BE Civil Engineering (IOE, TU) Engineering Drawing I (IOE, ME 401) Question Paper 2078 Nepal

Given data

• True length $TL = 80\text{ mm}$
• End $A$: 15 mm above HP, 20 mm in front of VP
• $\theta$ (inclination to HP) $= 30^\circ$, $\phi$ (inclination to VP) $= 45^\circ$

Step 1 — Length of the Top View (TV)

The top view is the projection on the HP. Its length is governed by the true inclination to the HP:

$$TV = TL \cos\theta = 80\cos 30^\circ = 80 \times 0.8660 = 69.28\text{ mm}$$

Step 2 — Length of the Front View (FV)

The front view is the projection on the VP. Its length is governed by the true inclination to the VP:

$$FV = TL \cos\phi = 80\cos 45^\circ = 80 \times 0.7071 = 56.57\text{ mm}$$

Step 3 — Vertical projector difference (for the second-stage rotation)

Difference in heights of the ends above HP (vertical distance between $a'$ and $b'$):

$$\Delta h = TL \sin\theta = 80\sin 30^\circ = 80 \times 0.5 = 40\text{ mm}$$

Difference in distances in front of VP (vertical distance between $a$ and $b$ in TV):

$$\Delta d = TL \sin\phi = 80\sin 45^\circ = 80 \times 0.7071 = 56.57\text{ mm}$$

Step 4 — Apparent angles (drafting construction)

The standard two-stage method:

1. Draw $a$ 20 mm below $xy$ and $a'$ 15 mm above $xy$ on the same projector.
2. Stage I (inclination to HP): draw $a'b_1'$ at $30^\circ$ to $xy$ with $a'b_1' = 80$. Project $b_1'$ down; mark $ab_1 = 69.28$ along $xy$ for the TV at this stage.
3. Stage I (inclination to VP): draw $ab_2$ at $45^\circ$ to $xy$ with $ab_2 = 80$. Project up; mark $a'b_2' = 56.57$ along the level for the FV at this stage.
4. Rotate: with locus lines (FV locus 40 mm above $a'$ level; TV locus 56.57 mm below $a$ level) intersect the swung arcs of radii $TV = 69.28$ (from $a$) and $FV = 56.57$ (from $a'$) to fix the final $b$ and $b'$.

The apparent angle of the FV, $\alpha$, satisfies $\tan\alpha = \dfrac{\Delta h}{FV} = \dfrac{40}{56.57} = 0.7071 \Rightarrow \alpha = 35.3^\circ$.

The apparent angle of the TV, $\beta$, satisfies $\tan\beta = \dfrac{\Delta d}{TV} = \dfrac{56.57}{69.28} = 0.8165 \Rightarrow \beta = 39.2^\circ$.

Step 5 — Traces

Produce the FV ($a'b'$) to meet $xy$ at $v'$; the projector through $v'$ meets the produced TV at the HT. Produce the TV ($ab$) to meet $xy$ at $h$; the projector through $h$ meets the produced FV at the VT.

Using the formulae for distances of traces from end $A$ along $xy$:

• Horizontal trace distance behind $a$ along $xy$: governed by where the front view crosses $xy$.
• Since $A$ is 15 mm above HP and the FV rises at apparent angle $35.3^\circ$, the HT lies on the side of $A$ at horizontal distance $\dfrac{15}{\tan 35.3^\circ} = \dfrac{15}{0.7071} = 21.2\text{ mm}$ from $a'$ measured along $xy$.
• Since $A$ is 20 mm in front of VP and the TV is inclined at apparent angle $39.2^\circ$, the VT lies at horizontal distance $\dfrac{20}{\tan 39.2^\circ} = \dfrac{20}{0.8165} = 24.5\text{ mm}$ from $a$ along $xy$.

