BE Civil Engineering (IOE, TU) Engineering Drawing I (IOE, ME 401) Question Paper 2077 Nepal

Given data

Distance between end projectors $= 60\text{ mm}$.

Both points are above HP and in front of VP, so the line lies in the first quadrant.

Step 1 — Locate the projections

Front view ($a'$, $b'$): heights above $xy$ are $15\text{ mm}$ and $55\text{ mm}$. Top view ($a$, $b$): distances below $xy$ are $20\text{ mm}$ and $50\text{ mm}$. Horizontal separation of projectors $= 60\text{ mm}$.

          b'
         /|
        / | (55-15)=40
  a'---+  |
   |   60 |
---+------+----------- xy
   |          
  a----+
    \  | (50-20)=30
     \ |
      \|
       b

Step 2 — True length (TL)

The true length is found by the trapezoidal / right-triangle method. Because both views are foreshortened, compute TL directly from the three coordinate differences:

• Difference along $xy$ (projector distance): $\Delta x = 60\text{ mm}$
• Difference in heights: $\Delta z = 55 - 15 = 40\text{ mm}$
• Difference in front distances: $\Delta y = 50 - 20 = 30\text{ mm}$

$$TL = \sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2} = \sqrt{60^2 + 30^2 + 40^2}$$ $$TL = \sqrt{3600 + 900 + 1600} = \sqrt{6100} = 78.10\text{ mm}$$

True length $\approx \mathbf{78.1\text{ mm}}$

Graphically: rotate the top view about a vertical projector (length $ab = \sqrt{60^2 + 30^2} = \sqrt{4500} = 67.08\text{ mm}$) to make it parallel to $xy$; project to the front view and join — the hypotenuse equals $78.1\text{ mm}$.

Step 3 — True inclinations

Inclination with HP ($\theta$): the vertical rise over the true length component. Using the right triangle whose base is the plan length and height is $\Delta z$:

$$\sin\theta = \frac{\Delta z}{TL} = \frac{40}{78.10} = 0.5122 \Rightarrow \theta = 30.8^\circ$$

Inclination with VP ($\phi$):

$$\sin\phi = \frac{\Delta y}{TL} = \frac{30}{78.10} = 0.3842 \Rightarrow \phi = 22.6^\circ$$

$\theta \approx \mathbf{30.8^\circ}$ (with HP), $\phi \approx \mathbf{22.6^\circ}$ (with VP)

(Check: $\sin^2\theta + \sin^2\phi = 0.262 + 0.148 = 0.410 \le 1$, consistent for a skew line.)

Step 4 — Traces

Extend the front view $a'b'$ to meet $xy$; the point where it crosses gives the Horizontal Trace (HT) projected down to the extended top view. Extend the top view $ab$ to meet $xy$; that crossing projected up to the extended front view gives the Vertical Trace (VT).

Using similar triangles on the projections (origin at $a$):

For HT, set height $= 0$ in the front view. Height varies $15 \to 55$ over projector distance $60$, i.e. slope $= 40/60$. To drop from $15$ to $0$ behind $a'$:

$$\text{distance back from } a' = \frac{15}{40}\times 60 = 22.5\text{ mm}$$

So HT lies $22.5\text{ mm}$ to the left of $a$'s projector, on the $xy$ line in the front view; its plan position is the corresponding point on the extended top view.

For VT, set front distance $= 0$ in the top view. Front distance varies $20 \to 50$ over projector $60$, slope $= 30/60$. To reduce from $20$ to $0$:

$$\text{distance back from } a = \frac{20}{30}\times 60 = 40\text{ mm}$$

So VT lies $40\text{ mm}$ to the left of $a$'s projector, projected up to $xy$.

HT is $22.5\text{ mm}$ beyond $A$; VT is $40\text{ mm}$ beyond $A$ (both measured along the projector direction, on the $A$ side).

Understanding the position

• Base is a regular pentagon, edge $a = 30\text{ mm}$.
• Axis horizontal, parallel to HP and VP $\Rightarrow$ axis runs left-to-right, perpendicular to the side view (profile plane).
• The prism lies on a rectangular face $\Rightarrow$ one face flat on HP.

