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Engineering Drawing and Engineering Graphics including engineering curves, projection of lines,

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    Engineering Drawing and Engineering Graphics including engineering curves, projection of lines,
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    • 1. Textbook of Engineering Drawing Second Edition K. Venkata Reddy Prof. & HOD of Mechanical Engineering Dept. C.R. Engineering College, Tirupati - 517 506 SSP BS Publications ;;;::::;;;;; 4-4-309, Giriraj Lane, Sultan Bazar, Hyderabad - 500 095 - A. P. Phone: 040-23445688
    • 2. Copyright © 2008, by Publisher AIl rights reserved iL No part of this book or parts thereof may be reproduced, stored in a retrieval system or I transmitted in any language or by any means, electronic, mechanical, photocopying, I recording or otherwise without the prior written permission of the publishers. ~ Published by : BSP BS Publications === 4-4-309, Giriraj Lane, -Sultan Bazar, Hyderabad - 500 095 A.P. Phone: 040-23445688 e-mail: contactus@bspublications.net www.bspublications.net Printed at Adithya Art Printers Hyderabad ISBN: 81-7800-149 '" !!!!:U BLOCK Fig. 2.4(b) Folding of drawing sheet for storing in filing cabinet 2.2.4 Lines (IS 10714 (part 20): 2001 and SP 46: 2003) Just as in English textbook the correct words are used for making correct sentences; in Engineering Graphics, the details of various objects are drawn by different types of lines. Each line has a defmite meaning and sense toconvey. IS 10714 (Pint 20): 2001 (General principles of presentation on technical drawings) and SP 46:2003 specify the following types oflines and their applications: • Visible Outlines, Visible .Edges : Type 01.2 (Continuous wide lines) The lines drawn to represent the visible outlines/ visible edges / surface boundary lines of objects should be outstanding in appearance. • Dimension Lines: Type 01.1 (Continuous narrow Lines) Dimension Lines are drawn to mark dimension. • Extension Lines: Type 01.1 (Continuous narrow Lines) • There are extended slightly beyond the respective dimension lines.
    • 20. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Lettering and Dimensioning Practices 2.5 • Construction Lines: Type 01.1 (Continuous narrow Lines) Construction Lines are drawn for constructing drawings and should not be erased after completion of the drawing. • Hatching / Section Lines: Type 01.1 (Continuous Narrow Lines) Hatching Lines are drawn for the sectioned portion of an object. These are drawn inclined at an angle of 45° to the axis or to the main outline of the section. • Guide Lines: Type 01.1 (Continuous Narrow Lines) Guide Lines are drawn for lettering and should not be erased after lettering. • Break Lines: Type 01.1 (Continuous Narrow Freehand Lines) Wavy continuous narrow line drawn freehand is used to represent bre~ of an object. • Break Lines : Type 01.1 (Continuous Narrow Lines With Zigzags) Straight continuous ~arrow line with zigzags is used to represent break of an object. • Dashed Narrow Lines: Type 02.1 (Dashed Narrow Lines) Hidden edges / Hidden outlines of objects are shown by dashed lines of short dashes of equal lengths of about 3 mm, spaced at equal distances of about 1 mm. the points of intersection of these lines with the outlines / another hidden line should be clearly shown. • Center Lines: Type 04.1 (Long-Dashed Dotted Narrow Lines) Center Lines are draWn at the center of the drawings symmetrical about an axis or both the axes. These are extended by a short distance beyond the outline of the drawing. • " Cutting Plane Lines: Type 04.1 and Type 04.2 Cutting Plane Line is drawn to show the location of a cutting plane. It is long-dashed dotted narrow line, made wide at the ends, bends and change of direction. The direction of viewing is shown by means of arrows resting on the cutting plane line. • Border Lines Border Lines are continuous wide lines of minimum thickness 0.7 mm Fig. 2.5 Types of Lines
    • 21. 2.6 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - - - ..c Fig. 2.6 Understanding the various types oflines used in drawing (i.e.,) their thickness, style of construction and appearance as per BIS and following them meticulously may be considered as the foundation of good drawing skills. Table 2.2 shows various types oflines with the recommended applications. Table 2.2 Types of Lines and their applications (IS 10714 (Part 20): 2001) and BIS: SP46 : 2003. Line description and Representation Continuous narrow line No. Ol.l B 01.1 Continuous narrow freehand line C ~ 01.1 Continuous narrow line with zigzags A~ 01.2 Continuous wide line 02.1 Dashed narrow line D 04.1 Long-dashed dotted narrow line E 04.2 ----- --.-_'_- Long-dashed dotted wide line F -_.--'-- Applications Dimension lines, Extension lines Leader lines, Reference lines Short centre lines Projection lines Hatching Construction lines, Guide lines Outlines of revolved sections Imaginary lines of intersection Preferably manually represented tenrunation of partIal interrupted views, cuts and sections, if the limit is not a line symmetry or a center line·. Preferably mechanically represented termination of partial interrupted vIews. cuts and sections, if the hmit is not a line symmetry or a center linea Visible edges, visible outlines or of or of Main representations in diagrams, ma~s. flow charts Hidden edges Hidden outlines Center lines / Axes. Lines of symmetry Cuttmg planes (Line 04.2 at ends and changes of direction) Cutting planes at the ends and changes of direction outlines of visible parts situated m front of cutting plane Line widths (IS 10714 : 2001) Line width means line thickness. Choose line widths according to the size of the drawing from the following range: 0.13,0.18, 0.25, 0.35, 0.5, 0.7 and 1 mm. BIS recommends two line widths on a drawing. Ratio between the thin and thick lines on a drawing shall not be less than 1 : 2.
    • 22. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Lettering and Dimensioning Practices 2.7 Precedence of Lines 1. When a Visible Line coincide with a Hidden Line or Center Line, draw the Visible Line. Also, extend the Center Line beyond the outlines of the view. 2. When a Hidden Line coincides with a Center Line, draw the Hidden Line. 3. When a Visible Line coincides with a Cutting Plane, draw the Visible Line. 4. When a Center line coincides with a Cutting Plane, draw the Center Line and show the Cutting Plane line outside the outlines of the view at the ends of the Center Line by thick dashes. 2.3 LETTERING [IS 9609 (PART 0) : 2001 AND SP 46 : 2003] Lettering is defined as writing of titles, sub-titles, dimensions, etc., on a drawing. 2.3.1 Importance of Lettering To undertake production work of an engineering components as per the drawing, the size and other details are indicated on the drawing. This is done in the fonn of notes and dimensions. Main Features of Lettering are legibility, unifonnity and rapidity of execution. Use of drawing instruments for lettering consumes more time. Lettering should be done freehand with speed. Practice accompanied by continuous efforts would improve the lettering skill and style. Poor lettering mars the appearance of an otherwise good drawing. BIS and ISO Conventions IS 9609 (Part 0) : 2001 and SP 46 : 2003 (Lettering for technical drawings) specifY lettering in technical product documentation. This BIS standard is based on ISO 3098-0: 1997. 2.3.2 Single Stroke Letters The word single-stroke should not be taken to mean that the lettering should be made in one stroke without lifting the pencil. It means that the thickness of the letter should be unifonn as if it is obtained in one stroke of the pencil. 2.3.3 Types of Single Stroke Letters 1. Lettering Type A: (i) Vertical and (ii) Sloped (~t 75 0 to the horizontal) 2. Lettering Type B : (i) Vertical and (ii) Sloped (at 75 0 to the horizontal) Type B Preferred In Type A, height of the capital letter is divided into 14 equal parts, while in Type B, height of the capital letter is divided into 10 equal parts. Type B is preferred for easy and fast execution, because of the division of height into 10 equal parts. Vertical Letters Preferred Vertical letters are preferred for easy and fast execution, instead of sloped letters.
