BE Civil Engineering (IOE, TU) Engineering Drawing I (IOE, ME 401) Question Paper 2080 Nepal

Given data

Distance between end projectors (plan/elevation length along $xy$) $= L = 60\text{ mm}$.

The point being above H.P. and in front of V.P. places the line in the first quadrant.

Step 1 — Lay out the projections

            b'
            |\
            | \
        a'  |  \           ELEVATION (above xy)
         \  |   \
  --------+--+----+-------- xy
         /   |    |
        a    |    b         PLAN (below xy)

• In elevation, $a'$ is $15\text{ mm}$ above $xy$, $b'$ is $55\text{ mm}$ above $xy$, with horizontal separation $60\text{ mm}$.
• In plan, $a$ is $20\text{ mm}$ below $xy$, $b$ is $50\text{ mm}$ below $xy$, with the same horizontal separation $60\text{ mm}$ (projectors are vertical, so $a'$ over $a$, $b'$ over $b$).

Step 2 — Difference quantities

• Difference in heights: $\Delta h = 55 - 15 = 40\text{ mm}$.
• Difference in distances from V.P.: $\Delta d = 50 - 20 = 30\text{ mm}$.

Step 3 — Apparent lengths of the projections

Elevation length $a'b'$ (rise $\Delta h$, run $L$):

$$a'b' = \sqrt{L^2 + \Delta h^2} = \sqrt{60^2 + 40^2} = \sqrt{3600+1600} = \sqrt{5200} = 72.11\text{ mm}$$

Plan length $ab$ (run $L$, the front difference $\Delta d$):

$$ab = \sqrt{L^2 + \Delta d^2} = \sqrt{60^2 + 30^2} = \sqrt{3600+900} = \sqrt{4500} = 67.08\text{ mm}$$

Step 4 — True length

The true length is found by rotating the plan to be parallel to $xy$ (rotate $ab$ about $a$) and projecting up to the locus of $b'$. Equivalently, the true length is the hypotenuse of a right triangle whose base is the plan length $ab = 67.08\text{ mm}$ and whose height is $\Delta h = 40\text{ mm}$:

$$TL = \sqrt{ab^2 + \Delta h^2} = \sqrt{67.08^2 + 40^2} = \sqrt{4500 + 1600} = \sqrt{6100} = 78.10\text{ mm}$$

Check using elevation: base $a'b' = 72.11\text{ mm}$, height $\Delta d = 30\text{ mm}$:

$$TL = \sqrt{72.11^2 + 30^2} = \sqrt{5200 + 900} = \sqrt{6100} = 78.10\text{ mm}\ \checkmark$$

True length $TL = 78.10\text{ mm}$.

Step 5 — Inclination with the H.P. ($\theta$)

Using the right triangle on the plan (base $= ab$, opposite $= \Delta h$):

$$\tan\theta = \frac{\Delta h}{ab} = \frac{40}{67.08} = 0.5963 \Rightarrow \theta = 30.8^{\circ}$$

$\theta \approx 30.8^{\circ}$ (inclination with H.P.).

Step 6 — Inclination with the V.P. ($\phi$)

Using the right triangle on the elevation (base $= a'b'$, opposite $= \Delta d$):

$$\tan\phi = \frac{\Delta d}{a'b'} = \frac{30}{72.11} = 0.4160 \Rightarrow \phi = 22.6^{\circ}$$

$\phi \approx 22.6^{\circ}$ (inclination with V.P.).

Step 7 — Traces

Vertical Trace (V.T.): Extend the elevation $a'b'$ to meet $xy$; from this meeting point drop a projector and extend the plan $ab$ to it; the V.T. lies on the extended elevation. The line rises from $A$ (15) to $B$ (55), so extending backwards beyond $A$ toward $xy$, the elevation reaches $xy$ where height $= 0$.

