Short questions answer based on the structural design and analysis by limit state method as per IS:456-2000.

1. STRUCTURAL DESIGN - I
GENERAL REQUIREMENTS OF IS:456-2000
LECTURE - 1
MADE BY:
Kavita Singh
Assistant Professor
ITM Vocational University, Vadodara
kavitas@itmvu.in; 9998935324

2. Q.1: What do you mean by “Limit State”?
aNS: aS per IS:456-2000, page No. 67:
The acceptable limit for the safety and serviceability requirements
before failure occurs is called Limit State.
Q.2: Define Characteristics Strength.
aNS: aS per IS: 456-2000, page No.67:
It means that value of strength of the material below which not
more than 5% of the tests results are expected to fall.

3. Q.3: Define Characteristic Load.
aNS: aS per IS: 456 - 2000, page No. 67:
It means that value of load which has a 95 % probability of not
being exceeded during the life of the structure.
Q.4: What is the Design strength of Concrete?
aNS: aS per IS: 456-2000, page No.68:

Fck
FactorSafetyPartial
ConcreteofstrengthsticCharacteri
concreteofstrengthDesign 

4. Q.5: What is the Design Strength of Steel?
aNS: aS per IS: 456-2000, page No. 68:
Q.6: Write down the formula for Design Load.
aNS: aS per IS: 456-2000, p.68:

Fy

FactorSafetyPartial
SteelofStrengthsticCharacteri
SteelforStrengthDesign

F
FdLoadDesign 

5. Q.7: As per IS:456-2000, write down the partial safety factor for
concrete and steel.
aNS: aS per IS:456-2000, page No. 68:
Partial Safety Factor for Concrete = 1.5
Partial Safety Factor for Steel = 1.15
Q.8: Which grade of concrete is used for PCC work?
aNS: M 15 or lower grade of concretes are used for the PCC
work.

6. Q.9: Write down the mix proportions for the different grades of
concrete.
aNS: M 5 = 1:5:10
M 7.5 = 1:4:8
M 10 = 1:3:6
M 15 = 1:2:4
M 20 = 1:1.5:3
M 25 = 1:1:2

7. Q.10: What to consider partial safety factor for loads?
aNS: (a) Error in load assessment.
(b) Possible Overloading.
(c) Load Combination.
(d) Repeated Loading.
(e) Modification in Loads.