    • 23. 2.8 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - - - Note: Lettering in drawing should be in CAPITALS (i.e., Upper-case letters). Lower-case (small) letters are used for abbreviations like mm, cm, etc. 2.3.4 Size of Letters • Size of Letters is measured by the height h of the CAPITAL letters as well as numerals. • Standard heights for CAPITAL letters and numerals recommended by BIS are given below : 1.8, 2.5, 3.5, 5, 6, 10, 14 and 20 mm Note: Size of the letters may be selected based upon the size of drawing. Guide Lines In order to obtain correct and uniform height ofletters and numerals, guide lines are drawn, using 2H pencil with light pressure. HB grade conical end pencil is used for lettering. 2.3.5 Procedure for Lettering 1. Thin horizontal guide lines are drawn first at a distance ' h' apart. 2. Lettering Technique: Horizontal lines of the letters are drawn from left to right. Vertical, inclined and curved lines are drawn from top to bottom. 3. After lettering has been completed, the guidelines are not erased. 2.3.6 Dimensioning of Type B Letters (Figs 2.5 and 2.6) BIS denotes the characteristics of lettering as : h (height of capita) letters), c i (height of lower-case letters), c2 (tail of lower-case letters), c3 (stem of lower-case letters), a (spacing between characters), b l & b2 (spacing between baselines), e (spacing between words) and d (line thickness), Table 2.3 Lettering Proportions Recommended Size (height h) of Letters I Numerals Main Title 5 mm, 7 mm, 10 mm Sub-Titles 3.5 mm, 5 mm Dimensions, Notes, etc. 2.5 mm, 3.5 mm, 5 mm
    • 24. ___________________ Lettering and Dimensioning Practices 2.9 2.3.7 Lettering practice Practice oflettering capital and lower case letters and numerals of type B are shown in Figs.2.7 and 2.8. 0 Base line Base line '" .D £ Base line Base line Fig. 2.7 Lettering Fig. 2.8 Vertical Lettering The following are some of the guide lines for lettering (Fig 2.9 & 2.10) 1. Drawing numbers, title block and letters denoting cutting planes, sections are written in 10 mrn size. 2. Drawing title is written in 7 mm size. 3. Hatching, sub-titles, materials, dimensions, notes, etc., are written in 3.5 mm size. 4. Space between lines = ~ h. 5. Space between words may be equal to the width of alphabet M or 3/5 h.
    • 25. 2.10 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - -_ _ _ _ __ Fig. 2.9 Inclined Lettering 6. Space between letters should be approximately equal to 115 h. Poor spacing will affect the visual effect. 7. The spacing between two characters may be reduced by half if th is gives a better visual effect, as for example LA, TV; over lapped in case of say LT, TA etc, and the space is increased for letters with adjoining stems. CAPITAL Letters • Ratio of height to width for most of the CAPITAL letters is approximately = 10:6 • However, for M and W, the ratio = 10:8 for I the ratio = 10:2 Lower-case Letters Height of lower-case letters with stem I tail (b, d, f, g, h, j, k, I, p, q, t, y) = Cz = c 3 = h • Ratio of height to width for lower-case letters with stem or tail = 10:5 • • Height of lower-case letters without stem or tail c 1 is approximately = (7/10) h • Ratio of height to width for most lower-case letters without stem or tail = 7: 5 • However, for m and w, the ratio = 7: 7. For I and I, the ratio = 10:2
    • 26. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Lettering and Dimensioning Practices Numerals • For numerals 0 to 9, the ratio of height to width = 10 : 5. For I, ratio = Spacing = a = (2/10)b e = (6/10)b • Spacing between characters • Spacing between words SMALL = SPACES SHOULD BE FOR GOOD LETTER USED SPACING Correct LETTER SPACING RES'U L T S FRO M SPA C E S BEING TOO BIG POOR In correct Ca) J. J. NIGHT NUMBERS Letters with adjoining Item. require more Ipacing VITAL ALTAR tt t Lett.r combin.tlonl with over I.pping len.,. (b) Fig. 2.10 Guide lines for lettering 10 : 2 2.11
    • 27. 2.12 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - Fig. 2.11 Vertical capital & Lowercase letters and numerals of type B EXAMPLE IN LETTERING PRACTICE Write freehand the following, using single stroke vertical CAPITAL letters of 5 mm (h) size .J- ENGINEERING GRAPHICS IS THE f :! LANGUAGE OF ENGINEERS Fig. 2.12 2.4 Dimensioning Drawing of a compo Inent, in addition to prividing complete shape description, must also furnish information regarding the size description. These are provided through the distances between the surfaces, location of holes, nature of surface finish, type of material, etc. The expression of these features on a drawing, using lines, symbols, figures and notes is called dimensioning.
    • 28. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Lettering and Dimensioning Practices IJ Rounds and Fillets R3 54 r 2.13 Extension line DimensIon General Note ~ I L_ ..., N ReferenceI-.Dl-im-e-nS-io-n-~'1I---;;'-=--D-i-m-e-n-1s~ion""'lIne ,c- Local Note DIA 28. DEEP 25 C' BORE DIA 20 D~E:.'::E:.':.P~37---r-"V'--_ _L R15 Centre Line used as an ExtensIon Lane 90 Dimensions in Millimetres ~ units of Measurements -e-.-$ Frojection Symbol ~ Fig.2.13 Elements of Dimensioning 2.4.1 Principles of Dimensioning Some of the basic principles of dimensioning are given below. I. All dimensional information necessary to describe a component clearly and completely shall be written directly on a drawing. 2. Each feature shall be dimensioned once only on a drawing, i.e., dimension marked in one view need not be repeated in another view. 3. Dimension should be placed on the view where the shape is best seen (Fig.2.14) 4. As far as possible, dimensions should be expressed in one unit only preferably in millimeters, without showing the unit symbol (mm). 5. As far as possible dimensions should be placed outside the view (Fig.2.15). 6. Dimensions should be taken from visible outlines rather than from hidden lines (Fig.2.16).
