Engineering Graphics (engineering Drawing) By P.h.jain

Please Refer Sample pages of the book.

Engineering Graphics Projections of Points and Straight Lines 6. Draw the projections of following points on the same reference line, keeping the projectors 30 mm apart. a) Point A, 40 mm above the H.P. and 25 mm in front of the V.P. b) Point B, in the H.P. and 20 mm behind the V.P. c) Point C, in the V.P. and 40 mm above the H.P. d) Point D, 25 mm below the H.P. and 25 mm behind the V.P. e) Point E, 15 mm above the H.P. and 40 mm behind the V.P. f) Point F, 40 mm below the H.P. and 25 mm in front of the V.P. g) Point G , in the H.P. and V.P. Also mention the quadrant in which they lie. Solution : Draw the projections of each point 30 mm apart as shown in fig. 1.9.

a′′

e

c′′ d

b

e′′ g′′ g

b′′ c

a

f

d′′

f′

A is in 1st Quad.

B is in 2nd & 3rd Quad.

C is in 1st & 2nd Quad.

D is in 3rd Quad.

E is in 2nd Quad.

F is in 4th Quad.

G is in All Four Quadrants

Fig. 1.9 7. A point 'P' is 40 mm from both the reference planes. Draw its projections in all possible positions. Solution : As there are four quadrants, there are four possible positions of point P. (i.e point P in 1st, 2nd, 3rd and 4th quadrant). Refer fig. 1.10. If Point P is in 1st quadrant; then p′′ will be 40 mm above x-y line and p will be 40 mm below x-y line. If Point P is in 2nd quadrant; then p′′ and p will be 40 mm above x-y line. If Point P is in 3rd quadrant; then p′′ will be 40 mm below x-y line and p will be 40 mm above x-y line. If Point P is in 4th quadrant; then p′′ and p will be 40 mm below x-y line.

p′′

p′′ p

p

p Point P in 1st Quad.

Point P in 2nd Quad.

Fig. 1.10

p′′

p′′ p

Point P in 3rd Quad.

Point P in 4th Quad.

1.9

1.14

Projections of Points and Straight Lines

Engineering Graphics

Problems on lines inclined (∠ ∠) to one reference plane and || to other : Important Notes : 1) When line is || to V.P., its F.V shows T.L. of line and true inclination of line with H.P. (θ θ) and its T.V is || to x-y line. 2) When line is || to H.P., its T.V shows T.L. of line and true inclination of line with V.P. (φ φ) and its F.V. is || to x-y line. 1. A line AB, 50 mm long has its end A 15 mm above H.P. and 20 mm in front of V.P. Draw the projections of line AB, when it is 30O ∠ to H.P. and || to V.P. Point B is in 1st quadrant. Solution : Refer fig. 1.27 (a) and (b).

Fig. 1.27 (a) Pictorial View of Line AB

Fig. 1.27 (b) Orthographic Projections of Line AB

Procedure : 1. Since A is 15 mm above H.P.; mark a′′ 15 mm above x-y line. Since A is 20 mm in front of V.P.; mark a 20 mm below x-y line. (a′′ is F.V. and a is T.V. of point A). 2. Since line AB of 50 mm length, is 30O ∠ to H.P.; through a′′, draw line a′′b′′ of 50 mm length and 30O ∠ to x-y line. [Since B is in 1st quadrant, b′′ should be above x-y line]. Through point b′′, draw projector of B vertically downward. 3. Since line AB is || to V.P.; through a, draw a line || to x-y line up to projector of B, and mark b. Join a-b. Line a′′b′′ and line ab are the required F.V. and T.V. of line AB respectively.

2. The F.V. of a line CD, 75 mm long, measures 55 mm. The line is || to H.P. and ∠ to V.P. Its end C is 20 mm above H.P. and in V.P. Draw its projections and determine its inclination with V.P. Point D is in 1st quadrant. Solution : Refer fig. 1.28 (a) and (b).

c′

d′

c d

d Fig. 1.28 (a) Pictorial View of Line CD

Fig. 1.28 (b) Orthographic Projections of Line CD

1. Since C is 20 mm above H.P.; mark c′′ 20 mm above x-y line. Since C is in V.P.; mark c on x-y line. (c′′ is F.V. and c is T.V. of point C). 2. Since F.V. of CD measures 55 mm and line is || to H.P.; through c′′, draw c′′d′′ 55 mm long and || to x-y line. Through d′′, draw projector of D. We know that when line is || to H.P., then its T.V shows T.L. of line. 3. Therefore, with center c and radius = T.L. = 75 mm, cut the projector of D at d. Join c-d and measure φ. [Since D is in 1st quadrant, d should be below x-y line]. Line c′′d′′ and line cd are the required F.V. and T.V. of line CD respectively.

