COVENANT UNIVERSITY

ALPHA SEMESTER TUTORIAL KIT (VOL. 2)

P R O G R A M M E : B U I L D I N G T EC H

200 LEVEL

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DISCLAIMER

The contents of this document are intended for practice and learning purposes at the undergraduate

level. The materials are from different sources including the internet and the contributors do not

in any way claim authorship or ownership of them. The materials are also not to be used for any

commercial purpose.

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LIST OF COURSES

*BLD211: Concrete & Workshop Practice Technology I

BLD212: Building Construction Practice I

BLD213: Building Science I

BLD214: Applied Mechanics for Builders

*BLD216: Statistics for Builder

BLD218: Building Drawing I

*SES211: Land Surveying I

*Not included

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COVENANT UNIVERSITY CANAANLAND, KM 10, IDIROKO ROAD

P.M.M 1023, OTA, OGUN STATE, NIGERIA TITLE OF EXAMINATION: B.Sc. EXAMINATION

COLLEGE: SCIENCE AND TECHNOLOGY

SCHOOL: ENVIRONMENTAL SCIENCES

DEPARTMENT: BUILDING TECHNOLOGY

SESSION: 2015/2016 SEMESTER: ALHPA

COURSE CODE: BLD 212 CREDIT UNIT: 3.0

COURSE TITLE: BUILDING CONSTRUCTION PRACTICE I

COURSE LECTURERS: DR. L. M. AMUSAN, MR. A. O. AFOLABI & MR. AJAO

MARKING GUIDE

Question 1

(a) discuss the types of excavation to be carried out?

Answer:

Types of Excavation

Over site Excavation: the removal of top soil, depth varies from site to site (usually in a 150 to 300

mm range), required since top soil often contains plant life, animal life and decaying matters making

soil compressible and thus unstable for supporting building.

Reduce Level ( R. L. )Excavation: required in irregular sites to form a level surface consists of both

cutting and filling operations, the level to which the ground is reduced is called the formation level.

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Trench Excavation: excavation of trench of required depth and width before laying out foundation

done both manually (with use of spade, pick axe, rammer, etc. and for small buildings) and

mechanically (with use of bulldozers, trench diggers, etc. and for large buildings)

(3 marks).

(b) problems to be encountered while excavating and the precautions to be put in place on the

construction site

Answer:

Collapse of Excavated walls – Timbering

Upsurge of water from water table into foundation – Dewatering using pumps (2 marks).

(c) discuss the functions of five (5) mechanical excavating plants that can be used on the

construction site and sketch two (2) out of the equipment discussed

Answer:

Bulldozer: This machine consists of a track or wheel mounted power unit with a mould blade infront

usually controlled by a hydraulic ram. Some bulldozer can tilt their mould blade to form an angle dozer.

The main function of the bulldozer are: for shallow excavations up to 300mm deep, clearance of shrubs

and trees, also as a towing tractor.

Dragline: This machine has a crane rigging which is attached to a drag bucket. It is designed for bulk

excavation in loose soil up to 3m below its own level. It can work in construction of swimming pools.

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Backacter: They are machines suitable for trench, foundation and basement excavation.

When used in trench operation, one can achieve a very accurate result.

Face Shovel: It is an excavating machine that excavates higher than its own level.

Skimmers: It is also an excavating machine that is rigged using a universal power unit for

surface stripping and shallow excavation wall up to 300mm deep. It is used where a higher

degree of accuracy is required.

Tractor Shovel (Loader): They are usually called payloader and their primary function is to

scoop up loose soil or materials into a bucket, elevate it and maneuver into a position to

deposit the loose material.

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(6½ marks).

(d) discuss the types of timbering that would be used for the soil condition

Answer: Open Timbering: This is also referred to as Stay Bracing. It is used for firm / stiff / rocky hard soil and for

excavation not exceeding about 2m in depth, it consists of placing vertical sheets called poling boards,

opposite to each other against the walls and holding them in position by one or two rows of struts, poling

boards are placed at an interval of 2 – 4m and extend to full height of trench poling board : 200 x 40 – 50

mm., struts : 100 x 100 mm for up to 2m wide excavation and 200 x 200 mm for up to 4m wide excavation.