Results (bold):

• Front view length FV $= 56.57$ mm
• Top view length TV $= 69.28$ mm
• Apparent angle of FV $\approx 35.3^\circ$; apparent angle of TV $\approx 39.2^\circ$
• HT lies $\approx 21.2$ mm from $a'$ along $xy$; VT lies $\approx 24.5$ mm from $a$ along $xy$

        b'
        /|
  FV   / |  Δh = 40
      /  |
 a'--+---+----------- (FV locus)
  x__|___|___________________ xy
 a---+---+
      \  |
  TV   \ |  Δd = 56.57
        \|
         b   (TV locus)

Given data

• Regular hexagonal prism, base side $s = 25\text{ mm}$, axis length $L = 70\text{ mm}$.
• Across-corners width of hexagon $= 2s = 50\text{ mm}$.
• Across-flats width of hexagon $= s\sqrt{3} = 25 \times 1.7320 = 43.30\text{ mm}$.

Stage 1 — Prism resting on a rectangular face, axis parallel to HP and VP

Because the axis is parallel to both planes, all three principal views show true shapes/lengths of the relevant features.

• Top view (TV): A rectangle of length 70 mm (axis seen as true length) and width 50 mm (across-corners), because the resting face puts the across-corners diagonal horizontal. Two internal lines show the longitudinal edges of the faces.
• Front view (FV): Because the prism rests on a face, the hexagonal end is seen edge-on partly; the FV is a rectangle 70 mm long and height equal to the across-flats distance $= 43.30\text{ mm}$.
• Side view (SV): A true hexagon of side 25 mm (the base), with one flat horizontal (resting face) — height 43.30 mm across flats, width 50 mm across corners.

 FV (70 x 43.30)            SV: true hexagon (s=25)
 +----------------+            __
 |   |    |   |   |           /  \
 +----------------+          |    |
         |                    \__/
        xy ------------------------
 TV (70 x 50)
 +----------------+
 |________________|
 +----------------+

Stage 2 — Axis turned to $40^\circ$ with VP, kept parallel to HP

When the axis stays parallel to HP but is inclined to VP, the work is done in the top view because that is where the inclination to VP appears.

1. Reproduce the Stage-1 top view (a 70 mm × 50 mm rectangle with internal edges).
2. Rotate the entire top view about a corner so the long axis makes $40^\circ$ with $xy$.
3. Project every corner of the rotated TV upward.
4. Heights of all points are unchanged from Stage 1 (axis parallel to HP), so carry the same heights (within the 43.30 mm band) from the Stage-1 FV across to the new projectors.
5. The new front view is obtained where the height lines meet the new vertical projectors — it becomes a foreshortened parallelogram-bounded figure; the apparent length of the axis in the FV is $70\cos 40^\circ = 70 \times 0.7660 = 53.62\text{ mm}$.

Key dimensions (bold):

• Across-flats $= 43.30$ mm; across-corners $= 50$ mm
• FV height (Stage 1) $= 43.30$ mm, length $= 70$ mm
• Apparent axis length in FV after tilt $= 53.62$ mm

The top view in Stage 2 retains the true 70 mm axis length (parallel to HP) but lies at $40^\circ$ to $xy$.

Given data

• Base diameter $D = 60\text{ mm}$ (radius $R = 30\text{ mm}$), axis $H = 75\text{ mm}$.
• Section plane $\perp$ VP, inclined $45^\circ$ to HP, cutting the axis 30 mm above base.

Step 1 — Cone geometry / semi-apex angle

The generator makes angle $\alpha$ with the axis where

$$\tan\alpha = \frac{R}{H} = \frac{30}{75} = 0.40 \Rightarrow \alpha = 21.80^\circ.$$

The generator makes angle with the base $= 90^\circ - \alpha = 68.20^\circ$.

Step 2 — Type of conic

The cutting plane is at $45^\circ$ to the base (HP). Compare with the base-angle of the generators, $68.20^\circ$.

• If the cutting-plane angle ($45^\circ$) is less than the generator's base-angle ($68.20^\circ$) and the plane is not parallel to a generator, the curve is an ellipse — the plane cuts all generators.

Since $45^\circ < 68.20^\circ$, the section is an ellipse (a complete closed curve passing through the cone all round).

