Engineering Graphics & Design (3110013) Semester 1 & 2 Winter 2025 GTU Solved Paper

Construct a plain scale of R.F. = 1:10 showing the feet and inches and long enough to read the distance of 5 feet. Show on it the distance of 4 feet and 10 inches.

1. Given Data

• Representative Fraction (R.F.): 1:10 = 1/10
• Maximum length to be measured (L_max): 5 feet
• Distance to show: 4 feet and 10 inches
• Unit relation: 1 foot = 12 inches; 1 inch = 2.54 cm = 25.4 mm; 1 foot = 304.8 mm = 30.48 cm.

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2. Calculation of Length of Scale (L_s)

• Formula:
  Length of Scale (Ls) = R.F. × Lmax
• Substitution:
  L_s = (1 / 10) × 5 feet
  Since 1 foot = 30.48 cm:
  L_s = (1 / 10) × 5 × 30.48 cm = 15.24 cm = 152.4 mm (or approximately 15.2 cm)

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3. Step-by-Step Construction Procedure

1. Draw a horizontal line of length 15.24 cm (152.4 mm) and create a rectangle of width 5 mm to 10 mm.
2. Divide into Main Units (Feet):

• Since maximum length is 5 feet, divide the 15.24 cm line into 5 equal parts. Each part represents 1 foot.
• Mark the zero point at the end of the first main division from the left.
• Number the divisions to the right of 0 as: 0, 1, 2, 3, 4 feet.

1. Divide into Sub-Units (Inches):

• The first part (to the left of 0) represents 1 foot (12 inches).
• Divide this leftmost division into 12 equal parts (or 6 equal parts of 2 inches each if space is limited).
• Number them to the left from 0 to 12 inches (mark 0, 3, 6, 9, 12 inches).

1. Marking the Distance (4 feet 10 inches):

• Locate 4 feet to the right of the 0 mark.
• Locate 10 inches to the left of the 0 mark.
• Draw extension lines from these two marks and draw a dimension line with arrowheads indicating 4 feet 10 inches.

1. Write the R.F. = 1:10 neatly below the scale.

   |<--------- 10 inches --------->|<-------------------- 4 feet ------------------->|
   |                               |                                                 |
  [12  9  6  3  0]-----------------[1]---------------[2]---------------[3]----------[4]
  <-- INCHES --->|                               FEET                                |
                                  R.F. = 1:10

(schematic, not to scale)

A circular disc of diameter AB = 90 mm, rotates with uniform angular velocity. The point P which is at A, moves with uniform linear velocity and reaches the point B, when the disc completes one revolution. Trace the locus of point P moving from A to B.

1. Concept

This problem generates a special spiral-like locus on a rotating plane (an Archimedean-type motion across a rotating diameter). As the circular disc rotates through 360° at a constant speed, the tracing point P travels linearly along diameter AB from end A to end B at a constant speed.

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2. Given Data

• Diameter of disc (AB): 90 mm
• Radius (R): 45 mm
• Initial position of P: At point A
• Final position of P: At point B
• Displacement of P: 90 mm along diameter AB during one complete revolution (360°).

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3. Step-by-Step Construction Procedure

1. Draw the Initial Geometry:

• Draw a circle with centre O and diameter AB = 90 mm horizontally.

1. Divide the Motion into Equal Intervals:

• Divide the angular rotation (one full revolution of 360°) into 12 equal parts of 30° each.
• Label the angular positions of diameter line AB after successive rotations as lines: 1-1', 2-2', 3-3', ..., 12-12' (where line 12-12' returns to AB).
• Divide the linear path AB (90 mm) into the same number of equal parts (12 equal divisions). Mark these points as 1, 2, 3, 4, ..., 12 from A towards B.

1. Locate the Successive Positions of Point P:

• At start (0°): Point P is at position A (point 0).
• When the disc rotates by 30° (position 1): Point P has moved linearly to division 1. With centre O and radius equal to distance O-1, draw an arc cutting the rotated radius line 1 at point P1.
• When the disc rotates by 60° (position 2): With centre O and radius O-2, draw an arc intersecting line 2 at point P2.
• Continue this process for all 12 positions:
• At 180° (position 6), the point reaches centre O (P6 = O).
• At position 7, with radius O-7, intersect line 7 to get P7.
• At 360° (position 12), point P reaches the opposite end B (P12 = B).

1. Draw the Curve:

• Join points P0 (A), P1, P2, P3, P4, P5, P6 (O), P7, ..., P12 (B) with a smooth, continuous curve using a French curve.
• This smooth curve represents the required locus of point P.

Construct an ellipse when the distance of its focus from its directrix is equal to 50 mm and the eccentricity is 2/3. Also, draw the tangent and normal to the ellipse at a point 70 mm away from the directrix.

1. Given Data

• Distance of focus (F) from directrix (AB): CF = 50 mm
• Eccentricity (e): 2/3 (Since e < 1, the conic is an ellipse)
• Point of Tangency (P): 70 mm away from the directrix

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2. Location of the Vertex (V)

• By definition of eccentricity: e = VF / VC = 2 / 3
• Total parts = 2 + 3 = 5 equal parts.
• Length of each part = 50 mm / 5 = 10 mm.
• Therefore:
• Distance of vertex V from directrix C: VC = 3 × 10 = 30 mm
• Distance of vertex V from focus F: VF = 2 × 10 = 20 mm

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3. Step-by-Step Construction (Directrix-Focus Method)

1. Draw a vertical directrix line AB.
2. Choose a point C on AB and draw the horizontal axis perpendicular to AB through C.
3. Mark focus F on the axis such that CF = 50 mm.
4. Mark vertex V on the axis such that CV = 30 mm and VF = 20 mm.
5. At vertex V, draw a vertical line and mark a point E such that VE = VF = 20 mm.
6. Draw a straight line from C through E and extend it. The slope of this line is VE / CV = 20 / 30 = 2/3 = e.
7. Mark several points 1, 2, 3, 4, 5... on the axis to the right of V (e.g., at intervals of 10 mm or 15 mm).
8. Through these points, draw vertical lines cutting the extended line CE at 1', 2', 3', 4', etc.
9. Taking focus F as centre and radius equal to 1-1', cut arcs above and below the axis on the vertical line through 1.
10. Repeat the step: with centre F and radius 2-2', cut arcs on the vertical line through 2; with radius 3-3', cut arcs on line 3, and so forth.
11. The curve closes at the other vertex V' when the arc radius from F intersects the corresponding vertical line on the axis.
12. Draw a smooth curve passing through V, all the plotted arc intersections, and V' to obtain the ellipse.

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4. Tangent and Normal at Point P (70 mm from Directrix)

1. Draw a line parallel to directrix AB at a distance of 70 mm to intersect the ellipse at point P.
2. Join point P with focus F to form line PF.
3. At focus F, draw a line perpendicular to PF (angle = 90°) to intersect the directrix AB at point T'.
4. Draw a straight line passing through T' and P. This line T-P-T is the required tangent to the ellipse.
5. Draw a line perpendicular to the tangent line at point P. This line N-P-N is the required normal.

Draw the projections of the following points on X-Y line: (1) A point ‘A’ 40 mm below H.P. and 40 mm in front of V.P. (2) A point ‘B’ 35 mm above H.P. and 45 mm in front of V.P. (3) A point ‘C’ on V.P. and 30 mm above H.P.

1. Projection Rules for Points

• Front View (denoted with a prime, e.g., a', b', c'): Measured with respect to the Horizontal Plane (H.P.). If above H.P., it lies above the XY line; if below H.P., it lies below the XY line.
• Top View (denoted without a prime, e.g., a, b, c): Measured with respect to the Vertical Plane (V.P.). If in front of V.P., it lies below the XY line; if behind V.P., it lies above the XY line.

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2. Analysis of Each Point

(1) Point A: 40 mm below H.P. and 40 mm in front of V.P. (Fourth Quadrant)

• Front view (a'): 40 mm below the XY line.
• Top view (a): 40 mm below the XY line.
• Result: Both views a' and a lie on the same projector, 40 mm below the XY line (coincident).

(2) Point B: 35 mm above H.P. and 45 mm in front of V.P. (First Quadrant)

• Front view (b'): 35 mm above the XY line.
• Top view (b): 45 mm below the XY line.
• Result: b' is 35 mm above XY, b is 45 mm below XY on the same vertical projector.

(3) Point C: On V.P. and 30 mm above H.P.

• Front view (c'): 30 mm above the XY line.
• Top view (c): Exactly on the XY line (distance from V.P. = 0 mm).

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3. Projection Schematic

                    b' (35 mm)      c' (30 mm)
                    |               |
  X ----------------+---------------+-------c------- Y
                    |               |
                    |               |
               a', a| (40 mm)       |
                    |               |
                    b (45 mm)       

(schematic, not to scale)

Classify solids based on their geometric properties.

1. Definition

A solid is a three-dimensional object having length, breadth, and thickness. In engineering drawing, solids are bounded by surfaces that can be planar, single-curved, or double-curved.

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2. Classification of Solids

Solids are broadly classified into two main groups:

                            SOLIDS
                              |
         +--------------------+--------------------+
         |                                         |
    POLYHEDRA                             SOLIDS OF REVOLUTION
         |
  +------+------+-------+
  |             |       |
Regular      Prisms  Pyramids
Polyhedra

Group 1: Polyhedra

A polyhedron is a solid bounded entirely by plane surfaces called faces. The lines of intersection of faces are called edges.

1. Regular Polyhedra:
   Solids bounded entirely by identical, regular plane polygons.

• Tetrahedron: 4 equal equilateral triangular faces.
• Hexahedron (Cube): 6 equal square faces.
• Octahedron: 8 equal equilateral triangular faces.
• Dodecahedron: 12 equal regular pentagonal faces.
• Icosahedron: 20 equal equilateral triangular faces.

1. Prisms:
   A prism has two identical and parallel plane faces called bases or ends, joined by side faces which are parallelograms or rectangles.

• Examples: Triangular prism, Square prism, Pentagonal prism, Hexagonal prism.

1. Pyramids:
   A pyramid has a plane polygon as its base and a number of triangular lateral faces meeting at a common vertex called the apex.

• Examples: Triangular pyramid, Square pyramid, Hexagonal pyramid.

Group 2: Solids of Revolution

These solids are generated by revolving a plane 2D geometric figure about a fixed straight axis.

1. Cylinder: Formed by the revolution of a rectangle about one of its sides.
2. Cone: Formed by the revolution of a right-angled triangle about one of its perpendicular sides.
3. Sphere: Formed by the revolution of a semicircle about its diameter.

Other Variations

• Frustum: Formed when a pyramid or cone is cut by a cutting plane parallel to its base and the top portion containing the apex is removed.
• Truncated Solid: Formed when a solid is cut by an inclined cutting plane (not parallel to the base) and the upper part is removed.

A straight line AB has its end A 10 mm above H.P. and end B 50 mm in front of the V.P. Draw the projections of line AB, if it is inclined to H.P. by 30° and to V.P. by 45° and the true length of the line is 50 mm.

1. Given Data

• True Length (TL): 50 mm
• True inclination to H.P. (θ): 30°
• True inclination to V.P. (Φ): 45°
• Position of end A:
• Above H.P. = 10 mm (a' is 10 mm above XY)
• Position of end B:
• In front of V.P. = 50 mm (locus of b is 50 mm below XY)

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2. Analytical Check & Dimensions

• Apparent length in Top View (Plan Length):
  TL × cos(θ) = 50 × cos(30°) = 50 × 0.8660 = 43.30 mm
• Vertical rise (difference in elevation between A and B):
  h = TL × sin(θ) = 50 × sin(30°) = 25 mm
  Therefore, locus of b' is at height = 10 + 25 = 35 mm above XY.
• Apparent length in Front View (Elevation Length):
  TL × cos(Φ) = 50 × cos(45°) = 50 × 0.7071 = 35.36 mm
• Horizontal shift (difference in plan distance between A and B):
  d = TL × sin(Φ) = 50 × sin(45°) = 35.36 mm
  Since end B is 50 mm in front of V.P., end A must be:
  y_a = 50 - 35.36 = 14.64 mm in front of V.P. (top view 'a' is 14.64 mm below XY).

