BE Civil Engineering (IOE, TU) Building Technology (IOE, CE 502) Question Paper 2077 Nepal

Given: $P_A = 800\ \text{kN}$, $P_B = 1200\ \text{kN}$, column spacing $= 4.0\ \text{m}$ c/c, column size $0.4\ \text{m}\times0.4\ \text{m}$, $\text{SBC} = 150\ \text{kN/m}^2$. Outer face of Column $A$ is on the property line.

Take the property line as origin and measure distances along the centre line of the footing.

• Centre of Column $A$: $x_A = 0.4/2 = 0.2\ \text{m}$ from the property line.
• Centre of Column $B$: $x_B = 0.2 + 4.0 = 4.2\ \text{m}$.

(a) Resultant load and its position

$$R = P_A + P_B = 800 + 1200 = 2000\ \text{kN}$$

Taking moments about the property line:

$$\bar{x} = \frac{P_A x_A + P_B x_B}{R} = \frac{800(0.2) + 1200(4.2)}{2000} = \frac{160 + 5040}{2000} = 2.6\ \text{m}$$

The resultant acts 2.6 m from the property line.

(b) Proportioning the rectangular footing For uniform soil pressure the centroid of the footing plan must lie under the resultant. For a rectangle the centroid is at mid-length, and since one end starts at the property line:

$$\frac{L}{2} = \bar{x} \;\Rightarrow\; L = 2 \times 2.6 = 5.2\ \text{m}$$

Required plan area for uniform pressure:

$$A_{req} = \frac{R}{\text{SBC}} = \frac{2000}{150} = 13.33\ \text{m}^2$$

Required width:

$$b_{req} = \frac{A_{req}}{L} = \frac{13.33}{5.2} = 2.56\ \text{m}$$

Adopt $b = 2.6\ \text{m}$.

Pressure check

$$A_{prov} = L\times b = 5.2 \times 2.6 = 13.52\ \text{m}^2$$ $$q = \frac{R}{A_{prov}} = \frac{2000}{13.52} = 147.93\ \text{kN/m}^2$$

Since $q = 147.93\ \text{kN/m}^2 \le \text{SBC} = 150\ \text{kN/m}^2$, the footing is safe and the pressure is essentially uniform (the small eccentricity is removed because the centroid coincides with the resultant).

Property line
   |<----------------- L = 5.2 m ----------------->|
   | A(800)            B(1200)                       |
   | x=0.2 m           x=4.2 m                        |
   |####|#############|####|##########################|  width b = 2.6 m
         R = 2000 kN at x = 2.6 m (centroid of plan)

Final answer: Resultant at 2.6 m from the property line; adopt a rectangular combined footing $5.2\ \text{m}\times2.6\ \text{m}$ giving uniform pressure $\mathbf{147.93\ kN/m^2 < 150\ kN/m^2}$ — safe.

Given: floor-to-floor height $H = 3.6\ \text{m} = 3600\ \text{mm}$, riser $R = 150\ \text{mm}$, tread $T = 300\ \text{mm}$.

(a) Check of riser and tread (comfort rules)

1. $2R + T = 2(150) + 300 = 600\ \text{mm}$ — lies within the recommended $550$–$650\ \text{mm}$ ✓
2. $R + T = 150 + 300 = 450\ \text{mm}$ — within the recommended $400$–$450\ \text{mm}$ ✓
3. $R \times T = 150 \times 300 = 45000\ \text{mm}^2$ — within the recommended $\approx 41000$–$45000\ \text{mm}^2$ ✓

All three rules are satisfied, so the chosen riser and tread give a comfortable stair.

(b) Number of risers, treads and going Number of risers:

$$n_R = \frac{H}{R} = \frac{3600}{150} = 24$$

Number of treads:

$$n_T = n_R - 1 = 24 - 1 = 23$$

For an open-well stair the $24$ risers are arranged in three flights around the central open well, e.g. $8 + 8 + 8 = 24$ risers per flight.

Going of one flight (a flight of $8$ risers has $8 - 1 = 7$ treads):

$$\text{Going} = (n_{R,\text{flight}} - 1)\times T = 7 \times 300 = 2100\ \text{mm} = 2.1\ \text{m}$$

        Flight 2 (8 risers)
      +------------------------+
      |                        |
  F   |     OPEN  WELL         |  F
  l   |                        |  l
  i 1 |                        |  i 3
  g   |                        |  g
  h   +------------------------+  h
  t          landings              t

Summary table

(c) Two situations where an open-well stair is preferred

1. When a wider stair hall is available and a more spacious, aesthetically pleasing layout with a central well (useful for a lift shaft) is required.
2. When a smoother, more gradual ascent with intermediate quarter-space landings is desired, reducing the abruptness of a single $180^\circ$ turn used in a dog-legged stair.

