BE Civil Engineering (IOE, TU) Engineering Drawing II (IOE, CE 451) Question Paper 2077 Nepal

Step 1 — Basic cone parameters

• Base radius $r = 30\ \text{mm}$, axis height $h = 80\ \text{mm}$.
• True slant length of a generator (apex to base circle):

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

Step 2 — Development of the FULL cone surface

The full lateral surface develops into a circular sector of radius $l$. The sector angle is:

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

Divide the base circle into 12 equal parts ($30^\circ$ each) giving generators $O\text{-}1,\ O\text{-}2,\dots,O\text{-}12$. In the development each chord-arc step subtends $\theta/12 = 126.4/12 \approx 10.53^\circ$ at the apex $O$.

Step 3 — Locate the cutting plane on each generator

The section plane is at $45^\circ$ and passes through the axis point $50\ \text{mm}$ above the base, i.e. $80-50 = 30\ \text{mm}$ below the apex on the axis. In the front view, project where the $45^\circ$ line crosses each generator. The cut point on a generator at horizontal distance $x$ from the axis lies at height:

$$z = 50 + x\,\tan 45^\circ = 50 + x \quad(\text{on the lower side of the axis, falling}\ z = 50 - x)$$

For each generator, measure the true distance $O\text{-}P$ from the apex to its cut point along the slant. Using similar triangles, the apex-to-cut true length on a generator whose cut point is at height $z$ above the base is:

$$OP = l\cdot\frac{h - z}{h} = 85.44\cdot\frac{80 - z}{80}$$

Project each cut point horizontally to the extreme (true-length) generator in the front view to read its true $OP$.

Step 4 — Tabulate true lengths (representative, by symmetry generators 1–7)

Take generator 1 at the near contour (front, $x=+30$), generator 7 at the far contour ($x=-30$), intermediate by $x = r\cos\phi$.

(Generator 7's cut point coincides with the apex side; the plane just reaches the contour near the top — values rounded.)

Step 5 — Construct the development

1. With centre $O$ and radius $l = 85.44\ \text{mm}$ draw an arc and step off 12 generator points so the total sector spans $126.4^\circ$. Join each to $O$: these are the developed generators $O\text{-}1'\dots O\text{-}12'$.
2. On each developed generator mark the true length $OP$ from the table measured from $O$.
3. Join the marked cut points $1',2',\dots$ with a smooth curve.

The region between this cut curve and the outer base arc is the development of the lower (retained) portion.

            O
           /|\
          / | \        each gen = 10.53 deg
   OP →  /  P  \      smooth curve through cut pts
        /__/‾\__\     
       1'  ...   12'   outer arc radius l=85.44, span 126.4 deg

Final Answer

Slant length $l = 85.44\ \text{mm}$; developed sector angle $\theta = 126.4^\circ$; the lower-portion development is the sector bounded by the outer base arc (radius $85.44\ \text{mm}$) and the inner true-length curve passing through cut points ranging from $OP = 0$ to $64.08\ \text{mm}$ as tabulated.

Step 1 — Given data

• Vertical cylinder: radius $R = 35\ \text{mm}$.
• Horizontal cylinder: radius $r = 25\ \text{mm}$, axis at $60\ \text{mm}$ above HP, intersecting and perpendicular to the vertical axis.

Since the horizontal cylinder ($\varnothing 50$) is smaller than the vertical one ($\varnothing 70$) and fully penetrates, there are two curves of intersection (entry and exit), which appear as a single symmetric curve pair in the front view.

Step 2 — Method (generator / cutting-plane)

Divide the top view circle of the smaller (horizontal) cylinder into 12 equal parts. Each division is a generator of the horizontal cylinder running front-to-back. The intersection point of each such generator with the surface of the vertical cylinder gives one point on the curve.

