A bar of cross-section area A is subjected to equal and opposite tensile forces at its end. Consider a plane section of the bar, whose normal makes an angle 
with axis (axis is along the length) of the bar. Match the Column I with Column II.
Column I | Column II | ||
A. | Shearing stress on plane section | I. | |
B. | Tensile stress on plane section | II. | |
C. | The Tensile stress is maximum for = | III. | sin cos |
D. | The Shearing stress is maximum for = | IV. |
Choose the correct answer from the options given below.
Text Solution
Verified by ExpertsB
The resolved part of F along the normal is the tensile stress on this plane and the resolved part
parallel to the plane is the shearing stress on the plane as shown below.
Given, cross-section area of 
Let cross-section area of plane 
, then,
From 
Area, 
Shearing stress 
Tensile stress 
(area of plane section 
)
Tensile stress or strength will be maximum when 
is maximum, i.e. 
or 
.
Shearing stress will be maximum when 
is maximum, i.e. or 
, i.e. 
.
Hence, A-III, B-I, C-II, D-IV
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