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Definition of normal forms when only primary keys are considered

First Normal Form: It states that the domain of an attribute must include only atomic values and that the values of any attribute in a tuple must be a single value from the domain of that attribute. In other words first normal form does not allow relations with in relation as attribute values within tuples.

Second Normal Form: It is based on concept of full functional dependency. A dependency X-> Y is full functional dependency if after removing any attribute A from X dependency does no hold any more. Else it is called partial dependency.

A relation schema R is said to be in 2NF f every nonprime attribute A in R is fully functional dependent on the primary key of R.

Third Normal Form: It is based on concept of transitive dependency. A functional dependency X->Y in a relation schema R is a transitive dependency if there is a set of attributes of Z, that are neither a candidate key nor a subset of any key of R, and both X->Z and Z->Y hold.

A relation schema is said to be in third normal form if it satisfies second normal form and no nonprime attribute of R is transitively dependent on the primary key.

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The general definitions of 2NF and 3NF are different from general definition because general definition takes into account candidate keys as well. As a general definition of prime atribute, an attribute that is part of any candidate key will be considered as prime. Partial and full functional dependencies and transitive dependencies will be considered with respect to all candidate keys of a relation.

General definition of 2NF: A relation schema R is in second normal form if every non-prime attribute A in R is not partially dependent on any key of R

General definition of 3NF: A relation schema is said to be in 3NF if, whenever a nontrivial functional dependency X->A holds in R, either (a) X is a super key of R, or (b) A is a prime attribute of R

A functional dependency is X-> Y trivial if X is superset of Y else dependency is non trivial.

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QUESTION 6 (a) The bar shown in Figure Q2(a) is subjected to tensile load of 150 Kn. If the stress in the middle portions is limited to 160 N/mm², determine the diameter of the middle portion. Find also the length of the middle portion if the total elongation of the bar is to be 0.25 mm. E E = 2.0 + 105N/mm². (12 marks) 150 KN 10 cm DIA 10 cm DIA 150 KN 45 cm Figure Q6(a) (b) A brass bar, having cross-section area of 900 mm², is subjected to axial forces as shown in Figure Q2(b), in which AB = 0.6 m, BC = 0.8 m, and CD = 1.0 m. Find the total elongation of the bar. E = 1.0 + 105N/mm2 . (8 marks) Page 4 of 5 B D 40 KN 70 KN 20 KN 10 KN Figure Q6(b) (TOTAL = 20 MARKS)

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Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Mechanical Engineering homework question answer, step 1, image 1 Mechanical Engineering homework question answer, step 1, image 2 Step 2 Mechanical Engineering homework question answer, step 2, image 1 Step 3 Mechanical Engineering homework question answer, step 3, image 1

Question: A 250-V, 4-pole, wave-wound d.c. series motor has 782 conductors on itsarmature. It has armature and series field resistance of 0.75 ohm. The motor takesa current of 40 A. Estimate its speed and gross torque developed if it has a flux per pole of 25 mWb Answer: Step 1 Step 2 Step 3

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