BE Civil Engineering (IOE, TU) Engineering Drawing II (IOE, CE 451) Question Paper 2078 Nepal

Given

• Base diameter $D = 60\text{ mm}$, so base radius $R = 30\text{ mm}$.
• Axis (vertical height) $h = 80\text{ mm}$.
• Section plane $\perp$ V.P., inclined $45^\circ$ to H.P., passing through the axis at height $40\text{ mm}$.

Step 1 — Slant height (true length of an extreme generator)

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

All generators of a right cone have the same true length $L$; this is the radius used for the development arc.

Step 2 — Development sector angle

The lateral surface develops into a sector of radius $L$. The sector angle is

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

Step 3 — Divide the base circle into 12 equal parts

Label generators $o\text{-}1,\ o\text{-}2,\dots,o\text{-}12$ ($o$ = apex). Each chord subtends $30^\circ$ at the base centre; on the development each generator is separated by

$$\frac{\theta}{12} = \frac{126.4^\circ}{12} = 10.53^\circ.$$

Step 4 — Locate the cut on each generator (true distances from apex)

The cutting plane is a straight line in the front view inclined $45^\circ$, crossing the axis at $y_0 = 40\text{ mm}$ above the base, i.e. $40\text{ mm}$ below the apex. Take the apex as origin; for a generator whose foot is at horizontal offset $x$ from the axis (positive toward the low side of the cut), the plane meets it at a vertical depth below apex of

$$d_v = 40 - x \quad(\text{since the }45^\circ\text{ plane drops }1:1).$$

The radius of the cone at vertical depth $d_v$ below apex is $r = \dfrac{R}{h}d_v = \dfrac{30}{80}d_v$. The horizontal position of a base point on generator $n$ projected to the V.P. is $x_n = R\cos\alpha_n$ where $\alpha_n$ is its angular position. Solving the intersection along each generator gives the true distance from apex along the generator, $\ell_n = \dfrac{L}{h}\,d_{v,n}$, found graphically by projecting the cut point horizontally onto the extreme (true-length) generator $o\text{-}1$.

Worked true lengths from apex (using $\ell = \dfrac{L}{h}d_v$, $\dfrac{L}{h}=\dfrac{85.44}{80}=1.068$):

All $\ell_n < L = 85.44\text{ mm}$, confirming the plane cuts every generator above the base (a complete elliptical section through the axis).

Step 5 — Draw the development of the lower portion

             o (apex)
            /|\
   sector  / | \   radius L = 85.44 mm
  angle   /  |  \  total angle 126.4 deg
         /   |   \
   base arc (full, radius L) at bottom = developed base circle
   cut curve: plot ell_n along each developed generator o-1'...o-7'...o-1

• With centre $o$ and radius $L = 85.44\text{ mm}$ strike the outer base arc; set off the 12 generators at $10.53^\circ$ each so the full sector spans $126.4^\circ$.
• On each developed generator, measure $\ell_n$ from $o$ and mark the cut point.
• Join the cut points with a smooth curve. The retained lower portion is bounded below by the base arc and above by this developed cut curve.

Final Answer

Slant height $L = 85.44\text{ mm}$, development sector angle $\theta = 126.4^\circ$, generator spacing $10.53^\circ$. The cut curve is plotted at apex distances $10.7,\,14.9,\,26.7,\,42.7,\,58.7,\,70.5,\,74.8\text{ mm}$ on generators 1→7; the lower frustum portion lies between the base arc (radius $85.44\text{ mm}$) and this curve.

Given

• Vertical cylinder: diameter $D_1 = 60\text{ mm}$ (radius $30\text{ mm}$), height $90\text{ mm}$, axis vertical.
• Horizontal cylinder: diameter $D_2 = 40\text{ mm}$ (radius $20\text{ mm}$), axis horizontal, $\parallel$ V.P., intersecting the vertical axis at $50\text{ mm}$ height.
• The smaller cylinder pierces the larger one (full penetration): there are two curves of intersection.