    • 29. 2.14 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - 13 26 CORRECT INCORRECT Fig. 2.14 Placing the Dimensions where the Shape is Best Shown I ~r$- - ~ 50 50 CORRECT INCORRECT Fig. 2.15 Placing Dimensions Outside the View 10 10 Correct 26 Incorrect Fig. 2.16 Marking the dimensions from the visible outlines 7. No gap should be left between the feature and the start of the extension line (Fig.2.I7). 8. Crossing of centre lines should be done by a long dash and not a short dash (Fig.2.I8).
    • 30. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Lettering and Dimensioning Practices 2.15 22 52 52 Incorrect Correct Fig. 2.17 Marking of Extension Lines Incorrect Correct Fig. 2.18 Crossing of Centre Lines 2.4.2 Execution of Dimensions 1. Prejection and dimension lines should be drawn as thin continuous lines. projection lines should extend slightly beyond the respective dimension line. Projection lines should be drawn perpendicular to the feature being dimensioned. If the space for dimensioning is insufficient, the arrow heads may be reversed and the adjacent arrow heads may be replaced by a dot (Fig.2.19). However, they may be drawn obliquely, but parallel to each other in special cases, such as on tapered feature (Fig.2.20). ~ 2 .1 °1. 20 1 30 .1 1 4=4= Fig. 2.19 Dimensioning in Narrow Spaces
    • 31. 2.16 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - Fig. 2.20 Dimensioning a Tapered Feature 2. A leader line is a line referring to a feature (object, outline, dimension). Leader lines should be inclined to the horizontal at an angle greater than 30°. Leader line should tenninate, (a) with a dot, if they end within the outline ofan object (Fig.2.21a). (b) with an arrow head, if they end on outside of the object (Fig.2.21b). (c) without a dot or arrow head, if they end on dimension line (Fig.2.21c). (b) (a) (c) Fig. 2.21 Termination of leader lines Dimension Termination and Origin Indication Dimension lines should show distinct tennination in the fonn of arrow heads or oblique strokes or where applicable an origin indication (Fig.2.22). The arrow head included angle is 15°. The origin indication is drawn as a small open circle of approximately 3 mm in diameter. The proportion lenght to depth 3 : 1 of arrow head is shown in Fig.2.23. -I -----------~o Fig. 2.22 Termination of Dimension Line ~..l. ...... _ ,.& __ A _ _ ... "' __ ...1 Fig. 2.23 Proportions of an Arrow Head
    • 32. __________________ Lettering and Dimensioning Practices 2.17 When a radius is dimensioned only one arrow head, with its point on the arc end of the dimension line should be used (Fig.2.24). The arrow head termination may be either on the inside or outside of the feature outline, depending on the size of the feature. Fig. 2.24 Dimensioning of Radii 2.4.3 Methods of Indicating Dimensions The dimensions are indicated on the drawings according to one of the following two methods. Method - 1 (Aligned method) Dimensions should be placed parallel to and above their dimension lines and preferably at the middle, and clear of the line. (Fig.2.25). 70 Fig. 2.25 Aligned Method Dimensions may be written so that they can be read from the bottom or from the right side of the drawing. Dinensions on oblique dimension lines should be oriented as shown in Fig.2.26a and except where unavoidable, they shall not be placed in the 30° zone. Angular dimensions are oriented as shown in Fig.2.26b Method - 2 (uni-directional method) Dimensions should be indicated so that they can be read from the bottom of the drawing only. Non-horizontal dimension lines are interrupted, preferably in the middle for insertion of the dimension (Fig.2.27a). Angular dimensions may be oriented as in Fig.2.27b Note: Horizontal dimensional lines are not broken to place the dimension in both cases.
    • 33. 2.18 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - (b) (a) Fig.2.26 Angular Dimensioning 70 f20 .30 +50 30 26 10 75 (b) (a) Fig.2.27 Uni-directional Method 2.4.4 Identification of Sbapes The following indications are used with dimensions to show applicable shape identification and to improve drawing interpretation. The diameter and square symbols may be omitted where the shape is clearly indicated. The applicable indication (symbol) shall precede the value for dimension (Fig. 2.28 to 2.32). o 1"' • Fig. 2.28 Fig. 2.29
    • 34. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Lettering and Dimensioning Practices 2.19 a ..:# o Fig. 2.30 Fig. 2.31 Fig. 2.32 2.5 Arrangement of Dimensions The arrangement of dimensions on a drawing must indicate clearly the purpose of the design ofthe object. They are arranged in three ways. 1. Chain dimensioning 2. Parallel dimensioning 3. Combined dimensioning. 1. Chain dimensioning Chain of single dimensioning should be used only where the possible accumulation oftolerances does not endanger the fundamental requirement of the component (Fig.2.33) 2. Parallel dimensioning In parallel dimensioning, a number of dimension lines parallel to one another and spaced out, are used. This method is used where a number of dimensions have a common datum feature (Fig.2.34).
    • 35. 2.20 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - ('I) ~ ('I) (Q 16{) 85 30 l!i Fig. 2.33 Chain Dimensioning , tv 64179 272 Fig. 2.34 Parallel Dimensioning I--- r--- - - - - - - - - r-- Fig. 2.35 Combined Dimensioning
    • 36. __________________ Lettering and Dimensioning Practices 2.21 Violation of some of the principles of drawing are indicated in Fig.2.36a. The corrected version of the same as per BIS SP 46-2003 is given is Fig.2.36b. The violations from 1 to 16 indicated in the figure are explained below. 1 I :. I I I , , , , ~1~ L~ I ,ko !.: I~t J FRONTVlEW TOPVl£W' (b) (a) Fig. 2.36 1. Dimension should follow the shape symbol. 2. 4. 5. 6. 7. S. 9. and 3. As far as possible, features should not be used as extension lines for dimensioning. Extension line should touch the feature. Extension line should project beyond the dimension line. Writing the dimension is not as per aligned method. Hidden lines should meet without a gap. Centre line representation is wrong. Dots should be replaced by small dashes. Horizontal dimension line should not be broken to insert the value ofdimension in both aligned and uni-direction methods. 10. Dimension should be placed above the dimension line. 11. Radius symbol should precede the dimension. 12. Centre line should cross with long dashes not short dashes. 13. Dimension should be written by symbol followed by its values and not abbreviation. 14. Note with dimensions should be written in capitals. 15. Elevation is not correct usage. 16. Plan is obsolete in graphic language
