Aircraft Design - Structures and Loads Quiz(MCQ)

1 .

What are the ground loads?

A)

Air loads

B)

Gust loads

C)

Lift and drag

D)

Loads acting due to aircraft motion on ground

Correct Answer : Loads acting due to aircraft motion on ground

Explanation : Loads that are acting due to aircraft motion on the ground is termed as ground loads. Air loads and gust loads are loads acting during the flight phase. Lift and drag are Aerodynamic forces acting on the aircraft. Lift is responsible for lifting the aircraft and drag is a resistive force.

2 .

Almost all the forces acting on aircraft can be categorized as _____

A)

body or surface forces

B)

body forces only

C)

only as surface force

D)

neither body nor surface force

Correct Answer : body or surface forces

Explanation : Typically, all aircraft forces are categorized as body forces or surface forces. Body forces are those forces which are acting through volume of the body. Typical example of body force is force due to gravity. Surface force is acting through surface of an object such as shear force.

3 .

Choose an appropriate incorrect option.

A)

At cruise, lift will be same as weight

B)

Lift is an example of Aerodynamic force

C)

At any given time, an aircraft can only experience any one type of load

D)

Aerodynamic centre is located at quarter chord point for typical subsonic a/c

Correct Answer : At any given time, an aircraft can only experience any one type of load

Explaination : At any given time, aircraft can experience the number of different types of loads. For example, at cruise aircraft experiences airloads, powerplant load etc. At cruise, lift and weight will be equal. Lift is an Aerodynamic force.

4 .

Which of the following is not categorized as powerplant load?

A)

Torque

B)

Taxing

C)

Thrust

D)

Duct pressure

Correct Answer : Taxing

Explanation : Taxing is not an example of power plant load. Powerplant load may include Thrust, torque, duct pressure, vibration etc. Thrust is propulsive force produced by the powerplant of an aircraft.

5 .

Design load is defined as _____

A)

lift and drag only

B)

the lowest load that structure has to withstand

C)

the Highest possible load that structure is designed to withstand without braking

D)

the highest normal stress when strain is only quarter

Correct Answer : the Highest possible load that structure is designed to withstand without braking

Explanation : Design load is the highest possible load that structure is designed to withstand without braking. The aircraft structure is always designed in order to withstand higher loads than the limit loads. Some factor of safety or safety margin is always used. Design load is also termed as the ultimate load.

6 .

The largest load that is expected to be experienced by aircraft is called _________

A)

limit load

B)

shear stress

C)

normal strain

D)

normal stress

Correct Answer : limit load

Explanation : Limit load is the largest load that is expected to be experienced by aircraft. Shear force will produce shear stress. Stress is force per unit area and strain is defined as change in dimensions divided by original dimension.

7 .

Which of the following is an example of air load?

A)

Taxi

B)

Gust

C)

Landing

D)

Wheel stress

Correct Answer : Gust

Explanation : Gust is primarily experienced during flight phase of an aircraft. Gust is experienced during flight in air. Hence, gust is an example of air load. Landing, taxi etc. are examples of the ground loads.

8 .

Load factor is defined as ____________

A)

thrust loading

B)

power to weight ratio

C)

aerodynamic efficiency

D)

ratio of lift produced to the weight of aircraft

Correct Answer : ratio of lift produced to the weight of aircraft

Explanation : Load factor is defined as the ratio of lift produced and the weight of the aircraft. For cruising flight lift is equal to weight. Hence, at such flight condition load factor is unity. Aerodynamic efficiency is defined as ratio of lift to drag. It is directly affected by aerodynamics of the aircraft. Thrust loading is ratio of thrust to weight ratio.

9 .

Following diagram represents _________

A)

drag polar

B)

load factor vs drag

C)

maneuver v-n diagram

D)

thrust required graph

Correct Answer : maneuver v-n diagram

Explaination : Above diagram represents a typical Maneuver characteristics. A typical Maneuver V-N diagram is shown in the above diagram. Drag polar is graphical representation of the aircraft drag properties. Typically it is used to show drag variation. Thrust required is used to oppose the drag produced.

10 .

An aircraft is cruising with lift of 150N. Find the weight of aircraft.

A)

0.150KN

B)

15KN

C)

150KN

D)

123N

Correct Answer : 0.150KN

Explanation : At cruise, Weight W = Lift = 150N = 0.150KN.

11 .

When an aircraft flies in air, it experiences _________

A)

only lift load

B)

only gust load

C)

only drag force

D)

lift, drag, thrust loads, etc

Correct Answer : lift, drag, thrust loads, etc

Explanation : When an aircraft flies in the air, it might experience various types of loads such as lift, drag Thrust, vibration, gust etc. Only single load cannot be experienced in real life. At a time there will be number of loads acting on the aircraft during flight.

12 .

Following diagram represents ________

A)

takeoff speed

B)

v-n diagram landing

C)

drag coefficient vs lift

D)

v-n diagram combined of gust and Maneuver

Correct Answer : v-n diagram combined of gust and Maneuver

Explaination : Above diagram is illustrating a typical V-n diagram. Above diagram is a typical combined V-n diagram of gust and Maneuver. Drag coefficient vs lift coefficient graph is nothing but the drag polar. Takeoff speed is based on takeoff loads.