Step 1 — Set up the views

• Front view (V.P.): vertical cylinder is a rectangle $60\text{ mm}$ wide, $90\text{ mm}$ high; horizontal cylinder axis is a horizontal line at $y = 50\text{ mm}$, the cylinder appearing as a rectangle $40\text{ mm}$ high crossing it.
• Top view (H.P.): vertical cylinder is a circle of $\varnothing 60\text{ mm}$; the horizontal cylinder is a rectangle of width $40\text{ mm}$ running across the circle.
• Side view: the horizontal cylinder appears as a circle of $\varnothing 40\text{ mm}$ — this is the auxiliary used to obtain generators.

Step 2 — Generator (line) method

Divide the circle of the penetrating (horizontal) cylinder in the side view into 12 equal parts; these define 12 generators running parallel to the horizontal axis. Number them $1\dots12$.

For each generator at side-view height $z_i$ above the horizontal axis (with $z_i = 20\cos\beta_i$, $\beta_i$ = angular position), the generator is a horizontal line in the front view at height

$$Y_i = 50 + 20\cos\beta_i.$$

Its intersection with the vertical cylinder happens where this generator meets the cylinder’s curved surface. In the top view, that surface is the circle of radius $30\text{ mm}$. The generator’s horizontal (depth) offset from the front plane is $u_i = 20\sin\beta_i$; it meets the circle where

$$x_i = \pm\sqrt{30^2 - u_i^2} = \pm\sqrt{900 - (20\sin\beta_i)^2}.$$

Project $x_i$ up to the front view at height $Y_i$ to obtain a point on the intersection curve.

Step 3 — Computed intersection points (front view coordinates)

Using $\beta = 0,30,60,90,120,150,180^\circ$:

Each $\pm x$ gives a point on the right/left curve; the two mirror-image curves form the intersection seen in the front view.

Step 4 — Drawing

   |======|  <- top of vertical cyl (z=90)
   |  ()  |   curve 1 (upper bulge around gen.1, x up to 30)
   |=)  (=|   horizontal cyl band (Y 30..70)
   |  ()  |   curve 2 (lower)
   |======|  <- base (z=0)
   30 wide each side

Plot the seven points (and their mirror partners) on the front view and join with a smooth curve. Because $D_2 < D_1$ the curve is a smooth ‘‘gothic-arch’’ shape, concave toward the vertical axis, on each side of the penetrating cylinder.

Final Answer

Two symmetric curves of intersection are obtained; key front-view points (height $Y$, half-width $x$): $(70,30),(67.3,28.3),(60,24.5),(50,22.4),(40,24.5),(32.7,28.3),(30,30)$ mm. The generator method projects each side-view division across to the top-view circle and up to the matching generator height.

Given

• Internal (clear) room: $4.0\text{ m} \times 3.5\text{ m}$.
• Wall thickness $t = 230\text{ mm} = 0.23\text{ m}$ all round.
• Clear height $3.0\text{ m}$; door $1.0\times2.1\text{ m}$; two windows $1.2\times1.2\text{ m}$, sill $0.9\text{ m}$.
• Suitable scale: 1:50 (1 cm = 0.5 m) for plan and elevations.

Step 1 — Overall (out-to-out) dimensions

$$L_{out} = 4.0 + 2(0.23) = 4.46\text{ m},\qquad B_{out} = 3.5 + 2(0.23) = 3.96\text{ m}.$$

Step 2 — Plinth area (built-up area at plinth level = out-to-out)

$$A_{plinth} = L_{out}\times B_{out} = 4.46 \times 3.96 = 17.66\text{ m}^2.$$

Step 3 — Carpet area (clear usable floor area inside walls)

$$A_{carpet} = 4.0 \times 3.5 = 14.00\text{ m}^2.$$

(For a single room with no internal partitions/columns, carpet area = internal floor area.)

Step 4 — Plan (drawn at 1:50)

  <------------- 4.46 m out-to-out ------------->
  +=====[ window 1.2 ]===[ window 1.2 ]=========+
  |                                             |  0.23 wall
  |        INTERNAL 4.0 x 3.5 (carpet)          | 3.96 m
  |                                             |  out-to-out
  +================[ door 1.0 ]=================+

• Show wall thickness $0.23\text{ m}$ hatched; mark internal $4.0\times3.5\text{ m}$.
• Door opening $1.0\text{ m}$ on one long wall with swing arc; two windows $1.2\text{ m}$ each on the opposite long wall.
• Dimension chains: internal, opening, and overall.