Distance behind $a'$ (along the run) where height becomes 0:

$$x_{VT} = \frac{15}{40}\times 60 = 22.5\text{ mm beyond } a' \text{ (on the lower side)}$$

The V.T. is on $xy$ in plan-projection and at $0$ height — it lies on $xy$. Its distance in front of V.P.:

$$d_{VT} = 20 - \frac{15}{40}\times 30 = 20 - 11.25 = 8.75\text{ mm in front of V.P.}$$

Since at V.T. the point is on the V.P.? No — V.T. is the point where the line (produced) meets the V.P., i.e. distance in front of V.P. $= 0$. Solve for where $\Delta d$-line gives $d=0$ starting from $A$ (20 in front), decreasing toward smaller $d$ requires going backward (toward $A$ and beyond):

$$\text{from } A:\; d = 20 - \frac{30}{60}\,s,\quad d=0 \Rightarrow s = 40\text{ mm behind } A.$$

At $s = 40\text{ mm}$ behind $A$, height $= 15 - \frac{40}{60}\times 40 = 15 - 26.67 = -11.67\text{ mm}$.

So the V.T. is $11.67\text{ mm below the H.P.}$ (the line produced backward dips below H.P. as it pierces the V.P.).

Horizontal Trace (H.T.): point where the line (produced) meets the H.P., i.e. height $= 0$. From $A$ (15 above H.P.), going backward height decreases:

$$h = 15 - \frac{40}{60}\,s,\quad h = 0 \Rightarrow s = 22.5\text{ mm behind } A.$$

At $s = 22.5\text{ mm}$: distance in front of V.P. $= 20 - \frac{30}{60}\times 22.5 = 20 - 11.25 = 8.75\text{ mm}$.

The H.T. lies $8.75\text{ mm in front of the V.P.}$

Summary of results

Given

• Hexagonal prism: base side $a = 25\text{ mm}$, axis (length) $H = 60\text{ mm}$.
• Final position: resting on a base corner, axis inclined $40^{\circ}$ to H.P., axis parallel to V.P.

We use the two-stage method: first draw the prism in the simple position (axis vertical, base on H.P.), then tilt.

Useful base dimensions of a regular hexagon (side 25 mm)

• Across corners (max diagonal) $= 2a = 2 \times 25 = 50\text{ mm}$.
• Across flats (distance between opposite sides) $= a\sqrt{3} = 25 \times 1.732 = 43.30\text{ mm}$.

Stage 1 — Simple position (axis vertical)

1. Draw the top view: a regular hexagon of side $25\text{ mm}$ with two sides horizontal. Label corners $1,2,3,4,5,6$. Across-corner span $1$–$4 = 50\text{ mm}$.
2. Project up to draw the front view: a rectangle of width equal to the across-corners projection ($50\text{ mm}$) and height $= 60\text{ mm}$ (axis vertical). The base edge $a'_1 d'_1$ lies on $xy$, top edge $60\text{ mm}$ above.

  Stage 1 front view (axis vertical)
   __________ top face (60 above xy)
  |          |
  |  axis    |   height = 60
  |    |     |
  |____|_____|  base on xy
  1'        4'

Stage 2 — Tilt the axis to 40° with H.P.

Keep the resting corner ($1$) on $xy$ in the front view. Redraw the Stage-1 front view rotated so the axis makes $40^{\circ}$ with $xy$ (the line of the axis goes up at $40^{\circ}$ from the resting corner).

• Reproduce the Stage-1 front view (same shape) but turned so that the axis line is at $40^{\circ}$ to $xy$; the corner $1'$ stays on $xy$.
• The far end of the axis (top face centre) is then at height

$$h_{top} = 60 \times \sin 40^{\circ} = 60 \times 0.6428 = 38.57\text{ mm above H.P.}$$

• Horizontal advance of the top face along $xy$

$$= 60 \times \cos 40^{\circ} = 60 \times 0.7660 = 45.96\text{ mm}.$$

Stage 2 — Obtain the new top view

Project every corner of the tilted front view downward and carry the width coordinates (perpendicular distances from $xy$ in plan) unchanged from the Stage-1 top view (because tilting about an axis parallel to V.P. preserves distances measured in front of V.P.).

For each corner: new plan point = (its tilted-front-view horizontal position) projected down, meeting the horizontal locus carried across from its Stage-1 top-view position.

   tilted FRONT VIEW (axis at 40 deg)
         o
        /  \        top face
       o    o
        \  /  axis
         X 40deg
   _____/__________________ xy
       1'
        \
  ........projectors down........
   NEW TOP VIEW (hexagonal outline
   foreshortened; resting corner 1 on the
   reference, top face hexagon shifted by
   45.96 mm along xy)

Key projected values to letter on the drawing

Final outcome

The front view shows the rectangular silhouette of the hexagonal prism with its long axis inclined at $40^{\circ}$ to $xy$, the lower base corner touching $xy$ and the top base centre $38.57\text{ mm}$ above $xy$. The top view is the foreshortened hexagonal outline of both end faces joined by the lateral edges, the top-face hexagon offset $45.96\text{ mm}$ along $xy$ from the bottom-face corner. Hidden lateral edges are shown dashed.

Interpretation of the model

• Slab: $90 \times 60 \times 20\text{ mm}$.
• Boss on top, centred: $40 \times 30 \times 35\text{ mm}$.
• Through hole $\varnothing 20\text{ mm}$, axis vertical, central.
• Overall height $= 20 + 35 = 55\text{ mm}$.

In first-angle: object is placed 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.

1. Front view in FULL SECTION (cutting plane through the hole axis)

The cutting plane removes the front half; we see the cut faces hatched and the hole as an open rectangular slot in section.

        |<------- 40 ------->|
        +---+           +---+   ___
        |///|           |///|    |
        |///|  (hole)   |///|    35   boss section
        |///|<-- 20 --> |///|    |     (hatched, hole void in middle)
   _____|///|___________|///|___ _|_
  |/////|///           ///|/////|  |
  |/////|              |/////|   20  slab section
  |/////|______________|/////|  _|_
  |<--------- 90 ----------->|

• Outer outline width $= 90\text{ mm}$, total height $= 55\text{ mm}$.
• Hole shown as a vertical void $20\text{ mm}$ wide running full height $55\text{ mm}$; its walls are visible cut edges (no hatching inside the void).
• All solid cut material is hatched at $45^{\circ}$, uniform spacing.

2. Top view (placed BELOW the front view in first-angle)

   +-----------------------------+   ___
   |         +---------+         |    |
   |         |  +---+  |         |    60   (slab width)
   |         |  |O  |<-- boss 40 x 30, hole O = dia 20 centred
   |         |  +---+  |         |    |
   |         +---------+         |   _|_
   +-----------------------------+
   |<----------- 90 ------------>|

• Outer rectangle $90 \times 60$ (slab).
• Inner rectangle $40 \times 30$ (boss), centred.
• Circle $\varnothing 20$ centred (the hole) with centre lines.

3. Left side view (placed to the RIGHT of the front view in first-angle)

        +-----+    ___
        |//   |     |
        |  ?  |    35   boss (width seen = 30)
        |     |     |
   _____|_____|___ _|_
  |             |    |
  |             |   20  slab (width seen = 60)
  |_____________|  _|_
  |<---- 60 ---->|

• Width seen $= 60\text{ mm}$ (slab depth along Y); boss width seen $= 30\text{ mm}$.
• Total height $= 55\text{ mm}$.
• The hole appears as two dashed (hidden) vertical lines $20\text{ mm}$ apart through boss and slab (this view is not sectioned).

Dimensions to mark

Notes on convention

• First-angle symbol (truncated cone) should be placed in the title block.
• Hatching only on cut solid material in the sectioned front view; the hole void is left blank.
• Centre lines (long-short-long chain) through hole in all three views.
• Hidden edges dashed only where required (side view hole).

Part 2 first — isometric scale (needed for the projection)

Isometric projection uses the isometric scale: actual lengths are foreshortened by the ratio

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

Isometric drawing ignores this foreshortening and uses true lengths directly along the isometric axes (simpler, slightly larger, but the standard for ordinary work).

Isometric length of a $50\text{ mm}$ true edge:

$$50 \times 0.8165 = 40.82\text{ mm}.$$

Isometric-projection length of the 50 mm edge $= 40.82\text{ mm}$.

(Similarly $30\text{ mm} \to 24.49\text{ mm}$, and height $45\text{ mm} \to 36.74\text{ mm}$.)

Part 1 — Isometric view (using an isometric drawing, i.e. true lengths)

Step 1 — Set up isometric axes

Draw the three isometric axes from an origin $O$: one vertical (Z), the other two at $30^{\circ}$ to the horizontal on each side (X to the right-up, Y to the left-up).

Step 2 — Construct the base rectangle

The base $50 \times 30$ becomes a rhombus in isometric:

• Along the $30^{\circ}$ right axis mark the $50\text{ mm}$ edge (longer edge, kept parallel to V.P.).
• Along the $30^{\circ}$ left axis mark the $30\text{ mm}$ edge.
• Complete the parallelogram $ABCD$.