Final answer: $\mathbf{24}$ risers, $\mathbf{23}$ treads, arranged $8+8+8$ in three flights, going per flight $= \mathbf{2.1\ m}$; adopted $R=150$ mm, $T=300$ mm satisfy all comfort rules.

(a) Classification of roofs and factors of choice

Roofs are broadly classified as:

1. Pitched (sloping) roofs — surface inclined more than $10^\circ$ to the horizontal; e.g. lean-to, couple, couple-close, collar-beam and trussed roofs. Suited to heavy rainfall/snow.
2. Flat roofs — slope less than about $10^\circ$ (just enough for drainage); e.g. RCC slab roofs. Provide usable terrace space.
3. Curved (shell / dome / vault) roofs — used for large column-free spans such as auditoria and assembly halls.

Factors governing the choice of a roof: climate (rainfall, snow, temperature), span to be covered, availability and cost of materials, required appearance, fire resistance, future use of roof space, and the type/importance of the building.

(b) King-post truss

A king-post truss is a triangular timber truss used for pitched roofs. Layout:

              Ridge / apex
                  /\
               P / |\ P     P = principal rafters
                /  |  \
        Strut /   K|   \ Strut   K = king post (central vertical)
              /    |    \
   ----------+-----+-----+----------
   Wall    Tie  king   Tie    Wall
   plate   beam  post  beam   plate

Components and their functions:

Suitable span: A king-post truss is economical for spans of about 5 m to 8 m. (For larger spans up to about 12 m, a queen-post truss with two vertical posts is used instead.)

(a) Cavity wall — definition and construction

A cavity wall (hollow wall) consists of two separate leaves (skins) of masonry — an outer leaf and an inner leaf — separated by a continuous air gap (cavity) of about $50\ \text{mm}$, the two leaves being connected at intervals by metal wall ties.

   Outer leaf       Cavity (≈50 mm)      Inner leaf
   (½-brick, 115)   air gap              (½-brick, 115)
   |======|         |     |              |======|
   |======|  <----- |  o  | wall tie --->|======|
   |======|         |     |              |======|
   |======|=========|  o  |==============|======|  <- DPC level
   ground floor     weep hole at base of cavity

Typical construction:

• Both leaves are usually a half-brick thick ($\approx 115\ \text{mm}$ each), giving an overall wall thickness of about $115 + 50 + 115 = 280\ \text{mm}$.
• Galvanised steel or non-corrodible wall ties are embedded in the bed joints of both leaves at about $900\ \text{mm}$ horizontally and $450\ \text{mm}$ vertically (staggered).
• The cavity is kept clean and is closed at the top, around openings, and at the base above the damp-proof course (DPC).
• Weep holes are left in the outer leaf just above the DPC to drain any water collected in the cavity.

(b) Advantages, wall ties and weep holes

Four advantages of a cavity wall over a solid wall:

1. Damp prevention — the air gap breaks the path of moisture so dampness cannot pass from the outer to the inner leaf.
2. Thermal insulation — the still air in the cavity is a poor conductor of heat, keeping interiors warmer in winter and cooler in summer.
3. Sound insulation — the discontinuous construction reduces transmission of airborne sound.
4. Economy and lighter walls — it gives the above benefits with less material and lighter dead load than an equally insulating solid wall.

Function of wall ties: They bond the two independent leaves together so that they act as one structural unit and share lateral loads, while their drip/twist shape prevents moisture from bridging across the cavity.

Function of weep holes: Small openings in the bed joints of the outer leaf, just above the DPC, that allow rainwater which has penetrated and run down the inside of the outer leaf to drain out of the cavity, keeping the inner leaf dry.

(a) Building byelaws — definition and objectives

Building byelaws are the legal rules and regulations framed by a local authority (municipality / metropolitan city) that control the design, construction, materials, and use of buildings within its jurisdiction.

Objectives (any three):

1. To ensure safety of occupants against structural failure, fire, and earthquakes.
2. To secure adequate light, air, ventilation and sanitation by controlling open spaces, set-backs and room sizes.
3. To regulate the density of development and orderly growth of the area through FAR, ground coverage and height limits.
4. To guide planned development and protect public health, traffic movement and the environment.

(b) Numerical computation

Given: plot area $A_p = 300\ \text{m}^2$, maximum $\text{FAR} = 2.5$, maximum ground coverage $= 65\% = 0.65$.