Step 3 — Front view construction

1. Draw the FV triangle: base 60 mm on $xy$, apex 75 mm above centre.
2. Mark the cutting line as a straight line in FV inclined at $45^\circ$ to $xy$, passing through the axis point 30 mm above base. Label where it crosses the two outline generators ($1'$ and $7'$, the major-axis ends in FV).
3. Draw the base circle as the TV (diameter 60 mm) and divide into 12 equal parts (generators $o1, o2,\dots o12$).

Step 4 — Section points by the generator (cutting-plane) method

For each generator, the cutting line in FV crosses it at a point; project that point down to the corresponding generator in TV to get the sectional TV point. Use horizontal sections to read the radius at each cut height.

The extreme points along the line of intersection in FV define the major axis of the ellipse. Its true length equals the FV chord length of the cut line between the two outline generators. By geometry the cut enters near the base on one side and exits 30 mm up on the axis side; measuring the FV chord gives the major axis.

Major axis (true length) $= \dfrac{\text{FV cut chord}}{\cos 45^\circ}$ when laid out true. Taking the FV chord of the cut $\approx 53\text{ mm}$ projected, the true major axis $\approx 53\text{ mm}$ along the inclined line itself.

Step 5 — True shape of the section

1. Draw a reference line parallel to the cutting line in the FV.
2. Project every section point perpendicular from the cut line onto this reference; the perpendicular offset of each point equals its half-width taken from the sectional TV (the horizontal distance of that generator point from the axis in TV).
3. Join the points with a smooth curve — the result is an ellipse.

• Major axis = length of cut line in FV between extreme generators.
• Minor axis = width of the cone at the mid-height of the cut, measured in the TV (the chord of the section at the centre of the cut). At the mid-height of the inclined cut the cone radius gives the half-minor-axis; doubling gives the minor axis.

Results (bold):

• Semi-apex angle $\alpha = 21.80^\circ$; generator base-angle $= 68.20^\circ$.
• Conic produced: ELLIPSE (because the cutting-plane angle $45^\circ <$ generator base-angle $68.20^\circ$).
• True shape obtained by transferring section-point offsets onto a line parallel to the cut.

 FV with 45° cut          True shape (ellipse)
     /\                       ____
    /  \  <-45° cut          /    \
   /----\                   (      )
  /______\                   \____/
 base 60       major axis = FV chord; minor = TV chord at mid-cut

Given data

• Block: length $l = 60\text{ mm}$, width $w = 40\text{ mm}$, height $h = 30\text{ mm}$.
• Central square through-hole: side 20 mm, depth 30 mm.

Part (a) — Isometric view (isometric drawing)

In an isometric drawing (view), true lengths are laid along the three isometric axes (two at $30^\circ$ to the horizontal, one vertical).

Construction steps:

1. Draw the isometric axes: vertical line, and two lines at $30^\circ$ to horizontal on each side.
2. Lay the box: 60 mm along the right $30^\circ$ axis, 40 mm along the left $30^\circ$ axis, 30 mm vertical. Complete the isometric box.
3. On the top $60\times40$ face, locate the square hole centrally: offsets $(60-20)/2 = 20\text{ mm}$ from the long edges and $(40-20)/2 = 10\text{ mm}$ from the short edges. Draw the 20 mm square in isometric.
4. Project the hole 30 mm down (vertical) through the block; show the inner vertical edges and the bottom opening where visible.
5. Darken visible edges; the back inside edges of the hole are partly hidden.

        ____________
       /  ___      /|
      /  |   |    / |
     /   |___|   /  |   30
    /___________/   /
    |    ___    |  /
    |   |   |   | /  40 (left axis)
    |___|___|___|/
       60 (right axis)

Part (b) — Isometric projection dimensions (isometric scale)

In a true isometric projection, lengths are foreshortened by the isometric scale:

$$\text{Isometric scale ratio} = \frac{\text{isometric length}}{\text{true length}} = \sqrt{\frac{2}{3}} = 0.8165 \approx 0.816.$$

(Equivalently $\cos 35.264^\circ = 0.8165$.)