Step 3 — Geometry of a generator

Let the horizontal cylinder axis be the $X$-axis; measure $y$ (distance of a generator from the axis, $-25\le y\le 25$) and $z$ (height, constant $=60$). A generator at offset $y$ meets the vertical cylinder (radius $R=35$, axis vertical) where $x^2 + y^2 = R^2$, so:

$$x = \pm\sqrt{R^2 - y^2} = \pm\sqrt{35^2 - y^2}$$

In the front view the relevant coordinate is $z$ (height of the generator on the horizontal cylinder) and the horizontal position is the projection $x$. The height of a generator on the horizontal cylinder at angular position $\phi$ is:

$$z = 60 + r\sin\phi,\qquad y = r\cos\phi$$

and it pierces the vertical cylinder at $x = \pm\sqrt{35^2 - (25\cos\phi)^2}$.

Step 4 — Tabulate (12 generators)

Step 5 — Plot the curve in the front view

For each generator plot the point $(x, z)$ on the right side and $(-x, z)$ on the left side. Join the points with a smooth curve on each side. The curve is widest ($x=35$) at $z = 85$ (top, $\phi=90°$) and $z=35$ (bottom, $\phi=270°$), and narrowest ($x=24.49$) at $z=60$ (the axis height). Two symmetric curves result — they bulge outward toward the top and bottom of the horizontal cylinder.

  z
 85 |        *  (x=±35)
 72 |      *   *
 60 |     *     *  (x=±24.5, narrowest at axis)
 47 |      *   *
 35 |        *  (x=±35)
    +----------------- x

Final Answer

The intersection appears as two symmetric closed curves in the front view; key piercing points are $(\pm35,85)$ and $(\pm35,35)$ at the extreme generators and $(\pm24.49,60)$ at the axis level, joined by smooth curves as tabulated.

Step 1 — Compute overall (out-to-out) plan dimensions

Overall length $= 4.0 + 2\times0.23 = 4.46\ \text{m}$. Overall width $= 3.5 + 2\times0.23 = 3.96\ \text{m}$.

At scale $1:50$: $4.46\ \text{m}\rightarrow 89.2\ \text{mm}$ on paper; $3.96\ \text{m}\rightarrow 79.2\ \text{mm}$.

Step 2 — PLAN

• Draw two concentric rectangles: outer $4.46\times3.96\ \text{m}$, inner (room) $4.0\times3.5\ \text{m}$, the gap being the $230\ \text{mm}$ wall.
• Door $1.0\ \text{m}$ wide on one wall, shown by a break in the wall hatch with a quarter-circle swing arc.
• Two windows $1.2\ \text{m}$ wide, shown by two thin lines across the wall thickness (window symbol).
• Hatch the cut walls at $45^\circ$ (brick convention: alternate hatch). Show centre-lines, the section line A–A through a window with arrows, and chain-dotted axes.
• Dimension three chains: overall, openings, and wall offsets.

  <------------- 4.46 m -------------->
  +======[ window ]==================+
  ||                                ||
  ||                                || 3.96 m
  || room 4.0 x 3.5                 ||
  ||                                ||
  +==[door]====[ window ]===========+
        wall = 0.23 m, hatched

Step 3 — SECTIONAL ELEVATION (section A–A through a window)

Vertical dimension chain from the foundation up:

• Show stepped foundation footing (e.g. $0.90\ \text{m}$ wide), DPC at plinth, floor slab/PCC, the window with sill and lintel ($230\ \text{mm}$ thick lintel over the opening), and the roof slab ($125\ \text{mm}$ RCC) at $2.85\ \text{m}$.
• Hatch the cut masonry; ground line hatched with earth symbol.

Step 4 — FRONT ELEVATION

• Outline $4.46\ \text{m}$ wide, total visible height $= 0.45\ (\text{plinth}) + 2.85\ (\text{wall}) + 0.125\ (\text{parapet/roof}) = 3.425\ \text{m}$ above GL.
• Show the door (centred or offset) $1.0\times2.1\ \text{m}$ from plinth level, and any window visible on the front face $1.2\times1.2\ \text{m}$ with sill at $0.90\ \text{m}$ above floor ($=1.35\ \text{m}$ above GL).
• No hatching (elevations show only visible edges); indicate material/finish notes.