(i) Maximum permissible total floor area

$$A_{\text{total}} = \text{FAR}\times A_p = 2.5 \times 300 = 750\ \text{m}^2$$

Maximum total floor area $= \mathbf{750\ m^2}$.

(ii) Maximum permissible ground (plinth) area of one floor

$$A_{\text{ground}} = 0.65 \times A_p = 0.65 \times 300 = 195\ \text{m}^2$$

Maximum ground coverage $= \mathbf{195\ m^2}$.

(iii) Number of floors If each floor is built to the full ground coverage of $195\ \text{m}^2$:

$$n = \frac{A_{\text{total}}}{A_{\text{ground}}} = \frac{750}{195} = 3.85$$

Since a fractional floor is not possible and the FAR must not be exceeded, round down:

$$n = 3\ \text{floors at full coverage } (3\times195 = 585\ \text{m}^2)$$

A partial fourth floor of $750 - 585 = 165\ \text{m}^2$ (less than the $195\ \text{m}^2$ coverage limit) may still be added within the FAR.

Final answer: Max total floor area $= \mathbf{750\ m^2}$; max ground area $= \mathbf{195\ m^2}$; 3 full floors (plus a possible reduced 4th floor of $165\ \text{m}^2$) are permissible within the FAR of $2.5$.

Dampness is the presence and movement of unwanted moisture (water) within the components of a building — walls, floors, roofs — which causes deterioration of the structure and an unhealthy interior environment.

Causes / sources of dampness (any four):

1. Rising damp — capillary rise of ground moisture through foundations, walls and floors.
2. Rain penetration — driving rain through exposed walls, parapets, copings and defective joints.
3. Faulty roof drainage / leaking pipes — defective gutters, downpipes and plumbing.
4. Condensation — warm moist indoor air condensing on cooler wall/ceiling surfaces.
5. Defective construction — poor workmanship, inadequate slopes, or absence/breakage of the DPC.

Requirements of an ideal damp-proofing material:

• It should be impervious (completely watertight).
• It should be durable and have a life equal to that of the building.
• It should be strong enough to carry the superimposed load without crushing.
• It should be dimensionally stable — should not crack or deform with movement or temperature.
• It should be flexible so as to accommodate small structural movements without rupture.
• It should be free from soluble salts and chemically inert.

Materials commonly used for DPC (any three):

1. Bituminous / asphalt felt (bitumen).
2. Mastic asphalt.
3. Dense cement mortar (1:2 or 1:3) with a waterproofing compound, or dense cement concrete. (Other acceptable: metal sheets such as lead/copper, plastic/polythene sheets, engineering bricks.)

Given: wall length $L = 5.0\ \text{m}$, height $h = 3.0\ \text{m}$, thickness $t = 0.23\ \text{m}$; nominal brick (brick + joint) $= 0.20\ \text{m}\times0.10\ \text{m}\times0.10\ \text{m}$.

Gross volume of wall

$$V_{wall} = L\times h\times t = 5.0 \times 3.0 \times 0.23 = 3.45\ \text{m}^3$$

(a) Number of bricks Volume occupied by one brick including its share of mortar joint (nominal size):

$$v_b = 0.20 \times 0.10 \times 0.10 = 0.002\ \text{m}^3$$

Number of bricks per cubic metre:

$$\frac{1}{0.002} = 500\ \text{bricks/m}^3$$

Total number of bricks:

$$N = 500 \times V_{wall} = 500 \times 3.45 = 1725\ \text{bricks}$$

Allowing $\approx 2\%$ for wastage $\Rightarrow$ provide about $1725\times1.02 \approx 1760$ bricks.

(b) Volume of mortar

$$V_{mortar} = 0.25 \times V_{wall} = 0.25 \times 3.45 = 0.8625\ \text{m}^3$$

(This is the wet volume of mortar in joints.)

Final answer: Number of bricks $\approx \mathbf{1725}$ (say $\mathbf{1760}$ with wastage); mortar volume $= \mathbf{0.86\ m^3}$.

Plastering is the process of applying a thin coat of mortar (cement/lime mortar) over masonry surfaces to give a smooth, durable, hard and finished surface.

Objectives: to conceal defective workmanship and joints, to protect the wall from rain/weather, to provide a smooth even base for painting/decoration, and to improve appearance, durability and hygiene.

Numerical estimate

Given: area $A = 6.0 \times 3.2 = 19.2\ \text{m}^2$, thickness $t = 12\ \text{mm} = 0.012\ \text{m}$, mix $1:6$, dry allowance $30\%$, cement bag $= 0.035\ \text{m}^3$.