          D________________ C
          /               /
         /   base rhombus /     (50 x 30)
        /_______________/
       A                 B

Step 3 — Locate the base centre and apex

• Draw the diagonals $AC$ and $BD$; their intersection is the base centre $P$.
• From $P$ draw a vertical line (true Z) of length equal to the height $= 45\text{ mm}$ to fix the apex $E$.

            E  (apex, 45 above P)
            |
          D | ____________ C
          / |.          /
         /  P          /
        /_____________/
       A               B

Step 4 — Complete the pyramid

Join the apex $E$ to the four base corners $A, B, C, D$. Show $EA, EB, EC$ as visible; the edge to the hidden rear corner is dashed if obscured. The base edges that are hidden ($DA$ region behind) are drawn dashed.

Summary table

Given

• Cone: base diameter $D = 60\text{ mm}$ (radius $30\text{ mm}$), axis $H = 70\text{ mm}$.
• Section plane $\perp$ V.P., inclined $45^{\circ}$ to H.P., cutting the axis $25\text{ mm}$ above the base.

Step 1 — Identify the type of section

Semi-cone (generator) angle with the base:

$$\alpha = \tan^{-1}\!\left(\frac{H}{R}\right) = \tan^{-1}\!\left(\frac{70}{30}\right) = \tan^{-1}(2.333) = 66.8^{\circ}.$$

The generator makes $66.8^{\circ}$ with the base; the section plane makes $45^{\circ}$ with the base.

Because the cutting plane is inclined to the axis at an angle greater than the semi-vertical angle but is not parallel to a generator (section angle $45^{\circ}$ to base $<$ generator angle $66.8^{\circ}$ to base, and the plane cuts all generators), the section is an ELLIPSE.

Step 2 — Front view & cutting line

1. Draw the front view: an isosceles triangle, base $60\text{ mm}$ on $xy$, apex $70\text{ mm}$ above centre.
2. Mark a point on the axis $25\text{ mm}$ above $xy$. Through it draw the section line (V.T. of the plane) at $45^{\circ}$ to $xy$. Label where it crosses the outline and several generators: points $1', 2', 3', \dots$

              apex
               /\
              /  \
         5'  / 45 \          section line at 45 deg
            X------\----     through axis-point 25 above base
           / \      \
          /   \      \
      ___/_____\______\___ xy  (base, dia 60)

Step 3 — Top view (sectioned)

1. Draw the base circle ($\varnothing 60$) as the top view, divided into $12$ equal generators $1,2,\dots,12$.
2. Project the cut points from the front view down onto the corresponding generators in the top view. At each cut height the cone radius is

$$r = R\left(1 - \frac{z}{H}\right) = 30\left(1 - \frac{z}{70}\right),$$

so each cut point sits at its proper radius along its generator. 3. Join the projected points with a smooth curve — this is the apparent (foreshortened) section in the top view.

Step 4 — True shape of the section

Project the cut points perpendicular to the section line onto an auxiliary reference parallel to the section line. Transfer the widths (half-ordinates) from the top view to each side of the new centre line. Join with a smooth curve.

• Major axis of the ellipse = true length of the cut line $1'5'$ on the front view (measured directly).
• Minor axis = chord width of the cone at the mid-height of the cut, taken from the top view.

Approximate major axis length: the cut runs from where the $45^{\circ}$ line meets the two outline generators. With the axis-point at $z=25$, the chord on the front view from the lower generator to the upper generator measured along the $45^{\circ}$ line gives the major axis; the minor axis equals the cone diameter at the height of the section centroid, $\;2\times 30(1-25/70)=2\times 30(0.643)=38.6\text{ mm}$ (approximate).

   TRUE SHAPE (auxiliary view) — an ellipse
          .-''''''-.
        .'          '.
       (    o   o     )   smooth elliptical curve
        '.          .'
          '-......-'

Result

The true shape of the section is an ELLIPSE (cutting plane inclined to the axis at an angle smaller than the generator's inclination to the axis, cutting all generators). The true shape is obtained by an auxiliary projection perpendicular to the section line.

