Saturday, May 2, 2020

MARKING SCHEM OF EXAM QUESTIONS FOR POST OF SCHOOL CONSTRUCTION ENGINEER


Question 1/ (10pts)
a) What do you understand by Bill of Quantities? Qu’est ce que vous entendez par devis
   quantitatif?
b) Civil engineers use standards codes to design structures; give two (2) of them. / Les
   ingénieurs civil utilisent des codes de normes pour concevoir les structures. Donnez deux
   d’entre elles.
c) Give the difference between a master builder and a master of construction work. /Donner
   la difference entre maitre d’oeuvre et maitre de l’ouvrage.
Answer1:
a) This is a statement of various items of works giving the description, quantities and unit
   rates. The column of rates is normally left unfilled to be filled by contractors. (4pts)
b) British standards, European, ISO standards, American standards, Chinese standards,
   Japanese standards, Australian and New Zealand standards. (2pts)
c) Master builder.: A person who has attained proficiency in one of the building crafts and
   is qualified or licensed to supervise building construction.
   Master of construction work is the client. (4pts)

Question 2/ (15pts)
The quantity of concrete needed in foundation work is 140 m3; considering the ratio 1:2:4 for the
mixture, Calculate the quantity of cement, fine aggregate and coarse aggregate needed/
La quantité du béton qu’on doit utiliser dans la foundation est de 140 m3 ; en considerant la
repartition des ingredients égale a 1:2:4 du mélange. Calculer la quantité du ciment, du sable et
du gravier

Answer 2:
The ratio is 1:2:4 = 7 parts.
Then the amounts required for concreting materials are worked out by first valuing 1 part,
i.e.: 140 m3 /7 = 20 m3
Therefore, cement is 1 part of 20 m3= 20 m3 (2pts)
And the fine aggregate is 2 part of 20 m3 = 40 m3 (2pts)
And the coarse aggregate is 4 part of 20 m3 =80 m3 (2pts)
                                         Check 140 m3
    a) Cement quantity:
        As the cement is approximately 1500 kg/m3,
        Hence, 1500 kg/m3 ×20 m3=30,000 kg of cement
        Add for decrease in volume and waste 42.5% =12,750
b) Fine aggregate:
   Fine aggregate required is40 m3
   Add 50% for decrease in volume and waste = 20 m3
                                                     60 m3
   As the aggregate is approximately 1500 kg/m3,
   Hence 60 m3 × 1500 kg/m3 = 90,000kg = 90 tons of fine aggregate required. (3pts)
c) Coarse aggregate:
        Coarse aggregate required is80 m3
        Add 50% for decrease in volume and waste = 40 m3
                                                        120 m3
        the aggregate is approximately 1500 kg/m3,
therefore, 120 m3 × 1500 kg/m3 = 180,000 kg = 180 tons of aggregate required (3pts)


Question 3:
 Give five types of flexible pavement failures/defects, their causes and
remedial measures / Donnez cinq types de défaillances ou défauts des chaussées asphalté,
leurs causes et les mesures correctives. (25pts)
Answer 3:
a) Corrugation/shoving:
These are in the form of plastic movements typified by ripples (corrugation)or an abrupt
wave (shoving) across the pavement surface. (2pt)
Causes: - mix contamination; (1pt)
- poor mix design (segregation). (1pt)
Repair: resurfacing (1pt)
b) Bleeding:
       It is a film of asphalt binder on the pavement surface which usually create a
shiny, glass-like reflecting surface that can become sticky when dry and slippery
when wet. (2pt)
Causes: - Excessive asphalt binder in the surface layer (poor mix design);
(0.5pt)
-Law surface layer, air void content. (0.5pt)
       Repair: - Minor bleeding can often be corrected by applying coarse sand to
blot up the excess asphalt binder; (1pt)
-Major bleeding can be corrected by cutting off excess asphalt and resurfacing. (1pt)
c) Polished aggregates:
       This is characterized by the areas of the wearing course with the portion of
the aggregate extending above the asphalt binder being either very small or they are
no rough angular aggregate particles. (2pt)
Causes: - Repeated traffic applications; (0.5pt)
- Use of aggregates susceptible to abrasion. (0.5pt)
Repair: - Slurry sealing; (0.5pt)
-Resurfacing; (0.5pt)
-Non-structural overlay. (1pt)
d) Alligator cracking:
These are series of interconnected cracks caused by fatigue of the surface layer under
repeated traffic loading. (2pt)
Causes: - Repeated traffic loading; (0.5pt)
- Stripping; (0.5pt)
- Poor construction (inadequate compaction). (0.5pt)
Repair: - Sealing; (0.5pt)
-Overlaying; (0.5pt)
-Patching. (0.5pt)

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