Stress on Inclined Section
Here, in this case, stress on the cross-section cut
by the plane mn can be easily determined as 𝜎𝑥
Stress on Inclined Section
• Here, in this case, stress on the cross-section
cut by the plane pq can not be determined
simply as 𝜎𝑥 .
• Rather both normal stress and shear stress act
on the surface
Stress on Inclined Section
• This shows that a surface can be subjected to
shear stress and normal stress at the same
time
Stress at a Point
• Stress over a surface is usually
determined by dividing the force acting on
the surface by the area.
• When the stress is constant over the
area, the surface is said to be at uniform
stress
• The state of stress at any point on such a
surface will be the same.
• If stress is not uniform, the state of stress
will vary on different points of the surface.
• Stress at any point is the stress
distribution over a differential area
enclosing the point of interest.
Plane Stress
• When the material is in plane stress in the
xy plane, only the x and y faces of the
element are subjected to stresses, and all
stresses act parallel to the x and y axes
• The stress 𝜎𝑥 acts on the 𝑥 face of the
element and the stress 𝜎𝑦 acts on the 𝑦
face of the element
• A shear stress 𝜏 has two subscripts—the
first subscript denotes the face on which
the stress acts, and the second gives the
direction.
❑ What’s the significance?
It is important to find the surface where the maximum stress is
acted upon and the corresponding stress values.
Plane Stress
𝜃
Plane Stress
𝐹𝑥1 = 0
𝐹𝑦1 = 0
Simplifying the above equations and putting 𝜏𝑥𝑦 = 𝜏𝑦𝑥
Plane Stress
By introducing the above trigonometric identities, the above equations become
Plane Stress
Using a Similar approach it can be shown that
Transformation Equations
The addition of the first two equations provide
Principal Stresses
• The maximum and minimum normal stresses are called principal stresses
• The planes on which principle stresses act are called the principal planes.
For Principle stresses:
𝐻𝑒𝑟𝑒, 𝜃𝑝 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑎𝑙 𝑝𝑙𝑎𝑛𝑒𝑠 𝑎𝑟𝑒 𝑙𝑜𝑐𝑎𝑡𝑒𝑑
• There are two values of 2𝜃𝑝
• These values differ by 180°
• Therefore, the angle 𝜃𝑝 has two values that differ by 90°
Principal Stresses
One of the
principal stresses
After simplification
This equation is shown in a previous slide
Principal Stresses
To obtain another principal stresses use the following formula
Now, 𝜎1 and 𝜎2 can be combined into a
single formula for the principal stresses
Maximum Shear Stress
For Maximum Shear stresses:
𝐻𝑒𝑟𝑒, 𝜃𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑠 𝑡ℎ𝑒 𝑜𝑟𝑖𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒
𝑝𝑙𝑎𝑛𝑒𝑠 𝑜𝑓 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑎𝑛𝑑
𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑠
• There are two values of 𝜃𝑠 This equation shows that the planes of
• These two values differ by 90° maximum shear stress occur at 45° to
the principal planes.
From a
previous slide
Maximum Shear Stresses
This equation was shown in a previous slide
Put these values
After
simplification By comparing with
the equations of
principle stresses
Mohr’s Circle
❑ The transformation equations for plane stress can be
represented in graphical form by a plot known as
Mohr’s circle.
………..(i)
By rearranging equation (i)
………..(ii)
………..(iii)
Where,
Squaring both sides of the equations and
then adding equations (i) and (ii)
Mohr’s Circle
Mohr’s Circle
Procedures of Drawing Mohr’s Circle
Mohr’s Circle
Mohr’s Circle-Problem
Using Mohr’s circle, determine the following quantities: (a) the stresses acting on an element inclined at an
angle 𝜃 = 400 , (b) the principal stresses, and (c) the maximum shear stresses. (Consider only the in-plane
stresses, and show all results on sketches of properly oriented elements.
Mohr’s Circle-Problem
Using Mohr’s circle, determine the following quantities: (a) the stresses acting on an element inclined at an
angle 𝜃 = 450 , (b) the principal stresses, and (c) the maximum shear stresses. (Consider only the in-plane
stresses, and show all results on sketches of properly oriented elements.
Combined Stress
❑ Types of Loading ❑ Possible Combinations
• Axial Loading :𝜎=
𝑃 1) Axial & Flexural
𝐴
2) Axial & Torsional
𝑀𝑦
• Flexural Loading :𝜎=
𝐼 3) Flexural & Torsional
𝑇𝜌 4) Axial, Flexural & Torsional
• Torsional Loading :𝜏=
𝐽
• Combines only normal
stresses
• Simplest
Combined Stress (Axial & Flexural)
Determine the resultant normal stresses at A and B at the wall.
𝑃 𝑀𝑦
𝜎=± ±
𝐴 𝐼
Combined Stress (Axial & Flexural)
To avoid interference, a link in a machine is designed so that its cross-sectional area is reduced one half at
section A-B as shown in the Fig. If the thickness of the link is 50 mm, compute the maximum force P that
can be applied if the maximum normal stress on section A-B is limited to 80 MPa.
𝑃 𝑀𝑦
𝜎=± ±
𝐴 𝐼
Combined Stress (Axial & Flexural)
A wooden beam 100 mm by 200 mm, supported as shown in the Figure, carries a load P. What is the
largest safe value of P, if the maximum stress is not to exceed 10 MPa?
𝑃 𝑀𝑦
𝜎=± ±
𝐴 𝐼
Combined Stress (Axial & Flexural)
Determine the largest load P that can be supported by the circular steel bracket shown in the Figure, if the
normal stress on section A-B is limited to 80 MPa.
𝑃 𝑀𝑦
𝜎=± ±
𝐴 𝐼
