CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
142
You visited us 142 times! Enjoying our articles? Unlock Full Access!
Question

For the cube of given dimensions which is fixed at the bottom and acted upon by the force as shown in figure, identify the correct statement(s).
[Poisson's ratio =0.8 and Young's modulus =2×1011 N/m2]


A
Normal stress is 253 N/m2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
Shear stress is 25 N/m2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Longitudinal strain is 12.53×1011
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Magnitude of lateral strain is 103×1011
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
A Normal stress is 253 N/m2
B Shear stress is 25 N/m2
C Longitudinal strain is 12.53×1011
D Magnitude of lateral strain is 103×1011
Given,
Poisson's ratio i.e ν=0.8
Young's modulus =2×1011 N/m2
According to question


Normal stress=Normal forceArea

Normal force is 253 N acting peprendicular to the surface of the cube.
σN=2531=253 N/m2 ...(1)
Shear stress=Shear forceArea
Similarly, 25 N is the shear force, acting parallel to the surface of cube.
σS=251=25 N/m2 ...(2)
Young's modulus has been defined as ratio of stress and strain for the normal direction, at the cross-sectional area where force is acting.
Y=σNεN
εN=2532×1011
εN=12.53×1011

εN is the strain along normal direction, also known as longitudinal strain.

Poisson ratio is defined as,
ν=Lateral strainLongitudinal strain
0.8=εS12.53×1011
εS=103×1011
Hence, magnitude of lateral strain is 103×1011

(a), (b), (c), (d) are correct choices.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon