Engineering Drawing Notes For 1st and 2nd sem (Engineering)
- 1. ADARSH SACHDEVA Presented By Engineering Graphics Projection of Planes
- 2. PROJECTIONS OF PLANES In this topic various plane figures are the objects. What will be given in the problem? 1. Description of the plane figure. 2. It’s position with HP and VP. In which manner it’s position with HP & VP will be described? 1.Inclination of it’s SURFACE with one of the reference planes will be given. 2. Inclination of one of it’s EDGES with other reference plane will be given (Hence this will be a case of an object inclined to both reference Planes.) To draw their projections means F.V, T.V. & S.V. What is usually asked in the problem? Study the illustration showing surface & side inclination given on next page.
- 3. HP VP VPVP a’ d’ c’b’ HP a b c d a1’ d1’ c1’ b1’ HP a1 b1 c1 d1 CASE OF A RECTANGLE – OBSERVE AND NOTE ALL STEPS. SURFACE PARALLEL TO HP PICTORIAL PRESENTATION SURFACE INCLINED TO HP PICTORIAL PRESENTATION ONE SMALL SIDE INCLINED TO VP PICTORIAL PRESENTATION ORTHOGRAPHIC TV-True Shape FV- Line // to xy ORTHOGRAPHIC FV- Inclined to XY TV- Reduced Shape ORTHOGRAPHIC FV- Apparent Shape TV-Previous Shape A B C
- 4. PROCEDURE OF SOLVING THE PROBLEM: IN THREE STEPS EACH PROBLEM CAN BE SOLVED:( As Shown In Previous Illustration ) STEP 1. Assume suitable conditions & draw Fv & Tv of initial position. STEP 2. Now consider surface inclination & draw 2nd Fv & Tv. STEP 3. After this,consider side/edge inclination and draw 3rd ( final) Fv & Tv. ASSUMPTIONS FOR INITIAL POSITION: (Initial Position means assuming surface // to HP or VP) 1.If in problem surface is inclined to HP – assume it // HP Or If surface is inclined to VP – assume it // to VP 2. Now if surface is assumed // to HP- It’s TV will show True Shape. And If surface is assumed // to VP – It’s FV will show True Shape. 3. Hence begin with drawing TV or FV as True Shape. 4. While drawing this True Shape – keep one side/edge ( which is making inclination) perpendicular to xy line ( similar to pair no. on previous page illustration ).A B Now Complete STEP 2. By making surface inclined to the resp plane & project it’s other view (Ref. 2nd pair on previous page illustration ) C Now Complete STEP 3. By making side inclined to the resp plane & project it’s other view. (Ref. 3nd pair on previous page illustration ) APPLY SAME STEPS TO SOLVE NEXT ELEVEN PROBLEMS
- 5. X Y a b c d a’ b’ c’d’ a1 b1 c1 d1 a’b’ d’c’ c’1 d’1 b’1 a’1450 300 Problem 1: Rectangle 30mm and 50mm sides is resting on HP on one small side which is 300 inclined to VP,while the surface of the plane makes 450 inclination with HP. Draw it’s projections. Read problem and answer following questions 1. Surface inclined to which plane? ------- HP 2. Assumption for initial position? ------// to HP 3. So which view will show True shape? --- TV 4. Which side will be vertical? ---One small side. Hence begin with TV, draw rectangle below X-Y drawing one small side vertical. Surface // to Hp Surface inclined to Hp Side Inclined to Vp
- 6. Problem 2: A 300 – 600 set square of longest side 100 mm long, is in VP and 300 inclined to HP while it’s surface is 450 inclined to VP.Draw it’s projections (Surface & Side inclinations directly given) Read problem and answer following questions 1 .Surface inclined to which plane? ------- VP 2. Assumption for initial position? ------// to VP 3. So which view will show True shape? --- FV 4. Which side will be vertical? ------longest side. c1 X Y 300 450 a’1 b’1 c’1 a c a’ a b1 b’ b a1b c a’1 b’1 c’1 c’ Hence begin with FV, draw triangle above X-Y keeping longest side vertical. Surface // to Vp Surface inclined to Vp side inclined to Hp
- 7. c c1 X Y 450 a’1 b’1 c’1 a c a’ a b1 b’ b a1b a’1 b’1 c’1 c’ 35 10 Problem 3: A 300 – 600 set square of longest side 100 mm long is in VP and it’s surface 450 inclined to VP. One end of longest side is 10 mm and other end is 35 mm above HP. Draw it’s projections (Surface inclination directly given. Side inclination indirectly given) Read problem and answer following questions 1 .Surface inclined to which plane? ------- VP 2. Assumption for initial position? ------// to VP 3. So which view will show True shape? --- FV 4. Which side will be vertical? ------longest side. Hence begin with FV, draw triangle above X- Y keeping longest side vertical. First TWO steps are similar to previous problem. Note the manner in which side inclination is given. End A 35 mm above Hp & End B is 10 mm above Hp. So redraw 2nd Fv as final Fv placing these ends as said.