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3. Step-by-Step Drawing Procedure

1. Draw reference line XY.
2. Draw horizontal locus lines:

• Locus of a': 10 mm above XY.
• Locus of b': 35 mm above XY (10 + 50 × sin 30°).
• Locus of b: 50 mm below XY.
• Locus of a: 14.64 mm below XY (obtained geometrically as shown below).

1. Geometric Construction:

• Mark a' anywhere on its locus (10 mm above XY).
• From a', draw a line inclined at θ = 30° to XY with length 50 mm to locate b1' on the locus of b'.
• Project b1' down to the horizontal line passing through a to locate the true top view length line of 43.3 mm.
• From the locus of b (50 mm below XY), draw a line backwards at angle Φ = 45° of length 50 mm to meet the locus of a. This establishes the exact horizontal position of top view point a.
• Mark projector of point A (passing through a' and a perpendicular to XY).
• With a as centre and radius equal to plan length (projected from b1'), rotate an arc to cut the locus of b at point b.
• Join a-b; this is the final Top View.
• With a' as centre and radius equal to elevation length (projected from true-length line in plan), rotate to cut the locus of b' at b'.
• Join a'-b'; this is the final Front View.

1. Check that b' and b lie on the same vertical projector.

The projections of the ends of a line AB are 50 mm apart. The end A is 20 mm above H.P. and 30 mm in front of the V.P. The end B is 10 mm below H.P. and 40 mm behind V.P. Determine the true length of the line AB and its inclination with H.P.

1. Given Data

• Distance between end projectors (D): 50 mm
• End A:
• 20 mm above H.P. -> a' is 20 mm above XY
• 30 mm in front of V.P. -> a is 30 mm below XY
• End B:
• 10 mm below H.P. -> b' is 10 mm below XY
• 40 mm behind V.P. -> b is 40 mm above XY

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2. Mathematical Calculations

• Difference in vertical height (Δy'):
  a' is +20 mm, b' is -10 mm.
  Vertical difference = 20 - (-10) = 30 mm
• Front View (Elevation) Length (EL):
  EL = √[(D)² + (Δy')²] = √[(50)² + (30)²] = √[2500 + 900] = √3400 ≈ 58.31 mm

• Difference in horizontal distance (from V.P.) (Δy):
  a is -30 mm (in front), b is +40 mm (behind).
  Total difference = 40 - (-30) = 70 mm
• Top View (Plan) Length (PL):
  PL = √[(D)² + (Δy)²] = √[(50)² + (70)²] = √[2500 + 4900] = √7400 ≈ 86.02 mm

• True Length (TL):
  TL = √[(D)² + (Δy')² + (Δy)²] = √[2500 + 900 + 4900] = √8300 ≈ 91.10 mm

• True Inclination with H.P. (θ):
  sin(θ) = Δy' / TL = 30 / 91.10 = 0.3293
  θ = arcsin(0.3293) ≈ 19.23° (approximately 19°)

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3. Step-by-Step Drawing Procedure

1. Draw the reference line XY.
2. Draw two vertical projectors at a distance of 50 mm apart.
3. On the first projector, mark:

• a': 20 mm above XY.
• a: 30 mm below XY.

1. On the second projector, mark:

• b': 10 mm below XY.
• b: 40 mm above XY.

1. Join a'-b' (Front View) and a-b (Top View).
2. Find True Length and θ by Rotating Line Method:

• Through a, draw a line parallel to XY.
• With centre a and radius = top view length ab, rotate point b until the line is parallel to XY at b1.
• Project b1 vertically to the locus line of b' (10 mm below XY) to get point b1'.
• Join a'-b1'.
• Line a'-b1' represents the True Length (TL).

1. Measure the length of a'-b1' and angle made with horizontal:

• True Length (TL) = 91.1 mm
• Inclination with H.P. (θ) = 19.2°

Explain A.I.P. and A.V.P.

1. Auxiliary Inclined Plane (A.I.P.)

• Definition: An Auxiliary Inclined Plane is an imaginary reference plane that is inclined to the Horizontal Plane (H.P.) and perpendicular to the Vertical Plane (V.P.).
• Trace:
• Its intersection with V.P. is called the Auxiliary Vertical Trace (A.V.T.) or simply the cutting plane line in the front view, which appears inclined to the XY line.
• It has no horizontal trace parallel to XY.
• Application: Used to find the true shape of sections of objects cut by planes inclined to H.P., and to project auxiliary top views showing true dimensions.

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2. Auxiliary Vertical Plane (A.V.P.)

• Definition: An Auxiliary Vertical Plane is an imaginary reference plane that is inclined to the Vertical Plane (V.P.) and perpendicular to the Horizontal Plane (H.P.).
• Trace:
• Its intersection with H.P. is called the Auxiliary Horizontal Trace (A.H.T.), which appears as an inclined line in the top view.
• It is perpendicular to H.P., so its projection in the front view is distributed.
• Application: Used to obtain auxiliary front views of objects, especially when features are inclined to the vertical plane of projection.

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Summary Table

| Parameter | Auxiliary Inclined Plane (A.I.P.) | Auxiliary Vertical Plane (A.V.P.) | |---|---|---| | Relation to H.P. | Inclined at an angle θ | Perpendicular (90°) | | Relation to V.P. | Perpendicular (90°) | Inclined at an angle Φ | | View showing line trace | Front View (shows inclination to H.P.) | Top View (shows inclination to V.P.) |

A square prism side of base 35 mm and height 50 mm, is resting on its base in the V.P. and axis perpendicular to the V.P. and 30 mm above the H.P. One of the edges of the base is inclined to 30° to the H.P. Draw the projection of the square prism.

