2 What will be given in the problem? PROJECTIONS OF PLANESIn this topic various plane figures are the objects.What will be given in the problem?Description of the plane figure.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 showingsurface & side inclination given on next page.
3 C A B CASE OF A RECTANGLE – OBSERVE AND NOTE ALL STEPS. SURFACE PARALLEL TO HPPICTORIAL PRESENTATIONSURFACE INCLINED TO HPPICTORIAL PRESENTATIONONE SMALL SIDE INCLINED TO VPPICTORIAL PRESENTATIONFor T.V.For T.V.For TvFV-1FV-2FV-3For FvFor F.V.For F.V.T V-1T V-2T V-3ORTHOGRAPHICTV-True ShapeFV- Line // to xyORTHOGRAPHICFV- Inclined to XYTV- Reduced ShapeORTHOGRAPHICFV- Apparent ShapeTV-Previous ShapeVPVPa’d’c’b’VPa’d’c’b’HPa1b1c1d1a1’d1’c1’b1’HPabcdHPa1b1c1d1CAB
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 // HPOr If surface is inclined to VP – assume it // to VP2. 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 ).ABNow Complete STEP 2. By making surface inclined to the resp plane & project it’s other view.(Ref. 2nd pair on previous page illustration )CNow Complete STEP 3. By making side inclined to the resp plane & project it’s other view.(Ref. 3nd pair on previous page illustration )
8 EG2111 Engineering Graphics Problem No.4.A circular lamina of diameter 60 mm is held vertical with its surface inclined at 45° to the VP. Its centre is 40 mm above the HP and 30 mm in front of the VP. Draw its top and front views.3’31’2’4’21’41’5’1’O’11’51’81’61’406’8’71’7’xy45°30o(81) (71) (61)(8) (7) (6)K.M.KumarEG2111 Engineering Graphics
9 EG2111 Engineering Graphics Problem No.5.A square lamina PQRS of side 40 mm rests on the ground on its corner P in such a way that the diagonal PR is inclined at 45° to the HP and apparently inclined at 30° to the VP. Draw its Projections.r2’p1’ q1’(s1’) r1’s2’q2’p’ q’(s’) r’45°xp2’y30°s1ss2r2prp1r1p2q2qq1K.M.KumarEG2111 Engineering Graphics
10 Problem No.6.A circular plate of diameter 70 mm has the end P of the diameter PQ in the HP and the plate is inclined at 40° to the HP. Draw its projections when the diameter PQ appears to be inclined at 45° to the VP in the top view.q2’p1’ r1’(s1’) q1’r2’s2’p’ r’(s’) q’40°p2’xys45°s1s2p2p1q1pqq2r2r1r
11 EG2111 Engineering Graphics Problem No.7.A rectangular plate 70 x 40 mm has one of its shorter edges in the VP inclined at 40° to the HP. Draw its top view if its front view is a square of side 40 mm.p2’p’q’p1’q1’s2’q2’s’r’s1’r1’r2’40°s2p2xp1(s1)yp(s) q(r)r2q2q1(r1)K.M.KumarEG2111 Engineering Graphics
12 Problem No.8.A hexagonal plate of side 20 mm rests on the HP on one of its sides inclined at 45° to the VP. The surface of the plate makes an angle of 30° with the HP. Draw the front and top views of the plate.s2’r2’u1’(p1’) t1’(q1’) s1’(r1’)t2’q2’u’(p’) t’(q’) s’(r’)30°xu2’p2’yqq145°p2p1prr1q2u2r2su1us1t2tt1s2