PROJ OF SOLIDS - Yola

SOLIDS To understand and remember various solids in this subject properly, those are classified & arranged in to two major groups. Group A Group B Solids having top and base of same shape Solids having base of some shape and just a point as a top, called apex. Cylinder Cone Prisms Triangular Cube Square ( A solid having six square faces) Pentagonal Hexagonal Pyramids Triangular Square Pentagonal Hexagonal Tetrahedron ( A solid having Four triangular faces) Laxmi Institute of Technology, sarigam SOLIDS Dimensional parameters of different solids. Square Prism Corner of base Cylinder Slant Edge Base Edge of Base Base Edge of Base Cone Apex Apex

Top Rectangular Face Longer Edge Square Pyramid Triangular Base Face Corner of base Sections of solids( top & base not parallel) Base Generators Imaginary lines generating curved surface of cylinder & cone. Frustum of cone & pyramids. ( top & base parallel to each other) STANDING ON H.P On its base. (Axis perpendicular to Hp And // to Vp.) F.V. X X RESTING ON H.P On one point of base circle. (Axis inclined to Hp And // to Vp) F.V. LYING ON H.P On one generator. (Axis inclined to Hp And // to Vp) F.V. While observing Fv, x-y line represents Horizontal Plane. (Hp) While observing Tv, x-y line represents Vertical Plane. (Vp) T.V. T.V. STANDING ON V.P RESTING ON V.P On its base. On one point of base circle. Axis perpendicular to Vp Axis inclined to Vp And // to Hp And // to Hp T.V. LYING ON V.P On one generator.

Axis inclined to Vp And // to Hp Y Y STEPS TO SOLVE PROBLEMS IN SOLIDS Problem is solved in three steps: STEP 1: ASSUME SOLID STANDING ON THE PLANE WITH WHICH IT IS MAKING INCLINATION. ( IF IT IS INCLINED TO HP, ASSUME IT STANDING ON HP) ( IF IT IS INCLINED TO VP, ASSUME IT STANDING ON VP) IF STANDING ON HP - ITS TV WILL BE TRUE SHAPE OF ITS BASE OR TOP: IF STANDING ON VP - ITS FV WILL BE TRUE SHAPE OF ITS BASE OR TOP. BEGIN WITH THIS VIEW: ITS OTHER VIEW WILL BE A RECTANGLE ( IF SOLID IS CYLINDER OR ONE OF THE PRISMS): ITS OTHER VIEW WILL BE A TRIANGLE ( IF SOLID IS CONE OR ONE OF THE PYRAMIDS): DRAW FV & TV OF THAT SOLID IN STANDING POSITION: STEP 2: CONSIDERING SOLIDS INCLINATION ( AXIS POSITION ) DRAW ITS FV & TV. STEP 3: IN LAST STEP, CONSIDERING REMAINING INCLINATION, DRAW ITS FINAL FV & TV. GENERAL PATTERN ( THREE STEPS ) OF SOLUTION: GROUP B SOLID. CONE AXIS AXIS VERTICAL INCLINED HP AXIS INCLINED VP Three steps If solid is inclined to Hp AXIS INCLINED HP AXIS AXIS VERTICAL INCLINED HP AXIS INCLINED VP Three steps If solid is inclined to Hp GROUP A SOLID. CYLINDER GROUP B SOLID. CONE GROUP A SOLID. CYLINDER AXIS er TO VP AXIS INCLINED VP Three steps If solid is inclined to Vp AXIS INCLINED HP AXIS

er TO VP AXIS INCLINED VP Three steps If solid is inclined to Vp Study Next Twelve Problems and Practice them separately !! CATEGORIES OF ILLUSTRATED PROBLEMS! PROBLEM NO.1, 2, 3, 4 GENERAL CASES OF SOLIDS INCLINED TO HP & VP PROBLEM NO. 5 & 6 CASES OF CUBE & TETRAHEDRON PROBLEM NO. 7 CASE OF FREELY SUSPENDED SOLID WITH SIDE VIEW. PROBLEM NO. 8 CASE OF CUBE ( WITH SIDE VIEW) PROBLEM NO. 9 CASE OF TRUE LENGTH INCLINATION WITH HP & VP. PROBLEM NO. 10 & 11 CASES OF COMPOSITE SOLIDS. (AUXILIARY PLANE) PROBLEM NO. 12 CASE OF A FRUSTUM (AUXILIARY PLANE) Problem 1. A square pyramid, 40 mm base sides and axis 60 mm long, has a triangular face on the ground and the vertical plane containing the axis makes an angle of 450 with the VP. Draw its projections. Take apex nearer to VP Solution Steps : Triangular face on Hp , means it is lying on Hp: 1.Assume it standing on Hp. 2.Its Tv will show True Shape of base( square) 3.Draw square of 40mm sides with one side vertical Tv & taking 50 mm axis project Fv. ( a triangle) 4.Name all points as shown in illustration. 5.Draw 2nd Fv in lying position I.e.ocd face on xy. And project its Tv. 6.Make visible lines dark and hidden dotted, as per the procedure. 7.Then construct remaining inclination with Vp ( Vp containing axis ic the center line of 2nd Tv.Make it 450 to xy as shown take apex near to xy, as it is nearer to Vp) & project final Fv.