Engineering Graphics

Projections of Points and Straight Lines

1.20

Problems on lines inclined (∠ ∠) to H.P. and V.P. : 1.

A line AB, 75 mm long is inclined at 45O with H.P. and 30O with V.P. Draw the projections of the line AB, when A is 20 mm above H.P. and 30 mm infront of V.P. and the end B is in second quadrant. Also find its H.T. and V.T.

Solution : Refer fig. 1.40.

Given data

Interpretation

AB = 75 mm θ = 45O φ = 30O A 20 mm above H.P., 30 mm infront of V.P. B is in 2nd quadrant

T.L. = a′b1′ = ab2 = 75 mm a′b1′ is 45O ∠ to x-y line ab2 is 30O ∠ to x-y line a′ 20 mm above x-y line, a 30 mm below x-y line. b′ and b are above x-y line

b′′

b1

Locus of b′

V.T.

a′′

b2 b2 Locus of b b′′ b

b h

v

b1

a

H.T Procedure :

Fig. 1.40

1. Since A is 20 mm above H.P.; mark its front view a′′, 20 mm above x-y line. 2. Since A is 30 mm in front of V.P.; mark its top view a, 30 mm below x-y line. Remember, since B is in 2nd quadrant; b′′ and b should be above x-y line. 3. Since line AB is 45O ∠ to H.P., draw a′′b1′ = T.L.= 75 mm, 45O ∠ to x-y line. Draw locus of b′ through b1′. 4. Since line AB is 30O ∠ to V.P., draw ab2 = T.L.= 75 mm, 30O ∠ to x-y line. Draw locus of b through b2. 5. Project b1′ on horizontal line through a and mark b1. Assuming point A fixed, with center a and radius = ab1, draw an arc to cut the locus of b at b. Join a-b, which is actual T.V. of the line AB. 6. Project b2 on horizontal line through a′′ and mark b2′. Assuming point A fixed, with center a′′ and radius = a′′b2′, draw an arc to cut the locus of b′ at b′′. Join a′′-b′′, which is actual F.V. of the line AB. Check : Points b′′ and b lies on the same projector. Traces : Mark h, where extension of a′′b′′ cuts x-y line. Mark H.T., where projector through h cuts the extension of ab. Mark v, where ab cuts x-y line. Mark V.T., where projector through v cuts a′′b′′. Note : Points a′′-b′′-h-V.T. lies on a straight line and Points a-b-v-H.T. lies on straight line.

Engineering Graphics

Projections of Points and Straight Lines

1.28

7. The line AB is in third quadrant. It is inclined at 30O to F.R.P. (i.e. V.P.). The end A is in H.P. It has its F.V. length of 70 mm and T.V. length of 60 mm. The H.T. is 25 mm behind V.P. Draw the projections and find : i) True length of the line; ii) Inclination of line with H.P.; iii) Locate V.T. Solution : Refer fig. 1.46.

Given data

Interpretation

F.V. is 40O ∠ with x-y line T.V. is 55O ∠ with x-y line

Line a′b′ is 40O ∠ with x-y line Line ab is 55O ∠ with x-y line b′ is above x-y line and b is below x-y line a′b1′ = ab2 = 75 mm

B is in 1st quadrant AB = 75 mm

b′′

b1′ Locus of b′

c′′ H.T .

