BE Civil Engineering (IOE, TU) Engineering Drawing II (IOE, CE 451) Question Paper 2076 Nepal

Given data

• Base radius $r = 30\text{ mm}$, axis height $h = 80\text{ mm}$.

Step 1 — True length of the slant generator (R)

The slant generator is the true radius used for development:

$$R = \sqrt{r^2 + h^2} = \sqrt{30^2 + 80^2} = \sqrt{900 + 6400} = \sqrt{7300} \approx 85.44\text{ mm}.$$

Step 2 — Included angle of the development sector

The development of a cone's lateral surface is a circular sector of radius $R$. The arc length equals the base circumference, $2\pi r$.

$$\theta = \frac{r}{R}\times 360^{\circ} = \frac{30}{85.44}\times 360^{\circ} \approx 0.3511\times 360^{\circ} \approx 126.4^{\circ}.$$

So the development is a sector of radius $85.44\text{ mm}$ and included angle $\approx 126.4^{\circ}$.

Construction (described):

1. Draw the front view (triangle, base $60$, apex $80$ above) and the top view (circle $\varnothing 60$ divided into 12 equal parts: $1,2,\dots,12$).
2. With centre $O$ (apex) and radius $R = 85.44$, draw an arc. From a starting radial line $O\text{-}1$, step off 12 chord-lengths equal to the chord of one $30^{\circ}$ division of the base circle ($\text{chord} = 2r\sin 15^{\circ} = 60\sin 15^{\circ} = 15.53\text{ mm}$) to mark $1',2',\dots,12',1'$. Join $O$ to each. The sector $1'\,O\,1'$ subtends $\approx 126.4^{\circ}$.

          O (apex)
         /|\
        / | \        R = 85.44 mm
       /  |  \
     1'---+---1'   arc (base, 2πr = 188.5 mm)
     sector angle ≈ 126.4°

Step 2 check: arc length $= R\theta_{rad} = 85.44\times(126.4\times\pi/180) = 85.44\times2.206 = 188.5\text{ mm}$, and $2\pi r = 2\pi(30)=188.5\text{ mm}$. ✓

(b) Locating point P

P is on the front generator (generator $O\text{-}1$ in V.P.) at base-height $30\text{ mm}$. Along this generator, true distance of P from the apex measured along the slant equals

$$OP = R\left(1 - \frac{30}{h}\right) = 85.44\left(1-\frac{30}{80}\right) = 85.44\times0.625 = 53.40\text{ mm}.$$

On the development, mark P on the radial line $O\text{-}1'$ at $53.4\text{ mm}$ from $O$. OP = 53.4 mm from apex.

(c) Truncation line of the $45^{\circ}$ section

The cutting plane is inclined $45^{\circ}$ to H.P. and passes through the axis at height $40\text{ mm}$. In the front view it appears as an edge (a $45^{\circ}$ line) crossing the axis at $40\text{ mm}$. It meets each generator $O\text{-}n$ at a point; transfer each cut-height to the corresponding radial line by true-length scaling along the slant (project the cut point on the front view horizontally to the extreme generator $O\text{-}1$ to read its true distance from $O$, then swing that radius onto $O\text{-}n'$ on the development). Joining the 12 transferred points with a smooth curve gives the truncation line on the development.

Mark distribution items: true length, sector angle, full development, point P, section curve.

Given: vertical cylinder $\varnothing 70$ (radius $R = 35$), horizontal penetrating cylinder $\varnothing 50$ (radius $r = 25$), axes intersecting at right angles.

(b) Method — cutting-plane (line) method

Use a series of horizontal cutting planes parallel to both axes' common plane, or more directly the generator/element method on the smaller cylinder:

1. In the side view, the smaller (horizontal) cylinder appears as a circle $\varnothing 50$. Divide it into 12 equal parts: $1,2,\dots,12$.
2. From each division draw a horizontal generator of the smaller cylinder into the front view.
3. The same divisions projected to the top view meet the larger cylinder's circle ($\varnothing 70$); project those intersection points up.
4. Each generator of the small cylinder meets the surface of the large cylinder at a point; the intersection of corresponding projectors gives a point on the curve.
5. Join the points with a smooth curve. By symmetry the curve is symmetric about both the vertical axis and the horizontal axis.

