

(i) Number of risers

(ii) Going of ht flight

(iii) Pitch of the stair.

Solution

Number of risers

= Total rise

Number of step

= 3.01

0.215

= 14m

(ii) Going of the flight

= 2r + g = 240mm

= 2(0.15) =g = 0.24m

= 0.3 +g = o.24m

G = 0.3 -024m

Going = 0.06

(iv) Pitch of the stair

= Total rise

Total going

=3.01

=-/24

= 12.54166667

=13

QUESTION FOUR ( A)

Data given

Length = 6.5m

Width = 4.5m

Height = 2.85m

1 roll of paper covers 6m

Required : To calculate the number of rolls of paper

Solution

Area = 1/2 (L +B ) x H

= (6.5m + 4.5m) x 2.85m

= 0.5 (11m) x 2.85m

=15.675m

Surface area =2(L + B) x H

= 2(6.5 +4.5) x 2.85

= 62. 7m

Area of room = 62 .7m

6m =1roll of paper

Number of roll of a paper

= Area of room

1 roll of paper

= 62.7m x 1 roll

6m

= 10.45 rolls

QUESTION FOUR (B)

Solution

Area of room

= L x B

=8. 5m x 10.5m

= 89. 25m

Area of skirling

0.012m x 0.075m

=0.0009m

Volume of timber

Area of room x Area of skirting

=89.25 x 0.00009m

=0.080325m

2.5% cutting waste

102.5 x 0.080325

100

= 0.082333125

QUESTION FOUR ( C)

Data given

Span = 2m

Mass = ?

Thickness = 215mm =0.215

Density = 2250kg/m

Required : To calculate the mass of brickwork

Solution

Mass = Volume x Density

Density = 2250kg/m

Volume = 2m x 0.215m

=0.43m

Mass =0.43m x 2250kg/m

=967.5kg/m

QUESTION FOUR

Solution

Cost of door =900,000.00

Labour cost at 0.5 hour =560.00

1 hour =?

0.5 x 560.00

100

=4.67

2.5 hour =850.00

= 2.5 x 850.00

60

35.42

Total cost

= 900,000 +67 +35.42

= 900,040.09

QUESTION FOUR (E)

Solution

Area = 470m

Area of roofing sheet

=2.4m x 1.2m

= 2.88m

Number of roofing sheet\

= Area

Area of roofing sheet

= 470m

2.88m

=163.194

= 163 roofing sheet

6% cutting waste

= 106 x 163

100

= 173.78

= 173 roofing sheets

(ii) Area = 4 70m

Area of laps

= 0.15m

Number of packets

= Area

Area of laps

= 470m

0.015m

=31.33

= 31packets

(ii) Cost of roofing sheet

=173 x 1,500,000.00

= 259,500,000.00

Cost of carpenter work

= 1 hour a day = 80,00.00

3 weeks = 21 days

= 1,680,000,00

Cost of Assistant work

=1 hour a d ay = 50,00.00

3 weeks = 21 days

= 21 x 50,000

= 1,050,000.00

QUESTION FIVE (A)

Slate size

o.45m 0.25m

Lap = 0.076m

Gauge x width Lap

=0.45m -00.76m

2

=0.374m

Area of slate

= Gauge + width

=o.37m +o.25m

=0. 624mm

Number of slate

= Area of Slate

= Area of Gauge

Area of slate

= 0.374m

0.624m

=0.5994m slate=

1m slate

(ii) Length of batten

=

Area of gauge

0.1m

==3.74m

Question five (B)

Length = 180m

Gradient = 1: 60

Fall =?

Slope Angle?

X =3

=0.0167

=tan -10.0167

=0.9568\

=1.0

=60 x=180

180

60

Fall = 3m

Mean slope

=Vertical interval

Horizontal equivalent

=3m/180 = 1/60= 1:60

=1.60 0.016666666

QUESTION FIVE (C )

Solution

Tile = 275mm

Gange = 150mm

Side lap = 25mm

=0.275 x 0. 275

=0.07562m

0.0.15x 0.15

=0.00225m

0.025 x 0.025

=0000.625

Total =000.10647653

=0000.1063 +76565m x 3000

2m

=1,595,214844

= 0.05 x 0.5

=000.25x 20

5m/2m

=2.5

= 2.5 x 1000.00

42

=5,952,38095

QUESTION

(i) sectional Area

=1/2 (a +b) x h

Where a = 1.0m

b = 1.5m

h =1.5m

= 05(1.0m +2.0m) x 1.5m

=0.5 (3.0m) x 1.5m

= 2.25m

Length of the slope side

C = a + b

C = 0.5 +1.5

C = 0.25 +2.25

C =2.5

C= 2.5

C= 1.58113883m

C 2 = a2 +62

C 2= 0.5 2+1.52

C2 = 0 .25+2.25

C2 =2.5

C = 1.58113883m2Total length of the slope

=1.58113883 + 1.58113883

=3.16227766m2(iii) Volume of trench excavation

= sectional area x length of trench

=2.25 x 40.50

= 91. 125m 220 % bulking

= 120 x 91.125 m2 100

= 109.35m3

(iv) Number of lorry loads with a capacity of 5m3.

