BE Civil Engineering (IOE, TU) Building Technology (IOE, CE 502) Question Paper 2078 Nepal

Functions of a foundation

• Distributes the superimposed load over a larger soil area so that the intensity of pressure on the soil stays within the safe bearing capacity.
• Provides a level, stable base for the superstructure and anchors it against sliding/overturning (wind, seismic).
• Reduces and equalises settlement, preventing differential settlement and cracking.

(a) Plan size of square footing

Total load on soil including footing/soil self-weight allowance:

$$P_{total} = 1200 + 0.10 \times 1200 = 1320\ \text{kN}$$

Required plan area:

$$A_{req} = \frac{P_{total}}{q_{safe}} = \frac{1320}{150} = 8.80\ \text{m}^2$$

Side of square footing:

$$B = \sqrt{8.80} = 2.966\ \text{m} \approx \mathbf{3.0\ m}$$

Provide a 3.0 m × 3.0 m square footing ($A_{prov} = 9.0\ \text{m}^2$).

(b) Net upward soil pressure for structural design

For structural (bending/shear) design, only the net column load is used (the soil self-weight allowance is not carried by the footing slab):

$$q_{net} = \frac{P_{column}}{A_{prov}} = \frac{1200}{9.0} = \mathbf{133.3\ kN/m^2}$$

Check gross pressure $\le$ safe value:

$$q_{gross} = \frac{1320}{9.0} = 146.7\ \text{kN/m}^2 < 150\ \text{kN/m}^2 \quad \checkmark$$

(c) When isolated footing is unsuitable

• When columns are so close together that individual footings would overlap → use a combined footing.
• When soil bearing capacity is very low / loads very heavy so footing area approaches half the plot area → use a raft (mat) foundation.
• (Also: a column near the property line where the footing cannot be placed concentrically.)

(a) Number of risers and treads

$$\text{No. of risers} = \frac{\text{floor height}}{R} = \frac{3300}{150} = 22\ \text{risers}$$

For a dog-legged stair split into two equal flights: 11 risers per flight.

Number of treads = risers − 1 per flight (top tread merges with landing):

$$\text{Treads per flight} = 11 - 1 = 10\ \text{treads}$$

Total treads in the run = $20$.

(b) Flight width, landing width, going check

Stair hall width = $2.6\ \text{m}$. The two flights run side by side with no central gap (dog-legged):

$$\text{Width of each flight} = \frac{2.6}{2} = \mathbf{1.30\ m}$$

Width (depth) of the mid-landing $\ge$ flight width; provide landing depth $= 1.30\ \text{m}$.

Going of one flight (10 treads of 250 mm):

$$\text{Going} = 10 \times 250 = 2500\ \text{mm} = 2.5\ \text{m}$$

Length required = going + landing depth = $2.5 + 1.30 = 3.80\ \text{m} < 5.0\ \text{m}$ available $\checkmark$ (remaining $1.20\ \text{m}$ accommodates the lower-floor approach/landing).

(c) Check $2R + T$

$$2R + T = 2(150) + 250 = 300 + 250 = 550\ \text{mm}$$

This lies within the comfortable range $550$–$600\ \text{mm}$, so the stair is comfortable. Pitch:

$$\theta = \tan^{-1}\!\left(\frac{150}{250}\right) = \tan^{-1}(0.60) = \mathbf{31.0^{\circ}}$$

which is within the recommended $25^{\circ}$–$40^{\circ}$ range for residential stairs $\checkmark$.

(d) Plan sketch (dog-legged)

  +-------------------------------+
  |   MID-LANDING (1.30 x 2.60)   |
  +---------------+---------------+
  | UP flight     | DOWN flight   |
  | 10 treads     | 10 treads     |   each 1.30 m wide
  | ->->->        |        <-<-<- |
  +---------------+---------------+
  |        ENTRY / lower landing   |
  +-------------------------------+
  Total run length used = 3.80 m of 5.0 m hall

(a) Load-bearing vs partition wall

(b) Number of bricks

Nominal brick size with joint = $(190+10)\times(90+10)\times(90+10) = 200 \times 100 \times 100\ \text{mm}$.

Volume of one brick with mortar:

$$V_{unit} = 0.200 \times 0.100 \times 0.100 = 0.002\ \text{m}^3$$

Volume of wall ($230\ \text{mm}$ thick = one-brick):

$$V_{wall} = 4.0 \times 3.0 \times 0.230 = 2.76\ \text{m}^3$$

Number of bricks (no wastage):

$$N = \frac{V_{wall}}{V_{unit}} = \frac{2.76}{0.002} = 1380\ \text{bricks}$$

Add $5\%$ wastage:

$$N_{total} = 1380 \times 1.05 = 1449\ \text{bricks} \approx \mathbf{1449\ bricks}$$

(c) Causes of cracks & prevention

1. Differential settlement of foundation → provide rigid, properly designed footing on uniform soil.
2. Thermal movement / expansion → provide expansion joints in long walls.
3. Moisture movement (shrinkage of mortar / sulphate attack) → use sound bricks, low water-cement mortar, sulphate-resisting cement.
4. Overloading / inadequate wall thickness → design wall thickness for actual loads; provide lintels over openings.