o ab cd o d1 a1 For dark and dotted lines c1 c1 d1 a1 o1 c o1 d o1 1 d1 b1 1.Draw proper outline of new view DARK. 2. Decide direction of an observer. 3. Select nearest point to observer and draw all lines starting from it-dark. 4. Select farthest point to observer and draw all lines (remaining)from it- dotted. Y a1 b d o a cd X ab b1 a1 c1 b 1(APEX c1 NEARER TO V.P). b 1 (APEX AWAY FROM V.P.) o

1 Q Draw the projections of a pentagonal prism , base 25 mm side and axis 50 mm long, resting on one of its rectangular faces on the H.P. with the axis inclined at 45 to the V.P. As the axis is to be inclined with the VP, in the first view it must be kept perpendicular to the VP i.e. true shape of the base will be drawn in the FV with one side on XY line b 2 b1 a 1 X a1 c 3 e 5 a e 25 b d 4 d 21 d1 e1 45 d 31 11 c1 c Y 41 51 b c e 50 a 4 2 5 1 5 2 4 3 1

3 Solution Steps: Resting on Hp on one generator, means lying on Hp: 1.Assume it standing on Hp. 2.Its Tv will show True Shape of base( circle ) 3.Draw 40mm dia. Circle as Tv & taking 50 mm axis project Fv. ( a triangle) 4.Name all points as shown in illustration. 5.Draw 2nd Fv in lying position I.e.oe on xy. And project its Tv below xy. 6.Make visible lines dark and hidden dotted, as per the procedure. 7.Then construct remaining inclination with Vp ( generator o1e1 300 to xy as shown) & project final Fv. Problem 2: A cone 40 mm diameter and 50 mm axis is resting on one generator on Hp which makes 300 inclination with Vp Draw its projections. For dark and dotted lines 1.Draw proper outline of new vie DARK. 2. Decide direction of an observer. 3. Select nearest point to observer and draw all lines starting from it-dark. 4. Select farthest point to observer and draw all lines (remaining) from it- dotted. o a hb c h1 g h f d e o e c g g g1 f f1 f1 e1 g1 h1 c d1 1 e e1 b

d c a1 b1 d1 c1 o1 a1 b1 e1 d1 Y o1 30 o1 h1 f1 a b1 g1 d f X a hb a1 c1 Solution Steps: Problem 3: Resting on Vp on one point of base, means inclined to Vp: 1.Assume it standing on Vp A cylinder 40 mm diameter and 50 mm Fv will show True Shape of base & top( circle ) axis is resting on one point of a base 2.Its 3.Draw 40mm dia. Circle as Fv & taking 50 mm axis project Tv. circle on Vp while its axis makes 450 ( a Rectangle) 4.Name all points as shown in illustration. with Vp and Fv of the axis 350 with Hp. 5.Draw 2nd Tv making axis 450 to xy And project its Fv above xy. Draw projections.. 6.Make visible lines dark and hidden dotted, as per the procedure. 7.Then construct remaining inclination with Hp ( Fv of axis I.e. center line of view to xy as shown) & project final Tv. 4 4d d 3 c a

1 a 4 c d 3 c 3 1 1 a 2 b bd c 450 2 350 c a b bd d1 Y b c1 b1 a1 3 3 a 1 24 24 4 3 1 X 2 1 2

