CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A mild steel bar of square cross-section 40 mm x 40 mm is 400 mm long. It is subjected to a longitudinal tensile stress of 440 N/mm2 and lateral compressive stress of 200 N/mm2 in perpendicular direction. E = 2 x 105 N/mm2, μ = 0.3. What is the approximate elongation of the bar in the longitudinal direction?

A
0.44 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0.88 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.22 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1 mm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 1 mm
The lateral compressive stress is applied on per
perpendicular direction (not directions). So considering 200 N/mm2lateral stress in one direction. Therefore, strain in longitudinal direction

ϵi=1E(σ1μσ2)

=12×105×(440+0.3×200)
=1400
Elongation of bar in longitudinal direction
=ϵi×400
=1 Mm

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Simple Stress and Strain-II
STRENGTH OF MATERIALS
Watch in App
Join BYJU'S Learning Program
CrossIcon