Closed Timbering: When the sides of the excavation is completely covered with timbering it is known

as close timbering unlike the partial covering in open timbering. It is a box like structure. It is

essentially good for loose / loamy / dry sandy soil. It consists of Poling boards placed very near to each

other / touching each other and keeping them in position by longitudinal rows (usually two) of

walings, struts are then provided across the walings.

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(4 marks).

(e) choice of choosing mechanical plants above manual excavation

Answer:

Mechanical Excavation: This type of excavation is used for big quantity and size of project, when

all soil is to be carried away from site and when time plays an important role to complete the job.

Accuracy is also paramount and the depth of excavation to be carried put must be considered. (2

marks)

Question 2

In preparing for the construction of the superstructure in Question 1;

(a). make a classification diagram of using an option of concrete floors in your superstructure

Answer:

(4 marks)

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(b). with neat sketches, distinguish between suspended timber ground floor and a hollow pot

upper floor

Answer:

Suspended timber ground floor

Hollow pot upper floor

(5 marks).

(c). draw a section of a typical ground floor and discuss three (3) components in a typical ground

floor in a building

Answer:

9

(4 marks).

(d). state four (4) technical requirements of an upper floor construction

Answer:

Technical requirements for an Upper floor

• The floor must be structurally stable to prevent deflection when a load is imposed on it.

• The floor should provide restraint for the external walls

• Provide suitable fire protection to delay fire spread.

• Provide good sound insulation (2 marks).

(e). discuss five (5) attributes of using concrete over steel sections

Answer:

10

. (2½ marks)

3. (a). Differentiate between a column and a beam

Answer:

Beam:

Column:

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(5 marks).

(b). State the steps involved in erecting formwork and list three (3) materials that can be used

Answer:

Erecting formwork

1. Props positioned and leveled thoroughly.

2. Soffit placed, leveled and position checked.

3. Side forms placed, their position checked before being fixed.

4. Final check before casting.

Suitable Formwork materials includes: - Timber, Steel and special plastics.

(3½ marks).

(c). With neat sketches, explain the term reinforced concrete upper floor and state the

construction sequence

Answer:

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(6 marks).

(d). State three (3) advantages of using precast concrete elements

(1½ marks).

(e). why should water in foundation be removed before constructing foundation

Answer:

Water in excavation should be removed since it can:

Undermine sides of excavation.

Make it impossible to adequately compact the bottom of excavation to receive foundation.

Bearing capacity of the soil is reduced with water stored in voids of the bottom of excavation.

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(1½ marks).

Question 4

(a). Explain the term “Finishes”

Answer:

Finishes are materials and item used to improve the service and decorative qualities that provides comfort and aesthetic satisfaction of buildings and structures, as well as to protect structural members from atmospheric weather conditions and other effects, this produces a final product fit for use and habitation.

(3 marks).

(b). State three (3) advantages of dry finishes over wet finishes

Answer:

The greatest advantage of dry finishes when compared to wet finishes is that no shrinkage will occur during the dry out process. A dry finish should be uniform in colour, size and be true to shape. The application of dry finishes does not prevent occupancy of the building (1½ marks).

(c). Give five (5) examples each of floor and wall finishes

Answer:

Wall finishes: Plaster, Paint, Alucobond, Wallpaper, Tiles, facing bricks

Floor finishes: Terrazo, Concrete Screed, Tiles, Interlocking blocks, timber boards

(5 marks).

(d). State five (5) factors affecting the choice of floor finish

Answer:

Type of Base

Room usage

Degree of comfort required (Sound control, freedom from slippery, warmth)

Maintenance(Cleaning, repairing)

Cost

Appearance

Safety

(2½ marks).

(e). Differentiate between Self finish and Applied finish and give two examples each

Answer:

Self finish - is a finish which is inherent in the material and does not have to be specially applied on site. E.g. Facing brick, Natural stone, Natural slate, Natural wood

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Applied finish - is a finish which is actually applied on site. E.g. Paint, Wallpaper, PVC tiles

(3 marks).

(f). Identify five (5) construction materials that can pollute the environment.

Answer:

Cement, grout, oil including fuel, asbestos, un-used plastics, metals, toxic paints, silt, dust,

chemicals, sewage and waste materials (2½ mark).

5. (a). Differentiate between Sustainable construction and Green construction.

Answer:

Sustainable Construction:

This explains the process of construction that involves meeting the needs of present generations

without compromising the ability of future generations to meet their needs.

