plane stress gere

Mechanics of Solids Plane Stress

Yatin Kumar Singh

Plane Stress:

To explain plane stress, we will consider the stress element shown in Fig. 7-1a. This element is infinitesimal in

size and can be sketched either as a cube or as a rectangular parallelepiped. The xyz axes are parallel to the

edges of the element, and the faces of the element are designated by the directions of their outward normals.

For instance, the right-hand face of the element is referred to as the positive x face, and the left-hand face

(hidden from the viewer) is referred to as the negative x face. Similarly, the top face is the positive y face, and

the front face is the positive z face.

FIG. 7-1 Elements in plane stress: (a) three-dimensional view of an element oriented to the xyz axes, (b) two-

dimensional view of the same element, and (c) two-dimensional view of an element oriented to the x1y1z1

axes.

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, as shown in Fig. 7-la. This stress condition is very

common because it exists at the surface of any stressed body, except at points where external loads act on the

surface. When the element shown in Fig. 7-1a is located at the free surface of a body, the z axis is normal to

the surface and the z face is in the plane of the surface.

The symbols for the stresses shown in Fig. 7-1a have the following meanings. A normal stress σ has a

subscript that identifies the face on which the stress acts; for instance, the stress σx acts on the x face of the

element and the stress σy acts on the y face of the element. Since the element is infinitesimal in size, equal

normal stresses act on the opposite faces.

The sign convention for normal stresses is the familiar one, namely, tension is positive and compression is

negative. A shear stress τ has two subscripts—the first subscript denotes the face on which the stress acts,

and the second gives the direction on that face. Thus, the stress τxy acts on the x face in the direction of the y

axis (Fig. 7-1a), and the stress τyx acts on the y face in the direction of the x axis.

The sign convention for shear stresses is as follows. A shear stress is positive when it acts on a positive face of

an element in the positive direction of an axis, and it is negative when it acts on a positive face of an element

in the negative direction of an axis. Therefore, the stresses τxy and τyx shown on the positive x and y faces in

Fig. 7-la are positive shear stresses. Similarly, on a negative face of the element, a shear stress is positive

when it acts in the negative direction of an axis. Hence, the stresses τxy and τyx shown on the negative x and y

faces of the element are also positive.

This sign convention for shear stresses is easy to remember if we state it as follows: A shear stress is positive

when the directions associated with its subscripts are plus-plus or minus-minus; the stress is negative when

the directions are plus-minus or minus-plus. The preceding sign convention for shear stresses is consistent

with the equilibrium of the element, because we know that shear stresses on opposite faces of an

infinitesimal element must be equal in magnitude and opposite in direction. Hence, according to our sign

convention, a positive stress τxy acts upward on the positive face (Fig. 7-1a) and downward on the negative

face. In a similar manner, the stresses τyx acting on the top and bottom faces of the element are positive

although they have opposite directions.


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We also know that shear stresses on perpendicular planes are equal in magnitude and have directions such

that both stresses point toward, or both point away from, the line of intersection of the faces. In as much as

τxy and τyx are positive in the directions shown in the figure, they are consistent with this observation.

Therefore, we note that

τxy = τyx (7-1)

For convenience in sketching plane-stress elements, we usually draw only a two-dimensional view of the

element, as shown in Fig. 7-1b. Although a figure of this kind is adequate for showing all stresses acting on the

element, we must still keep in mind that the element is a solid body with a thickness perpendicular to the

plane of the figure.

Stresses on Inclined Sections:

We are now ready to consider the stresses acting on inclined sections, assuming that the stresses σx, σy, and

τxy (Figs. 7-1a and b) are known. To portray the stresses acting on an inclined section, we consider a new

stress element (Fig. 7-1c) that is located at the same point in the material as the original element (Fig. 7-1b).

However, the new element has faces that are parallel and perpendicular to the inclined direction. Associated

with this new element are axes x1, y1, and z1, such that the z1 axis coincides with the z axis and the x1y1 axes

are rotated counterclockwise through an angle θ with respect to the xy axes.