1. Given Data

• Solid: Square prism
• Base side: 35 mm
• Height (length of axis): 50 mm
• Resting condition: Resting on its base in the V.P.
• Axis: Perpendicular to V.P. and 30 mm above H.P.
• Base orientation: One base edge is inclined at 30° to H.P.

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2. Visualization of Views

• Since the base rests in the V.P. and axis is perpendicular to V.P.:
• The Front View shows the true shape of the base, which is a square of side 35 mm.
• The Top View shows a rectangle of size 50 mm (length) × apparent width, with one face resting on XY.

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3. Step-by-Step Drawing Procedure

1. Draw the reference line XY.
2. Front View (Elevation):

• Draw a horizontal centre line 30 mm above the XY line.
• At this height, draw a square of side 35 mm such that one edge is inclined at 30° to the horizontal (H.P.).
• Label the corners of the front base as a'(1'), b'(2'), c'(3'), d'(4').

1. Top View (Plan):

• Since the base is in the V.P., the base line in the top view lies directly on the XY line.
• Project vertical lines downwards from all corners a', b', c', d' to the XY line.
• From the XY line, project perpendicular lines downwards of length 50 mm (height of prism) into the top view.
• Complete the rectangle of length 50 mm.
• Identify visible and hidden lines: edges facing the observer from the top are drawn as continuous dark lines, while edges hidden beneath are drawn as dashed lines.

A rhombus ABCD has its diagonal AC = 80 mm and BD = 50 mm. The side AD of the plane is in H.P. and the plane makes an angle of 30° with the H.P. Obtain the projections of the plane when the side CD is in V.P.

1. Given Data

• Surface shape: Rhombus ABCD
• Diagonals: AC = 80 mm, BD = 50 mm (diagonals bisect each other at right angles)
• Stage 1 (Simple position): Rhombus lying in H.P. to show true shape in Top View.
• Stage 2 (Inclination with H.P.): Plane inclined at 30° to H.P., with side AD remaining in H.P.
• Stage 3 (Inclination with V.P.): Side CD is placed in the V.P.

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2. Step-by-Step Construction Procedure

Stage 1: True Shape in H.P.

1. Draw reference line XY.
2. In the top view, construct the true shape of rhombus ABCD:

• Draw diagonal AC = 80 mm.
• Draw perpendicular bisector BD = 50 mm (25 mm on each side of AC).
• Connect A-B-C-D to form the rhombus.

1. Project points A, B, C, D upwards to the XY line to obtain the front view as a single straight line a'-d'-b'-c' lying on the XY line.

Stage 2: Plane Inclined at 30° to H.P.

1. Keep point a'(d') on the XY line.
2. Tilt the front view line at an angle of 30° to the XY line.
3. Transfer the distances of all points along this line so that the length equals the front view length from Stage 1.
4. Project points vertically down from this inclined front view and horizontally across from the Stage 1 top view.
5. The intersections give the new top view points a1, b1, c1, d1. Connect them to form the apparent top view (compressed rhombus).

Stage 3: Side CD in the V.P.

1. Measure the length of side c1-d1 in the Stage 2 top view.
2. Redraw the top view of Stage 2 such that the edge c1-d1 lies directly on the XY line (since it is in V.P.).
3. Project points vertically upwards from this new top view to meet the corresponding horizontal projectors drawn from the front view of Stage 2.
4. Mark the intersection points a2', b2', c2', d2' and join them with straight lines to get the final Front View.

What is the apparent shape and the true shape of the section in case of the section of solid?

1. Apparent Shape of Section

• Definition: The apparent shape of a section is the projection of the cut surface onto one of the principal planes of projection (Horizontal Plane or Vertical Plane).
• Characteristics:
• It represents how the cut surface looks when viewed perpendicular to the standard reference planes (i.e., directly from the top or front).
• When the cutting plane is inclined to the plane of projection, the dimensions in the direction of inclination are foreshortened.
• Therefore, the apparent shape does not reflect the real proportions or actual angles of the cut cross-section.

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2. True Shape of Section

• Definition: The true shape of a section is the actual, undistorted geometric shape of the cut surface obtained when viewed perpendicular (normal) to the cutting plane.
• Characteristics:
• It reveals the exact dimensions, true side lengths, and actual angles of the cross-section.
• It is obtained by projecting the cut points onto an Auxiliary Plane parallel to the cutting plane.

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Summary Comparison

| Feature | Apparent Shape | True Shape | |---|---|---| | Viewing Direction | Perpendicular to principal plane (H.P. or V.P.) | Perpendicular to the cutting plane | | Dimensions | Foreshortened / distorted | Exact true dimensions | | Projection Plane | Principal reference plane | Auxiliary plane parallel to cutting plane |

OBA is a simple slider crank chain. OB is a crank of 30 mm length. BA is a connecting rod of 90 mm length. Slider A is sliding on a straight path passing through point O. Draw the locus of the mid-point of the connecting rod AB for one complete revolution of the crank OB.

1. Given Data

• Crank length (OB): 30 mm
• Connecting rod length (BA): 90 mm
• Mid-point of connecting rod (M): Distance BM = MA = 45 mm
• Path of slider A: Horizontal straight line passing through fixed point O.

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2. Principle

As crank OB rotates 360° around fixed point O, slider A reciprocates back and forth along the horizontal line passing through O. The mid-point M of the connecting rod AB traces an ellipse (or an egg-shaped closed oval closely approximating an ellipse).

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3. Step-by-Step Construction Procedure

1. Mark the fixed crank centre O and draw a horizontal line through O representing the slider guide line.
2. With centre O and radius OB = 30 mm, draw a circle representing the crank path.
3. Divide the circle into 12 equal angular parts at intervals of 30°. Label the crank pin positions as B0, B1, B2, ..., B11.
4. Locate Slider Positions (A) and Mid-points (M):

• For position B0: Set compass to length of connecting rod BA = 90 mm. With centre at B0, strike an arc cutting the horizontal line through O to locate slider position A0.
• On line segment B0-A0, mark the mid-point M0 at a distance of 45 mm from B0.
• For position B1: With centre B1 and radius 90 mm, cut the horizontal line at A1. Mark mid-point M1 at 45 mm from B1.
• Repeat this process for all 12 positions: B2 -> A2 -> M2, ..., B11 -> A11 -> M11.

1. Draw the Curve:

• Join points M0, M1, M2, ..., M11, M0 smoothly using a French curve.
• The resulting closed curve is the required locus of mid-point M, which forms a smooth oval/ellipse.

Draw the projections of a circle, of 70 mm diameter, resting on the H.P. on a point ‘A’ of the circumference. Plane is inclined to the H.P. such that the plan of it is an ellipse of minor axis 40 mm. The plan of the diameter, through point ‘A’, is making an angle of 45° with the V.P. Measure the angle of the plane with the H.P.

1. Given Data

• Diameter of circle (D): 70 mm
• Resting point: Point 'A' on H.P.
• Top View (Plan) Shape: Ellipse with:
• Major axis = True diameter = 70 mm
• Minor axis = Apparent diameter = 40 mm
• Inclination of the plan diameter through 'A' with V.P. (β): 45°

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2. Calculation of Angle with H.P. (θ)

• The diameter parallel to H.P. remains unchanged in top view (Major axis = 70 mm).
• The diameter along the maximum slope is foreshortened in the top view to become the minor axis:
  Minor Axis = D × cos(θ)
• Substitution:
  40 = 70 × cos(θ)
  cos(θ) = 40 / 70 = 4 / 7 ≈ 0.5714
  θ = arccos(0.5714) ≈ 55.15° (approximately 55°)

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3. Step-by-Step Drawing Procedure

Stage 1: Circle Resting flat in H.P.

1. Draw reference line XY.
2. In the top view, draw a circle of diameter 70 mm.
3. Divide the circle into 8 (or 12) equal parts and mark points a, b, c, d, e, f, g, h. (Let diameter ae be parallel to the direction of tilt).
4. Project all points vertically to the XY line to obtain the front view as a single horizontal line of length 70 mm: a'-b'-c'-d'-e' on the XY line.

Stage 2: Plane Inclined to H.P.

1. At point a' on the XY line, tilt the front view line at angle θ = 55° until its length is 70 mm.
2. Project points vertically down from this inclined line and horizontally across from the Stage 1 top view.
3. The intersections yield an ellipse with major axis = 70 mm and minor axis = 40 mm.

Stage 3: Diameter through A Inclined at 45° to V.P.

1. Draw a line inclined at 45° to the XY line in the top view area.
2. Replicate the ellipse from Stage 2 such that its diameter line through point 'a' (the minor axis line) lies along this 45° line.
3. Project points vertically upwards to intersect horizontal projection lines from the front view of Stage 2.
4. Join the points with a smooth curve to get the final Front View.

• Measured Angle of plane with H.P.: θ ≈ 55°

Define ellipse, parabola and hyperbola based on eccentricity ratio.

1. General Definition of a Conic Section

A conic section is defined as the locus of a point moving in a plane such that the ratio of its distance from a fixed point (called the focus) to its perpendicular distance from a fixed straight line (called the directrix) is always constant. This constant ratio is known as the eccentricity (e):

Eccentricity (e) = (Distance of point from Focus / Perpendicular distance of point from Directrix)

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2. Definitions Based on Eccentricity Ratio

1. Ellipse (e < 1):

• An ellipse is the conic section formed when the eccentricity is strictly less than 1 (e < 1).
• The moving point is always closer to the focus than to the directrix.
• It forms a closed smooth oval curve.

1. Parabola (e = 1):

• A parabola is the conic section formed when the eccentricity is equal to 1 (e = 1).
• The moving point is always equidistant from the focus and the directrix.
• It forms an open, unbounded curve with a single branch.

1. Hyperbola (e > 1):

• A hyperbola is the conic section formed when the eccentricity is strictly greater than 1 (e > 1).
• The moving point is always farther from the focus than from the directrix.
• It forms an open curve having two symmetrical, separate branches.

Explain different types of lines with their illustrations and applications.

Types of Lines in Engineering Drawing (BIS / ISO Standards)

In technical drawings, standard line types and thicknesses are specified by Bureau of Indian Standards (BIS: SP 46) to convey specific information:

1. Continuous Thick Line (Type A)

• Description: Solid, dark, and thick line (approx. 0.5 mm).
• Illustration: ━━━━━━━━━━━━━━━━━━━━
• Applications:
• Visible outlines of the object.
• Visible edges and crests of threads.

2. Continuous Thin Line (Type B)

• Description: Solid, sharp, and thin line (approx. 0.25 mm).
• Illustration: ────────────────────
• Applications:
• Dimension lines and extension lines.
• Projection lines and leader lines.
• Hatching / section lines.
• Construction lines.

3. Continuous Thin Freehand Line (Type C)

• Description: Wavy, continuous thin line drawn freehand.
• Illustration: ~~~~~\/\/\/\/~~~~~~~
• Applications:
• Limits of partial or interrupted views and sections (short break lines).

4. Dashed Thin Line (Type E / F)

• Description: Series of short dashes of equal length (~3 mm) with small uniform gaps (~1 mm).
• Illustration: - - - - - - - - - -
• Applications:
• Hidden outlines and hidden edges of the object.

5. Long-Dashed Dotted Thin Line / Chain Thin (Type G)

• Description: Alternate long dash (~10–12 mm) and dot/short dash separated by small gaps.
• Illustration: ─── · ─── · ─── · ──
• Applications:
• Centre lines of circular features.
• Lines of symmetry.
• Pitch circles of gears and hole patterns.

6. Chain Thin with Thick Ends (Type H)

• Description: Thin chain line made thick at the ends and at changes of direction.
• Illustration: ━━─ · ─── · ─── · ━━
• Applications:
• Cutting plane lines indicating the sectioning path.