Green Construction:

This refers to a structure and using a process that is environmentally responsible and resource

efficient throughout a building’s life cycle: from siting to design, construction, operation,

maintenance, renovation and demolition.

(3 marks).

(b). State six (6) benefits of ICT to the construction industry

Answer:

Improves quality of work

Makes complex tasks easier to perform

Saves time

Improves productivity

Enhances public image

Saves cost

Facilitates decision making

(3 marks).

(c). State eight (8) constraints to the use of ICT in the construction industry

Answer:

Inadequate/ erratic power supply

High cost of hardware and software

Lack of sufficient jobs

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Fear of virus attack

High rate of obsolescence of hardware/software

Inadequate ICT content of construction education

High cost of employing computer professionals

Security fears

(4 marks).

(d). State three (3) softwares each under the following categories:

Answer:

(i) Word Processing and accounting softwares – MS Word, Word Perfect, MS Excel (1½ mark).

(ii) Presentation softwares – MS Powerpoint, Adobe Pagemaker, MS Outlook (1½ mark).

(iii) Architectural/Engineering Design and Drawing softwares – Corel Draw, Auto cad, Archicad, Autocad, StaadPro (1½ mark).

(iv) Q.S Measurement and Estimating softwares – WinQs, CatoPro, Master Bill, QS Elite (1½ mark).

(v) Project Planning softwares - MS Project, Primevera, SAP (1½ mark).

6. (a). Describe three (3) construction tools each used by five (5) specific specialty Trades in Building

Construction. Support your answers with neat diagrams. (5 marks)

(b). Explain in details the practical procedures involved in suspended concrete upper floor

construction and ensure you reach logical conclusion. (7½ marks)

(c). With annotated diagrams, clearly describe ten (10) symbols representing basic building

materials used in Construction works. (5 marks)

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COVENANT UNIVERSITY, OTA

CANAANLAND, KM 10, IDIROKO ROAD

P.M.M 1023, OTA, OGUN STATE, NIGERIA

TITLE OF EXAMINATION: B.Sc EXAMINATION

COLLEGE: SCIENCE AND TECHNOLOGY

SCHOOL: ENVIRONMENTAL SCIENCES

DEPARTMENT: BUILDING TECHNOLOGY

SEMESTER: ALHPA

COURSE CODE: BLD 214 CREDIT UNIT: 2.0

COURSE TITLE: APLIED MECHANICS FOR BUILDERS

INSTRUCTION: ANSWER ALL QUESTIONS IN SECTION A AND ANY 3

IN SECTION B TIME: 2 HOURS

SECTION A

1. …………………………… Is the single force which could replace for the given system without

altering the net effect? (a) Resultant of a single force system (b) Resultant of a force system

(c) concurrent system (d) force of a resultant single system

2. ……………… and ………………. are the quantities involved in expressing the value of a

moment of a force about a given point or fulcrum. (a) Moment and applied force (b)

Applied force and perpendicular distance (c) Resultant moment and applied force (d)

applied force and couples

3. A knife edge support has ……………. reactions component (a) 4 (b) 3 (c) 2 (d) 5

4. A structure is statically indeterminate if ………………………… (a) it can be analyzed by

equation of equilibrium (b) it cannot be analyzed by equation of equilibrium (c) All

structural members lies in one direction (d) structural members are well arranged.

5. Three (3) reactions component are sufficient for equilibrium in structures but the

arrangement of the reactions must not be …………… and ……….. (a) parallel to each other

and intersect at one point (b) aligned with each other and intersect at a point (c) intercept at

a point and parallel to each other (d) parallel to each other and horizontal at a point

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6. The formula for determining the determinacy of a structure is ………………. (a) D = M+R -2J

(b) D = M –R + 2J (c) M + 2R –J (d) D = 2R + M - J

7. A structure is said to be statically indeterminate externally when the reactions component is

more than (a) 3 (b) 4 (c) 5 (d) 6

8. A system of force may be classified into (a) concurrent and non- concurrent force (b)

vertical force and horizontal force (c) like force and unlike force (d) concurrent force and

horizontal force.