(a) Coordinates of the curve (analytic check)

Take origin at the intersection of axes. Vertical cylinder: $x^2 + y^2 = 35^2$ (axis along $z$). Horizontal cylinder: $y^2 + z^2 = 25^2$ (axis along $x$). On the intersection curve both hold, so projected onto the front view (the $xz$-plane, viewing along $y$):

$$x = \pm\sqrt{35^2 - y^2}, \qquad z = \pm\sqrt{25^2 - y^2}.$$

Eliminating $y^2$: $y^2 = 625 - z^2$, hence

$$x^2 = 1225 - (625 - z^2) = 600 + z^2.$$

So the front-view projection of the intersection is the hyperbola-like branch $x^2 - z^2 = 600$.

Sample points (front view, in mm):

At $z=25$, $x=\sqrt{600+625}=\sqrt{1225}=35$ — the curve reaches the extreme generator of the large cylinder, confirming full penetration. ✓

(c) Form of the curve

Because the smaller cylinder ($50$) is fully enclosed by the larger ($70$), there are two separate closed curves of intersection (one where it enters, one where it exits), each appearing in the front view as a curve that bulges toward the axis.

If the two diameters become equal ($\varnothing 70 = \varnothing 70$) with intersecting axes, the curve of intersection degenerates into two intersecting straight lines (ellipses seen edge-on) — the classic result that two equal cylinders with intersecting axes intersect in plane (elliptical) curves that appear as straight lines in the front view.

(a) Types of perspective

(b) Apparent length by similar triangles

When an object plane is parallel to the picture plane (PP), the perspective image is a uniform scale reduction governed by

$$\frac{\text{image size}}{\text{real size}} = \frac{\text{distance from SP to PP}}{\text{distance from SP to object}}.$$

• Distance SP → PP $= 6.0\text{ m}$.
• Distance SP → object $= 6.0 + 9.0 = 15.0\text{ m}$ (object is $9$ m behind the PP).

Scale factor:

$$k = \frac{6.0}{15.0} = 0.40.$$

Apparent (perspective) length:

$$L' = k\times L = 0.40\times 4.0\text{ m} = 1.60\text{ m}.$$

Apparent length on the picture plane = 1.60 m.

Scale relationship used: image-to-object ratio equals (SP-to-PP distance)/(SP-to-object distance), i.e. similar triangles formed by the visual rays converging at the station point.

Given: internal $4.50\times3.60$ m; wall thickness $t=0.23$ m; footing width $0.60$ m; plinth $0.45$ m; wall height (plinth top to ceiling) $H=3.0$ m.

(a) Centre-line and external dimensions

Centre line runs through the middle of each $0.23$ m wall, i.e. $t/2 = 0.115$ m outside the internal face on each side.

• Centre-line length (long direction) $= 4.50 + 2(0.115) = 4.50 + 0.23 = 4.73\text{ m}.$
• Centre-line length (short direction) $= 3.60 + 0.23 = 3.83\text{ m}.$
• External length $= 4.50 + 2t = 4.50 + 0.46 = 4.96\text{ m}.$
• External width $= 3.60 + 0.46 = 4.06\text{ m}.$

Centre-line plan = 4.73 m × 3.83 m; external plan = 4.96 m × 4.06 m.

(b) Centre-line wall length and masonry volume

Total centre-line length (perimeter through wall centres):

$$L_{cl} = 2(4.73 + 3.83) = 2(8.56) = 17.12\text{ m}.$$

Gross wall volume (above plinth) $= L_{cl}\times t\times H = 17.12\times0.23\times3.0 = 11.81\text{ m}^3.$

Deduct openings (wall thickness $\times$ opening area):

• Door: $1.0\times2.10 = 2.10\text{ m}^2 \Rightarrow 2.10\times0.23 = 0.483\text{ m}^3.$
• Window: $1.2\times1.2 = 1.44\text{ m}^2 \Rightarrow 1.44\times0.23 = 0.331\text{ m}^3.$
• Total deduction $= 0.483 + 0.331 = 0.814\text{ m}^3.$

Net superstructure masonry:

$$V = 11.81 - 0.814 = 11.00\text{ m}^3.$$

Net brick masonry (superstructure) ≈ 11.0 m³.

