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# Multiple choice question for engineering

## Set 1

1. Effective length of compression member is ________
a) distance between ends of members
b) distance between end point and midpoint of member
c) distance between points of contraflexure
d) distance between end point and centroid of member

Answer: c [Reason:] Effective length of compression member is distance between points of contraflexure. It should be derived from actual length and end conditions.

2. Magnitude of effective length depends upon
a) material of member
b) rotational restraint supplied at end of compression member
d) location where member is used

Answer: b [Reason:] Magnitude of effective length depends upon rotational restraint supplied at end of compression member and upon resistance to lateral movement provided.

3. Which of the following is true?
a) greater the effective length, greater the load carrying capacity
b) smaller the effective length, smaller the load carrying capacity
c) smaller the effective length, more the danger of lateral buckling
d) smaller the effective length, smaller the danger of lateral buckling

Answer: d [Reason:] Smaller the effective length of particular compression member, smaller is the danger of lateral buckling and greater is the load carrying capacity.

4. What is the effective length when both ends of compression member are fixed?
a) 0.65L
b) 0.8L
c) L
d) 2L

Answer: a [Reason:] The effective length of compression member when both ends of compression member are fixed is 0.65L (i.e. L/√2), where L is the length of the member.

5. What is the effective length when both ends of compression member are hinged?
a) 0.65L
b) 0.8L
c) L
d) 2L

Answer: c [Reason:] The effective length of compression member when both ends of compression member are hinged is L, where L is the length of the member.

6. What is the effective length when one end of compression member is fixed and other end is free?
a) 0.65L
b) 0.8L
c) L
d) 2L

Answer: d [Reason:] The effective length of compression member when one end is fixed and other end is free is 2L, where L is the length of the member.

7. What is the effective length when one end of compression member is fixed and other end is hinged?
a) 0.65L
b) 0.8L
c) L
d) 2L

Answer: b [Reason:] The effective length of compression member when one end is fixed and other end is hinged is 0.8L, where L is the length of the member.

8. What is slenderness ratio of compression member?
a) ratio of effective length to radius of gyration
b) ratio of radius of gyration to effective length
c) difference of radius of gyration and effective length
d) product of radius of gyration and effective length

Answer: a [Reason:] The tendency of member to buckle is usually measured by its slenderness ratio. Slenderness ratio of member is ratio of effective length to appropriate radius of gyration (λ = kL/r). This is valid only when column has equal unbraced heights for both axes and end condition is same for particular section.

9. Maximum radius of gyration (minimum slenderness ratio) can be obtained by
b) by spreading material of section towards its axis
c) by spreading material of section away from its axis
d) by spreading material of section at its axis

Answer: c [Reason:] Maximum radius of gyration is obtained when material of section is farthest from centroid i.e. away from its axis.

## Set 2

1. What is castellated beam?
a) beam with no openings in web
b) beam with number of regular openings in web and flange
c) beam with number of regular openings in web
d) beam with number of regular openings in flange

Answer: c [Reason:] A beam with number of regular openings in its web is called castellated beam. A castellated beam is formed by flame cutting a single rolled wide flange beam in a definite predetermined pattern and then rejoining the segments by welding to form a regular pattern of holes in the web.

2. The new rolled section of castellated beam will have depth
a) 50% more than original section
b) 50% less than original section
c) 25% less than original section
d) depth does not change

Answer: a [Reason:] The new rolled section of castellated beam will have depth at least 50% more and its section modulus increases by 2.25 times the original section. This allows the beam to span further than parent rolled section.

3. Castellated beams have ______ shear capacity than original beams
a) shear capacity does not change
b) twice
c) increased
d) reduced

Answer: d [Reason:] Castellated beams have reduced shear capacity. It has reduced shear capacity due to stress concentrations near the openings.

4. Which of the following measures can be taken to improve shear capacity of castellated beams?
a) openings can be made away from neutral axis
b) openings can be made close to neutral axis
c) making cuts in straight manner
d) by not using stiffenings

Answer: b [Reason:] Shear capacity of castellated beams can be improved by making openings close to neutral axis and making cuts in a wavy manner. Stiffening can be provided at load concentrations and reaction points to improve its shear carrying capacity.

5. Which of the following is not an advantage of castellated beam?
a) light in weight
b) can be assembled fast
c) cheaper
d) high fire resistance than original rolled section

Answer: d [Reason:] Castellated beams are light in weight, cheaper, they have relatively high resistance and can be assembled fast at the construction site. They are less fire resistant than normal rolled sections. Castellated beams can very easily be cambered and cranked.