The normal and shear stresses acting on this new element are denoted σx1, σy1, τx1y1, and τy1x1, using the same

subscript designations and sign conventions described previously for the stresses acting on the xy element.

The previous conclusions regarding the shear stresses still apply, so that

τx1y1 = τy1x1 (7.2)

From this equation and the equilibrium of the element, we see that the shear stresses acting on all four side

faces of an element in plane stress are known if we determine the shear stress acting on any one of those

faces. The stresses acting on the inclined x1y1 element (Fig. 7-1c) can be expressed in terms of the stresses on

the xy element (Fig. 7-1b) by using equations of equilibrium. For this purpose, we choose a wedge-shaped

stress element (Fig. 7-2a) having an inclined face that is the same as the x1 face of the inclined element

shown in Fig. 7-1c. The other two side faces of the wedge are parallel to the x and y axes.

In order to write equations of equilibrium for the wedge, we need to construct a free-body diagram showing

the forces acting on the faces.

FIG. 7-2 Wedge-shaped stress element in plane stress: (a) stresses acting on the element, and (b) forces acting on the element (free-body diagram)


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Let us denote the area of the left-hand side face (that is, the negative x face) as A0. Then the normal and shear

forces acting on that face are σx A0 and τxy A0, as shown in the free-body diagram of Fig. 7-2b. The area of the

bottom face (or negative y face) is A0 tanθ, and the area of the inclined face (or positive x1 face) is A0 secθ.

Thus, the normal and shear forces acting on these faces have the magnitudes and directions shown in Fig. 7-

2b. The forces acting on the left-hand and bottom faces can be resolved into orthogonal components acting in

the x1 and y1 directions. Then we can obtain two equations of equilibrium by summing forces in those

directions. The first equation, obtained by summing forces in the x1 direction, is

In the same manner, summation of forces in the y1 direction gives

Using the relationship τxy = τyx, and also simplifying and rearranging, we obtain the following two equations:

Equations (7-3a) and (7-3b) give the normal and shear stresses acting on the x1 plane in terms of the angle θ

and the stresses σx, σy, and τxy acting on the x and y planes.

For the special case when θ = 0, we note that Eqs. (7-3a) and (7-3b) give σx1 = σx and τx1y1 = τxy, as expected.

Also, when θ = 90°, the equations give σx1 = σy and τx1y1 = -τxy = -τyx. In the latter case, since the x1 axis is

vertical when θ = 90° , the stress τx1y1 will be positive when it acts to the left. However, the stress τyx acts to

the right, and therefore τx1y1 = τyx

Transformation Equations for Plane Stress:

Equations (7-3a) and (7-3b) for the stresses on an inclined section can be expressed in a more convenient

form by introducing the following trigonometric identities:

When these substitutions are made, the equations become

These equations are usually called the transformation equations for plane stress because they transform

the stress components from one set of axes to another. However, the intrinsic state of stress at the point

under consideration is the same whether represented by stresses acting on the xy element (Fig. 7-1b) or by

stresses acting on the inclined x1y1 element (Fig. 7-1c). Since the transformation equations were derived

solely from equilibrium of an element, they are applicable to stresses in any kind of material, whether linear

or nonlinear, elastic or inelastic.

An important observation concerning the normal stresses can be obtained from the transformation

equations. As a preliminary matter, we note that the normal stress σy1 acting on the y1 face of the inclined

element (Fig. 7-1c) can be obtained from Eq. (7-4a) by substituting θ + 90° for θ. The result is the following

equation for σy1:

Summing the expressions for σx1 and σy1 (Eqs. 7-4a and 7-5), we obtain the following equation for plane

stress:


Yatin Kumar Singh

This equation shows that the sum of the normal stresses acting on perpendicular faces of plane-stress

elements (at a given point in a stressed body) is constant and independent of the angle θ.