Step 5 — Sectional elevation (section through windows)

  ===== top of wall / lintel band =====   (height build-up below)
  ____ ceiling 3.0 m ____
  |  lintel @ 2.1 m (window head @ 0.9+1.2 = 2.1 m) |
  |  [   window 1.2 high   ]  head 2.1, sill 0.9    |
  |  sill 0.9 m                                     |
  |____ floor / plinth 0.0 (plinth ~0.45 above GL) _|
  foundation: footing below GL (stepped masonry)

• Floor-to-ceiling clear height $3.0\text{ m}$; window sill $0.9\text{ m}$, head $2.1\text{ m}$; lintel over openings.
• Show plinth ($\approx 0.45\text{ m}$ above ground level), DPC at plinth, and stepped foundation footing below GL.

Step 6 — Front elevation (door side)

   roof line / parapet
   +--------------------------------+
   |                                |
   |          [  door  ]            |  door 1.0 x 2.1
   |          |        |            |
   +----------+--------+------------+  ground line

Show door $1.0\times2.1\text{ m}$ centred, plinth band, and overall height to roof/parapet.

Final Answer

Scale 1:50. Out-to-out $4.46\text{ m}\times3.96\text{ m}$. Plinth area $= 17.66\text{ m}^2$; Carpet area $= 14.00\text{ m}^2$. Plan, sectional elevation (through windows, showing sill $0.9$, head $2.1$, ceiling $3.0\text{ m}$) and front elevation drawn with full dimensioning as above.

Definitions

• Picture Plane (PP): the transparent vertical plane on which the perspective is projected; the object is behind it, the observer in front.
• Station Point (SP): the position of the observer’s eye in the top view.
• Ground Line (GL): the line where the PP meets the ground (the horizontal trace of PP in the front view).
• Horizon Line (HL): a horizontal line on the PP at the eye level (height of SP above GL); all horizontal vanishing points lie on it.
• Vanishing Point (VP): the point on the HL where a set of parallel horizontal edges appears to converge.

Two-point perspective

The object is set with its vertical edges parallel to PP (so verticals stay vertical, no vertical VP) and its two sets of horizontal edges inclined to PP. Each set of horizontal edges converges to its own VP on the HL — hence two vanishing points.

Step 1 — Locate the vanishing points (visual-ray method)

The VP of a set of horizontal edges is found by drawing, from SP, a line parallel to that set of edges until it meets the PP; that piercing point is projected down to the HL.

Let SP be $90\text{ mm}$ behind PP (perpendicular distance $d = 90\text{ mm}$).

• For the longer face (edges at $30^\circ$ to PP): the ray from SP parallel to these edges meets PP at a horizontal distance from the foot of SP of

$$x_L = d\tan(30^\circ) = 90 \times 0.5774 = 51.96\text{ mm}\approx 52.0\text{ mm}.$$

• For the shorter face (edges at $60^\circ$ to PP):

$$x_S = d\tan(60^\circ) = 90 \times 1.7321 = 155.9\text{ mm}.$$

The two VPs lie on the HL on opposite sides of the point directly below SP. Their separation along HL is

$$x_L + x_S = 51.96 + 155.9 = 207.8\text{ mm}.$$

(Equivalently, because the two faces are perpendicular, the two rays from SP are at right angles, confirming the geometry: $\tan30^\circ\cdot\tan60^\circ$ relationship with $30^\circ+60^\circ=90^\circ$.)

Step 2 — Heights

• HL is drawn $35\text{ mm}$ above GL (eye level).
• The vertical edge touching PP is drawn true height $30\text{ mm}$ on the PP (a line on PP is in true size).
• All other heights are scaled down by projecting from SP through the object points onto PP.

Step 3 — Construction outline

1. Draw PP (a horizontal line in the top view) and place the object plan behind it with the front vertical edge on PP, longer side $30^\circ$ and shorter side $60^\circ$ to PP.
2. Mark SP $90\text{ mm}$ in front of PP.
3. From SP draw rays parallel to each set of edges → meet PP at $V_L$ (52 mm) and $V_S$ (156 mm); project these down to the HL to get the two VPs.
4. In the front view draw GL and HL ($35\text{ mm}$ apart). Erect the true-height vertical ($30\text{ mm}$) where the edge touches PP.
5. Join the top and bottom of this true-height line to $V_L$ and $V_S$.
6. From SP, draw visual rays to each plan corner; where they pierce PP, project down to the converging lines to fix the perspective corners. Join to complete the perspective view.