6. In which of the following cases are castellated beam desirable?
a) when more span to be covered than rolled section
b) when beam subjected to substantial concentrated loads
c) when beam to be used as continuous beam
d) when higher fire resistance than rolled section required

Answer: a [Reason:] The section of castellated beam will have more depth and section modulus than original rolled section. This allows the beam to span further than parent rolled section. Castellated beams may not be desirable when beam is subjected to substantial concentrated loads, or when castellated beam is used as a continuous beam across several supports. Castellated beams are less attractive when very high requirements for fire resistance are required because the fire resistant coating has to be around 20% thicker than for rolled sections in order to obtain the same fire resistance as rolled section.

7. Match the pairs

```   (A) Vierendeel mechanism		  (i) caused by heavy loading and short span
(B) Lateral torsional buckling of web  (ii) caused due to excessive horizontal shear
(C) Rupture of welded joint in web	  (iii) due to excessive deformation across openings in web
(D) Web Buckling			  (iv)caused by large shear```

a) A-i, B-ii, C-iii, D-iv
b) A-iv, B-iii, C-ii, D-i
c) A-iii, B-iv, C-ii, D-i
d) A-i, B-iv, C-iii, D-ii

Answer: c [Reason:] There are number of possible modes of failure for castellated beams. Some of them are as follows: (i) Vierendeel mechanism – occurs due to excessive deformation across one of the openings in web and formation of hinges in corners of castellation, (ii) Lateral torsional buckling of web – caused by large shear at welded joint, (iii) Rupture of welded joint in web – caused due to excessive horizontal shear at welded joint in the web, (iv) Web Buckling – caused by heavy loading and short span of beam, this may be avoided at support by filling firt castellation by welding plate in the hole.

8. What are lintels?
a) beams provided in foundation
b) beams on roof of building
c) columns above openings in wall
d) beams above openings in wall

Answer: d [Reason:] Beams provided above the openings in walls to support masonry that comes in between the opening and slab above are called as lintels. It is desirable that lintel is built flush from both the sides of the walls.

9. _____ section is suitable for small openings and _____ section is suitable for large openings
a) flat, I-section
b) I-section, flat
c) angles, flat
d) angles, angles

Answer: a [Reason:] Flats and plate sections are used for small openings. For openings of moderate dimension, back-to-back angles and inverted T-sections are best options. For large openings, channels, I-sections or built-up sections are preferred. If there is any doubt about lateral support from the wall, I-section with plates can be used.

10. Design of lintel is carried out for
a) weight of slab
c) small portion of masonry load above the opening
d) large portion of masonry load above the opening

Answer: c [Reason:] Design loads for lintels are not well defined because it is not certain as how much load from masonry will come over lintel. It is assumed that after setting of mortar, load from masonry is distributed by arch action. Design of lintel is carried out for small portion of masonry load above the opening.

11. When the slab over lintel is above apex of equilateral triangle formed on lintel, load of masonry is considered as

Answer: b [Reason:] When the slab over lintel is above apex of equilateral triangle formed on lintel, the load of masonry in the triangle thus formed is assumed to act over it. When the design load is from triangular portion of masonry , the maximum moment will be Wl/6, where W = triangular load from masonry and l = effective span of lintel.

12. When the slab over lintel is below apex of equilateral triangle formed on lintel, load of masonry is considered as

Answer: a [Reason:] When the slab over lintel is below apex of equilateral triangle formed on lintel, the load of masonry in the rectangle is considered. The load of masonry in the rectangle is assumed to act over by taking length equal to span of lintel and height equal to clear height of slab above the lintel.

## Set 3

1. Which IS code is used for calculating different loads on different structures?
a) IS 800
b) IS 200
c) IS 300
d) IS 875

Answer: d [Reason:] IS 875 (all 5 parts) is recommended by Bureau of Indian Standards for calculating various types of loads on the structure. Part 1 is for dead loads, part 2 for imposed loads, part 3 for wind load, part 4 for snow loads and part 5 for special loads and combinations.

2. Which of the following load is to be considered on liquid retaining structure?
c) earth pressure

Answer: a [Reason:] Hydrostatic load is considered on liquid retaining structures or hydraulic structures. Wave and current load is considered in marine and offshore structure. Earth pressure is considered in basements, retaining walls, column footings, etc. Dynamic load is due to earthquake and wind.

3. What is P-Δ effect?
b) second order moments arising from joint displaced
c) second order moments arising from member deflection
d) load due to shrinkage effect

Answer: b [Reason:] Second order moments arising from joint displaced is called P-Δ effect and second order moments arising from member deflection is called P-δ effect.

4. Match the pair :