Special Cases of Plane Stress:

The general case of plane stress reduces to simpler states of stress under special conditions. For instance, if

all stresses acting on the xy element (Fig. 7-1b) are zero except for the normal stress σx, then the element is in

uniaxial stress (Fig. 7-4). The corresponding transformation equations, obtained by setting σy and τxy equal

to zero in Eqs. (7-4a) and (7-4b), are

FIG. 7-4 Element in uniaxial stress FIG. 7-5 Element in pure shear FIG. 7-6 Element in biaxial stress

Another special case is pure shear (Fig. 7-5), for which the transformation equations are obtained by

substituting σx = 0 and σy = 0 into Eqs. (7-4a) and (7-4b):

Finally, we note the special case of biaxial stress, in which the xy element is subjected to normal stresses in

both the x and y directions but without any shear stresses (Fig. 7-6). The equations for biaxial stress are

obtained from Eqs. (7-4a) and (7-4b) simply by dropping the terms containing τxy, as follows:

Principal Stresses and Maximum Shear Stresses:

The transformation equations for plane stress show that the normal stresses and the shear stresses

vary continuously as the axes are rotated through the angle θ. This variation is pictured in Fig. 7-3 for a

particular combination of stresses. From the figure, we see that both the normal and shear stresses reach

maximum and minimum values at 90° intervals. Not surprisingly, these maximum and minimum values are

usually needed for design purposes. For instance, fatigue failures of structures such as machines and aircraft

are often associated with the maximum stresses, and hence their magnitudes and orientations should be

determined as part of the design process.

Principal Stresses:

The maximum and minimum normal stresses, called the principal stresses, can be found from the

transformation equation for the normal stress (Eq. 7-4a). By taking the derivative of with respect to θ

and setting it equal to zero, we obtain an equation from which we can find the values of θ at which is a

maximum or a minimum. The equation for the derivative is


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The subscript p indicates that the angle θp defines the orientation of the principal planes, that is, the planes

on which the principal stresses act. Two values of the angle 2θp in the range from 0 to 360° can be obtained

from Eq. (7-11). These values differ by 180°, with one value between 0 and 180° and the other between 180°

and 360°. Therefore, the angle θp has two values that differ by 90° , one value between 0 and 90° and the

other between 90° and 180° . The two values of θp are known as the principal angles. For one of these

angles, the normal stress σx1 is a maximum principal stress; for the other, it is a minimum principal stress.

Because the principal angles differ by 90°, we see that the principal stresses occur on mutually

perpendicular planes.

The principal stresses can be calculated by substituting each of the two values of θp into the first stress-

transformation equation (Eq. 7-4a) and solving for σx1. By determining the principal stresses in this manner,

we not only obtain the values of the principal stresses but we also learn which principal stress is associated

with which principal angle.

We can also obtain general formulas for the principal stresses. To do so, refer to the right triangle in Fig. 7-9,

which is constructed from Eq. (7-11). Note that the hypotenuse of the triangle, obtained from the

Pythagorean theorem, is

The quantity R is always a positive number and, like the other two sides of the triangle, has units of stress.

From the triangle we obtain two additional relations:

Now we substitute these expressions for cos 2θp and sin2θp into Eq. (7-4a) and obtain the algebraically

larger of the two principal stresses, denoted by σ1:

After substituting for R from Eq. (7-12) and performing some algebraic manipulations, we obtain

The smaller of the principal stresses, denoted by σ2, may be found from the condition that the sum of the

normal stresses on perpendicular planes is constant:

Substituting the expression for σ1 into Eq. (7-15) and solving for σ2, we get

This equation has the same form as the equation for σ1 but differs by the presence of the minus sign before

the square root. The preceding formulas for σ1 and σ2 can be combined into a single formula for the principal

stresses:


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The plus sign gives the algebraically larger principal stress and the minus sign gives the algebraically smaller

principal stress.

Principal Angles:

Let us now denote the two angles defining the principal planes as θp1 and θp2 corresponding to the principal

stresses σ1 and σ2, respectively. Both angles can be determined from the equation for tan2θp (Eq. 7-11).