13 .

My aircraft is flying at particular altitude. At that altitude ratio of actual airspeed to the equivalent airspeed is 1.4 then, what should be the desire value of density ratio?

A)

0.12

B)

0.189

C)

0.51

D)

1.89

Correct Answer : 0.51

Explaination : Density ratio = (1/ratio of actual airspeed and Equivalent airspeed)²
= (1/1.4)² = 0.51.

14 .

The dive speed in v-n diagram represents ______

A)

rate of climb

B)

maximum dynamic pressure

C)

hydraulic actuator speed limit

D)

maximum manifold pressure of supercharger

Correct Answer : maximum dynamic pressure

Explanation : Maximum dynamic pressure is being represented by the dive speed in a typical V-n diagram. Manifold pressure at intake is almost equal to the atmospheric. Supercharger is used to increase manifold pressure by large amounts. Rage of climb is vertical velocity.

15 .

A typical aircraft has level cruise speed of 0.6M. Find the dive speed for given subsonic aircraft.

A)

0.9M

B)

1.9M

C)

330m/s

D)

345m/s

Correct Answer : 0.9M

Explanation : Dive speed = cruise speed + 0.5*cruise speed
= 0.6 + 0.5*0.6 = 0.9M.

16 .

Following diagram represents ________

A)

drag polar

B)

lift curve slope

C)

drag curve slope

D)

typical stress strain diagram

Correct Answer : typical stress strain diagram

Explaination : Above diagram is showing typical relationship between stress and strain. As shown in diagram up to certain values of stress curve follows linear relationship. This is due to hook’s law. Lift curve slope is showing a typical variation in lift coefficient with respect to angle of attack.

17 .

Following diagram represents ____________

A)

shear

B)

typical bending

C)

tension

D)

twisted metal

Correct Answer : typical bending

Explaination : The above diagram is showing a typical bending concept. Bending as shown in the diagram is a combination of tension and compression. As shown in the diagram top portion is experiencing compression. Bottom portion is experiencing tension as seen in the diagram.

18 .

A material has elastic constant of 100MPa and Poisson’s ratio as 10. Find the shear modulus of the material.

A)

4.55MPa

B)

2.345MPa

C)

1.2Gpa

D)

345.23KPa

Correct Answer : 4.55MPa

Explanation : Shear modulus = Elastic constant / (2+2*Poisson’s ratio)
= 100 MPa / (2+2*10) = 4.55 MPa.

19 .

Determine Shear stress produced by material if shear force is 180Pa and area is 20m².

A)

9 Pa

B)

10.987 Pa

C)

12.55

D)

23.43

Correct Answer : 9 Pa

Explanation : Shear stress = shear force / area = 180/20 = 9 Pa.

20 .

If elastic constant of material is 6.6MPa then, find the value of direct stress. Given direct strain is 0.25.

A)

1.219MPa

B)

1.67MPa

C)

120KPa

D)

123678Pa

Correct Answer : 1.67MPa

Explanation : Stress = elastic constant*strain = 6.67 MPa*0.25 = 1.67MPa.

21 .

Consider a body is subjected to some force F which is acting on cross section area of 10m². If stress due to application of the force is 25Pa then, found the value of F.

A)

1500N

B)

890N

C)

250N

D)

0.89KN

Correct Answer : 250N

Explaination : Force or load = stress produced* cross section area = 25*10 = 250N.

22 .

An object is subjected to some force. As a result of the force object undergoes in the deformation. If deformation of length is 0.28mm and Actual length is 20 mm then, find simple strain.

A)

0.00215

B)

0.014

C)

10.0014

D)

23.002

Correct Answer : 0.014

Explaination : Strain = change in dimension / original dimension = 0.28/20 = 0.014.

23 .

An object has direct stress and direct strain of 250Pa and 0.015 respectively. Determine the Young's modulus.

A)

16.6 KPa

B)

177Pa

C)

2345Pa

D)

1265KPa

Correct Answer : 16.6 KPa

Explanation : Young’s modulus = stress/strain = 250/0.015 = 16.6 KPa.

24 .

What is creep?

A)

High lift device

B)

Anti ice device

C)

Under zero stress, it is permanent deformation

D)

Under low sustained stress, it is tendency of some materials to get deform slowly and permanently

Correct Answer : Under low sustained stress, it is tendency of some materials to get deform slowly and permanently

Explanation : Creep is nothing but the tendency of some materials to get deform slowly and permanently under low but sustained stress. For most of materials used in aerospace, creep is a problem at elevated temperatures. Although some titanium, plastics will exhibit creep at room temperatures.

25 .

Let us consider the ambient air temperature is 25°C and aircraft has Mach number of 1.5. Find the stagnation temperature.

A)

123.56K

B)

213.458K

C)

432.32K

D)

789K

Correct Answer : 432.32K

Explaination : Given, ambient air temperature T1 = 25°C = 25+273.15 = 298.15K, Mach number = 1.5
Stagnation temperature for air is given by,
Stagnation temperature = T1*(1+0.2*M²)
= (25+273.15)*(1+0.2*1.5²) = 432.32K.