Step 1 — Pentagon geometry (the true shape appears in the side view)

For a regular pentagon of edge $a = 30\text{ mm}$:

• Circumradius $R = \dfrac{a}{2\sin 36^\circ} = \dfrac{30}{2(0.5878)} = 25.52\text{ mm}$
• Inradius (apothem) $r = \dfrac{a}{2\tan 36^\circ} = \dfrac{30}{2(0.7265)} = 20.65\text{ mm}$
• Diagonal (across) $d = \dfrac{a}{2}\!\left(1+\sqrt5\right)= 15(3.236)=48.54\text{ mm}$
• Total height of pentagon (flat edge at bottom) $H = r + R = 20.65 + 25.52 = 46.17\text{ mm}$
• Width across the two upper sloping vertices $= d = 48.54\text{ mm}$

Step 2 — Side view (true shape of base pentagon)

Draw the pentagon with one edge $(30\text{ mm})$ resting on the ground line. Vertices (taking bottom-left base vertex as origin, $x$ right, $z$ up):

(The base edge spans $30\text{ mm}$; apex is centred at $x=15$.)

Step 3 — Front view

The axis is horizontal and parallel to VP, so the front view shows a rectangle:

• Length (axis) $= 70\text{ mm}$ horizontal
• Height $= H = 46.17\text{ mm}$ vertical (height of the pentagon)

Visible internal edges: the longitudinal edges of the prism corresponding to vertices 3, 4, 5 project as horizontal lines inside the rectangle at heights $28.53$, $46.17$, $28.53\text{ mm}$.

Step 4 — Top view

The top view is also a rectangle:

• Length (axis) $= 70\text{ mm}$
• Width $= 48.54\text{ mm}$ (max width $d$ of the pentagon)

The nearest end is $10\text{ mm}$ from VP, so the top view is drawn $10\text{ mm}$ below $xy$, extending to $10+70 = 80\text{ mm}$. Internal longitudinal edges appear at the projected widths of vertices.

 FV (rectangle 70 x 46.17)        SV (pentagon true shape)
 +----------------------+              4
 |   apex edge          |            /   \
 +----------------------+  3,5 edges 5     3
 |                      |            |     |
 +----------------------+            1-----2
 ===== xy (HP line) ======================
 TV (rectangle 70 x 48.54), 10 mm below xy

Final dimensions: FV $= 70 \times 46.17\text{ mm}$; TV $= 70 \times 48.54\text{ mm}$; SV $=$ true pentagon of edge $30\text{ mm}$ ($R=25.52$, $r=20.65$).

Step 1 — Identify the curve

For a right cone, the semi-cone (generator) angle $\alpha$ satisfies:

$$\tan\alpha = \frac{R}{h} = \frac{30}{70} \Rightarrow \alpha = 23.2^\circ$$

The generator makes $90^\circ - 23.2^\circ = 66.8^\circ$ with the base. The cutting plane is at $45^\circ$ to the base (HP). Since the plane inclination $45^\circ$ is less than the generator-to-base angle $66.8^\circ$ but greater than $0^\circ$, and it does not pass parallel to a generator, the cut crosses all generators on one nappe.

Check against a generator: a plane parallel to a generator (inclined at $66.8^\circ$ to base) gives a parabola. Here $45^\circ < 66.8^\circ$, so the plane cuts completely through the cone — the section is an ELLIPSE.

Step 2 — Geometry of the section line in the FV

In the front view the cone is a triangle: base $60\text{ mm}$ wide, apex $70\text{ mm}$ high. The section plane is a straight line at $45^\circ$ to $xy$ passing through axis point $P$ at height $30\text{ mm}$.

Find where the $45^\circ$ line meets the two extreme generators.

Left generator: from base-left $(-30, 0)$ to apex $(0,70)$. Parametrize $x = -30 + 30t,\; z = 70t$. The section line through $(0,30)$ with slope $\tan45^\circ = 1$ (rising to the right): $z - 30 = 1\cdot(x-0) \Rightarrow z = x + 30$.

Intersection with left generator: $70t = (-30+30t) + 30 = 30t \Rightarrow 70t = 30t$? That gives $t=0$ → meets base-left at $(-30,0)$, height $0$. Good — the cut starts at the base on the left side.

