Lecture 1 Basic Concept

Lecture 1Basic Concept

By

Dr. Lubna Salih Al-shammari

Mustansiriyah UniversityFaculty of Engineering1-What is Engineering Surveying?

Is the art and science of

• Locating the best positions for the

construction of bridges, tunnels, roads and

other structures

• Producing up-to-date maps and plans

• Setting out a site, so that a structure

is built to scale and in the right place

• Monitoring the construction process

• Providing control points so that

future movement of structures,

such as dams and bridges can

be monitored

Mustansiriyah UniversityFaculty of Engineering2-Types of Surveys:

1- Geodetic Surveying: The type of surveyingthat takes into account the true shape of theearth. These surveys are of high precision andextend over large areas.

2- Plane Surveying: The type of surveying in which the mean surface of the earth is considered as a plane, or in which its spheroidal shape is neglected, with regard to horizontal distances and directions.

Mustansiriyah UniversityFaculty of Engineering3-Operations in Surveying:

Control Survey: Made to establish thehorizontal and vertical positions of arbitrarypoints.

Construction Survey: Made to lay out, locate and monitor public and private engineering works.

Mining Survey: Made to control, locate and map underground and surface works related to mining operations.

Topographic Survey: Made to gather data to produce a topographic map showing the configuration of the terrain and the location of natural and man-made objects.

Boundary Survey: Made to determine the length and direction of land lines and to establish the position of these lines on the ground.

Hydrographic Survey: The survey of bodies of water made for the purpose of navigation, water supply, or sub-aqueous construction.

Route Survey: Refers to those control, topographic, and construction surveys necessary for the location and construction of highways, railroads, canals, transmission lines, and pipelines.

Photogrammetric Survey: Made to utilize the principles of aerial photogrammetry, in which measurements made on photographs are used to determine the positions of photographed objects.

Mustansiriyah UniversityFaculty of Engineering4-Units of Measurement

➢ Linear measurement units

1. Metric units

Length The basic unit of length in the

International System is the meter (m).

1 m = 1000 millimetres (mm)

1 m = 100 centimetres (cm)

1 m = 10 decimetres (dcm)

1 m = 10−3 kilometres (km)

Areas in the metric system are given is

square meters while larger measurements

are given in hectares.

1 m2 = 106 mm2

104 m2 = 1 hectare (ha)

106 m2 = 1 square kilometre (km2)

Volumes in the metric system are given in

cubic meters.

2. English units

1ft=12inch

1yard=3ft

1mile=1760yards=5280ft

1ft2=144inch2

1yard2=9ft2

1inch=2.54cm

1ft=30.48cm=0.3048m

1m=3.2808ft

1km=0.62137mile

Mustansiriyah UniversityFaculty of Engineering4-Units of Measurement

➢ Angular units of measurement

1. Degrees

In geometry, any horizontal or vertical angle ismeasured in degrees. These angles may be given indecimal degrees or degrees, minutes, and seconds.

1 circle = 360° = 21,600´ = 1,296,000˝

1° = 60´ = 3600˝

1´ = 60˝

Examples:

Degrees-Minutes-Seconds Decimal Degrees1- ? 35.37˚2- 231̊2̍ 18ʺ ?3- ? H.W 42.885˚4- ? H.W 63.545˚5- 875̊8̍ 48ʺ ? H.W

Solution:

1- 35+̊(0.37 x̊ 60̍) = 35 2̊2̍ + (0.2 x 60ʺ) = 35 2̊2̍ 12 ʺ

2- 23° + (12´ ÷ 60) + (18˝ ÷ 3600) = 23.205°

2. Radian

Is another unit of measure for angles. By definition, a

full circle has 2 π radians or 360 degrees.

Since 1 circle = 360° = 2πrad.,

then 1 rad. = 360°/2π = 57.29578...°

Examples:

Find the value of the angle 12°15 ̍ 26ʺ in radian.

Solution:

Rad=𝜋

180(12 +

15

60+

26

3600)

2. Grad

A grad is defined as 1/400 of a circle. The grad is widely used in much of the world as part of the metric system, even though the radian is the primary unit.