```	(A) Mass and gravitational effect		(i) wind load
(B) Mass and acceleration effect 		(ii) load due to settlement
(C) Environmental effects			(iii) imposed load```

a) A-i, B-ii, C-iii
b) A-iii, B-ii, C-i
c) A-iii, B-i, C-ii
d) A-ii, B-iii, C-i

Answer: c [Reason:] Load on structure may be due to following : 1) Mass and gravitational effect : examples of such loads are dead loads, imposed loads, snow and ice loads, earth loads, etc. 2) Mass and acceleration effect : examples of such loads are those caused by earthquake, wind, impact, blasts, etc. 3) Environmental effects : examples of such loads are due to temperature difference, settlement, shrinkage, etc.

5. The probability that a specific load will be exceeded during life of structure depends on _______
a) wind
b) factor of safety
c) partial factor of safety
d) period of exposure

Answer: d [Reason:] The probability that a specific load will be exceeded during life of structure depends on period of exposure. It also depends on magnitude of design load.

b) load which will be exceeded by certain probability during life of structure
c) load which will not be exceeded by certain probability during life of structure

7. Which of the following is not included in imposed load classification?

Answer: b [Reason:] Imposed loads are gravity loads other than dead load and cover factors such as occupancy by people, stored material etc. It is classified into following groups : (i)residential, (ii)educational, (iii)institutional, (iv)assembly halls, (v)office and business buildings, (vi)mercantile buildings, (vii)industrial, (viii)storage buildings.

8. What is the minimum imposed load on roof trusses as per IS code?
a) 0.5 kN/m2
b) 0.4 kN/m2
c) 0.9 kN/m2
d) 0.75 kN/m2

Answer: b [Reason:] As per IS 875, the minimum imposed load on roof truss should be 0.4 kN/m2. For sloping roof upto 10˚, the imposed load is taken as 0.5 kN/m2 if access is not provided and 0.75 kN/m2 if access is provided.

9. For roofs of slope greater than 10˚, the imposed load is reduced by ____ for every degree rise in slope.
a) 0.02 kN/m2
b) 0.05 kN/m2
c) 0.75 kN/m2
d) 0.5 kN/m2

Answer: a [Reason:] As per IS 875, for roofs of slope greater than 10o, the imposed load is taken as 0.75 kN/m2 and reduced by 0.02 kN/m2 for every degree rise in slope.

10. Calculate imposed load on roof truss of span 20m with slope of 20o.
a) 0.75 kN/m2
b) 0.95 kN/m2
c) 0.45 kN/m2
d) 0.55 kN/m2

Answer: d [Reason:] As per IS 875, for roofs of slope greater than 10o, the imposed load is reduced by 0.02 kN/m2 for every degree rise in slope. Therefore, Imposed load = 0.75 – 0.02*(20o-10o) = 0.55 kN/m2.

## Set 4

1. What is the condition for equilibrium in plastic analysis?
a) bending moment distribution defined by assumed plastic hinges must not be in static equilibrium with applied loads and reactions
b) shear force distribution defined by assumed plastic hinges must be in static equilibrium with applied loads and reactions
c) bending moment distribution defined by assumed plastic hinges must be in static equilibrium with applied loads and reactions
d) shear force distribution defined by assumed plastic hinges must not be in static equilibrium with applied loads and reactions

Answer: c [Reason:] The bending moment distribution defined by assumed plastic hinges must not be in static equilibrium with applied loads and reactions.

2. Which of the following is true?
a) ultimate load is reached when a mechanism is formed
b) ultimate load is not reached when a mechanism is formed
c) plastic hinges are not required for beam to form a mechanism
d) frictionless hinges are not required for beam to form a mechanism

Answer: a [Reason:] There must be sufficient number of plastic and frictionless hinges for the beam/structure to form a mechanism. The ultimate load is reached when a mechanism is formed.

3. Which of the following relation is correct?
a) -Mp ≥ M
b) M > Mp
c) M ≥ Mp
d) M ≤ Mp

Answer: d [Reason:] The bending moment in any section of a structure must be less than the plastic moment of the section -Mp ≤ M ≤ Mp.

