#### Transcript Construction of Mohr`s Circle - Mechanical and Aerospace

```Mechanics of Materials – MAE 243 (Section 002)
Spring 2008
Dr. Konstantinos A. Sierros
7.4: Mohr’s circle for plane stress
• The transformation equations for plane stress can be represented in
graphical form by a plot known as Mohr’s circle
• This graphical representation is extremely useful because it enables
you to visualize the relationships between the normal and shear stresses
acting on various inclined planes at a point in a stressed body
• Using Mohr’s circle you can also calculate principal stresses,
maximum shear stresses and stresses on inclined planes
• Mohr’s circle is also valid for strains and moments of inertia
* Mohr’s circle is named after the famous German civil engineer Otto Christian
Mohr (1835-1918), who developed the circle in 1882
Equations of Mohr’s circle
The equations of Mohr’s circle can
be derived from the transformation
equations for plane stress
This is the equation of a circle in standard algebraic form. The coordinates are
σx1 and τx1y1 the radius is R and the centre of circle has coordinates σx1 = σaver and
τx1y1 = 0
Two forms of Mohr’s circle
• Mohr's circle can be plotted from the previous equations in either of two
forms.
• In the first form of Mohr’s circle, we plot the normal stress σx1 positive to the
right and the shear stress τx1y1 positive downward (fig 7-14a). The advantage of
plotting shear stress positive downward is that the angle 2θ will be positive
when counterclockwise, which agrees with the positive direction of 2θ
• In the second form of Mohr’s circle, τx1y1 is plotted positive upward but the
angle 2θ is now positive clockwise (fig 7-14b), which is opposite to its usual
positive direction
Both forms of Mohr’s circle can be
used. However, it is easier to visualize
FIG. 7-14 Two forms of
the orientation of the stress element if
Mohr’s circle: (a) 
is positive downward
the positive direction of the angle 2θ is
and the angle 2 is
positive counterthe same in the Mohr’s circle as it is
clockwise, and (b) 
is positive upward and
for the element itself.
x1 y 1
x1 y 1
the angle 2 is positive
clockwise. (Note: The
first form is used in this
book.)
Therefore, we will choose the first
form of Mohr’s circle.
Construction of Mohr’s Circle
• Mohr’s circle can be constructed in a variety of ways, depending upon which
stresses are known and which are to be found
• Let us assume that we know the stresses σx , σy τxy acting on the x and y
planes of an element in plane stress (fig 7-15a)
• The above information is sufficient to construct the circle
• Then, with the circle drawn, we can determine the stresses σx1 , σy1 τx1y1
acting on an inclined element (fig 7-15b)
FIG. 7-15
Construction of
Mohr’s circle for
plane stress
Procedure for constructing Mohr’s Circle
1. Draw a set of coordinate axes
with σx1 as abscissa (positive
to the right) and τx1y1 as
ordinate (positive downward)
2. Locate the center C of the
circle at the point having
coordinates σx1= σaver and
τx1y1=0 (see eqs 7-31a and
7.32)
Procedure for constructing Mohr’s Circle
3. Locate point A, representing
the stress conditions on the x
face of the element shown in
fig 7-15a, by plotting its
coordinates σx1=σx and
τx1y1=τxy. Note that point A on
the circle corresponds to θ=0.
Also, note that the x face of
the element (fig 7-15a) is
labeled ‘A’ to show its
correspondence with point A
on the circle
Procedure for constructing Mohr’s Circle
4. Locate point B, representing
the stress conditions on the y
face of the element shown in
fig 7-15a, by plotting its
coordinates σx1=σy and
τx1y1=-τxy. Note that point B on
the circle corresponds to
θ =90. In addition, the y face
of the element (fig 7-15a) is
labeled ‘B’ to show its
correspondence with point B
on the circle
Procedure for constructing Mohr’s Circle
5. Draw a line from point A to
point B. This line is a
diameter of the circle and
passes through the center C.
Points A and B, representing
the stresses on planes at 90 to
each other (fig 7-15a), are at
opposite ends of the diameter
and therefore are 180 apart on
the circle
Procedure for constructing Mohr’s Circle
6. Using point C as the center,
draw Mohr’s circle through
points A and B. The circle
drawn in this manner has
Example 7-4
At a point on the surface of a pressurized cylinder, the material is subjected to
biaxial stresses σx = 90 MPa and σy = 20 MPa, as shown on the stress element
on fig 7-17a. Using Mohr’s circle, determine the stresses acting on an element
inclined at an angle θ = 30. (consider only the in-plane stresses, and show the
results on a sketch of a properly oriented element.)
Example 7-4. (a) Element in plane
stress, and (b) the corresponding Mohr’s
circle. (Note: All stresses on the circle have
units of MPa.)
FIG. 7-17
Solution
Construction of Mohr’s circle. We begin by setting up the axes for the normal
and shear stresses with σx1 positive to the right and τx1y1 positive downward.
Then we place the center C of the circle on the σx1 axis at the point where the
stress equals the average normal stress given by equation 7-31a;
Solution
Point A, representing the stresses on the x face of the element (θ=0), has
coordinates; σx1 = 90 MPa and τx1y1 = 0
Similarly, the coordinates of point B, representing the stresses on the y face
(θ=90), are; σx1 = 20 MPa and τx1y1 = 0
Solution
Now we can draw the circle through points A and B with center at C and radius
R (using eq. 7-31b) equal to;
Solution
Stresses on an element inclined at θ=30. The stresses acting on a plane oriented
at an angle θ=30 are given by the coordinates of point D, which is at an angle
2θ=60 from point A (fig 7-17b). By inspection of the circle we see that the
coordinates of point D are (using eqs. 7-33a,b);
Solution
In a similar manner, we can find the stresses represented by point D’, which
corresponds to an angle θ = 120 (or 2θ = 240);
Point D’ (using eqs. 7-33a,b);
Solution
These results are shown in the figure below on a sketch of an element oriented
at an angle θ = 30, with all stresses shown in their true directions. Note that the
sum of the normal stresses on the inclined element is equal to σx + σy = (90+20)
MPa = 110 MPa
Example 7-4 (continued). Stresses
acting on an element oriented at an angle
 = 30°
FIG. 7-18