26 .

What is the advantage of wood?

A)

Easy to fabricate

B)

Non availability

C)

Moisture sensitivity

D)

Complexity during fabrication

Correct Answer : Easy to fabricate

Explanation : Woods are easy to fabricate and repair. This is one of the advantages of the wood. Wood has some drawbacks as well such as moisture sensitivity. Today wood is largely used in homebuilt aircrafts.

27 .

Whiskers are _____

A)

isobaric

B)

isotropic

C)

isothermal

D)

isentropic compression

Correct Answer : isotropic

Explanation : Whiskers are Isotropic. Whiskers have same material properties in all directions and hence, termed as isotropic. Isothermal process has constant temperature. Isobaric process has constant pressure that means pressure will not change during such processes.

28 .

Following diagram represents _________

A)

fiber only

B)

matrix only

C)

aluminum alloy

D)

whisker reinforced composite

Correct Answer : whisker reinforced composite

Explaination : Above diagram is showing a typical whisker reinforced composite material. Short strand of the reinforcing materials are randomly located throughout the matrix. Chopped Fiberglass is an example of such arrangement.

29 .

Following diagram represents _______

A)

fiber reinforced polymer

B)

matrix arrangement only

C)

magnesium arrangement

D)

fiber reinforced nitrate only

Correct Answer : fiber reinforced polymer

Explaination : A typical Fiber reinforced polymer is shown in the diagram. It is also called filament reinforced composite. Most of the aircraft structure are made of the frp composite. They provide much better strength to weight ratio.

30 .

Magnesium is used for ________

A)

lofting

B)

drafting

C)

control hinges

D)

highest strength application

Correct Answer : control hinges

Explanation : Magnesium is used in control hinges, wheels, engine mounts etc. Magnesium has good strength to weight ratio. It can tolerate high temperatures and is easily formed. Magnesium can be easily formed by casting, forging etc.

31 .

For high strength applications ____ is widely used.

A)

1098 gallium

B)

7075 aluminum alloy

C)

7089 magnesium zinc

D)

Ge-Zn alloy

Correct Answer : 7075 aluminum alloy

Explanation : For high strength application 7075 aluminum alloy is widely used. 7075 is alloyed with zinc, copper and magnesium. Corrosion resistance is lessened by alloying. Aluminum sheet is frequently clad with a thin layer of pure aluminum.

32 .

Determine the actual length of the column which has slenderness ratio of 2.9 and radius of gyration as 1.6. Consider column is fixed at both ends.

A)

0.0045

B)

2.3674

C)

4.862

D)

9.28

Correct Answer : 9.28

Explanation : Given, column is fixed at both ends.
Actual length = slenderness ratio*radius of gyration/0.5 = 2.9*1.6/0.5 = 9.28.

33 .

If a column has slenderness ratio of 2.2 and 1.8 as the radius of gyration. Determine effective length of the column.

A)

1.09

B)

3.96

C)

4.5

D)

8.98

Correct Answer : 3.96

Explanation : Effective length = slenderness ratio*radius of gyration = 2.2*1.8 = 3.96.

34 .

When we step on an upright soda can, it would fail due to ______

A)

crippling

B)

torsion

C)

shearing

D)

bending

Correct Answer : crippling

Explanation : When we step on an upright soda can, it would fail due to crippling. Here, walls of cross section will collapse without warning and the ability to carry load will become virtually zero. This can be viewed as the short Column with very thin walls.

35 .

A rectangular section has b of 2m and h of 5m. Find centroid coordinates.

A)

(3, 0)

B)

(4, 2)

C)

(1, 2.5)

D)

(2, 4.8)

Correct Answer : (1, 2.5)

Explanation : For rectangle, centroid coordinates = (b/2, h/2)
= (2/2, 5/2) = (1, 2.5).

36 .

How do you define slenderness ratio?

A)

Ratio of lift to drag

B)

Ratio of Thrust to weight

C)

Ratio of slender body width to the bluff body height

D)

Ratio of column’s effective length to c/s radius of gyration

Correct Answer : Ratio of column’s effective length to c/s radius of gyration

Explanation : The ratio of column’s effective length to c/s radius of gyration can be termed as slenderness ratio. The effective length of the column is determined based on the end constraints such as pinned joints, fixed ends etc.

37 .

Column is supported by pin at both the ends and has length of 10 unit. If radius of gyration is 3.16 unit then find slenderness ratio.

A)

0.56

B)

2.13

C)

3.16

D)

5.67

Correct Answer : 3.16

Explanation : Given, column is supported by pin at both the end.
Now, slenderness ratio = effective length/ radius of gyration = 10/3.16 = 3.16.

38 .

The highest compression load which will not cause the elastic column buckling is called _________

A)

euler load

B)

divergence

C)

newtonian fluid

D)

bernoulli’s theorem

Correct Answer : euler load

Explanation : Euler load is defined as the highest compression load which will not cause the elastic column buckling. Euler load is also termed as critical load. Bernoulli’s theorem is used for fluids. Newtonian fluid are those fluids which follow the Newton’s law of viscosity.