4. Lowest plastic limit load is obtained when _____
a) only equilibrium condition of plastic analysis is satisfied
b) only equilibrium and mechanism condition of plastic analysis are satisfied
c) only mechanism condition of plastic analysis is satisfied
d) equilibrium, mechanism and plasticity condition of plastic analysis are satisfied

Answer: d [Reason:] The lowest plastic limit load is obtained when all the conditions of equilibrium, mechanism and plasticity of plastic analysis are satisfied.

5. Which load is obtained when equilibrium and mechanism conditions of plastic analysis are satisfied?
b) upper bound solution of true ultimate load
c) lower bound solution of true ultimate load
d) no solution

Answer: b [Reason:] When equilibrium and mechanism conditions of plastic analysis are satisfied, an upper bound solution of true ultimate load is obtained.

6. Which load is obtained when equilibrium and plasticity conditions of plastic analysis are satisfied?
b) upper bound solution of true ultimate load
c) lower bound solution of true ultimate load
d) no solution

Answer: c [Reason:] When equilibrium and plasticity conditions of plastic analysis are satisfied, an lower bound solution of true ultimate load is obtained.

7. What is principle of virtual work?
a) work done by external forces is greater than work done by internal forces
b) work done by external forces is less than work done by internal forces
c) work done by external forces is equal to work done by internal forces
d) work done by internal forces is greater than work done by external forces

Answer: c [Reason:] The principle of virtual work may be stated as if a system of forces in equilibrium is subjected to virtual displacement, the work done by external forces is equal to work done by internal forces.

8. Principle of virtual work is used to satisfy _____
a) mechanism condition
b) equilibrium condition
c) plasticity condition
d) no condition is satisfied

Answer: b [Reason:] The principle of virtual work is used to satisfy equilibrium condition of plastic analysis. If the work condition is formulated correctly the equilibrium condition is automatically satisfied.

9. Virtual work is used to determine _____

Answer: d [Reason:] Virtual work principle is used to determine collapse load. It is the application of principle of least action to study forces and movement of a system.

## Set 5

1. Which of the following factor is considered for classification of cross section?
a) location where member is used
b) width-to-thickness ratio
c) length of member
d) seismic force

Answer: b [Reason:] Cross section are classified into four behavioural groups depending upon the material yield strength, width-to-thickness ratio of individual components (e.g. webs and flanges) within the cross section, and the loading arrangement.

2. What is a plastic section?
a) cross section which can develop plastic moment
b) cross section which can resist seismic force
c) cross section in which buckling can occur
d) cross section which can develop plastic hinges

Answer: d [Reason:] Plastic or class I sections are cross sections which can develop plastic hinges and have a rotation capacity required for failure of structure by formation of plastic mechanism.

3. What is a compact section?
a) cross section which can develop plastic moment resistance
b) cross section which can resist seismic force
c) cross section in which buckling can occur
d) cross section which can develop plastic hinges

Answer: a [Reason:] Compact or class II sections are cross sections which can develop plastic moment resistance, but have inadequate plastic hinge rotation capacity because of local buckling.

4. What is a semi-compact section?
a) cross section which can develop plastic moment resistance
b) cross section which can resist seismic force
c) cross section in which elastically calculated stress in extreme compression fibre can reach yield strength
d) cross section which can develop plastic hinges

Answer: c [Reason:] Semi-compact or class III sections are cross sections in which elastically calculated stress in extreme compression fibre of steel can reach yield strength.

5. What is a slender section?
a) cross section which can develop plastic moment resistance
b) cross section which can resist seismic force
c) cross section in which elastically calculated stress in extreme compression fibre can reach yield strength
d) cross section in which local buckling will occur before yield stress

Answer: d [Reason:] Slender or class IV sections are cross section in which local buckling occurs even before the yield stress is attained in one or more parts of the cross section.

6. Which of the following is correct regarding class I section?
a) They are not fully effective under pure compression
b) They are capable of reaching and maintaining full plastic moment in bending
c) They are not capable of reaching and maintaining full plastic moment in bending
d) They does not exhibit sufficient ductility

Answer: b [Reason:] Class I sections are fully effective under pure compression, and capable of reaching and maintaining full plastic moment in bending and hence used in plastic design. These sections will exhibit sufficient ductility.

7. Match the following design moment capacity with the classes of cross section

```                    Cross section			Design moment capacity
i) Compact			A) Md = Zpfy
ii) Semi-Compact			B) Md < Zefy
iii) Plastic			C) Md = Zefy
iv) Slender```

a) i – B, ii – C, iii – A, iv – B
b) i – A, ii – B, iii – C, iv – B
c) i – A, ii – C, iii – A, iv – B
d) i – C, ii – C, iii – A, iv – B