(a) Describe various defects in clay bricks giving sources/causes of defects. 10 marks (b) Draw neat sketches, showing complete…

The five parts require material checks and geometric calculations.

(a) Brick defects. The common defects in clay bricks and their sources are: efflorescence, caused by soluble salts in the clay, water or lime migrating to the surface and crystallising; bloating, caused by over-firing or rapid heating which expands impurities or steam; laminations, caused by air pockets trapped during pugging, insufficient wedging, or organic matter in the clay; cracks, caused by uneven drying, rapid drying, shrinkage, poor curing, or thermal shock; warping, caused by improper stacking, uneven support, differential drying, or poor curing; black core, caused by under-firing, insufficient oxidation in the kiln, or impurities in the clay; spalling, caused by poor curing, frost action, or rapid heating; honeycombing, caused by poor clay, insufficient wedging, or excess sand; bulging, caused by over-firing or impurities in the clay; and glazing, caused by over-firing or fluxing impurities. These defects reduce strength, durability and appearance.

(b) English bond, 1½ brick wall. A 1½ brick wall is 13½ inch (345 mm) thick. In English bond, alternate courses of headers and stretchers are laid, giving a quarter-brick lap. For the plan of the header course, draw the two wall arms at a right angle using the standard brick 9 x 4.5 x 3 in and 10 mm mortar joints. The outer wythe consists of headers, i.e. brick ends exposed on the face; each header is 4.5 in wide on the face and 9 in deep across the wall. The inner wythe consists of stretchers, i.e. long faces parallel to the wall; each stretcher is 9 in long and 4.5 in wide. At the external corner, place a quoin header so that the header course is continuous; at the internal corner, place a queen closer to break the inner stretcher course. For the plan of the next stretcher course, the outer wythe consists of stretchers and the inner wythe consists of headers. At the external corner, a queen closer is used to break the outer stretcher course; at the internal corner, a quoin header is used. The vertical joints in the two courses are staggered, so no vertical joint is continuous from one course to the next. The two plans should be drawn one above the other, with the header course and the stretcher course clearly labelled. The lap is obtained because the header-course joints fall midway between the stretcher-course joints. The sketches should show the 13½ inch thickness, the two wythes, the header/stretcher arrangement, and the corner closers.

(c) Timber preservatives. Type 1 preservatives are tar-oil creosote and coal-tar based; they are highly toxic, dark brown, water resistant, and used for heavy timber, poles, sleepers, and ground-contact members by pressure treatment. They are not recommended for residential construction because of odour and toxicity. Type 2 preservatives are water-borne inorganic salts, especially CCA (chromated copper arsenate) and CCB; they are less toxic than creosote but still hazardous, give a green or brown stain, are used for general construction, outdoor timber, and pressure-treated for good retention. They should not be used where leaching into soil or water is likely. Type 3 preservatives are organic-solvent types, such as pentachlorophenol (penta); they are used for temporary or light-duty protection, indoor timber, and are applied by pressure, dipping, or brushing; they are toxic, less durable, and may darken or stain. They are not suitable for heavy ground contact. IS 401, the code of practice for preservative treatment of timber, classifies these types and prescribes retention, application methods, and toxicity considerations.

(d) Visibility. The dip of horizon is found from the right triangle formed by the eye, the horizon, and the Earth's centre. For height h above sea level and R = 6370 km, the geometrical distance is d = √(2Rh) (with h in km); with the usual refraction correction the working formula is d = 3.857√h km, h in metres. The line of sight just touches the Earth's surface at the horizon; if the sum of the two horizon distances is less than the separation, the top is hidden by the curvature of the Earth. For the lighthouse, h = 62 m, so dL = 3.857√62 = 30.4 km. For the observer, h = 10 m, so do = 3.857√10 = 12.2 km. The maximum distance at which the lighthouse top can be seen is dL + do = 42.6 km. Since the ship is 55 km away, the lighthouse top is not visible. The ship must travel 55 - 42.6 = 12.4 km towards the lighthouse.

(e) Stopping sight distance. The stopping sight distance is the sum of the distance travelled during reaction and the braking distance. On a downgrade, the grade reduces the effective friction, so f - S is used. The calculation assumes a level obstacle and that the driver applies brakes immediately after the reaction time. The speed is 60 km/h = 16.67 m/s. The reaction distance is vt = 16.67 x 0.75 = 12.5 m. For dry pavement, f = 0.4 and S = 0.025, so SSD = 12.5 + (16.67²)/(2 x 9.8 x (0.4 - 0.025)) = 12.5 + 277.8/7.35 = 12.5 + 37.8 = 50.3 m. Since 50.3 m is less than the available 55 m, the driver can stop in time on dry pavement, with about 4.7 m margin. For wet pavement, f = 0.32, so SSD = 12.5 + 277.8/(19.6 x 0.295) = 12.5 + 48.0 = 60.5 m. Since 60.5 m is greater than 55 m, the driver cannot stop in time on a rainy day; the available distance is short by about 5.5 m.

Thus, the lighthouse top becomes visible only after the ship has closed about 12.4 km, and the driver can stop safely only on dry pavement; on wet pavement the required SSD exceeds the available 55 m. These results show that material quality, correct bond, line-of-sight distance, and stopping sight distance are practical controls in construction and highway safety.