h

a′′ V.T . v

c2′ b′′

b

a

c

c2

Locus of c

Ans : θ = 25.7O φ = 47.5O

Locus of b

b

b2

Fig. 1.46 Procedure : 1. Since A is in H.P.; mark a′′ and h as a common point on x-y line. 2. Since H.T. is 25 mm behind V.P.; mark a and H.T. as a common point 25 mm above a′′, h. 3. Since F.V. length is 70 mm; through a′′, draw a′′b2′ 70 mm long on x-y line. Draw projector of B2 through b2′. 4. Since T.V. length is 60 mm; through a, draw ab1 60 mm long || to x-y line. Draw projector of B1 through b1. 5. Since line is 30O ∠ to V.P.; through a, draw a line 30O ∠ to x-y line to cut projector of B2 at b2. (Since line is in 3rd quadrant, b and b2 should be above x-y line). Measure T.L. Draw locus of b through b2. 6. With center a and radius = ab1, draw an arc to cut locus of b at b. Join a-b, which is the T.V. of the line. Draw projector of B through b. 7. With center a′′ and radius = a′′b2, draw an arc to cut the projector of B at b′′. Join a′′-b′′, which is the F.V. of the line. (Since line is in 3rd quadrant, b′′ should be below x-y line). Draw locus of b′ through b′′. 8. Mark b1′, where locus of b′ cuts the projector of B1. Join a′′-b1′. Measure θ. 9. Mark v, where ab extended cuts x-y line. Mark V.T., where projector through v cuts a′′b extended. Check : T.L. obtained a′′-b1′ = a-b2.

Engineering Graphics

Projections of Points and Straight Lines

1.48

27. The F.V. of a straight line ‘AB’ makes an angle of 30O with ‘X-Y’. The H.T. of the line is 45 mm behind V.P., while the V.T. is 30 mm above H.P. End A is 10 mm below H.P. and end B is in first quadrant. The line is 150 mm long. Draw projections of the line and find the true length of the portion of the line which is in second quadrant. Also determine the inclination of the line with H.P. and V.P. Solution : Refer fig. 1.66.

Given data

Interpretation

O

a′b′ is 30 O ∠ to x-y. H.T. is 45 mm above x-y and h V.T. is 30 mm above x-y and v T.L. = a′-b1′ = a-b2 = 150 mm

F.V. is 30 ∠ to x-y. H.T. is 45 mm behind V.P. V.T. is 30 mm above H.P. Line is 150 mm long

h1

a

b′ Locus of H.T.

Locus of b′

b1′

HT2 HT

Locus of V.T.

VT1′

VT

h h1′

v

v2

a′ Ans : T.L. in 2nd Quad. = h1′-VT1′ = v2-HT2 = 75 mm θ = 23.6O φ = 36.9O

Locus of b

b

b2

Fig. 1.66 Procedure : 1. Since A is 10 mm below H.P.; mark a′′, 10 mm below x-y line. 2. Since H.T. is 45 mm behind V.P.; draw locus of H.T., 45 mm above x-y line. 3. Since V.T. is 30 mm above H.P.; draw locus of V.T., 30 mm above x-y line. 4. Since F.V. of line is 30O ∠ to x-y; through a′′, draw a line 30O ∠ to x-y line [Initially assume some length]. (Remember that since B is in first quadrant; b′′ should be above x-y line). 5. Mark h, where F.V. of line cuts x-y line. Project h on locus of H.T. and mark HT. 6. Extend a′′-h upto locus of V.T. and mark VT. Project VT on x-y line and mark v. 7. Join v-HT and extend it upto projector of A and mark a. (Now, a-v is the line of T.V.) To determine θ and φ : 8. Consider a point on a-v (say HT). Now, rotate a-HT about a and make it || to x-y (a-h1). Project h1 on locus of h and mark h1′. 9. Join a′′-h1′ and extend it upto 150 mm and mark b1′, such that a′′-b1′ = T.L. = 150 mm. Measure θ. Through b1′, draw locus of b′. 10. Extend a′′-h-VT upto locus of b′ and mark b′′. Then a′′-b′′ is F.V. of line. Through b′′, draw projector of B. Extend a-HT-v upto projector of B and mark b. Then a-b is T.V. of line. Through b, draw locus of b. 11. With center a and radius = T.L. = 150 mm, cut the locus of b at b2. Join a-b2. Measure φ. 12. Measure T.L. of line in 2nd quadrant = h1′-VT1′ or v2-HT2 .

Engineering Graphics

Projections of Points and Straight Lines O

1.58

O

37. The plan of a straight line AB measures 80 mm. The line is inclined at 35 to V.P. and 45 to H.P., while the end B is 80 mm infront of V.P. Draw the projections of the line, when its V.T. is 15 mm above H.P. and the line is completely in first quadrant. Find the true length of line. Solution : Refer fig. 1.76.