Part 1 — Regular pentagon, side $40\text{ mm}$ (general method)

1. Draw the given side $AB = 40\text{ mm}$.
2. With $A$ as centre and radius $AB$, draw a semicircle on $AB$; divide the semicircle into $n = 5$ equal parts (same as number of sides) with a protractor or by trial, marking division points $1, 2, 3, 4$ from the end $B$.
3. Join $A$ to the second division point ($A2$); this gives the direction of the next side — point $2$ is the next vertex $C$ direction. (For an $n$-gon always join $A$ to the second division.)
4. With centre $B$ radius $40\text{ mm}$ cut an arc on line $A2$ to locate $C$.
5. From $C$ and successive vertices, strike arcs of radius $40\text{ mm}$; the perpendicular bisectors of $AB$ and $BC$ meet at the circumcentre $O$.
6. With $O$ as centre and $OA$ as radius draw the circumscribing circle; step the chord $40\text{ mm}$ around it to mark the remaining vertices $D$ and $E$.
7. Join $A$–$B$–$C$–$D$–$E$–$A$.

Check: interior angle of a regular pentagon $= \dfrac{(5-2)\times180^{\circ}}{5} = \dfrac{540^{\circ}}{5} = 108^{\circ}$.

Part 2 — Ellipse by the concentric-circles method (major $100$, minor $60$)

1. Draw the two axes intersecting at centre $O$: major axis $AB = 100\text{ mm}$ (so $OA = OB = 50\text{ mm}$), minor axis $CD = 60\text{ mm}$ ($OC = OD = 30\text{ mm}$), perpendicular to each other.
2. With $O$ as centre draw two concentric circles: outer radius $50\text{ mm}$, inner radius $30\text{ mm}$.
3. Divide both circles into the same number of equal angular parts (say $12$) by drawing radial lines from $O$; each radial line cuts the outer circle at point $P_o$ and the inner circle at point $P_i$.
4. From each outer point $P_o$ draw a line parallel to the minor axis (vertical), and from the corresponding inner point $P_i$ draw a line parallel to the major axis (horizontal).
5. Their intersection is a point on the ellipse.
6. Repeat for all radial lines and join the points with a smooth curve (French curve) to obtain the ellipse.

Relation used: a point is $(50\cos\theta,\;30\sin\theta)$, the standard parametric form of the ellipse $\dfrac{x^2}{50^2}+\dfrac{y^2}{30^2}=1$.

Concept

When a circle is tilted with respect to the H.P., its top view is an ellipse. The major axis of that ellipse equals the true diameter (the diameter that stays parallel to the H.P.), and the minor axis equals the diameter projected (foreshortened) by the tilt.

Given

• True diameter $d = 60\text{ mm}$.
• Inclination of plane to H.P. $\theta = 50^{\circ}$.

Step 1 — Stage 1 (lamina parallel to H.P.)

The top view is a true circle $\varnothing 60$. Divide it into $12$ equal parts $1,2,\dots,12$; project to the front view, which is a straight line ($60\text{ mm}$) on $xy$.

Step 2 — Stage 2 (tilt the front-view line to $50^{\circ}$)

Tilt the front-view edge so it makes $50^{\circ}$ with $xy$, keeping the resting point on $xy$. Project each of the $12$ points down; carry their horizontal coordinates from the Stage-1 top view to obtain the new top view.

Step 3 — Axes of the elliptical top view

• Major axis = true diameter (the diameter parallel to H.P. and perpendicular to the line of tilt):

$$\text{major} = d = \mathbf{60\text{ mm}}.$$

• Minor axis = projection of the diameter lying in the direction of tilt:

$$\text{minor} = d\cos\theta = 60\cos 50^{\circ} = 60 \times 0.6428 = \mathbf{38.57\text{ mm}}.$$

Result

   TOP VIEW = ellipse
        major = 60 mm
      .-------------------.
    (                     )  minor = 38.57 mm
      '-------------------'

The top view is an ellipse with major axis $60\text{ mm}$ and minor axis $38.57\text{ mm}$. (The front view is the inclined straight line of length $60\text{ mm}$ at $50^{\circ}$ to $xy$.)

Part 1 — Single-stroke vertical Gothic capitals, $h = 10\text{ mm}$

"Single-stroke" means each line of the letter is made by one stroke of the pencil/pen of uniform thickness.

(Letters $I$ and $W$/$M$ are the exceptions — $I$ is a single stroke, $W$ and $M$ are wider, about $h$ wide.)

Part 2 — Aligned vs unidirectional dimensioning

Summary: In the aligned system dimension figures follow the slope of the dimension line; in the unidirectional system every figure is horizontal. The unidirectional system is preferred for general engineering and CAD work because it never requires turning the drawing to read a value.