Step 5 — Conventions used

• Brick walls hatched at $45^\circ$ in section;
• DPC shown as a thick line at plinth;
• centre-lines chain-dotted;
• section plane marked A–A with direction arrows;
• dimension lines with arrowheads, three chains (overall / opening / detail);
• doors with swing arcs in plan, windows as double thin lines.

Final Answer

Overall plan $= 4.46\ \text{m}\times3.96\ \text{m}$; lintel level $= 2.10\ \text{m}$ above floor; floor-to-ceiling $= 2.85\ \text{m}$; total above-GL height $\approx 3.425\ \text{m}$; all drawn at $1:50$ with brick hatch, DPC, section A–A through a window, and three-chain dimensioning as detailed above.

Step 1 — Set up the reference lines

• Draw the ground line (GL) and the horizon line (HL) $40\ \text{mm}$ above it.
• The PP is represented by a horizontal line (in the plan setup, drawn above); the station point $SP$ is $70\ \text{mm}$ in front of (below) the PP.
• Place the plan of the block so the nearest vertical edge touches the PP, with the two faces inclined at $30^\circ$ and $60^\circ$.

Step 2 — Locate the Vanishing Points

From $SP$ draw lines parallel to each set of edges of the block until they meet the PP; project these intersections down to the HL.

• Visual ray parallel to the $30^\circ$ face: it hits the PP at a horizontal distance from the touching edge of:

$$d_1 = SP\ \text{distance}\times\tan(30^\circ)\ \text{component} \Rightarrow VP_1\ \text{at}\ 70\tan30^\circ = 70\times0.5774 = 40.4\ \text{mm}$$

• Visual ray parallel to the $60^\circ$ face:

$$VP_2\ \text{at}\ 70\tan60^\circ = 70\times1.7321 = 121.2\ \text{mm}$$

to the opposite side of the foot of $SP$ on the PP.

Project $VP_1$ and $VP_2$ vertically down onto the HL. (Note: the ray parallel to a face at $\theta$ to PP itself makes $90^\circ-\theta$ with the normal; the two VPs lie on opposite sides and the angle subtended at $SP$ between the two rays is $90^\circ$, confirming perpendicular faces.)

Step 3 — Heights and the touching edge

Because the front vertical edge touches the PP, it is shown in true height. On the GL mark the base of this edge; measure the true height $25\ \text{mm}$ up to get the top. This vertical line is the true-height line.

Step 4 — Draw the perspective

1. From the top and bottom of the true-height edge, draw lines to $VP_1$ and to $VP_2$. These give the perspective directions of the four top/bottom edges.
2. To fix the far corners, draw visual rays from $SP$ to the plan corners; where each ray crosses the PP, project vertically down and intersect with the corresponding edge-to-VP lines. This locates the receding edges' end-points.
3. Complete the cuboid by joining the located corners; the far vertical edges are drawn between the upper and lower receding lines.

  HL ----VP1(40.4)--------------touch-edge--------VP2(121.2)----
                       /|\
                      / | \   true height 25 (at PP edge)
  GL ________________/__|__\__________________________________
           SP 70 in front of PP; faces 30deg & 60deg

Final Answer

$VP_1$ lies $70\tan30^\circ = 40.4\ \text{mm}$ on one side and $VP_2$ lies $70\tan60^\circ = 121.2\ \text{mm}$ on the other side of the foot of the station point, both on the horizon. The vertical edge on the PP is shown in true height ($25\ \text{mm}$); all receding edges converge to the two VPs as constructed.