- 8. Read problem and answer following questions 1. Surface inclined to which plane? ------- HP 2. Assumption for initial position? ------ // to HP 3. So which view will show True shape? --- TV 4. Which side will be vertical? -------- any side. Hence begin with TV,draw pentagon below X-Y line, taking one side vertical. Problem 4: A regular pentagon of 30 mm sides is resting on HP on one of it’s sides with it’s surface 450 inclined to HP. Draw it’s projections when the side in HP makes 300 angle with VP a’b’ d’ b1 d c1 a c’e’ b c d1 b’1 a1 e’1 c’1 d’1 a1 b1 c1d1 d’ a’b’ c’e’ e1 e1 a’1 X Y450 300 e SURFACE AND SIDE INCLINATIONS ARE DIRECTLY GIVEN.
- 9. Problem 5: A regular pentagon of 30 mm sides is resting on HP on one of it’s sides while it’s opposite vertex (corner) is 30 mm above HP. Draw projections when side in HP is 300 inclined to VP. Read problem and answer following questions 1. Surface inclined to which plane? ------- HP 2. Assumption for initial position? ------ // to HP 3. So which view will show True shape? --- TV 4. Which side will be vertical? --------any side. Hence begin with TV,draw pentagon below X-Y line, taking one side vertical. b’ d’ a’ c’e’ a1 b1 c1d1 e1 b1 c1 d1 a1 e1 b’1 e’1 c’1 d’1 a’1 X Ya’b’ d’c’e’ 30 a b c d e 300 SURFACE INCLINATION INDIRECTLY GIVEN SIDE INCLINATION DIRECTLY GIVEN: ONLY CHANGE is the manner in which surface inclination is described: One side on Hp & it’s opposite corner 30 mm above Hp. Hence redraw 1st Fv as a 2nd Fv making above arrangement. Keep a’b’ on xy & d’ 30 mm above xy.
- 10. a d c b a’ b’ d’ c’ X Y a1 b1 d1 c1 450 300 a’1 b’1 c’1 d’1 a1 b1 d1 c1a d c b a’ b’ d’ c’ 300 a’1 b’1 c’1 d’1 Problem 7: A circle of 50 mm diameter is resting on Hp on end A of it’s diameter AC which is 300 inclined to Hp while it’s Tv is 450 inclined to Vp.Draw it’s projections. Problem 9: A circle of 50 mm diameter is resting on Hp on end A of it’s diameter AC which is 300 inclined to Hp while it makes 450 inclined to Vp. Draw it’s projections. Read problem and answer following questions 1. Surface inclined to which plane? ------- HP 2. Assumption for initial position? ------ // to HP 3. So which view will show True shape? --- TV 4. Which diameter horizontal? ---------- AC Hence begin with TV,draw rhombus below X-Y line, taking longer diagonal // to X-Y The difference in these two problems is in step 3 only. In problem no.8 inclination of Tv of that AC is given,It could be drawn directly as shown in 3rd step. While in no.9 angle of AC itself i.e. it’s TL, is given. Hence here angle of TL is taken,locus of c1 Is drawn and then LTV I.e. a1 c1 is marked and final TV was completed.Study illustration carefully. Note the difference in construction of 3rd step in both solutions.
- 11. As 3rd step redraw 2nd Tv keeping side DE on xy line. Because it is in VP as said in problem. X Y a b c d e f Problem 8: A hexagonal lamina has its one side in HP and Its apposite parallel side is 25mm above Hp and In Vp. Draw it’s projections. Take side of hexagon 30 mm long. ONLY CHANGE is the manner in which surface inclination is described: One side on Hp & it’s opposite side 25 mm above Hp. Hence redraw 1st Fv as a 2nd Fv making above arrangement. Keep a’b’ on xy & d’e’ 25 mm above xy. 25 f’ e’d’c’b’a’ a1 b1 c1 d1 e1 f1 c1 ’ b’1a’1 f’1 d’1e’1 f1 a1 c1 b1 d1e1 Read problem and answer following questions 1. Surface inclined to which plane? ------- HP 2. Assumption for initial position? ------ // to HP 3. So which view will show True shape? --- TV 4. Which diameter horizontal? ---------- AC Hence begin with TV,draw rhombus below X-Y line, taking longer diagonal // to X-Y