Applying the factor 0.8165:

Results (bold):

• Isometric scale ratio $= 0.8165$ (i.e. $\sqrt{2/3}$).
• Isometric-projection sizes: $48.99 \times 32.66 \times 24.50$ mm; hole side $16.33$ mm.
• The Part (a) isometric view uses the full true lengths $60 \times 40 \times 30$ with 20 mm hole.

Note the distinction: an isometric view/drawing uses true lengths; an isometric projection uses the foreshortened (0.816) lengths.

Given data (assembled bracket)

• Base plate: $80 \times 50 \times 12$ mm.
• Vertical plate: width 50, height 60, thickness 12, rising from the rear long edge of the base.
• Triangular rib: thickness 10, legs 40 (along base) and 40 (up vertical plate), centred.

Projection convention: First-angle (object between observer and plane). The top view is placed below the front view; the left side view is placed on the right of the front view.

Step 1 — Front view (looking along the 50 mm depth, seeing the 80 mm length)

• Outline: base plate seen as a rectangle 80 wide × 12 high.
• Vertical plate appears as a rectangle 50 wide × 60 tall sitting on the base, centred (since base 80, plate 50 → side margins $(80-50)/2 = 15$ mm each).
• The rib is hidden behind the vertical plate in this direction except its top sloping line where it projects forward; show as the inclined edge from base-top to plate at 40 mm height (hidden/visible per geometry).
• Total height $= 12 + 60 = 72\text{ mm}$.

Step 2 — Top view (placed below FV)

• Base plate true rectangle $80 \times 50$.
• Vertical plate seen as a 50-wide × 12-thick rectangle along the rear edge.
• Rib seen as a 10-wide strip, length 40 from the rear, centred on the 50 width → offsets $(50-10)/2 = 20$ mm... (centred on width of plate: rib centred → 20 mm from each side of the 50 width, leaving 10 wide rib).

Step 3 — Left side view (placed to the right of FV in first angle)

This is the most informative view for the L-profile:

• Base: 50 deep × 12 high rectangle.
• Vertical plate: 12 thick × 60 tall rising from the rear.
• Rib: triangle with horizontal leg 40 (along base, from the vertical plate forward) and vertical leg 40 (up the plate), hypotenuse joining them. The rib sits on top of the 12 mm base, so its horizontal leg starts at height 12 mm.
• Overall height $= 72$ mm; overall depth $= 50$ mm.

 FRONT VIEW (80 wide, 72 tall)        L.SIDE VIEW (50 deep)
   +------ 50 -----+                    | |\
   |              |                     | | \  rib (40x40)
   |  vert plate  | 60                  | |  \
   |              |                     | |   \___
  _|______________|_                   | |_______|
  |    base 80      | 12               |____________| base 12
  +----------------+                    <--- 50 --->

 TOP VIEW (80 x 50, below FV)
  +----------------------------+
  |==== vert plate (12 thick)==|  rear edge
  |        +--+ rib 10 wide     |
  |        |  | x40 long        |
  +----------------------------+

Principal dimensions to mark (bold):

• Base 80 × 50 × 12; vertical plate 50 × 60 × 12; rib 40 × 40 × 10 thick.
• Overall height 72 mm; vertical plate centred with 15 mm margins each side in FV.
• First-angle symbol to be drawn in the title block.

(a) Single-stroke vertical Gothic lettering

• The standard stroke thickness is about $\dfrac{1}{10}$ of the letter height (i.e. ratio of height to stroke width $\approx 10:1$), sometimes taken as $1/8$ for bolder text.
• For most capital letters the width is about $\dfrac{6}{10}$ to $\dfrac{7}{10}$ of the height (e.g. ratio height:width $\approx 10:6$).
• Two standard nominal lettering heights (per ISO/BIS series): 3.5 mm and 5 mm (the series being 2.5, 3.5, 5, 7, 10, 14, 20 mm).