Right generator: from base-right $(30,0)$ to apex $(0,70)$: $x = 30 - 30t,\; z = 70t$. Set $z = x+30$: $70t = (30-30t)+30 = 60 - 30t \Rightarrow 100t = 60 \Rightarrow t = 0.6$. So $z = 70(0.6) = 42\text{ mm}$, $x = 30 - 18 = 12\text{ mm}$.

Section line endpoints (FV): lower point $A'=(-30,\,0)$ on base; upper point $B'=(12,\,42)$ on the right generator.

Length of cut line in FV (this is the major axis of the ellipse):

$$AB = \sqrt{(12-(-30))^2 + (42-0)^2} = \sqrt{42^2 + 42^2} = 42\sqrt2 = 59.40\text{ mm}$$

Step 3 — Section points by cutting circles

The top view is a circle of $\varnothing 60\text{ mm}$ (radius $30$). At each height $z$ the cone radius is $\rho(z) = 30\left(1 - \dfrac{z}{70}\right)$. The section line crosses heights from $0$ to $42\text{ mm}$. Take horizontal cutting planes and mark where the section line lies, then project the chord half-width $y = \sqrt{\rho^2 - x_{\text{axis}}^2}$ where $x_{\text{axis}}$ is the section point's horizontal distance from the axis.

Sample stations along the section line ($z = x+30$, so $x = z-30$):

Plot these half-widths symmetrically about the trace in the top view to get the sectional top view (an ellipse-like closed curve, foreshortened).

Step 4 — True shape

Project the section points onto an auxiliary plane parallel to the $45^\circ$ cutting line. Distances along the cut are taken from the FV; the half-widths $y$ from the table are laid perpendicular to the new reference line. The result is the true ellipse:

• Major axis $= AB = 59.4\text{ mm}$ (the full length of the cut).
• Minor axis $=$ widest chord $=2\times 18.96 = 37.9\text{ mm}$ (at $z=20$, near mid-cut).

Curve of intersection: an ELLIPSE, major axis $\approx \mathbf{59.4\text{ mm}}$, minor axis $\approx \mathbf{37.9\text{ mm}}$.

Difference: isometric drawing vs isometric projection

• Isometric drawing: true (full) lengths are plotted directly.
• Isometric projection: lengths are reduced by the isometric scale; the ratio of isometric to true length is

$$\frac{\text{isometric length}}{\text{true length}} = \sqrt{\frac{2}{3}} = 0.8165.$$

This problem asks for isometric projection, so every dimension is multiplied by $0.8165$.

Step 1 — Isometric scale construction

Draw a line at $45^\circ$ (true lengths) and another at $30^\circ$ (isometric lengths) from a common point. Mark true lengths on the $45^\circ$ line; drop verticals to the $30^\circ$ line to read isometric lengths. Equivalently multiply by $0.8165$:

Step 2 — Square prism (bottom)

The isometric axes are vertical and two lines at $30^\circ$ to the horizontal. Construct the bottom square base as a rhombus of side $65.32\text{ mm}$ (isometric). Raise vertical edges of $24.50\text{ mm}$ to form the top face (also a rhombus of side $65.32\text{ mm}$).

Step 3 — Cylinder (top), centred on prism top

The cylinder base is a circle of true $\varnothing 50\text{ mm}$. In isometric it becomes an ellipse inscribed in a rhombus of side $= 40.83\text{ mm}$ (isometric diameter). Use the four-centre method:

• Draw the rhombus (side $40.83$) centred on the prism top face.
• Major axis of ellipse $= 0.8165 \times 50 \times 1.0 \approx$ (graphically the rhombus long diagonal) $= 49.99\text{ mm}$; minor axis $\approx 28.87\text{ mm}$. (Standard isometric ellipse ratio: major $= 1.0$, minor $= 0.577$ of the isometric diameter on the four-centre construction.)
• The four centres: two obtuse-angle vertices of the rhombus, and two points found from the perpendicular bisectors of the rhombus sides.