Final Answer

$V_L$ is $52.0\text{ mm}$ and $V_S$ is $155.9\text{ mm}$ from the point below SP (opposite sides), so the two vanishing points are $207.8\text{ mm}$ apart on the HL, which sits $35\text{ mm}$ above GL; the true-height vertical of $30\text{ mm}$ is set on the PP.

Given

• Clear span $L = 5.0\text{ m} = 5000\text{ mm}$; section $b\times D = 300\times450\text{ mm}$.
• Bottom steel: $4\,\varnothing16$ (2 straight + 2 bent up at $45^\circ$ near supports).
• Top steel: $2\,\varnothing12$ (hanger/anchor bars).
• Stirrups: $\varnothing8$ two-legged @ $150\text{ mm}$ c/c; clear cover $25\text{ mm}$.

Step 1 — Longitudinal section (drawn ~1:25)

  support                                         support
  |====================================================|  <- 2-12phi top
  |  \____/  bent-up bars rise to top near supports     |
  |  ||||||||||||  stirrups @150 c/c  ||||||||||||||||  |
  |====================================================|  <- 4-16phi bottom
  |<--------------- 5000 clear span ------------------->|

• 2 of the 4 bottom $\varnothing16$ bars are cranked (bent up at $45^\circ$) at about $L/7\approx750\text{ mm}$ from each support face to resist hogging/shear.
• 2 straight $\varnothing16$ run full length; $2\,\varnothing12$ hangers at top tie the stirrups.

Step 2 — Cross-section

   <-------- 300 -------->
   +---------------------+   25 cover
   |  o            o     |  <- 2-12 phi (top)
   |   [ stirrup 8phi ]  |
   |                     |  450
   |  o   o    o    o    |  <- 4-16 phi (bottom)
   +---------------------+

• Effective depth $d = 450 - 25 - 8 - 16/2 = 450 - 25 - 8 - 8 = 409\text{ mm}$.

Step 3 — Number of stirrups

Provide stirrups over the clear span. Effective stirrup length for spacing = span $-$ allowance; using clear span with end stirrups:

$$N = \frac{L}{s} + 1 = \frac{5000}{150} + 1 = 33.33 + 1 = 34.33 \Rightarrow \textbf{35 stirrups}.$$

(Round up to 34 spaces → 35 bars.)

Step 4 — Bar bending schedule (summary)

Cut length of one stirrup (perimeter of $8\,\varnothing$ stirrup with $25\text{ mm}$ cover, two $90^\circ$ hooks of $10\,d$ each):

• Stirrup width $= 300 - 2(25) - 8 = 242\text{ mm}$ (centre-line ~$250$).
• Stirrup depth $= 450 - 2(25) - 8 = 392\text{ mm}$ (centre-line ~$400$).
• Perimeter $\approx 2(242 + 392) = 1268\text{ mm}$; add 2 hooks $\approx 2(10\times8)=160\text{ mm}$ → cut length $\approx 1428\text{ mm}$ per stirrup.

(Crank extra length per bend $= 0.42\times$ vertical rise $\approx 0.42\times392 \approx 165\text{ mm}$ each.)

Final Answer

Effective depth $d = 409\text{ mm}$; number of stirrups $= 35$ (8 mm two-legged @ 150 c/c); stirrup cut length $\approx 1428\text{ mm}$. Longitudinal section shows 2 straight + 2 bent-up $\varnothing16$ bottom bars and $2\,\varnothing12$ top hangers, with the full BBS summarized above.

Given

• Square prism, base side $a = 30\text{ mm}$, height $70\text{ mm}$.
• Section plane inclined $30^\circ$ to H.P., meeting one edge (edge 1) at $40\text{ mm}$.

Step 1 — Development is a rectangle (parallel-line method)

Total developed width $= 4a = 4\times30 = 120\text{ mm}$ (four faces laid flat), height up to the cut on each edge.

Step 2 — Heights of the cut on each edge

The plane rises at $30^\circ$ across the prism. The horizontal travel between adjacent edges is the face width $a = 30\text{ mm}$. Rise per face $= 30\tan(30^\circ) = 30\times0.5774 = 17.32\text{ mm}$.