Solution Steps : 1.Assume it standing on Hp but as said on apex.( inverted ). 2.Its Tv will show True Shape of base( square) 3.Draw a corner case square of 30 mm sides as Tv(as shown) Showing all slant edges dotted, as those will not be visible from top. 4.taking 50 mm axis project Fv. ( a triangle) 5.Name all points as shown in illustration. 6.Draw 2nd Fv keeping oa slant edge vertical & project its Tv 7.Make visible lines dark and hidden dotted, as per the procedure. 8.Then redrew 2nd Tv as final Tv keeping a1o1d1 triangular face perpendicular to Vp I.e.xy. Then as usual project final Fv. Problem 4:A square pyramid 30 mm base side and 50 mm long axis is resting on its apex on Hp, such that its one slant edge is vertical and a triangular face through it is perpendicular to Vp. Draw its projections. a bd X c a o o d a bo bd c ao1 1 c a1 d1 b1 c1 o1 d1 b1 Y d1 c1 c1 a1 1 o b1 Solution Steps: Problem 5: A cube of 50 mm long edges is so placed on Hp on one

corner that a body diagonal is parallel to Hp and perpendicular to Vp Draw its projections. 1.Assuming standing on Hp, begin with Tv,a square with all sides equally inclined to xy.Project Fv and name all points of FV & TV. 2.Draw a body-diagonal joining c with 3( This can become // to xy) 3.From 1 drop a perpendicular on this and name it p 4.Draw 2nd Fv in which 1-p line is vertical means c-3 diagonal must be horizontal. .Now as usual project Tv.. 6.In final Tv draw same diagonal is perpendicular to Vp as said in problem. Then as usual project final FV. a a1 bd c p c1 1 3 1 d d1 d1 a c a1 c1 a1 X c p 3 b b1 1 Y b1 bd d1 c1 a d1

Problem 6:A tetrahedron of 50 mm long edges is resting on one edge on Hp while one triangular face containing this edge is vertical and 450 inclined to Vp. Draw projections. IMPORTANT: Tetrahedron is a special type of triangular pyramid in which base sides & slant edges are equal in length. Solid of four faces. Like cube it is also described by One X dimension only.. Axis length generally not given. Solution Steps As it is resting assume it standing on Hp. Begin with Tv , an equilateral triangle as side case as shown: First project base points of Fv on xy, name those & axis line. From a with TL of edge, 50 mm, cut on axis line & mark o (as axis is not known, o is finalized by slant edge length) Then complete Fv. In 2nd Fv make face obc vertical as said in problem. And like all previous problems solve completely. o1 o o TL a a a 90 b c b c c c1 o 0 a1 b1 c1 450 a1 c1 o1 o1

b1 b b1 a1 Y FREELY SUSPENDED SOLIDS: Positions of CG, on axis, from base, for different solids are shown below. CG H/2 H CG H/4 GROUP A SOLIDS ( Cylinder & Prisms) GROUP B SOLIDS ( Cone & Pyramids) Problem 7: A pentagonal pyramid 30 mm base sides & 60 mm long axis, is freely suspended from one corner of base so that a plane containing its axis remains parallel to Vp. Draw its three views. Solution Steps: In all suspended cases axis shows inclination with Hp. 1.Hence assuming it standing on Hp, drew Tv - a regular pentagon,corner case. 2.Project Fv & locate CG position on axis ( H from base.) and name g and Join it with corner d 3.As 2nd Fv, redraw first keeping line gd vertical. 4.As usual project corresponding Tv and then Side View looking from. dg VERTICAL d LINE o d ce FOR SIDE VIEW g H e ab g IMPORTANT: When a solid is freely suspended from a corner, then line joining point of contact & C.G. remains vertical. ( Here axis shows

inclination with Hp.) So in all such cases, assume solid standing on Hp initially.) X H/4 c e a b e1 e a1 do 1 o d1 b b1 c a b o d a c c1 Y Solution Steps: 1.Assuming it standing on Hp begin with Tv, a square of corner case. 2.Project corresponding Fv.& name all points as usual in both views. 3.Join a1 as body diagonal and draw 2nd Fv making it vertical (I on xy) 4.Project its Tv drawing dark and dotted lines as per the procedure. 5.With standard method construct Left-hand side view. a d bd a bd A cube of 50 mm long edges is so placed on Hp on one corner that a body diagonal through this corner is perpendicular to Hp and parallel to Vp Draw its three views. a ( Draw a 450 inclined Line in Tv region ( below xy). Project horizontally all points of Tv on this line and

reflect vertically upward, above xy.After this, draw horizontal lines, from all points of Fv, to meet these lines. Name points of intersections and join properly. For dark & dotted lines locate observer on left side of Fv as shown.) Problem 8: c c X d a c b 1 1 d1 a1 c 1 c1 b b Y Problem 9: A right circular cone, 40 mm base diameter and 60 mm long axis is resting on Hp on one point of base circle such that its axis makes 450 inclination with Hp and 400 inclination with Vp. Draw its projections. This case resembles to problem no.7 & 9 from projections of planes topic. In previous all cases 2nd inclination was done by a parameter not showing TL.Like Tv of axis is inclined to Vp etc. But here it is clearly said that the axis is 40 0 inclined to Vp. Means here TL inclination is expected. So the same construction done in those Problems is done here also. See carefully the final Tv and inclination taken there. So assuming it standing on HP begin as usual. o o o1 a1 a h1 h b g1 g c

f d e 450 g1 h1 f a e b d c f1 a1 e g h c1 d1 c g d f X a hb b1 o1 400 e1 y Axis True Length f1 1 b1 e1 e1 o1 f1 d1 g1 c1 Axis Tv Length d1

