Jawaharlal Nehru Technological University, Hyderabad
Jawaharlal Nehru Technological University, Hyderabad B.Tech I ENGINEERING GRAPHICS Previous Question Paper (August - 2018 - R16)
Answer all five questions. All questions carry equal marks.
Main
Show by means of drawing that when the diameter of the directing circle is twice that of the generating circle, the hypocycloid is a straight line. Take the diameter of the generating circle is equal to 50 mm.
Answer
When the diameter of the generating circle is exactly half the diameter of the directing circle, the hypocycloid generated by a point on the circumference of the generating circle is a straight line coinciding with the diameter of the directing circle. Given the generating circle diameter (d) = 50 mm, the directing circle diameter (D) must be 100 mm.
- Draw the directing circle with radius R = 50 mm and mark its center as O.
- Draw the generating circle of radius r = 25 mm such that it is tangent to the directing circle at a point P.
- Divide the generating circle into 12 equal parts and label them 1, 2, ..., 12, starting from the point of contact P.
- As the generating circle rolls inside the directing circle, the center of the generating circle moves along an arc of radius (R - r) = 25 mm. Divide this path into 12 equal divisions.
- For each position of the center, draw the corresponding position of the generating circle.
- Locate the path of point P by marking the intersection of the arcs drawn from the center positions with the radii of the generating circles.
- It will be observed that all points P1, P2, ..., P12 lie on the diameter of the directing circle passing through the initial point P. Thus, the locus of the point is a straight line.
Draw a vernier scale of R.F.= 4 to read 1/5 cm and 1/25 cm and to measure up to 5 cm. Mark on the scale distance of 3.36 cm.
Answer
To construct the vernier scale, we first determine the necessary parameters:
- Representative Fraction (R.F.): 4
- Maximum Length (L): 5 cm
- Length of Scale (L.O.S.): R.F. × Maximum Length = 4 × 5 cm = 20 cm
- Least Count (L.C.): 1/25 cm = 0.04 cm
The main scale is divided into 5 equal divisions, each representing 1 cm. Each 1 cm division is further subdivided into 5 parts, where each part equals 1/5 cm (0.2 cm). To find the vernier scale length, we use the relation: 1 vernier division (V.D.) = (n-1)/n × (smallest division of main scale), where n is the number of divisions on the vernier scale. Here, 5 V.D. = 4 main scale divisions (each 0.2 cm). Thus, 1 V.D. = (4/5) × 0.2 cm = 0.16 cm. The difference between 1 main scale division (0.2 cm) and 1 V.D. (0.16 cm) is 0.04 cm, which is the required Least Count.
To mark 3.36 cm:
- Identify the main scale reading: 3.2 cm (which is 3 cm + 1 division of 0.2 cm).
- The remaining distance is 3.36 - 3.2 = 0.16 cm.
- Since 0.16 cm corresponds to exactly 1 vernier division, align the 1st mark of the vernier scale with the 3.2 cm mark on the main scale.
- The distance from the zero of the main scale to the 1st vernier mark represents 3.36 cm.
A regular Hexagon of 40 mm side has a corner in the H.P. Its surface is inclined at 45° to the H.P. and the top view of the diagonal through the corner which is in the H.P. makes an angle of 60° with the V.P. Draw its projections.
Answer
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The projection drawn from H.T and V.T of a straight line AB are 80mm apart while those drawn from its ends are 50mm apart. The H.T is 35mm in front of the V.P, the V.T is 55mm above the H.P. and the end A is 10mm above the H.P. Draw the projections of AB and determine its true length and inclinations with the reference planes.
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A square pyramid, base 40mm side and axis 90mm long, has a triangular face on the ground and the vertical plane containing the axis makes an angle of 45° with the VP. Draw its projections.
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A Tetrahedron of 75 mm long edges has one edge parallel to the H.P. and inclined at an angle of 45° to the V.P. While a face containing that edge is vertical. Draw its projections.
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A Cube of 65 mm long edges has its vertical faces equally inclined to the V.P. It is cut by a section plane, perpendicular to the V.P., so that true shape of the section is a regular Hexagon. Determine the inclination of the cutting plane with the H.P. and draw the sectional top view and true shape of the section.
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A hexagonal pyramid, base 50mm side and axis 100mm long, is lying on the H.P on one of its triangular faces with the axis parallel to the V.P. A vertical sectional plane the H.T of which makes an angle of 30° with the reference line passes through the center of the base and cuts the pyramid, the apex being retained. Draw the top view, sectional front view, true shape of the section.
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Draw the isometric view for the given figure 1. (All dimensions are in mm).
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Draw the front view, side view from the right, and top view of the block as shown in figure 2. (All dimensions are in mm).
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