9. Linear relations between the component of stress and strain are generally known as

………………. (a) Hooke’s law (b) Euler load (c) modulus (d) homogenous

10. Isotropic material properties are …….. (a) different in all direction (b) same in all direction

(c) parallel in all direction (d) are in the same angle

11. ……………….. and …………………… are examples of anisotropic material (a) wood and fabric

(b) fabric and steel (c) wood and iron (d) iron and steel

12. …………….. and ………………… are examples of isotropic material (a) copper and tin (b) steel

and aluminum (c) gold and diamond (d) wood and iron

13. The bending of a beam depend upon 4 parameters, they are …………(a) type of load, length

of beam, elasticity of the beam and type of beam (b) dimension of the beam, type of load,

load position and type of beam (c) load position, type of load, length of the beam and type

of beam (d) load orientation, beam dimension, type of load and type of beam

14. Unstable framework is also called a ……………… (a) mechanism (b) determinate (c)

redundant (d) indeterminate

(17 ½ marks)

SECTION B

1(i) Calculate the moment of inertial of a hollow circular section of external and internal diameters

100mm and 80mm respectively about an axis passing through its centroid.

(ii) A hollow triangular section show below is symmetrical about its vertical axis, find the moment

of inertial of the section about BC.

(iii) Find the moment of inertial of the T-section below about X-X and Y-Y axes through the center

of gravity of the section.

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(17 ½ marks)

2(i) Define shear force and bending moment

(ii) For the beams shown below, calculate the reactions and sketch the shear force and bending

moment diagram.

Fig 1

Fig2 (17 ½ marks)

3(i) Distinguish between statically determinate and statically indeterminate structure

(ii) Highlights the methods of constructing stable framework

(iii) Determine the determinacy of the frames shown below and comment on them

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(iv) Calculate the reactions at A and B in the diagram below.

(17 ½ marks)

4. (i) Identify three methods of plane truss analysis

(ii) Determine the force in each member of the truss and state if the members are in tension or

compression.

(17 ½ marks)

5 (i) Define kinematics and list motions of rigid bodies

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(ii) A body rotates with θ= 5t3

+ 15t2

+20, where θ is in radians and t is in seconds

(a) Find angular velocity and angular acceleration when t = 0 and t = 2.5s

(b) Find the angular distance and acceleration when the angular velocity is equal to zero

(iii) The geometry of the particle motions are given by the equation XA = 5t3

-2t2

+ 5m and VB = 7t

– 8t2

+ 9t3

m, when t = 0, XB was 20m. Find

(a) The distance at B relative to A

(b) The speed of A relative to B

(c) The velocity of B relative to A

(d) The magnitude of the acceleration of B relative to A when the time lapsed is 10s (17 ½ marks)

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COVENANT UNIVERSITY CANAANLAND, KM 10, IDIROKO ROAD

P.M.B 1023, OTA, OGUN STATE

TITLE OF EXAMINATION: B.Sc. DEGREE EXAMINATION

COLLEGE: SCIENCE AND TECHNOLOGY

DEPARTMENT: BUILDING TECHNOLOGY

SESSION: 2014/2015 SEMESTER: ALPHA

COURSE CODE: BLD 214 CREDIT UNIT: 2.0

COURSE TITLE: APPLIED MECHANICS FOR BUILDERS

COURSE COORDINATOR: MR. O. JOSHUA

COURSE LECTURERS: MR. O. JOSHUA and MR. R. A. OJELABI

MARKING GUIDES

1. (a) External Diameter = 80mm, Internal Diameter = 60mm

Applying the formula

Ixx = ∏/64(D4-d4) when ∏= 3.142

Substituting the known parameters into the equation

Ixx = 3.142/64(1004-804)

Ixx = 2.898 x 106mm4

(b) Main triangle, B= 180mm, H=100mm

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Cut out triangle, b=120mm, h=60mm

Moment of inertial of a triangular section about BC is given by the formula below

IBC= BH3/12 – bh3/12

Substituting the known parameters into the equation

IBC= 180 x (100)3/12 – 120 x (60)3

IBC = (15 x 106) – (2.16 x 106)

IBC = (15-2.16) x 106

IBC= 12.84 x 106mm4

(c)

Centroid of the given diagram =

Ў = (A1Y1 + A2Y2)/ A1+ A2

Rectangle 1

A1 = 150mm x 50mm = 7500mm2

Y1 = 150mm + 50mm/2 = 175mm

Rectangle 2

A2 = 150mm x 50mm = 7500mm2

Y2 = 150mm/2 = 75mm

Therefore;