(Check: gross $17.12\times0.23 = 3.9376\text{ m}^2$ per metre height $\times 3.0 = 11.813\text{ m}^3$. ✓)

(c) Components of a typical cross-section (bottom to top)

1. Natural/firm soil (founding stratum).
2. Lean concrete / soling bed under footing.
3. Strip footing ($600\text{ mm}$ wide, often stepped).
4. Foundation masonry up to plinth.
5. Damp-proof course (DPC) at plinth level ($0.45\text{ m}$).
6. Plinth beam (if RCC framing) / plinth masonry.
7. Ground-floor finish over compacted fill + PCC bed.
8. Superstructure brick wall ($230\text{ mm}$), with door/window openings.
9. Sill below window; lintel over door and window.
10. Roof slab / truss with eaves and parapet or fascia.
11. Plaster (internal & external), flooring, and roof finish.

Given: square side $a = 200$ mm (so half-side $=100$); circle radius $r = 80$ mm; vertical height $H = 150$ mm; concentric.

(a) Triangulation method

A square-to-round transition cannot be unrolled like a cone or cylinder, so its surface is approximated by triangles:

1. Divide the top circle into equal parts (commonly 12). Number them.
2. Each corner of the square connects to the nearest arc of the circle. The surface between one square corner and the adjacent set of circle points forms a part of a cone (warped triangular strip) — approximated by small flat triangles.
3. The surface between a side of the square and the circle point directly above its mid-region forms a plane triangle.
4. Find the true length of every element line (corner→circle point) using the right-triangle rule $\text{TL}=\sqrt{(\text{plan length})^2 + H^2}$.
5. Lay the triangles side by side, sharing edges, to build the flat pattern; add seam allowance.

(b) True lengths of representative lines

Place origin at centre. Square corner at $(100, 100)$. Top circle points at angle $\phi$: $(80\cos\phi, 80\sin\phi)$.

Element 1 — corner to the nearest circle point (the $45^{\circ}$ point, $\phi=45^{\circ}$): circle point $=(80\cos45^{\circ}, 80\sin45^{\circ}) = (56.57, 56.57)$.

• Plan length $d_1 = \sqrt{(100-56.57)^2 + (100-56.57)^2} = \sqrt{43.43^2 + 43.43^2} = \sqrt{2}\times43.43 = 61.42\text{ mm}.$
• True length $\text{TL}_1 = \sqrt{61.42^2 + 150^2} = \sqrt{3772 + 22500} = \sqrt{26272} = 162.1\text{ mm}.$

Element 2 — corner to the circle point on the axis of the adjacent side ($\phi = 0^{\circ}$, point $(80,0)$):

• Plan length $d_2 = \sqrt{(100-80)^2 + (100-0)^2} = \sqrt{20^2 + 100^2} = \sqrt{400 + 10000} = \sqrt{10400} = 101.98\text{ mm}.$
• True length $\text{TL}_2 = \sqrt{101.98^2 + 150^2} = \sqrt{10400 + 22500} = \sqrt{32900} = 181.4\text{ mm}.$

Element 3 — mid-side of square to nearest circle point (mid-side $(100,0)$ to $\phi=0$ point $(80,0)$):

• Plan length $d_3 = 100 - 80 = 20\text{ mm}.$
• True length $\text{TL}_3 = \sqrt{20^2 + 150^2} = \sqrt{400 + 22500} = \sqrt{22900} = 151.3\text{ mm}.$

(c) Composition of the piece

The square-to-round transition consists of:

• 4 plane (flat) triangles, one rising from the middle of each side of the square to the nearest circle point.
• 4 warped/conical (warped triangular) corner elements, one at each square corner, each spanning an arc of the circle (subdivided into small triangles in the development).