Axis Tv Length b1 1 h1 c1 a1 Locus of Center 1 Problem 10: A triangular prism, 40 mm base side 60 mm axis is lying on Hp on one rectangular face with axis perpendicular to Vp. One square pyramid is leaning on its face F.V. centrally with axis // to vp. Its base side is 30 mm & axis is 60 mm long resting on Hp on one edge of base.Draw FV & TV of both solids.Project another FV X on an AVP 450 inclined to VP. 1 y 45 0 to Vp ) 450 (A VP T.V. 1 Aux.F.V. X Steps : Draw Fv of lying prism ( an equilateral Triangle) And Fv of a leaning pyramid. Project Tv of both solids. Draw x1y1 450 inclined to xy and project aux.Fv on it. Mark the distances of first FV from first xy for the distances of aux. Fv from x1y1 line. Note the observers directions Shown by arrows and further steps carefully. Y Problem 11:A hexagonal prism of base side 30 mm longand axis 40 mm long, is standing on Hp on its base with one base edge // to Vp. A tetrahedron is placed centrally

on the top of it.The base of tetrahedron is a triangle formed by joining alternate corners of top of prism..Draw projections of both solids. Project an auxiliary Tv on AIP 450 inclined to Hp. b f d c e Fv X Aux.Tv Y e Tva o o1 e1 450 f Y1 to a Hp ) TL (A IP 45 0 STEPS: Draw a regular hexagon as Tv of standing prism With one side // to xy and name the top points.Project its Fv a rectangle and name its top. Now join its alternate corners a-c-e and the triangle formed is base of a tetrahedron as said. Locate center of this triangle & locate apex o Extending its axis line upward mark apex o By cutting TL of edge of tetrahedron equal to a-c. and complete Fv of tetrahedron. Draw an AIP ( x1y1) 450 inclined to xy And project Aux.Tv on it by using similar Steps like previous problem. o f1 d d1 a1

c1 b1 b c X1 Problem 12: A frustum of regular hexagonal pyrami is standing on its larger base On Hp with one base side perpendicular to Vp.Draw its Fv & Tv. Project its Aux.Tv on an AIP parallel to one of the slant edges showing TL. Base side is 50 mm long , top side is 30 mm long and 50 mm is height of frustum. Fv 1 25 AIP // to slant edge Showing true length i.e. a- 1 Y1 34 4 TL 5 3 1 X a c d Y b e Aux.Tv e a d1 d Tv 5 4 2 e1 X1 1 b 2 3 c

a1 c1 b1 ab ab a1 b1 cd cd 11 12 c1 d1 21 450 12 b2 c3 41 34 34 21 31 30 b1 21 0 31 c1 11 31 b1 41 a1 d4 11 41 a1 d1 a1 c1

d1 Q13.22: A hexagonal pyramid base 25 mm side and axis 55 mm long has one of its slant edge on the ground. A plane containing that edge and the axis is perpendicular to the H.P. and inclined at 45 to the V.P. Draw its projections when the apex is nearer to the V.P. than the base. The inclination of the axis is given indirectly in this problem. When the slant edge of a pyramid rests The plane containing theHP slant edge deciding on the HP is the seen in the on thevertical HP its axis is inclined with the so while firstand viewthe the axis axis of solid mustTV be as operpendicular auxiliary draw plane 45Secondly from d1owhen kept to HP i.e. trueFV shape of an the auxiliary base will be seenXin drawing 1d1 for drawing 1Ythe 1 atTV. 1 extended. hexagon in the TV we have to keep corners thef1extreme ends. to X1Y1 and mark the Then draw projectors from eachthe point i.e. aat1 to perpendicular points measuring their distances in the FV from old XY line. o f1 a

X1 b f b1 c e X c e b f a d e f a d o b o f1 d1 a1 o1 c c1 c1 d1 d e1 a1 e1 b1 Y o1 45 Y1

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