Given data

Interpretation

Plan of AB is 80 mm long Line is 45O ∠ to H.P. Line is 35O ∠ to V.P. B is 80 mm infront of V.P. V.T. is 15 mm above H.P.

b-a1 = 80 mm θ = 45O φ = 35O b is 80 mm below x-y line V.T. is 15 mm above x-y and v. Projector of A

b′ a1′ Locus of a0′ a0′ Projector of B Projector of A1

a′ b0′

V.T.

v Locus of a

a2

Ans : T.L. = 113.1 mm

a

a1

b

Fig. 1.76 Procedure : 1. Since B is 80 mm infront of V.P.; mark b 80 mm below x-y line. Through b, draw projector of B. 2. Since plan of line measures 80 mm; through b, draw b-a1 80 mm long and || to x-y line. Through a1, draw projector of A1. 3. Mark b0′ (b0′ is the dummy of b′′ to be located) any where on projector of B. Since line is 45O ∠ to H.P.; through b0′, draw a line 45O ∠ to x-y line to cut projector of A1 at a1′. Measure T.L. = b0′-a1′ (= 113.1 mm). Through a1′, draw locus of a0′. (a0′ is the dummy of a′′ to be located). 4. Since line is 35O ∠ to V.P.; through b, draw b-a2, 35O ∠ to x-y line and 113.1 mm long. Through a2, draw locus of a. 5. With center b and radius = b-a1, draw an arc to cut the locus of a at a. Join a-b, which is T.V. of line. Through a, draw projector of A. 6. Mark a0′, where projector of A cuts locus of a0′. Join a0′- b0′, which is dummy F.V. of the line. Measure angle α (= 59.7O). 7. Extend a-b up to x-y line and mark v. Since V.T. is 15 mm above H.P.; mark V.T. 15 mm above v. 8. We know that points a′′-b′′-V.T. are co-linear, therefore, through V.T. draw a line αO (= 59.7O) ∠ to x-y line to cut the projectors of A and B at a′′ and b′′. Then a′′-b′′ is actual F.V. of the line.

Engineering Graphics

Projections of Points and Straight Lines

1.76

55. A line AB, 75 mm long, makes an angle of 30O with the V.P. and has its V.T. 15 mm above the H.P. The end A is 40 mm above H.P. and 25 mm in front of V.P. Find the angle made by the line with the H.P., if line is in first quadrant. Solution : Refer fig. 1.94. Given data

Interpretation

Line AB is 75 mm long

T.L. = a′-b1′ = a-b2 = 75 mm

O

φ = 30O

Line is 30 ∠ to V.P. V.T. is 15 mm above H.P.

V.T. is 15 mm above x-y.

A is 40 mm above H.P.

a′ is 40 mm above x-y line

B is 25 mm infront of V.P.

b is 25 mm below x-y line

b′′

v 2′

b1 ′

Locus of b′

b2 ′

a′′

Locus of V.T.

V.T. v v2

a Ans : θ = 28.5O Locus of b

b

b2

Fig. 1.94 Procedure : 1.

Since A is 40 mm above H.P.; mark a′′, 40 mm above x-y line.

2.

Since A is 25 mm infront of V.P.; mark a, 25 mm below x-y line.

3.

Since V.T. is 15 mm above H.P.; draw locus of V.T., 15 mm above x-y line.

4.

Since line AB is 75 mm long and makes 30O with V.P.; through a, draw a-b2 75 mm long and 30O ∠ to x-y line. [Since line is in first quadrant, b and b2 should be below x-y line]. Through b2, draw locus of b.

5.

Project b2 on horizontal line through a′′ and mark b2′. With center a′′ and radius a′′-b2′ draw an arc.

6.

Extend a-b2 upto x-y line and mark v2. Project v2 on horizontal line through a′′ and mark v2′. With center a′′ and radius a′′-v2′ draw an arc to cut the locus of V.T. at V.T.

7.

Join V.T. - a′′ and extend it upto the arc drawn with radius a′′-b2′ and mark b′′. Then a′′-b′′ is F.V. of line. Through b′′, draw locus of b′.

8.

With center a′′ and radius = T.L. = 75 mm, cut the locus of b′ at b1′. Join a′′-b1′ and measure θ.

9.

Project V.T. on x-y line and mark v. Join v–a and extend it upto locus of b and mark b. Then a-b is T.V. of line.

Check : Points b′′ and b lies on same projector.