Rule recap

• Front view (elevation) is the projection on the V.P.: distance above H.P. is plotted above $xy$, below H.P. below $xy$.
• Top view (plan) is the projection on the H.P.: distance in front of V.P. is plotted below $xy$, behind V.P. above $xy$.

Point-by-point

Diagram (schematic, single $xy$)

            q  (25 above xy : plan of Q, behind VP)
   r (45 above xy : plan of R)
            p' (30 above xy : elevation of P)   s' (20 above xy)
  ----------+--------+--------+--------+-------- xy
            p (40 below xy : plan of P)
            q' (35 below xy : elevation of Q)
   (R elevation r' on xy ; S plan s on xy)

Quadrant conclusions

1. $P$ — above H.P., in front of V.P. $\Rightarrow$ First quadrant.
2. $Q$ — below H.P., behind V.P. $\Rightarrow$ Third quadrant.
3. $R$ — on H.P. and behind V.P. $\Rightarrow$ lies on the H.P. in the second-quadrant region (plan above $xy$, elevation on $xy$).
4. $S$ — above H.P. and on V.P. $\Rightarrow$ lies on the V.P. between the first and second quadrants (elevation above $xy$, plan on $xy$).

Part 1 — Representative Fraction

Convert to the same unit. Ground distance $5\text{ km} = 5 \times 1000 \times 100 = 500{,}000\text{ cm}$.

$$\text{R.F.} = \frac{\text{drawing length}}{\text{actual length}} = \frac{20\text{ cm}}{500{,}000\text{ cm}} = \frac{20}{500000} = \frac{1}{25000}.$$

R.F. $= \dfrac{1}{25000}$.

Part 2 — Actual length of a $7.5\text{ cm}$ plot

$$\text{Actual length} = \frac{\text{map length}}{\text{R.F.}} = 7.5\text{ cm} \times 25000 = 187{,}500\text{ cm}.$$

Convert: $187{,}500\text{ cm} = 1875\text{ m} = \mathbf{1.875\text{ km}}$.

Part 3 — Scale length for $6\text{ km}$ and scale type

Drawing length needed for a maximum of $6\text{ km}$:

$$6\text{ km} = 600{,}000\text{ cm};\quad \text{drawing length} = 600{,}000 \times \text{R.F.} = 600{,}000 \times \frac{1}{25000} = 24\text{ cm}.$$

Length of scale to construct $= 24\text{ cm}$.

Scale type: To read kilometres and tenths of a kilometre ($0.1\text{ km}$) only, a plain scale is sufficient — main divisions for km, the first main division sub-divided into $10$ equal parts each $= 0.1\text{ km}$. If readings to hundredths ($0.01\text{ km}$) were required, a diagonal scale would be constructed instead.

Part 1 — First-angle vs third-angle projection

Symbols (truncated cone)

  FIRST-ANGLE                 THIRD-ANGLE
   ___                              ___
  /   |====|                   |====|   \
  \___|====|                   |====|___/
  (small circle on left)     (small circle on right)

The truncated cone's smaller end (apex side) points toward the larger circle differently: in first-angle the small-diameter (front) view sits to the left; in third-angle to the right.

Part 2 — Procedure to develop the missing right side view (first-angle)

1. Set up reference lines. Draw $xy$ between front and top views; for the side view draw a vertical reference line to the left of the front view (right side view goes on the left in first-angle).
2. Use a 45° mitre line. From the corner where $xy$ and the side-view vertical reference meet, draw a $45^{\circ}$ mitre line. This transfers the depth (width) dimensions from the top view to the side view.
3. Transfer heights. Project all horizontal edges (heights) horizontally from the front view across to the side-view region — heights are common to front and side views.
4. Transfer depths. Project each depth from the top view down to $xy$, across the $45^{\circ}$ mitre line, then turn it up/horizontal into the side view. (Depths in the top view = widths in the side view.)
5. Locate points. The intersection of each height projector (from front view) with each depth projector (via mitre line from top view) fixes a corner of the side view.
6. Join and finalise. Join the points to outline each step; show hidden edges dashed, add centre lines, and check that the side view aligns in height with the front view and in depth with the top view.

Key alignment rule: Front and side views share heights; top and side views share depths (transferred through the 45° mitre line).