Step 4 — Build the cylinder

Draw the base ellipse on the prism top. Raise vertical generators of $53.07\text{ mm}$ at the extreme points (ends of the major axis and the tangent points). Draw the identical top ellipse $53.07\text{ mm}$ above. Join the outer tangent generators to complete the cylinder.

        ___________            <- top ellipse of cylinder
      (           )
       |         |  h=53.07
      (___________)            <- base ellipse on prism top
      /          /|
     /  prism   / |  height 24.50
    /__________/  |
    |          |  /
    |  65.32   | /
    |__________|/

Final isometric projection dimensions: prism rhombus side $65.32\text{ mm}$, height $24.50\text{ mm}$; cylinder base ellipse from isometric $\varnothing 40.83\text{ mm}$, generator height $53.07\text{ mm}$, centred on the prism top. Total isometric height $= 24.50 + 53.07 = \mathbf{77.57\text{ mm}}$.

Step 1 — Set up first-angle layout

In first-angle projection the object lies between observer and plane; the top view goes below the front view, and the left side view goes to the right of the front view.

Step 2 — Front view (looking along the 60 mm width)

We see the length $\times$ height silhouette: an inverted-L step.

• Base block: rectangle $90\text{ mm}$ long $\times\, 15\text{ mm}$ high.
• Vertical wall at the left end: rectangle $15\text{ mm}$ wide $\times\, 50\text{ mm}$ high, sitting on the base, so the wall top is at $15+50 = 65\text{ mm}$.
• The $\varnothing20\text{ mm}$ hole is in the base, $30\text{ mm}$ from the right end → its centre is at $90-30 = 60\text{ mm}$ from the left. Through the base it shows as two hidden vertical lines $20\text{ mm}$ apart, between heights $0$ and $15\text{ mm}$.

 |15|
 +--+                              <- wall top at 65
 |  |
 |  | 50
 |  |
 +--+---------------------------+   <- 15 (top of base)
 |  |        : :                |
 +--+--------:.:----------------+   <- 0  (HP)
            hole (hidden), 20 wide
  |<----- 90 overall ----->|

Step 3 — Top view (placed below the front view)

Length $\times$ width plan = $90 \times 60\text{ mm}$ rectangle.

• The vertical wall appears at the left end as a $15\text{ mm} \times 60\text{ mm}$ region; its rear face is flush with the base back edge. Its front edge (the $15\text{ mm}$ thickness line) is drawn solid.
• The $\varnothing20\text{ mm}$ hole appears as a full circle (visible) centred $30\text{ mm}$ from the right edge and $30\text{ mm}$ from each long edge (centred on the $60\text{ mm}$ width).

 +--+--------------------------+
 |  |                          |
 |  |            (O) hole 20    | 60
 |  |                          |
 +--+--------------------------+
 |15|        |<- 30 ->|
  |<-------- 90 ------------->|

Step 4 — Left side view (placed to the right of the front view)

Width $\times$ height profile = looking from the left, we see the wall in full plus the base.

• Base: $60\text{ mm}$ (width) $\times\, 15\text{ mm}$ (height).
• Wall: rises full height to $65\text{ mm}$ over the full $60\text{ mm}$ width and $15\text{ mm}$ thickness (the wall is the nearest feature on the left, so it shows as a solid rectangle $60\text{ mm} \times 65\text{ mm}$ for the wall portion).
• The hole is to the right of the wall in reality; in the left side view it lies behind, shown as a hidden circle/lines $20\text{ mm}$ within the $0$–$15\text{ mm}$ base band.

 +--------------+
 |              |
 |    wall      | 65
 |              |
 |              |
 +--------------+   <-15
 |  : :  hidden |
 +--:.:---------+   <-0
 |<--- 60 --->|

Step 5 — Dimensioning

Overall: length $90$, width $60$, height $65$. Base thickness $15$; wall height $50$, wall thickness $15$. Hole $\varnothing20$, located $30$ from right and centred on width ($30$ from each side).

All three views are aligned: front and top share the length; front and side share the height; top and side share the width (via a $45^\circ$ mitre line if used).

Single-stroke lettering means the thickness of a line of the letter is obtained in one stroke of the pencil/pen (the stem width equals the natural width of a single stroke); it does not mean one stroke per letter.