However, we cannot tell from that equation which angle is θp1 and which is θp2 A simple procedure for

making this determination is to take one of the values and substitute it into the equation for σx1 (Eq. 7-4a).

The resulting value of σx1 will be recognized as either σ1 or σ2 (assuming we have already found σ1 and σ2

from Eq. 7-17), thus correlating the two principal angles with the two principal stresses.

Another method for correlating the principal angles and principal stresses is to use Eqs. (7-13a) and (7-13b)

to find θp since the only angle that satisfies both of those equations is θp1. Thus, we can rewrite those

equations as follows:

Only one angle exists between 0 and 360° that satisfies both of these equations. Thus, the value of θp1 can be

determined uniquely from Eqs. (7-18a) and (7-18b). The angle θp2, corresponding to σ2, defines a plane that

is perpendicular to the plane defined by θp1 Therefore, θp2 can be taken as 90° larger or 90° smaller than θp1.

Shear Stresses on the Principal Planes:

An important characteristic of the principal planes can be obtained from the transformation equation for the

shear stresses (Eq. 7-4b). If we set the shear stress τx1y1 equal to zero, we get an equation that is the same as

Eq. (7-10). Therefore, if we solve that equation for the angle 2θ, we get the same expression for tan2θ as

before (Eq. 7-11). In other words, the angles to the planes of zero shear stress are the same as the angles to

the principal planes. Thus, we can make the following important observation: The shear stresses are zero on

the principal planes.

FIG. 7-10 Elements in uniaxial and biaxial stress

Special Cases:

The principal planes for elements in uniaxial stress and biaxial stress are the x and y planes themselves

(Fig. 7-10), because tan2θp = 0 (see Eq. 7-11) and the two values of θp are 0 and 90°. We also know that the x

and y planes are the principal planes from the fact that the shear stresses are zero on those planes.

For an element in pure shear (Fig. 7-11a), the principal planes are oriented at 45° to the x axis (Fig. 7-11b),

because tan2θp is infinite and the two values of θp are 45° and 135° . If τxy is positive, the principal stresses

are σ1 = τxy and σ2 = τxy


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FIG. 7-11 (a) Element in pure shear, and (b) principal stresses.

The Third Principal Stress:

The preceding discussion of principal stresses refers only to rotation of axes in the xy plane, that is, rotation

about the z axis (Fig. 7-12a). Therefore, the two principal stresses determined from Eq. (7-17) are called the

in-plane principal stresses. However, we must not overlook the fact that the stress element is actually

three-dimensional and has three (not two) principal stresses acting on three mutually perpendicular planes.

By making a more complete three-dimensional analysis, it can be shown that the three principal planes for a

plane-stress element are the two principal planes already described plus the z face of the element. These

principal planes are shown in Fig. 7-12b, where a stress element has been oriented at the principal angle θp1,

which corresponds to the principal stress σ1. The principal stresses σ1 and σ2 are given by Eq. (7-17), and the

third principal stress (σ3) equals zero. By definition, σ1 is algebraically larger than σ2 but σ3 may be

algebraically larger than, between, or smaller than σ1 and σ2 Of course, it is also possible for some or all of the

principal stresses to be equal. Note again that there are no shear stresses on any of the principal planes.*

FIG. 7-12 Elements in plane stress: (a) original element, and (b) element oriented to the three principal

planes and three principal stresses.

Maximum Shear Stresses:

Having found the principal stresses and their directions for an element in plane stress, we now consider the

determination of the maximum shear stresses and the planes on which they act. The shear stresses τx1y1

acting on inclined planes are given by the second transformation equation (Eq. 7-4b). Taking the derivative of