Ў = {(7500mm2

x 175mm) + (7500mm2

+ 75mm)}/ 7500mm2

+ 7500mm2

Ў = 125m

Moment of inertial about x-x axis

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M.I of rectangle 1 about x-x axis through its center of gravity and parallel to x-x axis

IG1 = bd3

/12 b = 150mm, d = 50mm

IG1 = 150 X 503

/12 = 1562500mm4

= 1.5625 x 106

Distance between center of gravity of rectangle (1) and x-x axis

h1 = 175mm – 125mm = 50mm

Therefore moment of inertial of rectangle 1 about x-x axis is given by the equation below

IG1+ a1h1

2

= (1.5625 x 106

) + (7500 x 502

)

= 20.3125 x 106

mm4

Similarly, moment of inertial of rectangle 2 about an axis through its center of gravity parallel to x-x

axis

IG2 = bd3

/12 = 50 x (150)3

/12 = 14.0625 x 106

mm4

And distance between the center of gravity of rectangle 2 and x-x axis

h2 = 125mm – 75mm = 50mm

Therefore moment of inertial of rectangle 2 about x-x axis is given by the equation below

IG2 + a2h2

2

= (14.0625 x 106

mm4

) + (7500 x (50)3

= 32.8125 x 106

Moment of the whole section about x-x axis is given by

Ixx = (IG1 + A1h1

2

) + (IG2 + A2h2

2

)

Substitute the known parameters into the equation to solve for second moment area of inertial (I)

I = (1562500mm4

+ 7500mm2

x502

mm2

) + (14062500mm4 + 7500mm2

x502

mm2

)

Ixx = 53.125 X 106

mm4

Moment of inertial about y-y axis

M.I of rectangle 1 about y-y axis is given by the formula

Iyy = db3

/12, where d=150mm and b=50mm

Substitute the known parameters unto the equation

Iyy1= 50 x 1503

/12 = 14.0625 x 106

mm4

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And moment of inertial of rectangle 2 about y-y axis

Iyy2= db3/12, where d = 150mm and b =50mm

Substitute the known parameters into the equation

Iyy2 = 150 x 503

/12 = 1.5625 x 106

mm4

Now moment of inertial of the whole section about y-y axis

Iyy = Iyy1 + Iyy2

= (14.0625 x 106

+ 1.5625 x 106

)

= 15.625 x 106

mm4

2. (a) Shear force at the section of a beam may be defined as the algebraic sum of the loads to the

left or to the right of a point

Bending moment at the section of a beam may be defined as the algebraic sum of the

moments of the forces to the left or to the right of the section.

b. (i)

∑MD = 0

Ra x 4 – 3x3 – 2x1 = 0

4Ra – 9 – 2 = 0

4Ra = 9 +2

4Ra = 11

Ra = 11/2 = 2.75KN

∑MA= 0

-Rd x 4 + 2x3 + 3x1 = 0

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-4Rd +6 + 3 = 0

4Rd = 6+3

4Rd = 9

Rd = 9/4

Rd = 2.25KN

SHEAR FORCE CACULATION

S.F @ A = 2.75KN

S.F @ B = 2.75KN – 3KN = -0.25KN

S.F @ C = 2.75KN – 3KN – 2KN = -2.25KN

S.F @ D = 2.75KN – 3KN – 2KN + 2.25KN = 0

BENDING MOMENT CACULATION

B.M @ A = 0

B.M @ B = 2.75KN X 1M = 2.75KNM

B.M @ C = 2.75 X 3 – 3 X 2

= 8.25KNM – 6KNM =2.25KNM

B.M @D = 2.75 X4 – 3X3 -2X1 = 0

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(ii).