    • 37. 2.22 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - -_ _ 120 ~100 100 H-120 8 40$ (a) Incorrect (b) Correct Fig. 2.37 3 HOLES DIA 10 ~.., 7f~-------------+~~.., 10 90 (b) Correct (a) Incorrect Fig. 2.38 30 40 1 + - - - 40 - - - - + I (b) Correct (a) Incorrect Fig. 2.39
    • 38. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Lettering and Dimensioning Practices (b) Correct Fig. 2.40 -:.0 ~ o '" 15 20 I Fig. 2.41 o '" 35 Fig. 2.42 o ..... '" 40 1+---....-:20 Fig. 2.43 2.23
    • 39. 2.24 Textbook of Enginnering D r a w i n g - - - - - - - - - - -_ _ _ _ _ __ 1l1li 15 5 30 ~_ _ ,I 40 Fig. 2.44 I -+I I I 20 ~ 14- ~ I _.J _ _ _ .1_ 40 Fig. 2.45 o --------- ------ .... ~20 20 Fig. 2.46 Drill ~ 10, C Bore
      li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-40-638.jpg?cb=1398399596" title="_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Lettering and Dimensioning ..." target="_blank" 40. /a _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Lettering and Dimensioning Practices, 2.25 4CMd r t SCM i 5CM LI--_ _----I SCM t Fig. 2.48 T f co '" + o o 25 ~ ~ 25 -+ 25 Fig. 2.49 Fig. 2.50 EXERCISE Write freehand the following, using single stroke vertical (CAPITAL and lower-case) letters: 1. Alphabets (Upper-case & Lower-case) and Numerals 0 to 9 (h = 5 and 7 nun) 2. PRACTICE MAKES A PERSON PERFECT (h = 3.5 and 5) 3. BE A LEADER NOT A FOLLOWER (h = 5) 4. LETTERlNG SHOULD BE DONE FREEHAND WITH SPEED (h = 5) /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-41-638.jpg?cb=1398399596" title="3 CHAPTER Scales 3.1 Introduction It is not possible a..." target="_blank" 41. /a 3 CHAPTER Scales 3.1 Introduction It is not possible always to make drawings of an object to its actual size. If the actual linear dimensions of an object are shown in its drawing, the scale used is said to be a full size scale. Wherever possible, it is desirable to make drawings to full size. 3.2 Reducing and Enlarging Scales Objects which are very big in size can not be represented in drawing to full size. In such cases the object is represented in reduced size by making use of reducing scales. Reducing scales are used to represent objects such as large machine parts, buildings, town plans etc. A reducing scale, say 1: 10 means that 10 units length on the object is represented by 1 unit length on the drawing. Similarly, for drawing small objects such as watch parts, instrument components etc., use offull scale may not be useful to represent the object clearly. In those cases enlarging scales are used. An enlarging scale, say 10: 1 means one unit length on the object is represented by 10 units on the drawing. The designation of a scale consists of the word. SCALE, followed by the indication of its ratio as follows. (Standard scales are shown in Fig. 3.1) Scale 1: 1 for full size scale Scale 1: x for reducing scales (x = 10,20 ...... etc.,) Scale x: 1 for enlarging scales. Note: For all drawings the scale has to be mentioned without fail. r r 111111111111111111111111111111111111111111111111111 1:1 0 10 20 30 40 50 111111111111111111111111111111111111111111111111111 1.2 0 20 40 60 80 100 r IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIITIIIIIIIII 1:5 0 100 200 r 1IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIImllllllllllili 1.100 Fig.3.1 Scales 200 400 /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-42-638.jpg?cb=1398399596" title="3.2 Textbook of Enginnering D r a w i n g - - - - - - - - ..." target="_blank" 42. /a 3.2 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - 3.3 Representative Fraction The ratio of the dimension of the object shown on the drawing to its actual size is called the Representative Fraction (RF). RF = Drawing size of an object (. . ) m same umts Its actual size For example, if an actual length of3 metres of an object is represented by a line of 15mm length on the drawing RF = 15mm = lSmm =_1_ orl:200 3m (3 x 1000)mm 200 If the desired scale is not available in the set of scales it may be constructed and then used. Metric Measurements 10 millimetres (mm) = 1 centimetre(cm) 10 centimetres (cm) = 1 decimetre(dm) 10 decimetre (dm) = 1 metre(m) 10 metres (m) = 1 decametre (dam) 10 decametre (dam) = 1 hectometre (bm) 10 hectometres (bm) = 1 kilometre (km) 1 hectare = 10,000 m2 3.4 Types of Scales The types of scales normally used are: 1. Plain scales. 2. Diagonal Scales. 3. Vernier Scales. 3.4.1 Plain Scales A plain scale is simply a line which is divided into a suitable number of equal parts, the fIrst of which is further sub-divided into small parts. It is used to represent either two units or a unit and its fraction such as km and bm, m and dm, cm and mm etc. Problem 1 : On a survey map the distance between two places 1km apart is 5 cm. Construct the scale to read 4.6 km. Solution: (Fig 3.2) RF= Scm 1 =-1x 1000 x 100cm 20000 /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-43-638.jpg?cb=1398399596" title="----------------------------------------------------Scru~ ..." target="_blank" 43. /a ----------------------------------------------------Scru~ 3.3 1 Ifx is the drawing size required x = 5(1000)(100) x 20000 Therefore, x = 25 cm Note: If 4.6 km itself were to be taken x = 23 cm. To get 1 km divisions this length has to be divided into 4.6 parts which is difficult. Therefore, the nearest round figure 5 km is considered. When this length is divided into 5 equal parts each part will be 1 km. 1. Draw a line of length 25 cm. 2. Divide this into 5 equal parts. Now each part is 1 km. 3. Divide the first part into 10 equal divisions. Each division is 0.1 km. 4. Mark on the scale the required distance 4.6 km. 46km I III IIIIII 10 I I I I I I I 1 2 3 5 I 0 HECfOMETRE I LENGTH OFTHE SCALE 4 KILOMETRE SCALE .1.20000 Fig. 3.2 Plain Scale Problem 2 : Construct a scale of 1:50 to read metres and decimetres and long enough to measure 6 m. Mark on it a distance of 5.5 m. Construction (Fig. 3.3) 1. Obtairrthe length of the scale as: RF x 6m = _1 x 6 x 100 =12 cm 50 2. Draw a rectangle strip oflength 12 cm and width 0.5 cm. 3. Divide the length into 6 equal parts, by geometrical method each part representing 1m. 4. Mark O(zero) after the first division and continue 1,2,3 etc., to the right of the scale. 5. Divide the first division into 10 equal parts (secondary divisions), each representing 1 cm. 6. Mark the above division points from right to left. 7. Write the units at the bottom of the scale in their respective positions. 8. Indicate RF at the bottom of the figure. 9. Mark the distance 5.5 m as shown. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-44-638.jpg?cb=1398399596" title="3.4 Textbook of Enginnering D r a w i n g - - - - - - - - ..." target="_blank" 44. /a 3.4 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - -'I 5.5m ;(1 j /, t if I I I I 10 o 5 I I I I 2 3 I I 5 4 RF=l/50 DECIMETRE METRES Fig. 3.3 Problem 3 : The distance between two towns is 250 km and is represented by a line of length 50mm on a map. Construct a scale to read 600 km and indicate a distance of 530 km on it. Solution: (F ig 3.4) 50mm 250km 1. Determine the RF value as - - 2. Obtain the length of the scale as: 3. 