Total: 4 plane triangles + 4 conical corner pieces.

(a) Coordinate entry modes

(b) Resulting absolute coordinates from $(40,30)$

Relative $(@30,20)$:

$$x = 40 + 30 = 70,\qquad y = 30 + 20 = 50 \;\Rightarrow\; (70,\,50).$$

Polar $(@50 < 30^{\circ})$:

$$\Delta x = 50\cos30^{\circ} = 50\times0.8660 = 43.30,$$ $$\Delta y = 50\sin30^{\circ} = 50\times0.5000 = 25.00,$$ $$x = 40 + 43.30 = 83.30,\qquad y = 30 + 25.00 = 55.00 \;\Rightarrow\; (83.30,\,55.00).$$

Relative result = (70, 50); polar result = (83.30, 55.00).

Given: floor height $= 3.30\text{ m} = 3300\text{ mm}$; trial riser $R \approx 165$ mm; tread $T = 250$ mm.

(a) Number of risers and treads

Number of risers:

$$N = \frac{3300}{165} = 20\text{ risers}.$$

Actual riser $= 3300/20 = 165\text{ mm}$ (exact). Number of treads $= N - 1 = 19$ (the top tread coincides with the landing/floor).

20 risers of 165 mm; 19 treads of 250 mm.

(b) Proportion check and going

Rule check:

$$2R + T = 2(165) + 250 = 330 + 250 = 580\text{ mm}.$$

This is within the comfortable range $550\text{–}600\text{ mm}$, so the proportion is satisfactory (also $R\times T = 165\times250 = 41250 \approx 40000\text{–}45000\text{ mm}^2$, acceptable).

Dog-legged stair → two equal flights of $20/2 = 10$ risers each, i.e. $9$ treads per flight. Going (horizontal run) of one flight:

$$\text{Going} = (\text{treads per flight})\times T = 9\times250 = 2250\text{ mm} = 2.25\text{ m}.$$

Each flight: 10 risers, 9 treads, going = 2.25 m; 2R+T = 580 mm (OK).

Line (edge) method — intersection of two prisms

When two prisms penetrate, the line of intersection is a series of straight-line segments (since all faces are planes); each segment lies where one edge of a prism pierces a face of the other.

Steps:

1. Draw the views. Draw the front, top and (where helpful) side views of both prisms in their given positions, with the vertical square prism standing on H.P. and the horizontal triangular prism penetrating it.
2. Identify the piercing edges. In the view where the horizontal (triangular) prism shows its end as a true shape (side view), its three edges appear as points/lines. Number these edges $1,2,3$.
3. Locate piercing points. Project each edge of the penetrating triangular prism onto the faces of the square prism. Where an edge meets a face, mark the piercing point. Also find where edges of the square prism (if any) pierce the faces of the triangular prism.
4. Transfer points between views. Use projectors to carry each piercing point from the view where it is found to the other views, maintaining alignment.
5. Join in correct sequence. Connect the piercing points by straight lines, joining only those points that lie on the same face of both solids. Maintain visibility — use continuous lines for visible segments and dashed for hidden ones.
6. Repeat for entry and exit. Because the prism penetrates fully, obtain the intersection on both the near and far sides; complete both closed figures.

Result: a closed polygon (or two polygons for full penetration) of straight segments forming the line of intersection.

(b) Why straight segments; entry vs exit

Every face of both prisms is a plane. The intersection of two planes is always a straight line, so each portion of the intersection lying on one face of the square prism and one face of the triangular prism is a straight segment; the complete intersection is therefore a polygon of straight lines (no curves, unlike solids with curved surfaces).

Because the triangular prism (side $40$ mm) is fully enclosed within the square prism (side $50$ mm), it both enters and exits, giving two separate closed polygons. The entry figure (front face of square prism) and the exit figure (rear face) are mirror images about the vertical axis; in the front view they overlap, with the exit polygon drawn as hidden (dashed) lines and the entry polygon as visible (continuous) lines.