(b) Two instruments and their uses

(Other acceptable answers: T-square — drawing horizontal lines and a guide for set squares; divider — transferring/stepping off equal distances; protractor — measuring angles.)

Key results (bold): Stroke = height/10; widths follow the 10:6 ratio; standard heights 3.5 mm and 5 mm.

Given: Regular pentagon, side $a = 35\text{ mm}$.

Step 1 — Interior angle

For a regular $n$-gon, interior angle $= \dfrac{(n-2)\times 180^\circ}{n}$. For $n=5$:

$$\frac{(5-2)\times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ.$$

Step 2 — Circumradius $R$

$$R = \frac{a}{2\sin(180^\circ/n)} = \frac{35}{2\sin 36^\circ} = \frac{35}{2 \times 0.5878} = \frac{35}{1.1756} = 29.77\text{ mm}.$$

Step 3 — General construction method (side AB given)

1. Draw line $AB = 35\text{ mm}$.
2. At $A$ and $B$, erect perpendiculars; with centre the midpoint $M$ of $AB$ and radius $MA$, draw a semicircle... (alternatively the standard 5-division method below).
3. Standard method: Bisect $AB$ at $M$. Draw $BP \perp AB$ at $B$ with $BP = AB$. Join $M$ to $P$. With centre $M$, radius $MP$, draw an arc to cut $AB$ produced at $Q$. Then $AQ$ is the diagonal length. With centre $A$ and radius $AB$, and centre $B$ and radius $AQ$, intersect to locate the opposite vertex region; step the side 35 mm around to get the five vertices.
4. More simply, draw the circumscribing circle of radius $R = 29.77\text{ mm}$ and step the chord $35\text{ mm}$ five times around it, OR mark central angles of $72^\circ$ ($=360^\circ/5$).
5. Join consecutive points to complete the regular pentagon.

Results (bold):

• Interior angle $= 108^\circ$.
• Central angle $= 72^\circ$.
• Circumradius $R = 29.77$ mm.

Recall the four-quadrant convention

Point $P$: 25 mm above HP, 35 mm behind VP

• Above HP + behind VP ⇒ Second quadrant (II).
• Front view $p'$: above $xy$ by the height above HP ⇒ 25 mm above $xy$.
• Top view $p$: behind VP, so on rotation of HP it falls above $xy$ as well, 35 mm above $xy$ (in the second quadrant both views lie above $xy$, with $p$ farther typically). So $p'$ is 25 mm above $xy$ and $p$ is 35 mm above $xy$.

Point $Q$: 30 mm below HP, 20 mm in front of VP

• Below HP + in front of VP ⇒ Fourth quadrant (IV).
• Front view $q'$: below HP ⇒ $q'$ lies 30 mm below $xy$.
• Top view $q$: in front of VP ⇒ after rotating HP downward, $q$ lies below $xy$ as well, 20 mm below $xy$ (in the fourth quadrant both views lie below $xy$).

Results (bold):

• $P$ is in the 2nd quadrant: $p'$ 25 mm above $xy$, $p$ 35 mm above $xy$.
• $Q$ is in the 4th quadrant: $q'$ 30 mm below $xy$, $q$ 20 mm below $xy$.

            p (35 above)
            p' (25 above)
  __________________________ xy
            q (20 below)
            q' (30 below)

Given: Circular lamina, diameter $d = 60\text{ mm}$; inclined at $\theta = 50^\circ$ to HP; one diameter parallel to VP.

Step 1 — Top view becomes an ellipse

When a circle is inclined to the HP, its top view is an ellipse:

• Major axis = the diameter that stays parallel to HP = true diameter = 60 mm (the diameter perpendicular to the direction of tilt does not foreshorten in the top view).
• Minor axis = projection of the diameter along the tilt direction = $d\cos\theta$.

$$\text{Minor axis} = 60\cos 50^\circ = 60 \times 0.6428 = 38.57\text{ mm}.$$

Step 2 — Two-stage procedure

Stage 1 (lamina parallel to HP):

1. Draw the true-shape top view as a circle of 60 mm diameter.
2. Divide it into 12 equal parts (points 1–12).
3. Project up to get the front view as a straight horizontal line (edge view) of length 60 mm on $xy$.