4. 5. 6. 50 1 250x 1000x 1000 = 5 X 10 6 1 x 600km =120mm . 5xl06 Draw a rectangular strip oflength 120 mm and width 5 mm. Divide the length into 6 eq1}AI parts, each part representing 10 km. Repeat the steps 4 to 8 of construction in Fig 3.2. suitably. Mark the distance 530 km as shown. / II ~ if j. I I I II 530km I I I 100 50 0 I I I I I I I I 300 100 Fig. 3.4 400 500 /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-45-638.jpg?cb=1398399596" title="---------------------------Scales 3.5 Problem 4: Construc..." target="_blank" 45. /a ---------------------------Scales 3.5 Problem 4: Construct a plain scale of convenient length to measure a distance of 1 cm and mark on it a distance of 0.94 em. Solution: (Fig 3.5) This is a problem of enlarged scale. 1. Take the length of the scale as 10 cm 2. RF = 1011, scale is 10:1 3. The construction is shown in Fig 3.5 094cm 1111 1 0 1 Z 3 4 5 6 7 89 LENG1H OFTHE SCAlE 100mm SCAlE: 1:1 Fig. 3.5 3.4.2 Diagonal Scales . Plain scales are used to read lengths in two units such as metres and decimetres, centimetres and millimetres etc., or to read to the accuracy correct to first decimal. Diagonal scales are used to represent either three units of measurements such as metres, decimetres, centimetres or to read to the accuracy correct to two decimals. Principle of Diagonal Scale (Fig 3.6) 1. Draw a line AB and errect a perperrdicular at B. 2. Mark 10 equi-distant points (1,2,3, etc) of any suitable length along this perpendicular and mark C. 3. Complete the rectangle ABCD 4. Draw the diagonal BD. 5. Draw horizontals through the division points to meet BD at l' , 2' , 3' etc. Considering the similar triangles say BCD and B44' B4' B4 4 I 4 , - = _ . =-xBCx-=-· 44 =O.4CD CD BC' 10 BC 10' /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-46-638.jpg?cb=1398399596" title="3.6 Textbook of Enginnering Drawing 0 C 9 8 7 9' 7' 6 ..." target="_blank" 46. /a 3.6 Textbook of Enginnering Drawing 0 C 9 8 7 9' 7' 6 6' 5 5' 4 4' 3 3' 2 1 2' A B Fig. 3.6 Principle ofDiagonal Scale Thus, the lines 1-1',2 - 2', 3 - 3' etc., measure O.lCD, 0.2CD, 0.3CD etc. respectively. Thus, CD is divided into 1110 the divisions by the diagonal BD, i.e., each horizontal line is a multiple of 11 10 CD. This principle is used in the construction of diagonal scales. Note: B C must be divided into the same number of parts as there are units of the third dimension in one unit of the secondary division. Problem 5 : on a plan, a line of 22 em long represents a distance of 440 metres. Draw a diagonal scale for the plan to read upto a single metre. Measure and mark a distance of 187 m on the scale. Solution: (Fig 3.7) 187m DEC I5ETRE 10 1 5 o t 1 I 40 211 0 40 80 LENGTH OF SCALE 100mm 160 1211 MeIJ& SOUE.1:2000 Fig. 3.7 Diagonal Scale. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-47-638.jpg?cb=1398399596" title="---------------------------SCales 1. RF = 3.7 22 1 440x..." target="_blank" 47. /a ---------------------------SCales 1. RF = 3.7 22 1 440xl00 = 2000 2. As 187 m are required consider 200 m. 1 Therefore drawing size = R F x actual size = 2000 x 200 x 100 =10 em When a length of 10 cm representing 200 m is divided into 5 equal parts, each part represents 40 m as marked in the figure. 3. The first part is sub-divided into 4 divisions so that each division is 10 cm 4. On the diagonal portion 10 divisions are taken to get 1 m. 5. Mark on it 187 m as shown. Problem 6 : An area of 144 sq cm on a map represents an area of 36 sq /an on the field. Find the RF of the scale of the map and draw a diagonal scale to show Km, hectometres and decametres and to measure upto 10 /an. Indicate on the scale a distance 7 /an, 5 hectometres and 6 decemetres. Solution: (Fig. 3.8) 1. 144 sq cm represents 36 sq km or 12 cm represent 6 km 12 RF Drawing size x = 1 6xl000xl00 = 50000 = R F x actual size = 10xl000xl00 50000 = 20 cm 7.56Icm DECA Mem: to 1111 0 5 o I 10 5 0 HECTOMETRE 1 3 5 LENGTH Of THE SCALE Fig. 3.8 7 9 KILOMETRE /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-48-638.jpg?cb=1398399596" title="3.8 Textbook of Enginnering D r a w i n g - - - - - - - - ..." target="_blank" 48. /a 3.8 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - 2. Draw a length of 20 cm and divide it into 10 equal parts. Each part represents 1 km. 3. Divide the first part into 10 equal subdivisions. Each secondary division represents 1 hecometre 4. On the diagonal scale portion take 10 eqal divisions so that 1110 ofhectometre = 1 decametre is obtained. 5. Mark on it 7.56 km. as shown. Problem 7 : Construct a diagonal scale 1/50, showing metres, decimetres and centimetres, to measure upto 5 metres. Mark a length 4. 75 m on it. Solution: (Fig 3.9) 1. Obatin the length of the scale as _1 x 5 x 100 =10 cm 50 2. Draw a line A B, 10 cm long and divide it into 5 equal parts, each representing 1 m. 3-. -Divide the fIrst part into 10 equal parts, to represent decimetres. 4. Choosing any convenient length, draw 10 equi-distant parallel lines alJove AB and complete the rectangle ABC D. 5. Erect perpendiculars to the line A B, through 0, 1,2,3 etc., to meet the line C D. 6. Join D to 9, the fIrst sub-division from A on the main scale AB, forming the fIrst diagonal. 7. Draw the remaining diagonals, parallel to the fIrst. Thus, each decimetre is divided into II 10th division by diagonals. 8. Mark the length 4.75m as shown. 4.75m 0 c 10 8 ~ w .... w 6 ~ w 4 :i o 2 A1()~ J;!fCIMETEB Q 2 1 FtF = 1:50 Fig. 3.9 3 4 METRES 5 /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-49-638.jpg?cb=1398399596" title="---------------------------Scales 3.9 3.4.3 Vernier Scale..." target="_blank" 49. /a ---------------------------Scales 3.9 3.4.3 Vernier Scales The vernier scale is a short auxiliary scale constructed along the plain or main scale, which can read upto two decimal places. The smallest division on the main scale and vernier scale are 1 msd or 1 vsd repectively. Generally (n+ 1) or (n-l) divisions on the main scale is divided into n equal parts on the vernier scale. (1) (n -1) Thus, 1 vsd = -n-msd or 1-; msd When 1 vsd 1 it is called forward or direct vernier. The vernier divisions are numbered in the same direction as those on the main scale. When 1 vsd 1 or (1 + lin), It is called backward or retrograde vernier. The vernier divisions are numbered in the opposite direction compared to those on the main scale. The least count (LC) is the smallest dimension correct to which a measurement can be made with a vernier. For forward vernier, L C = (1 msd - 1 vsd) For backward viermier, LC = (1 vsd - 1 msd) Problem 8 : Construct a forward reading vernier scale to read distance correct to decametre on a map in which the actual distances are reduced in the ratio of 1 : 40,000. The scale should be long enough to measure upto 6 km. Mark on the scale a length of 3.34 km and 0.59 km. Solution: (Fig. 3.10) 1. RF = 1140000; length of drawing = 6xl000x 100 40000 = 15 em 2. 