A hexagonal pyramid base 30 mm and height 60 mm is resting, on H.P. on its base with two sides of base perpendicular to V.P. It is cut by an Auxiliary Inclined Plane (A.I.P.), inclined to H.P. by 45°, passing to a point 25 mm from the apex on the axis. Draw the front view, sectional top view and true shape of the section.

1. Given Data

• Solid: Regular hexagonal pyramid
• Base side: 30 mm
• Axis height: 60 mm
• Base condition: Resting on H.P., two base sides perpendicular to V.P.
• Cutting Plane: Auxiliary Inclined Plane (A.I.P.) inclined at 45° to H.P.
• Cutting plane location: Passes through the axis at a distance of 25 mm from the apex (i.e., 35 mm above base).

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2. Step-by-Step Drawing Procedure

Step 1: Draw Projections of the Uncut Solid

1. Top View (Initial):

• Draw a regular hexagon of side 30 mm with two opposite sides perpendicular to XY.
• Mark base corners as a, b, c, d, e, f and centre apex as o.
• Join all corners to apex o (representing the slant edges).

1. Front View (Initial):

• Project base corners to the XY line to form the base line.
• Draw the vertical axis of height 60 mm to locate the apex o'.
• Connect o' to the base points to complete the triangular elevation.

Step 2: Introduce the Cutting Plane

1. On the axis in the front view, mark a point K at a distance of 25 mm below the apex o'.
2. Through point K, draw a cutting plane line V-T inclined at 45° to the XY line.
3. Identify the cutting points where the line V-T cuts the slant edges:

• Point 1' on slant edge o'-a'
• Point 2' on slant edge o'-b'
• Point 3' on slant edge o'-c'
• Point 4' on slant edge o'-d'
• Point 5' on slant edge o'-e'
• Point 6' on slant edge o'-f'

Step 3: Draw the Sectional Top View

1. Project the cutting points 1', 2', 3', 4', 5', 6' vertically downwards to the corresponding slant edges in the top view (oa, ob, oc, od, oe, of).
   (Note: For edges nearly vertical or overlapping, transfer points via horizontal projection to the true length generator and then rotate down).
2. Join points 1, 2, 3, 4, 5, 6 with straight lines to form the closed cut boundary.
3. Hatch the interior of this cut area with thin parallel lines at 45°.
4. Retain the remaining bottom portion of the pyramid with firm dark lines.

Step 4: Draw the True Shape of the Section

1. Draw a reference line X1-Y1 parallel to the cutting plane line V-T.
2. From cutting points 1', 2', 3', 4', 5', 6', draw projection lines perpendicular to V-T (and X1-Y1).
3. Measure the perpendicular distances of points 1, 2, 3, 4, 5, 6 from the XY line in the top view.
4. Transfer these distances onto the corresponding perpendicular projectors from X1-Y1.
5. Join the plotted points in sequence to obtain the True Shape of the Section and hatch the region at 45°.

Explain first angle and third angle projection system with their symbols.

1. Comparison of First Angle and Third Angle Projection Systems

| Parameter | First Angle Projection | Third Angle Projection | |---|---|---| | Location of Object | Placed in the 1st Quadrant (between observer and plane) | Placed in the 3rd Quadrant (plane is between observer and object) | | Nature of Planes | Assumed to be opaque | Assumed to be transparent | | Top View Position | Placed below the Front View | Placed above the Front View | | Left Side View Position | Drawn on the right side of the Front View | Drawn on the left side of the Front View | | Right Side View Position| Drawn on the left side of the Front View | Drawn on the right side of the Front View | | Standard Adoption | Widely used in India (BIS) and Europe (ISO) | Widely used in USA, Canada, and Japan (ASME/ANSI) |

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2. Standard Projection Symbols

Both symbols are based on the front view and side view of a frustum of a right circular cone.

(a) First Angle Projection Symbol

• In First Angle, the side view is projected on the opposite side. When looking from the left at the smaller end, the two concentric circles appear on the right side of the trapezoid.

          +---------+
         /           \            ( O )   (Concentric circles
        /             \                     on the right)
       +---------------+        
       <-- Trapezoid -->

(b) Third Angle Projection Symbol

• In Third Angle, the side view remains on the viewing side. The two concentric circles appear on the left side of the trapezoid.

                                  +---------+
            ( O )                /           \
     (Concentric circles        /             \
        on the left)           +---------------+
                               <-- Trapezoid -->

(schematic, not to scale)

What are the advantages and disadvantages of using AUTOCAD?

1. Advantages of AutoCAD

1. High Accuracy and Precision: Drawings are created to exact mathematical dimensions with negligible tolerances, eliminating manual drafting inaccuracies.
2. Speed and Time Savings: Standard shapes, symbols, and entire assemblies can be quickly generated, copied, patterned, and scaled.
3. Easy Editing and Revision: Modifications can be made instantly using commands like Erase, Trim, Move, and Stretch without redrawing from scratch.
4. Layer Management: Different drawing elements (outlines, dimensions, centre lines, hatching) can be assigned to different layers, colours, and line weights for clear organization.
5. Standard Symbol Libraries (Blocks): Frequently used components (fasteners, electrical symbols, valves) can be saved as blocks and inserted repeatedly.
6. Safe Storage and Portability: Files are saved digitally, reducing physical storage space and allowing instant transfer via email or cloud systems.

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2. Disadvantages of AutoCAD

1. High Initial Cost: Professional software licenses and compatible high-performance computer hardware require significant investment.
2. Steep Learning Curve: Mastering advanced drafting, shortcut keys, 3D modelling, and parametric tools requires dedicated training.
3. Risk of Data Loss: Digital drawing files may be lost, corrupted, or infected by malware if regular backups are not maintained.
4. System Dependency: Operates exclusively on computer systems; power outages or hardware malfunctions can interrupt work immediately.
5. Limited Sculpting/Freeform Capabilities: Unlike specialized organic modelling software, purely parametric/vector CAD tools are less suited for complex artistic sculpting.

The development of a cone is a semicircle of 80 mm radius having a circular hole of 80 mm diameter. Draw the plan and elevation of the cone along with periphery of a circular hole shown on them.