Standard height ratios (take capital height $= h$):

• Capital letters & numerals height $= h$ (the nominal lettering size, e.g. $h = 5\text{ mm}$, $7\text{ mm}$, $10\text{ mm}$ from the standard range $2.5, 3.5, 5, 7, 10, 14, 20\text{ mm}$).
• Lower-case letter body height (x-height) $= \dfrac{5}{7}h \approx 0.7h$ (ascenders/descenders reach the full $h$).
• Line (stroke) thickness $d = \dfrac{h}{10}$ for type B (or $\dfrac{h}{14}$ for type A).
• Spacing between letters $\approx \dfrac{2}{10}h$; minimum line spacing $\approx \dfrac{14}{10}h$.

Worked ratio (for $h = 7\text{ mm}$, type B): capitals/numerals $=7\text{ mm}$; lower-case body $= 0.7\times7 = 4.9\approx 5\text{ mm}$; stroke thickness $= 7/10 = 0.7\text{ mm}$.

Construction steps (compass method)

1. Draw a circle of radius $R$ equal to the side length, i.e. $R = 35\text{ mm}$ (for a regular hexagon, circumradius $=$ side).
2. With the same radius $35\text{ mm}$, step off arcs around the circle from any starting point on the circumference; the compass divides the circle into exactly 6 equal arcs.
3. Join the six successive division points with straight lines to obtain the regular hexagon of side $35\text{ mm}$.

(Alternative: draw a horizontal line $= 35\text{ mm}$, and at each end construct $120^\circ$ interior angles, repeating around.)

Step — Circumscribed circle (passes through the vertices)

Circumradius $R = a = 35\text{ mm}$ (a defining property of a regular hexagon).

$$\boxed{D_{\text{circumscribed}} = 2R = 2(35) = 70\text{ mm}}$$

Step — Inscribed circle (tangent to the sides)

Apothem (inradius) $r = \dfrac{a\sqrt3}{2} = \dfrac{35 \times 1.7320}{2} = \dfrac{60.62}{2} = 30.31\text{ mm}$

$$\boxed{D_{\text{inscribed}} = 2r = 2(30.31) = 60.62\text{ mm}}$$

Summary: circumscribed circle diameter $= \mathbf{70\text{ mm}}$; inscribed circle diameter $= \mathbf{60.62\text{ mm}}$.

Point $P$: $25\text{ mm}$ below HP, $40\text{ mm}$ behind VP

Below HP $\Rightarrow$ in the lower half; behind VP $\Rightarrow$ behind. This is the third quadrant.

In the rabatment (rotating HP about $xy$):

• The front view $p'$ lies below $xy$ at $25\text{ mm}$ (object below HP → FV below $xy$).
• The top view $p$ lies above $xy$ at $40\text{ mm}$ (object behind VP → TV above $xy$).

So for $P$ both $p$ and $p'$ are on opposite sides crossing over: $p$ above $xy$ by $40\text{ mm}$ and $p'$ below $xy$ by $25\text{ mm}$. (Characteristic of the 3rd quadrant: top view above, front view below.)

Point $Q$: on the HP, $35\text{ mm}$ in front of VP

On HP $\Rightarrow$ height $= 0$; in front of VP $\Rightarrow$ first-quadrant boundary.

• The front view $q'$ lies on the $xy$ line (height above HP is zero).
• The top view $q$ lies below $xy$ at $35\text{ mm}$ (in front of VP → TV below $xy$).

Summary table

Concept

When a circle is tilted to the HP but kept perpendicular to the VP, its top view is an ellipse. The diameter parallel to the $xy$ line keeps its true length (major axis); the diameter along the line of steepest slope is foreshortened by $\cos\theta$ (minor axis).

Step 1 — Stage 1 (circle parallel to HP)

Assume the plate first lies flat on the HP.

• Top view: a true circle, $\varnothing 70\text{ mm}$.
• Front view: a horizontal line of length $70\text{ mm}$ on $xy$. Divide the circle into 12 equal parts and number them; project to the front view line.