Mustansiriyah UniversityFaculty of Engineering

5-Rounding NumbersRound the numbers below to two decimal places

38.157 38.1638.132 38.1338.125 38.1238.135 38.1438.5 38.50

6-Accuracy and Precision

7-Errors in measurement

(1) Natural errors

(2) Instrumental errors

(3) Personal errors

8-Classification of errors

1- Mistakes: are sometimes called gross errors,

but should not be classified as errors at all.

2- Systematic errors

3- Random errors


9-Map Scale

Map scale is the relationship between a unit of

length on a map and the corresponding length

on the ground. It's also an expression of how

much the area represented has been reduced

on the map.

𝑆𝑐𝑎𝑙𝑒 =𝑚𝑎𝑝 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑔𝑟𝑜𝑢𝑛𝑑 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

10-Types of Map Scales

• Verbal scale1 inch represents 16 miles

• Graphic scale, or bar scale

• Representative fraction, or ratio scale

Example

map with a scale of 1:50,000. The distance along a

property boundary as 1.7 cm. What is the length

in the real world?

Solution:

𝑆𝑐𝑎𝑙𝑒 =𝑚𝑎𝑝 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑔𝑟𝑜𝑢𝑛𝑑 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

1

50000=1.7𝑐𝑚

𝑋

𝑋 = 85000 𝑐𝑚 = 0.85 𝑘𝑚


1-Distance measurements

➢ Direct measurement by taping ➢ Electromagnetic measurement

by total station.

In engineering surveying, three types of distance are used: slope distance, horizontal distance and vertical distance (or height difference)

Linear Measurements

2-Types of tapes used in surveying

➢ Steel tape➢ Fibreglass tape

3-Measuring distance using tapes

➢ For short distances: measurements are carried out by unwinding and positioning the tape along the straight line between the points

➢ For long distances: When the distance to be measured exceeds the length of the tape, some form of alignment of the tape is necessary. This is known as ranging and is achieved using ranging poles or rods, survey or marking arrows.

Mustansiriyah UniversityFaculty of EngineeringLinear Measurements

4-Types of error in taping

1-Mistakes: The two most common sources of error in taping

are misreading the tape and incorrectly identifying where the

zero is.

2-Systematic errors: taping is subject to a series of systematic

errors that must be accounted for in order to improve the

precision of a measured distance. Also, it is important to know

the techniques involved and corrections that have to be applied

to remove each of these errors. This includes:

• Standardisation

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑠𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 =)𝐿(𝑙′ − 𝑙

𝑙

𝐿 = 𝑟𝑒𝑐𝑜𝑟𝑑𝑒𝑑 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑎 𝑙𝑖𝑛𝑒

𝑙 = 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑓𝑖𝑒𝑙𝑑 𝑡𝑎𝑝𝑒 (𝑠𝑎𝑦 30 𝑚)

𝑙′ = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑠𝑒𝑑 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑓𝑖𝑒𝑙𝑑 𝑡𝑎𝑝𝑒 (𝑠𝑎𝑦 30.011 𝑚)

• Tension

𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 =)𝐿(𝑇𝐹 − 𝑇𝑆

𝐴𝐸𝑇𝐹 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒 (𝑁)

𝑇𝑆 = 𝑐𝑎𝑙𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 (𝑁)

𝐴 = 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒 (𝑚𝑚2)

𝐸 = 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 (𝑁 𝑚𝑚−2)

(𝑓𝑜𝑟 𝑠𝑡𝑒𝑒𝑙 𝑡𝑎𝑝𝑒𝑠, 𝑡𝑦𝑝𝑖𝑐𝑎𝑙𝑙𝑦 200,000 𝑁 𝑚𝑚−2)

• Temperature variations

𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = 𝛼𝐿(𝑡𝑓 − 𝑡𝑠)

Where

𝛼 = 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙

(𝑓𝑜𝑟 𝑒𝑥𝑎𝑚𝑝𝑙𝑒 0.0000112 𝑝𝑒𝑟 °𝐶 𝑓𝑜𝑟 𝑠𝑡𝑒𝑒𝑙)

𝑡𝑓 = 𝑚𝑒𝑎𝑛 𝑓𝑖𝑒𝑙𝑑 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 (°𝐶)

𝑡𝑠 = 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑐𝑎𝑙𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛

The sign of this correction is given by the magnitudes of 𝑡𝑓 and 𝑡𝑠.