15 em is divided into 6 parts and each part is 1 km 3. This is further divided into 10 divitions and each division is equal to 0.1 km = 1 hectometre. Ims d = 0.1 km = 1 hectometre L.C expressed in terms of m s d = (111 0) m s d L C is 1 decametre = 1 m s d - 1 v s d 1 v s d = 1 - 1110 = 9110 m s d = 0.09 km 4. 9 m sd are taken and divided into 10 divisions as shown. Thus 1 vsd = 9110 = 0.09 km 5. Mark on itbytaking6vsd=6x 0.9 = 0.54km, 28msd(27 + 1 on the LHS of 1) =2.8 kmand Tota12.8 + 0.54 = 3.34 km. 6. Mark on it 5 msd = 0.5 km and add to it one vsd = 0.09, total 0.59 km as marked. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-50-638.jpg?cb=1398399596" title="3.10 Textbook of Enginnering D r a w i n g - - - - - - - -..." target="_blank" 50. /a 3.10 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - O.54+2.S=3.34lcm O.SSkm o 5 10 LENGTH OF THE SCALE 150 mm SCAlE:1:40000 Fig. 3.10 Forward Reading Vernier Scale Problem 9 : construct a vernier scale to read metres, decimetres and centimetres and long enough to measure upto 4m. The RF of the scale in 1120. Mark on it a distance of 2.28 m. Solution: (Fig 3.11) Backward or Retrograde Vernier scale 1. The smallest measurement in the scale is cm. Therefore LC = O.Olm 2. Length of the scale = RF x Max. Distance to be measured =-1 x 4m =-1 x 400 =20 em 20 20 3 LeastCO.llt= O.01m Rf" 1120 Fig. 3.11 Backward or Retrograde Vernier Scale /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-51-638.jpg?cb=1398399596" title="--------------------------Scales 3.11 3. Draw a line of20..." target="_blank" 51. /a --------------------------Scales 3.11 3. Draw a line of20 em length. Complete the rectangle of20 em x 0.5 em and divide it into 4 equal parts each representing 1 metre. Sub divide all into 10 main scale divisions. 1 msd = Im/l0 = Idm. 4. Take 10+ 1 = 11 divisions on the main scale and divide it into 10 equal parts on the vernier scale by geometrical construction. Thus Ivsd= llmsd/lO= 1.1dm= llcm 5. Mark 0,55, 110 towards the left from 0 (zero) on the vernier scale as shown. 6. Name the units of the divisions as shown. 7. 2.28m = (8 x vsd) + 14msd) = (8 x O.llm) + (14 x O.lm) = 0.88 + 1.4 = 2.28m. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-52-638.jpg?cb=1398399596" title="3.12 Textbook of Enginnering D r a w i n g - - - - - - - -..." target="_blank" 52. /a 3.12 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - EXERCISES 1. Construct a plain scale of 1:50 to measure a distance of 7 meters. Mark a distance of 3.6 metres on it. 2. The length of a scale with a RF of2:3 is 20 cm. Construct this scale and mark a distance of 16.5 cm on it. 3. Construct a scale of 2 cm = 1 decimetre to read upto 1 metre and mark on it a length of 0.67 metre. 4. Construct a plain scale of RF = 1:50,000 to show kilometres and hectometres and long enough to measure upto 7 krn. Mark a distance of 5:3 kilometres on the scale. 5. On a map, the distance between two places 5 krn apart is 10 cm. Construct the scale to read 8 krn. What is the RF of the scale? 6. Construct a diagonal scale ofRF = 1150, to read metres, decimetres and centimetres. Mark a distance of 4.35 krn on it. 7. Construct a diagonal scale of five times full size, to read accurately upto 0.2 mm and mark a distance of 3 .65 cm on it. 8. Construct a diagonal scale to read upto 0.1 mm and mark on it a 'distance of 1.63 cm and 6.77 cm. Take the scale as 3: 1. 9. Draw a diagonal scale of 1 cm = 2.5krn and mark on the scale a length of26.7 krn. 10. Construct a diagonal scale to read 2krn when its RF=I:20,000. Mark on it a distance of 1:15 km. 11. Draw a venier scale of metres when Imm represents 25cm and mark on it a length of 24.4 cm and 23.1 mm. What is the RF? 12. The LC of a forward reading vernier scale is 1 cm. Its vernier scale division represents 9 cm. There are 40 msd on the scale. It is drawn to 1:25 scale. Construct the scale and mark on it a distance ofO.91m. 13. 15cm of a vernier scale represents 1cm. Construct a backward reading vernier scale of RF 1:4.8 to show decimetres cm and mm. The scale should be capable of reading upto 12 decimeters. Mark on the scale 2.69 decimetres and 5.57 decimetres. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-53-638.jpg?cb=1398399596" title="CHAPTER 4 Geometrical Constructions 4.1 Introduction Engi..." target="_blank" 53. /a CHAPTER 4 Geometrical Constructions 4.1 Introduction Engineering drawing consists of a number of geometrical constructions. A few methods are illustrated here without mathematical proofs. 1. To divide a straight line into a given number of equal parts say 5. construction (Fig.4.1) A " 4' 3' 2' B 2 I 3 4 5 c Fig. 4.1 Dividing a line 1. Draw AC at any angle e to AB. 2. Construct the required number of equal parts of convenient length on AC like 1,2,3. 3. Join the last point 5 to B 4. Through 4, 3, 2, 1 draw lines parallel to 5B to intersect AB at 4',3',2' and 1'. 2. To divide a line in the ratio 1 : 3 : 4. construction (Fig.4.2) /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-54-638.jpg?cb=1398399596" title="4.2 Textbook of Enginnering D r a w i n g - - - - - - - - ..." target="_blank" 54. /a 4.2 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - As the line is to be divided in the ratio 1:3:4 it has to be divided into 8 equal divisions. By following the previous example divide AC into 8 equal parts and obtain P and Q to divide the lineAB in the ratio 1:3:4. A~~----------~----------~ K c Fig. 4.2 3. To bisect a given angle. construction (Fig.4.3) .. Fig. 4.3 1. Draw a line AB and AC making the given angle. 2. With centre A and any convenient radius R draw an arc intersecting the sides at D and E. 3. With centres D and E and radius larger than half the chord length DE, draw arcs intersecting at F 4. JoinAF, BAF = PAC. 4. To inscribe a square in a given circle. construction (Fig. 4.4) 1. With centre 0, draw a circle of diameter D. 2. Through the centre 0, drwaw two diameters, say AC and BD at right angle to each other. 3. ,(oin A-B, B-C, C- D, and D-A. ABCD is the required square. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-55-638.jpg?cb=1398399596" title="_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ GeometricaIContru..." target="_blank" 55. /a _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ GeometricaIContructions 4.3 D c 8 Fig. 4.4 5. To inscribe a regular polygon of any number of sides in a given circle. construction (Fig. 4.5) G f / A D Fig. 4.5 1. 2. 3. 