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∑MD = 0

Ra x 8 – 60 x 6 – 50 x 2 = 0

8Ra -360 -100 = 0

8Ra = 460

Ra = 460/8

Ra = 57.5KN

∑MA = 0

-Rd x 8 + 50 x 6 + 60 x 2 = 0

-8Rd + 300 + 120 = 0

420 = 8Rd

Rd = 420/8

Rd = 52.5KN

SHEAR FORCE CACULATION

S.F @ A = 0

S.F @ B = 57.5KN – 60 = -2.5KN

S.F @ C = 57.5KN – 60KN – 50KN = -52.5KN

S.F @ D = 57.5KN – 60KN – 50KN – 52.5KN = 0

BENDING MOMENT CACULATION

B.M @ A = 0

B.M @ B = 57.5 X 6 – 60 X 2

230KNM – 120KNM = 110KNM

B.M @ C = 57.5 X 6 – 60 X 4

345KNM – 240KNM = 105KNM

B.M @ D = 57.5 X 8 – 60 X 6 – 50 X 2 = 0

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3. (a) A statically determinate structure is such that after application of equation of determinacy D

= M+R-2J give D = 0

While statically indeterminate structures are such that after analysis with the equation of

determinacy D = M+R – 2J gives a value of D greater than 0.

(b) Methods of constructing stable structure

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(i). By starting with a triangular configuration and successive addition of two new members and a

joint

(ii). By connecting together by means of 3 pins or 3 hinges all of which do not lie in the same

straight line.

(iii). By connecting together by means of 3 bars all of which not parallel to each other nor

intersect at one point

(iv). By connecting together by means of a pin or a hinge and a bar, the axis of which does not

pass through the pin or hinge.

b. (i)

To solve for the determinacy D

D = M + R – 2J

M = 11

J = 6

R = 2

Substitute the known parameters into the equation

D = 11 + 2 – 2(6)

13 – 12 = 1

Comment: The structure is statically indeterminate externally and internally

(ii)

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D = M + R – 2J

M = 12

R = 3

J = 7

Substitute the known parameters

D = 12 + 3 – 2(7)

D = 15 – 14

D = 1

Comment: The structure is statically determinate externally, but statically indeterminate internally

(c) To solve for the reactions of the diagram below

Resolving the rectangular uniform load distribution first

To solve for point load

L = 4m

B = 3KN/M

Point load = 4M x 3KN/M

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= 12KN

It will be acting at the midpoint of the rectangular section

Resolving the triangular section with distribution load into a point load

B = 3M

H = 3KN/M

Point load = ½ b x h

Substitute the known parameters into the equation

Point load = ½ x 3m x 3KN/m

= 4.5KN

It will be acting at L/3 to the right of the triangular section and 2L/3 to the left hand of the

triangular section

L/3 of 3m = 3/3 = 1m to the right

2L/3 = of 3m = 2 x 3 /3 = 2m to the left

To solve for the reactions

∑MB = 0

Ra x 7 – 4.5 x 5 – 12 x 2 =0

7Ra – 22.5 – 24 = 0

7Ra = 46.5KN

Ra = 46.5/ 7

32

Ra = 6.643KN

∑MA = 0

-7Rb x 12 x 5 + 4.5 x 2 = 0

-7Rb + 60 + 9 = 0

7Rb = 69

Rb = 69/7

Rb = 9.86KN.

4. (a)(i). By joint resolution method

(ii).By sectioning method

(iii).By combination of both methods

(b)

Solving for reactions

∑MD = 0

Ra x 9M – 10KN x 6M – 10KN x 3M = 0

9Ra – 60KNM – 30KNM = 0

9Ra = 90KNM

Ra = 90KNM/9

Ra = 10KN

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∑MA = 0

-Rd X 9M + 10KN X 6M + 10KN X 3M = 0

-9Rd + 60KNM + 30KNM = 0

9MRd = 90KNM

Rd = 90KNM/9M

Rd = 10KN

∑MD = 0

By Pythagoras theorem to solve for X

X2

= 42

+ 32

= 16 + 9

X = √25

X = 5M

Resolving using joint resolution method

Joint A Sin θ = 4/5

Cos θ = 3/5

∑fy = 0

10KN + SAB Sin θ = 0

10KN + SAB x 4/5 = 0

10KN = - SAB x 4/5

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10KN X 5 = -4SAB

SAB = -50KN/4

SAB = - 12.5 KN (compression)

∑fx = 0

SAF + SAB Cos θ = 0

SAF + (-12.5KN) X 3/5 = 0

SAF = 12.5 X 3/5

SAF = 7.5KN (Tension)

JOINT F

∑fy = 0

SFB – 10KN = 0

SFB = 10KN (Tension)