• Sag (catenary)

𝑆𝑎𝑔 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = −𝑤2𝐿3𝑐𝑜𝑠2𝜃

24𝑇𝐹2 = −

𝑊2𝐿 𝑐𝑜𝑠2𝜃

24𝑇𝐹2

Where

𝜃 = 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑠𝑙𝑜𝑝𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑎𝑝𝑒 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑠

𝑤 = 𝑡ℎ𝑒 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒 𝑝𝑒𝑟 𝑚𝑒𝑡𝑟𝑒 𝑙𝑒𝑛𝑔𝑡ℎ (𝑁 𝑚–1)

𝑊 = 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒 (𝑁)

𝑇𝐹 = 𝑡ℎ𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒 (𝑁)

Question

A steel tape of nominal length 30 m was used to

measure the distance between two points A and B on

a structure. The following measurements were

recorded with the tape suspended between A and B:

Line Length measured Slope angle Mean temperature Tension appliedAB 29.872 m 3°40' 5 °C 120 N

The standardised length of the tape against a reference tape is 30.014

m at 20 °C and 50 N tension. The tape weighs 0.17 N m–1 and has a

cross-sectional area of 2 mm2. Calculate the horizontal length of AB.

Solution

𝑆𝑙𝑜𝑝 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = −𝐿(1 − 𝑐𝑜𝑠 𝜃) =– 29.872 (1 – 𝑐𝑜𝑠 3°40′) = – 0.0611 𝑚

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑠𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 =𝐿(𝑙′−𝑙)

𝑙=

29.872 (30.014−30)

30= +0.0139

𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 =𝐿(𝑇𝐹−𝑇𝑆)

𝐴𝐸=

29 .872 (120−50)

2∗200000= +0.0052

𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = 𝛼𝐿 𝑡𝑓 − 𝑡𝑠 = 0. 0000112 ∗ 29. 872 (5 − 20 )

= –0.0050 𝑚

𝑆𝑎𝑔 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑜𝑛 = −𝑤2𝐿3𝑐𝑜𝑠2𝜃

24𝑇𝐹2 = −

0.172∗29.8723∗𝑐𝑜𝑠23°40′

24∗1202= −0.0022𝑚

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝐴𝐵

= 29.872 − 0.0611 + 0.0139 + 0.0052 − 0.0050 − 0.0022 = 29.8228

= 29.823𝑚(𝑡𝑜 𝑡ℎ𝑒 𝑛𝑒𝑎𝑟𝑠𝑡 𝑚𝑚)


5-Types of obstacles in direct measurement of a line

➢ Vision obstructed

𝑇ℎ𝑒𝑛 𝐴𝐵 = ൯(𝐴𝐵′ 2

+ 𝐴𝐵′ 2

➢ Direct measurement obstructed, vision free:

𝑇ℎ𝑒𝑛 𝐴𝐵 = 𝐵𝐶 2 − 𝐴𝐶 2

1. When the obstacle can be measured around, e.g. a pond.


➢ Direct measurement obstructed, vision free:

2. Any one of the following methods may be

employed to find the width of the river along the

direction of the line

ቇ∴𝐴𝐵

𝐴𝐷=𝐷𝐹

𝐹𝐸(𝑏𝑢𝑡 𝐷𝐹 = 𝐴𝐶, 𝑎𝑛𝑑 𝐹𝐸 = 𝐶𝐸 − 𝐶𝐹 = 𝐶𝐸 − 𝐴𝐷

3. Direct measurement and vision both obstructed

BG = DE

Good luck

Lecture 1 Basic Concept

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