Given: $b = 230$ mm, $D = 400$ mm, cover $c = 25$ mm, bar $\varnothing = 16$ mm.

(a) Effective depth

Effective depth is measured to the centroid of tension steel:

$$d = D - c - \frac{\varnothing}{2} = 400 - 25 - \frac{16}{2} = 400 - 25 - 8 = 367\text{ mm}.$$

Effective depth d = 367 mm.

(b) Area of steel and percentage

Area of one $16$ mm bar:

$$a_1 = \frac{\pi}{4}(16)^2 = 0.7854\times256 = 201.06\text{ mm}^2.$$

For 3 bars:

$$A_{st} = 3\times201.06 = 603.2\text{ mm}^2.$$

Percentage of steel on $b\times d$:

$$p = \frac{A_{st}}{b\,d}\times100 = \frac{603.2}{230\times367}\times100 = \frac{603.2}{84410}\times100 = 0.715\%.$$

$A_{st} = 603.2\text{ mm}^2$; $p \approx 0.72\%$.

(c) Two other detailing items

1. Stirrups / shear links (e.g. $\varnothing 8$ two-legged at given spacing) for shear and to hold bars.
2. Anchorage / development length & hooks at supports (and top hanger bars). (Bar bending schedule, curtailment points, or cover blocks are also acceptable.)

Given: diameter $d = 50$ mm (radius $r = 25$), height $h_{max} = 90$ mm; cut is a single oblique plane from $0$ on one generator to $90$ mm on the diametrically opposite generator.

Step 1 — Development width

The lateral surface unrolls into a rectangle of width equal to the base circumference:

$$W = \pi d = \pi\times50 = 157.08\text{ mm}.$$

Divide the circumference into 12 equal parts; spacing $= W/12 = 13.09\text{ mm}$, generators numbered $1,2,\dots,12$ starting from the lowest-cut generator.

Step 2 — Cut height at each generator

For an oblique plane on a cylinder, the cut height varies as a cosine of the angular position. With generator $1$ at height $0$ (one end of the diameter) and generator $7$ (opposite) at $90$ mm, the height at angle $\theta$ (measured from generator 1) is

$$\boxed{\,h(\theta) = \frac{h_{max}}{2}\,(1 - \cos\theta)\,} = 45\,(1 - \cos\theta)\text{ mm}.$$

Heights at the 12 generators ($\theta = 0^{\circ},30^{\circ},\dots$):

Step 3 — Development

Plot the 12 generator lines at $13.09$ mm spacing on the rectangle; mark the above heights on each; join the tops with a smooth curve (a sinusoid). The lateral development is bounded below by the base line and above by this curve, from $0$ at gen 1 rising to $90$ mm at gen 7.

h(mm)
90 ┤                  ╭─7
   │              ╭───╯
45 ┤      ╭───────╯ (gen 4)
   │  ╭───╯
 0 ┤──╯ 1   2   3   4   5   6   7  ...
   └──────────────────────────────► W = 157.08 mm

Width 157.08 mm; cut height $h = 45(1-\cos\theta)$ mm.

(a) Common standard scales

• Site / layout plans: 1:500 and 1:200 (also 1:1000 for large sites).
• Detailed construction drawings: 1:50 and 1:20 (also 1:10, 1:5 for fine details).

(b) Scale conversions

Wall at 1:50, drawn 92 mm:

$$\text{Actual} = 92\text{ mm}\times50 = 4600\text{ mm} = 4.60\text{ m}.$$

Beam 4.5 m at 1:25, drawn length:

$$\text{Drawn} = \frac{4500\text{ mm}}{25} = 180\text{ mm} = 18.0\text{ cm}.$$

Actual wall = 4.60 m; beam drawn at 180 mm.

(c) Conventional symbols (described)

Brick:  ////////      Earth:  - - -  /  /      Door (plan):
        ////////              - - -                 ┌───┐ wall
                              (random dashes)        │   .─ leaf
                                                     │  ( arc swing