Step 1 — Effective dimensions

• Width $b = 250\ \text{mm}$, overall depth $D = 450\ \text{mm}$, cover $= 25\ \text{mm}$.
• Effective depth $d = D - \text{cover} - \frac{\varnothing_{\text{main}}}{2} = 450 - 25 - 8 = 417\ \text{mm}$.

Step 2 — Stirrup count

Assume the beam length includes $230\ \text{mm}$ bearing each end; take the total beam length over which stirrups run $\approx$ clear span $4.0\ \text{m} = 4000\ \text{mm}$ (stirrups within the clear span).

• End zones: $1.0\ \text{m}$ each end at $150\ \text{mm}$ c/c. Stirrups per end $= \frac{1000}{150} + 1 = 6.67 + 1 \approx 8$ (round up, including end stirrup). For two ends $= 2\times8 = 16$. More precisely each 1.0 m end zone: $\lceil 1000/150\rceil = 7$ spaces $\Rightarrow 8$ stirrups per end including both boundary positions; counting non-overlapping, take $7$ stirrups per end interior + boundaries shared.
• Middle zone: length $= 4000 - 2\times1000 = 2000\ \text{mm}$ at $250\ \text{mm}$ c/c. Spaces $= 2000/250 = 8 \Rightarrow 9$ positions, but the two boundary positions are shared with end zones, so $9 - 2 = 7$ additional stirrups in the middle.

Total stirrups $= (\text{end zones } 2\times7 = 14) + (\text{middle } 7) + (\text{2 shared boundaries counted once each, already included}) + 2\ \text{end boundary stirrups} = 14 + 7 + 2 = 23$ stirrups.

(A clean practical count: end zone $1000/150\to 8$ stirrups each $=16$; middle $2000/250\to 8$ spaces $=7$ interior; total $\approx 23$ stirrups.)

Number of stirrups $\approx 23$.

Step 3 — Cutting length of one stirrup (8 mm two-legged)

Inner dimensions of stirrup: width $= b - 2\,\text{cover} = 250 - 50 = 200\ \text{mm}$ depth $= D - 2\,\text{cover} = 450 - 50 = 400\ \text{mm}$ Perimeter $= 2(200+400) = 1200\ \text{mm}$. Add 2 hooks ($\approx 10\,\varnothing = 10\times8 = 80\ \text{mm}$ each, $=160$) and subtract bend allowances ($\approx 3\times 2\varnothing \approx$ negligible for schedule): Cutting length $\approx 1200 + 160 = 1360\ \text{mm} \approx 1.36\ \text{m}$ per stirrup.

Step 4 — Bar Bending Schedule (main bars)

Main bottom bars $4{-}16\varnothing$: provide $L_d$-type end hooks/bends. Length $=$ clear span $+ 2\times$ bearing $- 2\times$cover $+ 2\times$ 90° bend.

Take bearing $230\ \text{mm}$ each end: gross length $= 4000 + 2\times230 = 4460\ \text{mm}$; deduct end cover $2\times25 = 50$; add two 90° bends of $\approx 9\varnothing = 9\times16 = 144\ \text{mm}$ each $(=288)$: Cutting length $\approx 4460 - 50 + 288 = 4698\ \text{mm} \approx 4.70\ \text{m}$ per bar.

Top anchor bar length $\approx 4000 + 2\times230 - 50 + 2\times(9\times12)= 4460 - 50 + 216 = 4626 \approx 4.40\ \text{m}$ (after deducting laps/practical) — value rounded for schedule.

Step 5 — Drawing notes

• Cross-section: $250\times450$ rectangle; 4–16φ at bottom corners + intermediate, 2–12φ at top corners, stirrup outline at $25\ \text{mm}$ cover.
• Longitudinal section: show closer stirrups (150 c/c) at supports, wider (250 c/c) at mid-span, main bars at bottom, hangers at top, dimension the zones $1.0\ \text{m}+2.0\ \text{m}+1.0\ \text{m}$.