(a) Lean-to roof vs Couple roof

LEAN-TO (single slope)        COUPLE ROOF (two slopes, no tie)
      /|                              /\
     / |                             /  \
    /  | higher wall               /    \
   /___|                          /______\
  low wall                       two opposite walls equal height

• Lean-to: single sloping surface, one wall higher than the other; used for verandahs/sheds.
• Couple roof: a pair of rafters sloping from a ridge to two equal walls, with NO tie at feet; suitable only for small spans (≤ ~3.5 m) as rafters tend to spread.

(b) Slope length, pitch and roofing area

Half span (horizontal run of one slope):

$$\text{run} = \frac{8.0}{2} = 4.0\ \text{m}, \qquad \text{rise} = 2.0\ \text{m}$$

Slope (rafter) length:

$$L_s = \sqrt{(\text{run})^2 + (\text{rise})^2} = \sqrt{4.0^2 + 2.0^2} = \sqrt{16 + 4} = \sqrt{20} = 4.472\ \text{m}$$

Pitch angle:

$$\theta = \tan^{-1}\!\left(\frac{\text{rise}}{\text{run}}\right) = \tan^{-1}\!\left(\frac{2.0}{4.0}\right) = \tan^{-1}(0.5) = \mathbf{26.57^{\circ}}$$

Pitch as a ratio = rise / span = $2.0 / 8.0 = \mathbf{1/4}$.

Total roof covering area (two slopes, each $L_s \times$ length $12\ \text{m}$):

$$A = 2 \times (L_s \times 12.0) = 2 \times (4.472 \times 12.0) = 2 \times 53.67 = \mathbf{107.3\ m^2}$$

(c) Functional requirements of a good roof

1. Adequate strength and stability to carry dead, live, wind and snow loads.
2. Weather resistance — watertight against rain and proper drainage.
3. Thermal insulation and sound insulation for indoor comfort.
4. Durability, fire resistance and economy / ease of maintenance.

Plot area

$$A_{plot} = 15.0 \times 20.0 = 300\ \text{m}^2$$

(a) Maximum ground coverage and total floor area

Max ground coverage (from GCR):

$$A_{GCR} = 0.60 \times 300 = \mathbf{180\ m^2}$$

Max total floor area (from FAR):

$$A_{floor} = \text{FAR} \times A_{plot} = 2.5 \times 300 = \mathbf{750\ m^2}$$

(b) Buildable area after setbacks

Buildable width = plot width − (both side setbacks):

$$15.0 - (1.5 + 1.5) = 12.0\ \text{m}$$

Buildable length = plot length − (front + rear setback):

$$20.0 - (3.0 + 2.0) = 15.0\ \text{m}$$

Buildable (setback-limited) footprint:

$$A_{setback} = 12.0 \times 15.0 = 180\ \text{m}^2$$

Compare: $A_{setback} = 180\ \text{m}^2$ and $A_{GCR} = 180\ \text{m}^2$ — they are equal, so both limits give the same governing footprint of $\mathbf{180\ m^2}$ (neither is more restrictive here).

(c) Maximum number of storeys from FAR

Governing floor plate = $180\ \text{m}^2$.

$$n = \frac{A_{floor}}{\text{floor plate}} = \frac{750}{180} = 4.17$$

Since a fractional storey is not allowed, maximum 4 storeys (built floor area $= 4 \times 180 = 720\ \text{m}^2 \le 750\ \text{m}^2$ $\checkmark$).

(d) Purpose of setbacks and FAR

• Setbacks: provide light, ventilation, fire safety/access, privacy between buildings, space for utilities and future road widening.
• FAR: controls the total built-up intensity on a plot, thereby regulating population density, infrastructure load (water, sewer, traffic) and skyline of the area.

(a) Dampness and its ill-effects

Dampness is the unwanted entry, presence and travel of moisture into the building components (walls, floors, roof) through capillary action, leakage, condensation or rising ground water.

Three ill-effects:

1. Deterioration of plaster, paint and finishes (efflorescence, blistering, peeling).
2. Growth of mould/fungus causing unhealthy living conditions and bad odour.
3. Corrosion of embedded metals and decay of timber, weakening the structure.

(b) Damp Proof Course (DPC)

A DPC is a continuous impervious layer/membrane provided in walls and floors to prevent the entry and upward/downward travel of moisture.