1. Given Data

• Development of Cone: Semicircle of radius R_slant = 80 mm
• Slant height of cone (L): 80 mm
• Subtended sector angle (θ): 180° (semicircle)
• Hole inside development: Circular hole of diameter 80 mm (radius = 40 mm)

---

2. Dimensions of the Cone

• Sector angle formula for development of a cone:
  θ = (r / L) × 360°
• Substituting given values:
  180° = (r / 80) × 360° implies r = (180° / 360°) × 80 = 40 mm
• Base diameter of cone (2r): 80 mm
• Vertical height of cone (H):
  H = √(L^2 - r^2) = √(80^2 - 40^2) = √(6400 - 1600) = √(4800) ≈ 69.28 mm

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3. Step-by-Step Drawing Procedure

Step 1: Draw the Development and the Hole

1. Draw a semicircle of radius L = 80 mm with centre O. This represents the lateral development of the cone.
2. Divide the semicircle into 12 equal sectors of 15° each (generators 1 to 12).
3. Draw the given 80 mm diameter circular hole (radius 40 mm) at its specified position within this development sheet.
4. The circular boundary of the hole intersects the various radial generators at points marked as p1, p2, p3, etc.
5. With centre O, swing arcs through each intersection point to the extreme outer generator line (true-length generator) to obtain their true radial distances from apex O.

Step 2: Draw the Elevation (Front View)

1. Draw the reference line XY.
2. Draw the front view of the uncut cone: an isosceles triangle with base = 80 mm and vertical axis height H = 69.28 mm.
3. Transfer the true radial distances of hole points from apex O onto the outer slant contour line of the front view.
4. From these points on the outer slant edge, project horizontal lines to intersect the corresponding generators in the front view.
5. Connect these plotted points with a smooth curve using a French curve to show the periphery of the hole in the elevation.

Step 3: Draw the Plan (Top View)

1. Below XY, draw a circle of diameter 80 mm representing the top view of the cone base.
2. Divide the circle into the same 12 radial generators corresponding to those in the development.
3. Project the points from the front view vertically downwards to intersect their respective radial generator lines in the top view.
4. Join these points with a smooth curve to show the periphery of the hole in the plan.

Differentiate between aligned system and unidirectional system of dimensioning using sketches.

1. Comparison of Dimensioning Systems

| Parameter | Aligned System | Unidirectional System | |---|---|---| | Orientation of Text | Dimension text is aligned parallel to the dimension line. | Dimension text is always placed horizontally regardless of line direction. | | Reading Direction | Read from the bottom or from the right-hand edge of the drawing sheet. | Read strictly from the bottom of the drawing sheet only. | | Placement of Text | Placed above and clear of unbroken dimension lines. | Dimension line is usually broken near the centre to insert the horizontal text. | | Applications | Traditional engineering drawings, civil and structural drawings. | Large engineering drawings, automotive and aerospace drafting. |

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2. Explanatory Sketches

(a) Aligned System

• Horizontal dimensions are read from the bottom.
• Vertical and inclined dimensions are read from the right-hand side.

             50
      |<------------>|
      +--------------+
      |              |   ^
      |              |   | 30
      |              |   v
      +--------------+

(b) Unidirectional System

• All dimensions are oriented horizontally and read exclusively from the bottom edge.

             50
      |<------------>|
      +--------------+
      |              |   ^
      |              |   | 30
      |              |   v
      +--------------+

(schematic, not to scale)

Draw isometric circle on the three sides of cube of 60 mm dimension using four centre method.

1. Concept of Isometric Circle (Four-Centre Method)

In isometric projection, a circle appears as an ellipse. The Four-Centre Method is an approximate standard geometric technique where the ellipse is drawn using four circular arcs struck from four specific centres inside an isometric square (rhombus).

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2. Step-by-Step Construction Procedure for a Cube of 60 mm

Step 1: Draw the Isometric View of the Cube

1. From an origin point, draw three isometric axes:

• One vertical line.
• Two receding lines at 30° to the horizontal on either side.

1. Measure 60 mm along each axis.
2. Complete the isometric cube of 60 mm × 60 mm × 60 mm to form three visible faces: Top face, Left-hand vertical face, and Right-hand vertical face.

Step 2: Draw the Four-Centre Ellipse on the Top Face (Horizontal Plane)

1. The top face is a rhombus ABCD.
2. Locate the mid-points of all four sides: 1, 2, 3, 4.
3. Identify the two obtuse angle corners (corners with angle = 120°).
4. From obtuse corner A, draw lines to the opposite side mid-points (e.g., A-2 and A-3).
5. From the opposite obtuse corner C, draw lines to the remaining mid-points (C-1 and C-4).
6. These lines intersect each other at two points, say c1 and c2.
7. The four centres are: A, C, c1, and c2.
8. Draw the Four Arcs:

• With centre A and radius A-2, draw an arc connecting mid-points 2 and 3.
• With centre C and radius C-1, draw an arc connecting mid-points 1 and 4.
• With centre c1 and radius c1-1, draw a small arc connecting mid-points 1 and 2.
• With centre c2 and radius c2-3, draw a small arc connecting mid-points 3 and 4.
• Together, these four smooth arcs form the isometric circle on the top face.

Step 3: Repeat for Left and Right Vertical Faces

1. On the Left Vertical Face, repeat the identical four-centre construction using its two obtuse corners and side mid-points.
2. On the Right Vertical Face, construct the four centres and draw the four blending arcs.
3. The result is three seamlessly inscribed isometric circles on all three visible faces of the 60 mm cube.

Fig. 1 shows pictorial view of an object. Draw following views: (1) Front view looking in the direction X, (2) Left Hand Side View. Use first angle projection method.

1. Principle of First Angle Orthographic Projection

In First Angle Projection:

• Front View (FV): Drawn looking in the direction of arrow X, positioned above the reference line XY.
• Left Hand Side View (LHSV): Placed to the right of the Front View.
• Visible outlines and intersections are drawn as thick continuous lines.
• Internal features and hidden edges are drawn with dashed lines.