Final Answer

Effective depth $d = 417\ \text{mm}$; total stirrups $\approx 23$; stirrup cutting length $\approx 1.36\ \text{m}$; main bar cutting length $\approx 4.70\ \text{m}$ (4 nos, total $18.80\ \text{m}$), as tabulated in the bar bending schedule.

Step 1 — Why parallel-line method?

A prism has lateral edges that are all parallel and of equal true length (perpendicular to the base). The parallel-line method develops such prismatic/cylindrical surfaces by laying the edges side by side as parallel lines spaced by the true edge-lengths of the base.

Step 2 — Lateral surface dimensions

• Each of the 5 rectangular faces has width = base side $= 25\ \text{mm}$ and height $= 60\ \text{mm}$.
• Total developed width $= 5\times25 = 125\ \text{mm}$; height $= 60\ \text{mm}$.

Step 3 — Construction

1. Draw a horizontal base line and step off 5 equal segments of $25\ \text{mm}$: marks $A,B,C,D,E,A$.
2. At each mark erect a vertical line of height $60\ \text{mm}$ (the lateral edges).
3. Join the tops to form the top edge line.

  +----+----+----+----+----+
  |    |    |    |    |    | 60 mm
  A    B    C    D    E    A
  <-25-> each, total 125 mm

Final Answer

The development is a rectangle $125\ \text{mm}\times60\ \text{mm}$ divided into five $25\ \text{mm}$ panels; the parallel-line method applies because all lateral edges are parallel and of equal true length.

General procedure

1. Draw the projections (FV, TV) of both solids in the intersecting position.
2. Select a series of points common to both surfaces using an auxiliary construction (generators or cutting planes).
3. Project each common point to all views.
4. Join the points in correct sequence with a smooth curve (curved surfaces) or straight lines (plane faces).

Line (generator) method

• Uses the straight-line generators / edges of one solid (cylinder, cone, prism, pyramid).
• Each generator is intersected with the surface of the other solid; the piercing points lie on the intersection.
• Suitable example: a cylinder penetrating a prism (or cone penetrating a prism) — divide the cylinder/cone surface into generators and find where each pierces the prism faces.

Cutting-plane method

• A series of auxiliary section planes (usually parallel to a principal plane or containing the apex) cut both solids.
• Each plane produces a section line on each solid; their intersection gives points on the curve of intersection.
• Suitable example: cone and cylinder with non-intersecting/eccentric axes, or any case where generators are hard to use — use horizontal cutting planes giving circles on the cone and rectangles/lines on the cylinder.

Comparison table

Final Answer

Both methods generate a set of common surface points joined into the intersection curve; the line method pierces generators through the other solid (e.g. cylinder–prism), while the cutting-plane method intersects section outlines from auxiliary planes (e.g. cone–cylinder with offset axes).

Coordinate systems

• Absolute X,Y — measured from the origin (0,0). Example: 80,20.
• Relative / incremental @dX,dY — measured from the last point entered. Example: @60,0 moves 60 right.
• Polar @dist<angle — a distance at an angle from the last point. Example: @40<90 moves 40 up.

Triangle vertices

$A(20,20)$, $B(80,20)$, $C(80,60)$. Side $AB = 60$ (horizontal), $BC = 40$ (vertical), $CA = \sqrt{60^2+40^2} = \sqrt{3600+1600} = \sqrt{5200} \approx 72.11$ at angle $\arctan(40/60)=33.69^\circ$ above horizontal (so back to $A$ the polar direction is $180^\circ+33.69^\circ = 213.69^\circ$).

