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# MCQs on Structures Fundamentals

Following diagram represents ____________

a) typical stress strain diagram

b) lift curve slope

c) drag curve slope

d) drag polar

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a

Explanation: 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.

Determine the value of stress produced if body is subjected to a force of hundred Newton and cross sectional area of 10 m2.

a) 10Pa

b) 15MPa

c) 20GPa

d) 27Pa

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a

Explanation: Stress = force / c/s area = 100/10 = 10Pa.

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) 250N

b) 1500N

c) 0.89KN

d) 890N

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a

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

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) 0.014

b) 10.0014

c) 23.002

d) 0.00215

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a

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

If stain in body is 0.01 in longitudinal direction and has length of 52 mm then, find the deformation in length of the body.

a) 0.52mm

b) 4m

c) 2m

d) 22mm

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a

Explanation: Change in length = original length*strain = 52*0.01 = 0.52mm.

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

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a

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

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

a) 1.67MPa

b) 123678Pa

c) 120KPa

d) 1.219MPa

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a

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

Following diagram represents _______________

a) typical bending

b) twisted metal

c) shear

d) tension

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a

Explanation: 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.

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

a) 9 Pa

b) 10.987 Pa

c) 23.43

d) 12.55

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a

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

Which of the following is correct?

a) G = 0.5*E / (1+μ)

b) G = 0.5*E

c) G = 0.5*E*(1+μ)

d) G = 0.5 / (1+μ)

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a

Explanation: Above equation is showing a typical relationship between Shear modulus G, elastic constant E and Poisson’s ratio µ. Correct relationship is given as follows: G = 0.5*E / (1+μ) where, G = shear modulus, E = elastic constant or Young’s modulus, μ = Poisson’s ratio.

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

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a

Explanation: Shear modulus = Elastic constant / (2+2*Poisson’s ratio)

= 100 MPa / (2+2*10) = 4.55 MPa.