Taking edge 1 at $40\text{ mm}$ and moving to the diagonally opposite edge through edge 2 (the section line crosses the prism along the direction perpendicular to the plane’s line of intersection). For a plane sloping from edge 1 up toward edge 3 (opposite corner), the horizontal projection across the diagonal $= a\sqrt2$... but along the edge sequence the cut heights are:

Since computed $74.64\text{ mm} > 70\text{ mm}$, edge 3 is cut at the top face ($70\text{ mm}$); the plane exits through the top. Practical developed heights: edge 1 $= 40.0$, edges 2 and 4 $= 57.3\text{ mm}$, edge 3 $= 70\text{ mm}$ (top).

Step 3 — Development

  |    /\          |   top of cut
  |   /  \___      |
  |40 57.3 70 57.3|  heights along edges 1-2-3-4-1
  +---------------+  base line, width 120 mm

Mark the four edge verticals at $0,30,60,90,120\text{ mm}$; set the heights above; join cut points by straight lines (flat faces → straight cut lines).

Final Answer

Developed rectangle width $120\text{ mm}$; rise per face $17.32\text{ mm}$; edge heights: edge 1 $=40.0\text{ mm}$, edges 2 & 4 $=57.3\text{ mm}$, edge 3 $=70\text{ mm}$.

Given

• Vertical square prism: side $50\text{ mm}$.
• Horizontal square prism: side $30\text{ mm}$, axis $\perp$ vertical axis and intersecting it; its faces at $45^\circ$ to V.P.

Step 1 — Number of intersection lines

A square prism has 4 lateral edges. The penetrating prism completely passes through, so each of its 4 edges enters and exits the vertical prism → it cuts the vertical prism’s faces at points along each edge. The intersection appears as two closed figures (one where it enters, one where it exits). Because both are prisms (flat faces), the curves of intersection are made of straight line segments, not curves.

Each closed figure is a polygon formed where the 4 edges of the horizontal prism pierce the faces of the vertical prism — giving up to 8 corner points total (4 per side), joined by straight lines.

Step 2 — Method (edge/line method)

1. Draw the three views: front view (both prisms as rectangles), top view (vertical prism as $50\text{ mm}$ square, horizontal prism crossing it), side view (horizontal prism as a $45^\circ$-tilted square).
2. Number the 4 edges of the penetrating (horizontal) prism in the side view (where it shows as a true square at $45^\circ$).
3. Project each edge as a line across the front view at its known height.
4. In the top view, locate where each such edge meets the faces of the vertical prism (straight $50\text{ mm}$ square sides).
5. Project these piercing points up to the corresponding edge line in the front view → intersection points.
6. Join points lying on the same face of the vertical prism by straight lines (key rule: only join points on the same pair of faces). This yields the two intersection figures.

Step 3 — Result

   front view of vertical prism (50 wide)
   +---------------------------+
   |    /\        /\           |  entry & exit polygons
   |   /  \      /  \          |  (straight-line segments)
   |   \  /      \  /          |
   |    \/        \/           |
   +---------------------------+

Final Answer

The penetration produces two closed polygonal figures of intersection (entry and exit), composed of straight-line segments meeting at up to 8 points; they are found by the edge/line method—numbering the 4 edges of the smaller prism, finding their piercing points on the vertical prism faces in the top view, and projecting to the front view.

Given

• Floor-to-floor height $H = 3.30\text{ m} = 3300\text{ mm}$.
• Trial riser $R \approx 165\text{ mm}$, tread $T = 250\text{ mm}$.

Step 1 — Number of risers

$$N_R = \frac{H}{R} = \frac{3300}{165} = 20 \text{ risers}.$$

Exact riser $= 3300/20 = 165\text{ mm}$ (clean).

Step 2 — Number of treads

Number of treads $= N_R - 1 = 20 - 1 = 19$ treads total. For a dog-legged stair split into two equal flights:

• Risers per flight $= 20/2 = 10$.
• Treads per flight $= 10 - 1 = 9$ (the landing replaces the last tread of each flight).

Step 3 — Check comfort rule $2R + T$

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

The recommended range is $580$–$600\text{ mm}$ → satisfies the rule. (Also $R\times T = 165\times250 = 41{,}250\text{ mm}^2$, within the $40{,}000$–$45{,}000$ range.)