∑fx = 0

SFE – SFA = 0

SFE – 7.5KN = 0

SFE = 7.5KN

JOINT B

Sin θ = 3/5

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Cos θ = 4/5

∑fy = 0

-SBF – SBA .Cos θ – SBE. Cos θ = 0

- 10KN – (-12.5KN) X 4/5 – SBE X 4/5 = 0

- 10KN + 12.5 X 4/5 – SBE X 4/5 = 0

Multiply through by 5

-10KN X 5 + 12.5 X 4 – 4SBE = 0

-50 KN + 50KN = 4SBE

0 = 4SBE

SBE = 0/ 4

SBE = 0

∑fx = 0

SBC + SBE x Sin θ – SBA x Sin θ = 0

SBC + 0 x 3/5 – (-12.5) x 3/5 =0

SBC + 0 +12.5 x 3/5 = 0

SBC + 12.5KN x 3/5 = 0

SBC = -7.5KN

JOINT E

Sin θ = 4/5

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Cos θ = 3/5

∑fy = 0

SEB x Sin θ + SEC – 10KN = 0

0 x Sin θ + SEC – 10KN = 0

SEC = 10KN (Tension)

∑fx = 0

SED - SEF – SEB Cos θ = 0

SED – 7.5KN – 0 Cos θ = 0

SED = 7.5KN (Tension)

JOINT DCos θ = 3/5

Sin θ = 4/5,

∑fy = 0

10KN + SDC x Sin θ = 0

10KN + SDC x 4/5 = 0

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Multiply through by 5

50KN + 4SDC = 0

4SDC = -50KN

SDC= - 50KN/4

SDC = -12.5KN (Compression)

∑fy = 0

-SDE – SDC Cos θ = 0

-SDE – SDC x 3/5 = 0

-7.5KN – (-12.5KN) x 3/5 = 0

7.5KN + 12.5KN x 3/5 = 0

-7.5KN + 7.5KN = 0

5. (a (i) Kinematics is the study of a body in motion without reference to the force producing it. it

can also be defined as the study of the geometry of motion.

Types of rigid bodies

Translational

Rotation about a fixed axis

General plane motion

Motion about a fixed point

(b (i) Θ = 5t3 + 15t2 + 20

Angular velocity is given as

W = dθ/dt = 5t2 + 30t

When t = 0; W = 15(0)2 + 30(0) = 0

When t = 2.5s; W = 15(2.5)2 + 30(2.5) =168.7rad/s

Angular acceleration is given as

α = dw/dt = 30t + 30

38

When t = 0; α = 30(0) + 30 = 30rad/s2

When t = 2.5s; α = 30(2.5) + 30 = 105rad/s2

(ii). When W = 0

W = dθ/dt = 15t2 + 30t = 0

15t (t+2) = 0

Thus t = 0 or t = -2s

Therefore, angular distance is given as

Θ = 5t3 + 15t3 + 20

When t = o; θ = 5(0)3 + 15(0)2 +20 = 20rad

When t = -2s; θ = 5(-2)3 + 15(-2)2 + 20

-40 + 60 + 20 = 40rad

Angular acceleration is given as

α = 30t + 20

When t = 0; α = 30(0) + 30 = 30rad/s2

When t = -2s; α = 30(-2) + 30 = -30rad/s2

(c) XA = 5t3 – 2t2 +5m, VB = 7t – 8t2 +9t3, XB = 20m

(i) Distance at B relative to A XB/A

XB/A = XB – XA = 20 -5t3 – 2t2 + 5m

When t = 0

XB/A = 20 – 5(0)3 – 2(0)2 + 5

XB/A = 20 – 5

XB/A = 15m

(ii) Speed at A relative to B (VA/B)

VA/B = VA – VB

VB = 7t – 8t2 + 9t3

39

DXA/dt = VA = 15t2 – 4t

Therefore VA/B = 15t2 – 4t – 7t + 8t2 – 9t3

VA/B = 23t2

– 11t – 9t3

When t = 0; VA/B = 23(0)2

– 11(0) – 9(0)3

= 0

(iii) Velocity of B relative to A (VB/A)

VB/A = VB – VA

dXA/dt = VA = 15t3

– 4t, VB = 7t – 8t2

+ 9t3

Therefore, VB/A = 7t – 8t2 + 9t3 – (15t2

– 4t)

VB/A = 11t – 23t2

+ 9t3

When t = 0; VB/A = 11(0) – 23(0)2

+ 9(0)3

= 0