Wet volume of mortar

$$V_{wet} = A\times t = 19.2 \times 0.012 = 0.2304\ \text{m}^3$$

Dry volume of mortar (add $30\%$)

$$V_{dry} = 1.30 \times V_{wet} = 1.30 \times 0.2304 = 0.2995\ \text{m}^3$$

Proportioning ($1:6$, sum of parts $= 1 + 6 = 7$)

Cement volume:

$$V_c = \frac{1}{7}\times V_{dry} = \frac{0.2995}{7} = 0.0428\ \text{m}^3$$

Cement in bags:

$$N_{bags} = \frac{V_c}{0.035} = \frac{0.0428}{0.035} = 1.22\ \text{bags}\;\Rightarrow\; \text{say } 1.25\ \text{bags (≈ 61 kg)}$$

Sand volume:

$$V_s = \frac{6}{7}\times V_{dry} = \frac{6}{7}\times 0.2995 = 0.257\ \text{m}^3$$

Final answer: Cement $= \mathbf{1.22\ bags}$ (provide ≈ $1.25$ bags); sand $= \mathbf{0.257\ m^3}$.

Difference between a lintel and an arch

Terms of an arch

• Springing line: the imaginary horizontal line from which the arch curve springs (the level at which the arch starts from the supports/skewbacks).
• Crown: the highest point of the extrados (top) of the arch.
• Voussoir: each of the individual wedge-shaped units (blocks) that make up the arch ring.
• Keystone: the central, topmost voussoir at the crown that locks the other voussoirs in place.

Rise of the semicircular arch For a semicircular arch the curve is a half-circle, so the radius equals half the span and the rise equals the radius:

$$\text{rise} = \frac{\text{span}}{2} = \frac{1.2}{2} = 0.6\ \text{m}$$

Rise $= 0.6\ \text{m}$.

Why an odd number of voussoirs? An odd number of voussoirs allows a single keystone to be placed exactly at the crown (centre) of the arch. This gives a symmetrical arrangement about the centre line, locks the two halves together neatly, and is more pleasing in appearance; an even number would place a joint (instead of a solid block) at the crown, which is structurally and aesthetically less desirable.

Purpose of a plate load test: It is a field test in which a rigid steel plate is loaded in increments at the proposed founding level and the settlement is measured for each load. From the load–settlement curve, the ultimate bearing capacity and the safe bearing capacity of the soil and the probable settlement of the actual footing are estimated. It is a quick in-situ assessment of the soil's behaviour under load.

Estimate of footing settlement

Given: plate width $B_p = 0.30\ \text{m}$, plate settlement $S_p = 8\ \text{mm}$, footing width $B_f = 1.50\ \text{m}$, same pressure intensity, cohesionless soil.

Using the given relation:

$$S_f = S_p\left[\frac{B_f\,(B_p + 0.3)}{B_p\,(B_f + 0.3)}\right]^2$$

Substitute the values:

$$S_f = 8\left[\frac{1.50\,(0.30 + 0.30)}{0.30\,(1.50 + 0.30)}\right]^2 = 8\left[\frac{1.50 \times 0.60}{0.30 \times 1.80}\right]^2$$ $$= 8\left[\frac{0.90}{0.54}\right]^2 = 8\,(1.6667)^2 = 8 \times 2.7778 = 22.22\ \text{mm}$$

Final answer: The probable settlement of the $1.5\ \text{m}$ footing $\approx \mathbf{22.2\ mm}$.

Note: because settlement of footings on sand increases with footing size, the larger footing settles considerably more than the small test plate even under the same pressure intensity.

(a) Requirements of a good window and types

Requirements of a well-designed window:

• It should admit sufficient natural light and provide adequate ventilation (air movement).
• Its size and position should suit the function of the room (rule of thumb: glazed area $\approx$ $\tfrac{1}{8}$ to $\tfrac{1}{10}$ of the floor area).
• It should give a good outside view and privacy where needed.
• It should be weather-tight, durable, easy to operate and clean, and offer security.

Four common types of windows: casement window, sliding window, pivoted window, and bay window. (Also acceptable: sash, louvered/ventilator, dormer, corner windows.)

(b) Components of a panelled door

   +=====================+  <- Head / top rail
   ||  stile |  stile   ||
   ||  +----+   +----+  ||
   ||  |PANE|   |PANE|  ||  <- upper panels (in panels)
   ||  +----+   +----+  ||
   ||========= lock rail ||  <- middle (lock) rail
   ||  +----+   +----+  ||
   ||  |PANE|   |PANE|  ||  <- lower panels
   ||  +----+   +----+  ||
   +=====================+  <- bottom rail
   ^left stile      ^right stile

Principal components:

The shutter so framed is hung on hinges to the door frame (which itself has a head, two posts and a sill/threshold).