---

2. General Step-by-Step Projection Procedure

1. Determine Overall Dimensions:

• Total Length (L) of the object along direction X.
• Total Width / Depth (W) perpendicular to direction X.
• Total Height (H) from the base to the highest point.

1. Layout and Positioning:

• Draw horizontal reference line XY.
• Draw vertical reference line X1-Y1 to separate the Front View from the Side View.
• Set up bounding boxes:
• Front View box: Length (L) × Height (H) above XY.
• Left Hand Side View box: Width (W) × Height (H) to the right of X1-Y1.

1. Front View (Direction X):

• Draw the base outline and step/profile transitions visible directly from direction X.
• Show any cut-outs, slots, or holes. If visible, draw with solid lines; if hidden behind a solid face, represent with dashed lines.

1. Left Hand Side View (LHSV):

• Project heights horizontally from the Front View to ensure perfect alignment.
• Transfer width dimensions using a 45° mitre line or direct measurement.
• Draw visible stepped contours and represent internal passages/recesses with hidden dashed lines.

1. Finishing:

• Add centre lines for all circular features and holes.
• Apply overall dimensions using the aligned or unidirectional system.
• Indicate "All dimensions are in mm" and "First Angle Projection".

Draw isometric scale showing 100 mm normal scale.

1. Theory & Principle of Isometric Scale

• In isometric projection, all dimensions along the isometric axes are foreshortened due to the tilt of the object.
• Ratio of Isometric Length to True Length:
  (Isometric Length / True Length) = (cos 45° / cos 30°) = frac{1/√(2)}{√(3)/2} = √((2 / 3)) ≈ 0.8165 (approx. 82%)

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2. Step-by-Step Construction Procedure (Up to 100 mm)

1. Draw a horizontal reference line from origin O.
2. Draw a line OA inclined at 30° to the horizontal. This line represents the Isometric Scale.
3. Draw another line OB inclined at 45° to the horizontal. This line represents the True (Normal) Scale.
4. On the 45° line (True Scale), mark true lengths from 0 to 100 mm using a standard ruler:

• Mark 0, 10, 20, 30, ..., 100 mm.

1. From each marked division point on the 45° line, draw vertical projection lines (perpendicular to the horizontal baseline) downwards to intersect the 30° line (OA).
2. The points of intersection on the 30° line provide the corresponding isometric lengths (e.g., true 100 mm produces an isometric length of 81.65 mm).
3. Subdivide the first division (0 to 10 mm) into 1 mm units on the true scale and project down to obtain single-millimetre isometric divisions.

                     B (True Scale, 45°)
                    /  |  |  |  |  | (100 mm true)
                   /   |  |  |  |  |
   Isometric Scale/    |  |  |  |  |
      (30°)      /     |  |  |  |  |
                A------+--+--+--+--+ (81.65 mm iso)
               /       |  |  |  |  |
              O---------------------- Horizontal Base

(schematic, not to scale)

Explain the following AUTOCAD commands: (i) Trim, (ii) Arc, (iii) Mirror, (iv) Array

(i) TRIM Command

• Purpose: Trims (cuts away) unwanted portions of objects that intersect with other cutting edges.
• Shortcut: TR
• Working Method:

1. Type TRIM or TR and press Enter.
2. Select the cutting edges (or press Enter for default "Select All").
3. Click on the segments of lines, arcs, or circles that need to be removed.

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(ii) ARC Command

• Purpose: Creates a circular arc curve without drawing a full circle.
• Shortcut: A
• Working Method:

1. Type ARC or A and press Enter.
2. Specify points based on chosen method. Standard default is the 3-Point method (Start point, Second point on arc, End point).
3. Other options include Start-Center-End, Start-Center-Angle, and Start-End-Radius.

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(iii) MIRROR Command

• Purpose: Creates a reverse (mirror-image) duplicate of selected objects across a specified mirror line.
• Shortcut: MI
• Working Method:

1. Type MIRROR or MI and press Enter.
2. Select the objects to mirror.
3. Specify the first and second points of the mirror line.
4. Choose whether to erase the source object (Y/N, default is No).

---

(iv) ARRAY Command

• Purpose: Creates copies of selected objects arranged in a systematic regular pattern.
• Shortcut: AR
• Types of Array:

1. Rectangular Array: Distributes copies into a specified number of rows and columns.
2. Polar Array: Distributes copies circularly around a specified central point over a given angle (e.g., bolt patterns on a flange).
3. Path Array: Evenly spaces copies along a selected curve or polyline.

Fig. 2 shows F.V. and T.V. of an object. Draw isometric view.

1. Concept of Isometric View

An isometric view is a pictorial 3D representation where all three principal axes are drawn at 120° to each other (two axes at 30° to the horizontal and one vertical axis). All dimensions along the isometric axes are drawn to full scale (true scale).

---

2. Step-by-Step Procedure to Construct Isometric View from F.V. & T.V.

1. Analyze Orthographic Views:

• Obtain the maximum overall dimensions from the given views:
• Length (L) from the Front View.
• Width / Depth (W) from the Top View.
• Height (H) from the Front View.

1. Construct the Isometric Enclosing Box:

• Draw a horizontal reference line and establish an origin point.
• Draw three isometric axes: one vertical, and two at 30° to the horizontal line.
• Measure Length along the length axis, Width along the depth axis, and Height along the vertical axis.
• Complete the isometric block (enclosing box) using faint construction lines.

1. Map the Visible Features:

• Start with prominent primary surfaces (e.g., base plate, vertical uprights, or webs).
• Transfer distances of steps, cuts, and slots directly from the F.V. and T.V. onto the corresponding isometric faces.

1. Carve out Cut-outs and Slopes:

• Locate the terminal points of inclined/slanted surfaces on the isometric box edges and join them with straight non-isometric lines.
• If circular holes or cylinders exist, construct them using the Four-Centre Method.

1. Finishing and Line Weight:

• Darken all final visible edges with firm, uniform thick lines.
• Erase or dim the construction lines.
• Omit hidden lines in isometric views unless strictly required for clarity.