=109.35m3

5m3

=21.87

=22 lorries

QUESTION FIVE (E)

Solution

=01m x 10,000

2m

=500

5% waste =105 x 500

100

525

=0.1 x 2,500

2

=125

5% waste =105 x 125

100

=131.25

=0.1 x 1700

2

=85

5% waste =105 x 85

100

=89.25

1 hour = 3600

0.5 hour

= 0.5 x 3600

60

= 30

15% profit and cost

= 15 x 30

100

= 34.5

QUESTION FIVE (F)

Data given

Opening = 950 mm x 2250mm

= 0.95 m x 2.25m

Lining = 150mm x 40mm

= 0.15m x 0.04m

Required: to calculate the volume of the timber.

Solution

Area of room

= LXB

= 0.95m x 2.25m

= 2.1375m2Timber size

150mm x 40mm

= 0.15m x 0.04m

Area of timber

= 0.15m x 0.0 4m

= 0.00 6m 2Volume of timber requited

= Area of door x Area of timber living

= 2.1375m2 x 0.006m2= 0.012 825m3

2.5% cutting waste

= 102.5 x 0. 1282

100

= 0.013145625625m2QUESTION SIX (A)

Data

With = 450mm = 0.45m

Depth = 1.000m

Thickness = 650mm x 225mm

= 0. 65m x 0.225m

Required: to calculate the volume of

(i) Trench excavations allow 15% bulking

(ii) Concrete allow 4.5% residual waste/

(iii) Gravel as backfilling allow 20% compaction

Solution

2L X 2 r

= 25 x with of foundation

= 25- x 0.45m

= 24- 0.225m

= 24.775m

= (2 x 110) + (2x3.142 x 24775)

= 375.69m

Depth of trench = 1.000m

Width of trench = 0.45m

Area = depth x width

= 1m x 0.45m

= 0.45m2(i) volume of trench excavation

= Area x mean girth

= 0.45m2 x 375.69m

= 169.06m3

15% bulking

= 115x 169. 06m2 100

= 194.419m3

(ii) concrete used

Depth = 650mm = 0.65m

Width = 450mm = 0.45m

Area = Depth x width

= 0. 65m x 0.45m

= 0.2925m2

Volume of concrete used

= 375.69 x 0. 2925m2

= 109.89325m3

4.5% residual waste

= 104.5 x 109.889325

100

= 114.8343446m3

(iii) Backing filing

= volume of excavation volume of block work + volume of concrete.

Back height = 1m 0.65m

= 0.35m

Thickness = 225mm = 0.225m

Area

= 0.225m x 0.35= 0.0 7875m2

Volume = Area x mean girth

= 0.0 7 875m2 x 375.69

= 29.586m3

Back filling

= 169.06 (109.889325 + 29.586) m3

= 29. 584675m3

20% compaction

= 120 x 29. 584675m3

100

= 35.50161m3

QUESTION SIX (B)

Solution

Numbers of ceiling joist

= 0.05 x 0.05

= 000.25m24.2x 3

= 12.6m2Number of ceiling joist

= Area of room

Area of ceiling joist

= 12.6m2000.252m

= 50.4 ceiling joist

5% cutting waste

= 105.x 50.4

100

= 52.92 ceiling joist

Joist

= 52. 92 x 8,000

4.2

= 100.800

Carpenter

= 1 hour = 10.000

= 3 weeks = 21 days

= 21 x 10,000

= 210, 000

Assistant

= 1 hour = 8,000

= 3 weeks = 21 days

= 21 x 8000

= 168, 000

QUESTION SIX (C)

SOH CAH TOA

Sin = 32= opposite

Hypo

= 0.57.x x

0.45

= 0.57 x 0 .45

= 0.2565m.

3m

8m

5m

8m

8m

15mm

3m

8m

8m

5m

20m

15m

AB = BC = AC

AB = 5 +8

AB = 25 +64

AB 89

AB = 9.4m

5

x

8

a

b

c

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