Command sequence

Command: LINE
Specify first point: 20,20            (absolute -> A)
Specify next point: @60,0             (relative -> B at 80,20)
Specify next point: @40<90            (polar  -> C at 80,60)
Specify next point: @72.11<213.69     (polar  -> back to A)  
        ... or simply:  C  (CLOSE)    (closes to start point A)
Specify next point: <Enter>           (end)

Verification:

• A→B: @60,0 gives $(20+60,20+0)=(80,20)=B$. ✓
• B→C: @40<90 gives $(80+40\cos90°,\ 20+40\sin90°)=(80,60)=C$. ✓
• C→A: @72.11<213.69 gives $(80+72.11\cos213.69°,\ 60+72.11\sin213.69°)=(80-60,60-40)=(20,20)=A$. ✓

Final Answer

Absolute = from origin (20,20); relative = from last point (@60,0); polar = distance<angle from last point (@40<90). The triangle closes with @72.11<213.69 or the CLOSE option, returning exactly to $A(20,20)$.

Step 1 — Number of risers

$$\text{No. of risers} = \frac{\text{floor-to-floor height}}{\text{riser}} = \frac{3000}{150} = 20\ \text{risers}$$

Step 2 — Number of treads

In a stair, number of treads (goings) $=$ risers $-1$ per continuous flight; for a dog-legged stair in two equal flights of 10 risers each:

• Risers per flight $= 20/2 = 10$.
• Treads per flight $= 10 - 1 = 9$.
• Total treads $= 2\times9 = 18$.

Step 3 — Comfort rule check

Rule 1: $2R + T = 2\times150 + 300 = 600\ \text{mm}$ — should lie between $550$ and $700\ \text{mm}$. 600 mm is within range ✓. Rule 2 (product): $R\times T = 150\times300 = 45000\ \text{mm}^2$ — should be about $40000\text{–}45000$. 45000 ✓ (at upper limit). Rule 3 (sum): $R + T = 150 + 300 = 450\ \text{mm}$ — should be about $400\text{–}450$. 450 ✓.

The stair satisfies the comfort criteria.

Step 4 — Going length of one flight

$$\text{Going of one flight} = (\text{treads per flight})\times T = 9\times300 = 2700\ \text{mm} = 2.70\ \text{m}$$

Final Answer

Total risers $= 20$ (10 per flight); total treads $= 18$ (9 per flight); comfort rule $2R+T = 600\ \text{mm}$ (within $550$–$700$) ✓; going length per flight $= 2.70\ \text{m}$.

Definitions

• (a) Picture Plane (PP): the transparent vertical plane placed between the observer and the object onto which the perspective view is projected (the 'screen'). It is usually perpendicular to the line of sight.
• (b) Station Point (SP): the position of the observer's eye in space from which the object is viewed; all visual rays originate here.
• (c) Horizon Line (HL): the horizontal line on the PP at the level of the observer's eye (eye level). All horizontal-receding parallel edges vanish on it.
• (d) Vanishing Point (VP): the point on the HL where a set of mutually parallel horizontal lines (receding from the observer) appears to converge.

        eye HL ---- VP1 -------------- VP2 ----  (horizon, eye level)
        SP •  \        \            /
               \        PP        /
   GL ___________\______|________/_______________ (ground line)

One-point vs Two-point perspective

Final Answer

PP = projection screen; SP = eye position; HL = eye-level horizontal line; VP = convergence point of receding parallels on HL. One-point perspective has one face parallel to the PP and a single VP, whereas two-point perspective has the object turned so two sets of edges recede to two VPs.

Purpose of layers

Layers organise drawing objects into logical groups so they can be controlled collectively — switched on/off, frozen, locked, or assigned a common colour, linetype, and lineweight. This keeps complex civil drawings (plans with structure, plumbing, electrical, dimensions) manageable, supports drawing standards, and allows selective plotting.

Two civil-engineering layer examples

(Any two suffice, e.g. WALLS = continuous white; CENTRE-LINE = CENTER red.)

Final Answer

Layers group related objects for collective control (visibility, colour, linetype, plotting). Example: WALLS = continuous white line; CENTRE-LINE = CENTER linetype, red.