Step 4 — Going (horizontal length) of one flight

With 9 treads per flight:

$$\text{Going} = 9 \times 250 = 2250\text{ mm} = 2.25\text{ m}.$$

Final Answer

20 risers (165 mm each), 19 treads total — 10 risers and 9 treads per flight; $2R+T = 580\text{ mm}$ (acceptable); going of one flight $= 2.25\text{ m}$.

Coordinate systems in AutoCAD

• Absolute coordinates X,Y — measured from the fixed origin $(0,0)$. Every point is given by its true position.
• Relative (incremental) coordinates @dX,dY — measured from the last point entered; the @ symbol means ‘‘relative to previous point.’’
• Polar coordinates @dist<angle — measured from the last point by a distance and an angle (degrees, CCW from positive X).

Drawing a 50 mm horizontal line from (20, 30)

Start command: LINE (or L), then:

All three produce the same horizontal segment from $(20,30)$ to $(70,30)$.

Four common 2D drafting commands

(Others: FILLET, ARRAY, MIRROR, EXTEND, DIMLINEAR.)

Final Answer

Absolute = from origin (70,30); Relative = from last point (@50,0); Polar = distance<angle (@50<0). Four commands: LINE, CIRCLE, TRIM, OFFSET as described.

One-point vs two-point perspective

Suitability

• Interior of a room → one-point perspective. The back wall is set parallel to the PP, so it is drawn in true shape; the side walls, floor and ceiling recede to a single central VP, giving a natural, symmetric view down into the room.
• Exterior corner of a building → two-point perspective. A corner shows two adjacent faces at angles to the observer; with no face parallel to PP, the two wall sets converge to two VPs on the HL, realistically depicting the projecting corner and depth on both sides.

Measuring Point (MP)

The measuring point is a special point on the HL used to transfer true (real) lengths from the PP/ground line onto the receding perspective lines. For a set of edges with vanishing point V, the MP is located on the HL at a distance from V equal to the distance from V to the SP (i.e. MP is found by swinging the SP–V distance onto the HL). True measurements marked on the measuring line are projected through the MP to scale receding distances correctly without separate plan projection.

Final Answer

One-point has a single VP and is best for room interiors (back wall parallel to PP); two-point has two VPs and is best for an exterior building corner (two faces inclined to PP). The measuring point (MP) is a point on the HL used to lay off true lengths along receding (vanishing) lines.

Given

• Column $300\times300\text{ mm}$; axial service load $P = 600\text{ kN}$.
• Safe bearing capacity (SBC) $q = 150\text{ kN/m}^2$.
• Allow $10\%$ extra for footing self-weight.

Step 1 — Total load on soil

$$P_{total} = 1.10 \times 600 = 660\text{ kN}.$$

Step 2 — Required base area

$$A_{req} = \frac{P_{total}}{q} = \frac{660}{150} = 4.40\text{ m}^2.$$

Step 3 — Square footing size

$$B = \sqrt{A_{req}} = \sqrt{4.40} = 2.098\text{ m} \approx \textbf{2.10 m} \times 2.10\text{ m}.$$

Provided area $= 2.10^2 = 4.41\text{ m}^2 > 4.40\text{ m}^2$ → OK.

Step 4 — Soil pressure check

$$q_{actual} = \frac{660}{4.41} = 149.7\text{ kN/m}^2 < 150\text{ kN/m}^2 \;\checkmark$$

Step 5 — Stepped footing detail (sketch)

   |  300x300 column  |
   |==================|
   +--- step 1 (upper)---+      smaller step over the column
  +----- step 2 (lower) -----+  full 2.10 x 2.10 base on soil
  |  reinforcement mesh both ways near bottom  |
  PCC (lean concrete) bedding below

• Two concrete steps reduce depth at edges while keeping required thickness near the column (resists punching & bending).
• Bottom mesh: main bars both ways with $\approx 50\text{ mm}$ clear cover; $75\text{ mm}$ PCC (1:4:8) levelling course below.

Final Answer

Total load $660\text{ kN}$; required area $4.40\text{ m}^2$; provide a square stepped footing $2.10\text{ m}\times2.10\text{ m}$ (actual soil pressure $149.7\text{ kN/m}^2 < 150\text{ kN/m}^2$, safe).