Step 2 — Stage 2 (tilt the front view to $30^\circ$)

Tilt the front-view line so it makes $30^\circ$ with $xy$ (the plane is perpendicular to VP, so the front view stays a straight line of length $70\text{ mm}$, now inclined at $30^\circ$). Project the 12 points down; from the first-stage top view carry the horizontal distances across. The resulting top view is the ellipse.

Step 3 — Axes of the top-view ellipse

• Major axis $=$ the diameter that is parallel to the $xy$ line (horizontal axis of tilt) $=$ true diameter $= 70\text{ mm}$.
• Minor axis $=$ true diameter $\times \cos\theta = 70 \times \cos30^\circ = 70 \times 0.8660 = 60.62\text{ mm}$.

Front view: a straight line of length $70\text{ mm}$ inclined at $30^\circ$ to $xy$.

Result: Top view is an ellipse with major axis $= 70\text{ mm}$ and minor axis $\approx 60.62\text{ mm}$.

Given: major axis $2a = 100\text{ mm} \Rightarrow a = 50\text{ mm}$; minor axis $2b = 60\text{ mm} \Rightarrow b = 30\text{ mm}$.

Concentric-circles method — steps

1. Draw the two axes $AB = 100\text{ mm}$ (horizontal) and $CD = 60\text{ mm}$ (vertical), intersecting at centre $O$.
2. With centre $O$, draw two concentric circles of diameters equal to the major axis ($\varnothing100$, radius $50$) and the minor axis ($\varnothing60$, radius $30$).
3. Divide both circles into the same number of equal angular parts (say $12$), drawing radial lines through $O$ cutting both circles.
4. From each point on the outer (major) circle, draw a line parallel to the minor axis (vertical).
5. From the corresponding point on the inner (minor) circle, draw a line parallel to the major axis (horizontal).
6. The intersection of each such pair of lines is a point on the ellipse.
7. Join all the points with a smooth curve (French curve) to complete the ellipse.

Foci location

The distance of each focus from the centre is $c$, where

$$c = \sqrt{a^2 - b^2} = \sqrt{50^2 - 30^2} = \sqrt{2500 - 900} = \sqrt{1600} = 40\text{ mm}.$$

The two foci lie on the major axis, each $40\text{ mm}$ from the centre $O$ (i.e. $10\text{ mm}$ inside each vertex $A$, $B$). Check: $a^2 = b^2 + c^2 \Rightarrow 2500 = 900 + 1600$ ✓.

Foci distance from centre $= \mathbf{40\text{ mm}}$ each.

Purpose of sectional views

A sectional view is obtained by imagining the object cut by a cutting plane and removing the portion between the observer and the plane, so that the interior features (holes, bores, cavities, ribs) are revealed clearly without a clutter of hidden (dashed) lines. It improves clarity and makes dimensioning of internal features unambiguous.

Five conventions/rules for section lines (hatching)

1. Angle: section lines are thin continuous lines drawn at $45^\circ$ to the principal outline (or to the axis of symmetry). If the outline is itself at $45^\circ$, use $30^\circ$ or $60^\circ$ instead.
2. Spacing: hatching lines are evenly spaced (typically $1$–$3\text{ mm}$ apart depending on the area), uniform throughout one cut surface.
3. Adjacent parts: different parts in an assembly are hatched in opposite directions or with different spacing so each part is distinguishable.
4. Same part: all sectioned areas of the same part are hatched in the same direction and spacing across all views.
5. Exceptions (not sectioned even when cut): ribs, webs, shafts, bolts, nuts, rivets, keys, pins, and spokes are shown un-hatched when the cutting plane passes longitudinally through them.

(Additional: the cutting plane is shown by a long chain line, thick at the ends and at changes of direction, with arrows indicating the viewing direction and labelled e.g. A–A.)

Half-section

A half-section is produced by cutting the object with two perpendicular cutting planes that remove one quarter of the object. The view then shows one half in section (revealing interior) and the other half as an external (outside) view, separated by a centre line (not a solid line).

When preferred: for symmetrical objects (e.g. flanges, pulleys, cylindrical parts). It conveniently shows both internal and external features in a single view, saving a separate exterior view, while keeping the drawing uncluttered. Hidden lines are usually omitted on both halves.