Common materials for horizontal DPC:

• Cement concrete (rich M15/M20) with waterproofing compound — 1:2:4 or richer, often a 40–75 mm layer.
• Bituminous / mastic asphalt or bitumen felt — flexible, fully impervious membrane.

Position: a horizontal DPC is laid in the external wall at the plinth level, about $150\ \text{mm}$ above the finished ground level and below the floor finish, continuous through the full wall thickness so rising damp from the foundation cannot pass above it.

(a) Objectives of plastering & defects

Objectives:

• Provide a smooth, even, hard finished surface that can take paint/decoration.
• Protect masonry from rain, atmospheric agents and improve weather/abrasion resistance.
• Conceal defective workmanship and joints; improve appearance and hygiene.

Two common defects: cracking (crazing / hair cracks) and blistering / peeling (also efflorescence, popping).

(b) Cement and sand quantity for plaster

Wet volume of plaster:

$$V_{wet} = 5.0 \times 3.0 \times 0.012 = 0.18\ \text{m}^3$$

Dry volume (×1.35):

$$V_{dry} = 0.18 \times 1.35 = 0.243\ \text{m}^3$$

Mix $1:6$, total parts $= 1 + 6 = 7$.

Volume of cement:

$$V_{cement} = \frac{1}{7} \times 0.243 = 0.0347\ \text{m}^3$$

Number of cement bags ($1$ bag $= 0.0347\ \text{m}^3$):

$$N = \frac{0.0347}{0.0347} = 1.0\ \text{bag}$$

Cement required ≈ 1 bag ($0.0347\ \text{m}^3$, $\approx 50\ \text{kg}$).

Volume of sand:

$$V_{sand} = \frac{6}{7} \times 0.243 = \mathbf{0.208\ m^3}$$

(a) Components

• (i) Frame: the fixed outer assembly (jambs + head, and sill for windows) fixed to the wall opening that supports the shutters.
• (ii) Shutter: the movable (openable) part of a door/window that swings or slides on hinges/runners.
• (iii) Sill: the bottom horizontal member of a window frame (the lowest member resting on the wall) that throws off water.
• (iv) Mullion: a vertical intermediate member that divides a window/door opening into two or more panes/lights.

(b) Panelled door vs Flush door

(a) Components of a ground floor (bottom → top)

1. Natural / compacted earth (subgrade), well rammed.
2. Sub-base / hard core — broken brick or stone soling.
3. Base concrete (lean cement concrete, e.g. 1:4:8), with DPC / damp-proofing where required.
4. Bedding / screed layer (cement mortar) for the topping.
5. Floor finish / topping (the wearing surface — concrete, tile, terrazzo, etc.).

(b) Comparison of flooring materials

(a) Lintel and its function

A lintel is a horizontal structural member (beam) placed across the top of a door/window opening to support the masonry and loads above it and transfer them to the walls on either side.

Lintel vs Arch:

(b) Length and effective span of lintel

Clear span (opening width) $= 1.2\ \text{m} = 1200\ \text{mm}$.

Total length of lintel = clear span + 2 × bearing:

$$L = 1200 + 2(200) = 1600\ \text{mm} = \mathbf{1.60\ m}$$

Effective span = lesser of:

• (i) clear span + effective depth $= 1200 + 150 = 1350\ \text{mm} = 1.35\ \text{m}$
• (ii) centre-to-centre of bearings $= 1200 + \dfrac{200}{2} + \dfrac{200}{2} = 1200 + 200 = 1400\ \text{mm} = 1.40\ \text{m}$

Effective span $= \min(1.35, 1.40) = \mathbf{1.35\ m}$.

(a) Volume of earthwork in excavation

$$V_{exc} = \text{width} \times \text{depth} \times \text{length} = 0.90 \times 1.20 \times 46$$ $$V_{exc} = 1.08 \times 46 = \mathbf{49.68\ m^3}$$

(b) Volume carted away and retained

$25\%$ removed off-site:

$$V_{away} = 0.25 \times 49.68 = \mathbf{12.42\ m^3}$$

Retained soil ($75\%$):

$$V_{retained} = 0.75 \times 49.68 = \mathbf{37.26\ m^3}$$

(c) Net volume to be back-filled and check

Volume occupied by foundation concrete + below-ground masonry:

$$V_{found} = 0.55 \times 46 = 25.30\ \text{m}^3$$

Net volume of trench available for back-filling:

$$V_{backfill} = V_{exc} - V_{found} = 49.68 - 25.30 = \mathbf{24.38\ m^3}$$

Check: retained soil $= 37.26\ \text{m}^3 > 24.38\ \text{m}^3$ required.

So the $75\%$ retained soil is more than sufficient for back-filling; a surplus of $37.26 - 24.38 = 12.88\ \text{m}^3$ would also need to be disposed of or used in levelling the plinth.