4. Draw the given circle with AD as diameter. Divide the diameter AD into N equal parts say 6. With AD as radius and A and D as centres, draw arcs intersecting each other at G Join G-2 and extend to intesect the circle at B. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-56-638.jpg?cb=1398399596" title="4.4 Textbook of Enginnering D r a w i n g - - - - - - - - ..." target="_blank" 56. /a 4.4 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - 5. loinA-B which is the length of the side of the required polygon. 6. Set the compass to the length AB and strating from B mark off on the circuference of the circles, obtaining the points C, D, etc. The figure obtained by joing the points A,B, C etc., is the required polygon. 6. To inscribe a hexagon in a given circle. (a) Construction (Fig. 4.6) by using a set-square or mini-draughter o 2 2 E 21 2~, A 0 60' 60' A Fig. 4.6 1. With centre 0 and radius R draw the given crcle. 2. Draw any diameter AD to the circle. 3. Using 30° - 60° set-square and through the point A draw lines AI, A2 at an angle 60° with AD, intesecting the circle at B and F respectively. 4. Using 30° - 60° and through the point D draw lines Dl, D2 at an angle 60° with DA, intersecting the circle at C and E respectively. By joining A,B,C,D,E,F, and A the required hexagon is obtained. (b) Construction (Fig.4.7) By using campass 1. With centre 0 and radius R draw the given circle. 2. Draw any diameter AD to the circle. 3. With centres A and D and radius equal to the radius of the circle draw arcs intesecting the circles at B, F, C and E respectively. 4. ABC D E F is the required hexagon. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-57-638.jpg?cb=1398399596" title="- - - - - - - - - - - -_ _ _ _ _ _ _ _ _ GeometricaIContruc..." target="_blank" 57. /a - - - - - - - - - - - -_ _ _ _ _ _ _ _ _ GeometricaIContructions ...------f...-o''------. A 4.5 :J 8 Fig. 4.7 7. To circumscribe a hexagon on a given circle of radius R construction (Fig. 4.8) / / / R/ . / I At-+---1-- I ./ +0 o c B Fig. 4.8 I. With centre 0 and radius R draw the given circle. 2. Using 60° position of the mini draughter or 300-600set square, circumscribe the hexagon as shown. 8. To construct a hexagon, given the length of the side. (a) contruction (Fig. 4.9) Using set square 1. Draw a line AB equal to the side of the hexagon. 2. Using 30° - 60° set-square draw lines AI, A2, and BI, B2. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-58-638.jpg?cb=1398399596" title="4.6 Textbook of Enginnering D r a w i n g - - - - - - - - ..." target="_blank" 58. /a 4.6 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - 1 2 1 . A ..---'---+---'----t' A B Fig. 4.9 3. Through 0, the point of intesection between the lines A2 at D and B2 at E. 4. loinD,E 5. ABC D E F is the required hexagon. (b) By using compass (Fig.4.10) D E c x a B A Fig. 4.10 1. Draw a line AB equal to the of side of the hexagon. 2. With centres A and B and radius AB, draw arcs intersecting at 0, the centre of the hexagon. 3. With centres 0 and B and radius OB (=AB) draw arcs intersecting at C. 4. Obtain points D, E and F in a sinilar manner. 9. To construct a regular polygon (say a pentagon) given the length of the side. construction (Fig.4.11) 1. Draw a line AB equal to the side and extend to P such that AB = BP 2. Draw a semicircle on AP and divide it into 5 equal parts by trial and error. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-59-638.jpg?cb=1398399596" title="- - - - - - - - - - - - - - - - - - - - ' - - - G e o m e t..." target="_blank" 59. /a - - - - - - - - - - - - - - - - - - - - ' - - - G e o m e t r i c a l Contructions 4.7 Fig. 4.11 3. Join B to second division 2. Irrespective of the number of sides of the polygon B is always joined to the second division. 4. Draw the perpendicular bisectors of AB and B2 to intersect at O. 5. Draw a circle with 0 as centre and OB as radius. 6. WithAB as radius intersect the circle successively at D and E. Thenjoin CD. DE and EA. 10. To construct a regular polygon (say a hexagon) given the side AB - alternate method. construction (Fig.4.12) E F A B Fig. 4.12 1. Steps 1 to 3 are same as above 2. Join B- 3, B-4, B-5 and produce them. 3. With 2 as centre and radius AB intersect the line B, 3 produced at D. Similarly get the point E and F. 4. Join 2- D, D-E, E-F and F-A to get the required hexagon. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-60-638.jpg?cb=1398399596" title="4.8 Textbook of Enginnering D r a w i n g - - - - - - - - ..." target="_blank" 60. /a 4.8 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - 11. To construct a pentagon, given the length of side. (a) Construction (Fig.4.13a) 1. Draw a line AB equal to the given length of side. 2. Bisect AB at P. 3. Draw a line BQ equal to AB in length and perpendicular to AB. 4. With centre P and radius PQ, draw an arc intersecting AB produced at R. AR is equal to the diagonal length of the pentagon. S. With centres A and B and radii AR and AB respectively draw arcs intersecting at C. 6. With centres A and B and radius AR draw arcs intersecting at D. 7. With centres A and B and radii AB and AR respectively draw arcs intersecting at E. ABCDE is the required pentagon. 2 1 F / / ,I' C R A Fig.4.13a Fig.4.13b (b)By included angle method 1. 2. 3. 4. Draw a line AB equal to the length of the given side. Draw a line B 1 such that AB 1 = 108° (included angle) Mark Con Bl such that BC = AB Repeat steps 2 and 3 and complete the pentagon ABCDE 12. To construct a regular figure of given side length and of N sides on a straight line. construction (Fig 4.14) 1. 2. 3. 4. S. Draw the given straight line AB. At B erect a perpendicular BC equal in length to AB. Join AC and where it cuts the perpendicular bisector of AB, number the point 4. Complete the square ABeD of which AC is the diagonal. With radius AB and centre B describe arc AC as shown. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-61-638.jpg?cb=1398399596" title="- _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Geometrical Contr..." target="_blank" 61. /a - _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Geometrical Contructions 4.9 Fig. 4.14 6. 7. 8. 9. Where this arc cuts the vertical centre line number the point 6. This is the centre of a circle inside which a hexagon of side AB can now be drawn. Bisect the distance 4-6 on the vertical centre line. Mark this bisection 5. This is the centre in which a regular pentagon of side AB can now be drawn. 10. On the vertical centre line step off from point 6 a distance equal in length to the distance 5-6. This is the centre of a circle in which a regular heptagon of side AB can now be drawn. 11. If further distances 5-6 are now stepped off along the vertical centre line and are numbered consecutively, each will be the centre of a circle in which a regular polygon can be inscribed with sice of length AB and with a number of sides denoted by the number against the centre. 