Stage 2 (tilt to 50°): 4. Tilt the edge-view (front view line) to $50^\circ$ with $xy$, keeping the resting point on $xy$. 5. Project each of the 12 points up from the Stage-1 top view and across from the tilted front view. 6. The intersections trace the ellipse: major axis 60 mm (horizontal), minor axis 38.57 mm (along tilt).

Results (bold):

• Major axis of top view $= 60$ mm.
• Minor axis of top view $= 38.57$ mm.

 Stage1 TV (circle 60)   Stage2 FV at 50°        Stage2 TV (ellipse)
     ___                    /                        ____
   (     )                 /  50°                   (      )  major 60
    \___/        ---->     /_________ xy             (______)  minor 38.57

Full section vs Half section

Four conventions for section (hatching) lines

1. Section lines are thin continuous lines drawn at $45^\circ$ to the principal outline (or to the axis), unless $45^\circ$ coincides with an outline — then use $30^\circ$ or $60^\circ$.
2. Hatching lines are uniformly spaced (typically 1–3 mm apart depending on the size of the hatched area).
3. Adjacent parts in an assembly are hatched in opposite directions or with different spacing to distinguish them.
4. The same part is hatched in the same direction and spacing across all sectioned views; large areas may be hatched only along the boundary.
5. (Additional) Certain features are never sectioned/hatched even when the plane passes through them — ribs, webs, shafts, bolts, nuts, keys, rivets (shown in full).

Key results (bold): Full section = one half removed (whole view sectioned); Half section = one quarter removed (half sectioned, half external). Hatching: thin lines at 45°, uniform spacing, opposite directions for adjacent parts, consistent for the same part.

Given: Pentagonal pyramid, base side $a = 30\text{ mm}$, axis $h = 65\text{ mm}$; rests on one base edge with base at $30^\circ$ to HP; axis parallel to VP.

Step 1 — Base geometry (circumradius and apothem)

Circumradius of base (centre to corner):

$$R = \frac{a}{2\sin 36^\circ} = \frac{30}{2 \times 0.5878} = \frac{30}{1.1756} = 25.52\text{ mm}.$$

Apothem (centre to mid-side), the inradius:

$$r = \frac{a}{2\tan 36^\circ} = \frac{30}{2 \times 0.7265} = \frac{30}{1.4531} = 20.65\text{ mm}.$$

Step 2 — Slant edge (apex to base corner)

$$\text{Slant edge} = \sqrt{h^2 + R^2} = \sqrt{65^2 + 25.52^2} = \sqrt{4225 + 651.3} = \sqrt{4876.3} = 69.83\text{ mm}.$$

Step 3 — Slant height (apex to mid-point of base side)

$$\text{Slant height} = \sqrt{h^2 + r^2} = \sqrt{65^2 + 20.65^2} = \sqrt{4225 + 426.4} = \sqrt{4651.4} = 68.20\text{ mm}.$$

Step 4 — Two-stage projection method

Stage 1 (base on HP, axis vertical):

1. Draw the true-shape top view: a regular pentagon of side 30 mm (one edge placed front-parallel as required for the later tilt).
2. Mark the centre and join to corners; project the apex.
3. Draw the front view: base as a line on $xy$, apex 65 mm above the base centre; join apex to the base corners.

Stage 2 (tilt base to 30° on one edge): 4. Redraw (reproduce) the Stage-1 front view tilted so the base line makes $30^\circ$ with $xy$, with the resting edge point on $xy$. 5. Project all corners and the apex of the tilted front view downward. 6. Project the corresponding points horizontally from the Stage-1 top view; intersections give the new top view. 7. Join to complete the projections. Since the axis is parallel to VP, no third rotation is needed.

Results (bold):

• Circumradius $R = 25.52$ mm; apothem $r = 20.65$ mm.
• Slant edge $= 69.83$ mm.
• Slant height (to mid-side) $= 68.20$ mm.