13. To inscribe a square in a triangle. construction (Fig. 4.15) 1. Draw the given triangle ABC. 2. From C drop a perpendicular to cut the base AB at D. 3. From C draw CE parallel toAB and equal in length to CD. 4. Draw AE and where it cuts the line CB mark F. 5. From F draw FG parallel to AB. 6. From F draw FJ parallel to CD. 7. From G draw GH parallel to CD. 8. Join H to 1. Then HJFG is the required square. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-62-638.jpg?cb=1398399596" title="4.10 Textbook of Enginnering D r a w i n g - - - - - - - -..." target="_blank" 62. /a 4.10 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - E C / '"" .. ,/ ' /~ / ,F "-"-, ! "" "-, I A /' H 0 J B Fig. 4.15 14. To inscribe within a given square ABCD, another square, one angle of the required square to touch a side of the given square at a given point construction (Fig 4.16) Fig. 4.16 1. Draw the given square ABeD. 2. Draw the diagonals and where they intersect mark the point O. 3. Mark the given point E on the line AB. 4. With centre 0 and radius OE, draw a circle. S. Where the circle cuts the given square mark the points G, H, and F. 6. Join the points GHFE. Then GHFE is the required square. 15. To draw an arc of given radius touching two straight lines at right 8rngles to each _ other. construction (Fig 4.17) Let r be the given radius and AB and AC the given straight lines. With A as centre and radius equal to r draw arcs cutting AB at P and Q. With P and Q as centres draw arcs to meet at O. With 0 as centre and radius equal to r draw the required arc. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-63-638.jpg?cb=1398399596" title="- - - - -_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _Geometrical Contruc..." target="_blank" 63. /a - - - - -_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _Geometrical Contructions 4.11 B p c Q Fig. 4.17 16. To draw an arc of a given radius, touching two given straight lines making an angle between them. construction (Fig 4.18) Let AB and CD be the two straight lines and r, the radius. Draw a line PQ parallel to AB at a distance r from AB. Similarly, draw a line RS parallel to CD. Extend them to meet at O. With 0 as centre and radius equal to r draw the arc to the two given lines. 8 c Fig. 4.18 17. To draw a tangent to a circle construction (Fig 4.19 a and b) (a) At any point P on the circle. (a) Fig. 4.19 /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-64-638.jpg?cb=1398399596" title="4.12 Textbook of Enginnering D r a w i n g - - - - - - - -..." target="_blank" 64. /a 4.12 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - 1. With 0 as centre, draw the given circle. P is any point on the circle at which tangent to be drawn (Fig 4.l6a) 2. Join 0 with P and produce it to pI so that OP = ppl 3. With 0 and pI as centres and a length greater than OP as radius, draw arcs intersecting each other at Q. 4. Draw a line through P and Q. This line is the required tangent that will be perpendicular to OP at P. (b) From any point outside the circle. 1. With 0 as centre, draw the given circle. P is the point outside the circle from which tangent is to be drawn to the circle (F ig 4 .16b). 2. Join 0 with P. With OP as diameter, draw a semi-circle intersecting the given circle at M. Then, the line drawn through P and M is the required tangent. 3. If the semi-circle is drawn on the other side, it will cut the given circle at MI. Then the line through P and MI will also be a tangent to the circle from P. 4.2 Conic Sections e Cone is formed when a right angled triangle with an apex and angle is rotated about its altitude as the axis. The length or height of the cone is equal to the altitude of the triangle and the radius of the base of the cone is equal to the base of the triangle. The apex angle of the cone is 2 (Fig.4.20a). e When a cone is cut by a plane, the curve formed along the section is known as a conic. For this purpose, the cone may be cut by different section planes (Fig.4.20b) and the conic sections obtained are shown in Fig.4.20c, d, and e. r Apex r End generator rCutting plane / perpendicular --+_;---4--_1. to the axis Base Circle Fig.4.20a&b /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-65-638.jpg?cb=1398399596" title="_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Geometrical Con..." target="_blank" 65. /a _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Geometrical Contructions section p section plane 8-B 4.13 c-c c- Ellipse d- Parabola section plane D-D e- Hyperbola Fig. 4.20c,d&e 4.2.1 Circle When a cone is cut by a section plane A-A making an angle a = 90° with the axis, the section obtained is a circle. (Fig 4.20a) 4.2.2 Ellipse e When a cone is cut by a section plane B-B at an angle, a more than half of the apex angle i.e., and less than 90°, the curve of the section is an ellipse. Its size depends on the angle a and the distance of the section plane from the apex of the cone. 4.2.3 Parabola e If the angle a is equal to i.e., when the section plane C-C is parallel to the slant side of the cone. the curve at the section is a parobola. This is not a closed figure like circle or ellipse. The size of the parabola depends upon the distance of the section plane from the slant side of the cone. 4.2.4 Hyperbola e(section plane D-D), the curve at the section is hyperbola. The curve of intersection is hyperbola, even if a = e, provided the section plane is not passing through the If the angle a is less than apex of the cone. However if the section plane passes through the apex, the section produced is an isosceles triangle. /li li a href="http://image.slidesharecdn.com/5905...engineering-graphics-66-638.jpg?cb=1398399596" title="4.14 Textbook of Enginnering D r a w i n g - - - - - - - -..." target="_blank" 66. /a 4.14 Textbook of Enginnering D r a w i n g - - - - - - - - - - - - - - - - - - - - 4.2.5 Conic Sections as Loci of a Moving Point A conic section may be defined as the locus of a point moving in a plane such that the ratio of its distance from a fixed point (Focus) and fixed straight line (Directrix) is always a constant. The ratio is called eccentricity. The line passing through the focus and perpendicular to the directrix is the axis of the curve. The point at which the conic section intersects the axis is called the vertex or apex of the curve. The eccentricity value is less than 1 for ellipse, equal to I for parabola and greater than 1 for hyperbola (F ig.4.21). ~_ _ _ Q,f-C_--i./ p, I HYPERBOLA e = P, F,IP,Q,> 1 ~ V, . V2. V3 - VERTICES F" F2 - FOCI AXIS Q2t--+--. f.P 2 O Q3t--t--+-;~P3 A~~~~~~~~-~~------B V2 V3~ ' V, - - - - ELLIPSE ~~P3F'/P3Q3
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