τx1y1 with respect to θ and setting it equal to zero, we obtain

from which


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The subscript s indicates that the angle θs defines the orientation of the planes of maximum positive and

negative shear stresses. Equation (7-20) yields one value of θs between 0 and 90° and another between 90°

and 180°. Furthermore, these two values differ by 90°, and therefore the maximum shear stresses occur on

perpendicular planes. Because shear stresses on perpendicular planes are equal in absolute value, the

maximum positive and negative shear stresses differ only in sign. Comparing Eq. (7-20) for θs with Eq. (7-11)

for θp shows that

From this equation we can obtain a relationship between the angles θs and θp. First, we rewrite the preceding

equation in the form

Multiplying by the terms in the denominator, we get

which is equivalent to the following expression

Therefore,

This equation shows that the planes of maximum shear stress occur at 45° to the principal planes. The

plane of the maximum positive shear stress τmax is defined by the angle θs1 for which the following equations

apply:

in which R is given by Eq. (7-12). Also, the angle is related to the angle as follows:

The corresponding maximum shear stress is obtained by substituting the expressions for and

into the second transformation equation (Eq. 7-4b), yielding

The maximum negative shear stress τmin has the same magnitude but opposite sign. Another expression for

the maximum shear stress can be obtained from the principal stresses σ1 and σ2 both of which are given by

Eq. (7-17). Subtracting the expression for σ2 from that for σ1 and then and comparing with Eq. (7-25), we see

that

Thus, the maximum shear stress is equal to one-half the difference of the principal stresses.

The planes of maximum shear stress also contain normal stresses. The normal stress acting on the planes of

maximum positive shear stress can be determined by substituting the expressions for the angle θs1 (Eqs. 7-

23a and 7-23b) into the equation for σx1 (Eq. 7-4a). The resulting stress is equal to the average of the normal

stresses on the x and y planes:

This same normal stress acts on the planes of maximum negative shear stress.

In the particular cases of uniaxial stress and biaxial stress (Fig. 7-10), the planes of maximum shear stress

occur at 45° to the x and y axes. In the case of pure shear (Fig. 7-11), the maximum shear stresses occur on

the x and y planes.


Yatin Kumar Singh

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. It also provides a means for calculating principal stresses, maximum shear stresses, and stresses on

Inclined planes. Furthermore, Mohr’s circle is valid not only for stresses But also for other quantities of a

similar mathematical nature, including Strains and moments of inertia.

Equations of Mohr’s Circle:

The equations of Mohr’s circle can be derived from the transformation equations for plane stress (Eqs. 7-4a

and 7-4b). The two equations are repeated here, but with a slight rearrangement of the first equation:

From analytic geometry, we might recognize that these two equations are the equations of a circle in

parametric form. The angle 2θ is the parameter and the stresses σx1 and τx1y1 are the coordinates. However, it

is not necessary to recognize the nature of the equations at this stage—if we eliminate the parameter, the

significance of the equations will become apparent. To eliminate the parameter 2θ, we square both sides of

each equation and then add the two equations. The equation that results is

Equation (7-30) now becomes

which is the equation of a circle in standard algebraic form. The coordinates are σx1 and τx1y1 , the radius is R,

and the center of the circle has coordinates σx1 = σaver and τx1y1 = 0.

FIG. 7-14 Two forms of Mohr’s circle: (a) τx1y1 is positive downward and the angle 2θ is positive

counterclockwise, and (b) τx1y1 is positive upward and the angle 2θ is positive clockwise.

Two Forms of Mohr’s Circle:

Mohr’s circle can be plotted from Eqs. (7-29) and (7-32) 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, as

shown in Fig. 7-14a. The advantage of plotting shear stresses positive downward is that the angle 2θ on

Mohr’s circle will be positive when counterclockwise, which agrees with the positive direction of 2θ in the

derivation of the transformation equations (see Figs. 7-1 and 7-2).


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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 are mathematically

correct, and either one can be used. However, it is easier to visualize the orientation of the stress element if

the positive direction of the angle 2θ is the same in Mohr’s circle as it is for the element itself. Furthermore, a

counterclockwise rotation agrees with the customary right-hand rule for rotation. Therefore, we will choose

the first form of Mohr’s circle (Fig. 7-14a) in which positive shear stress is plotted downward and a positive

angle 2θ is plotted counterclockwise.

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. For our immediate purpose, which is to show the basic properties of the circle, let us assume

that we know the stresses σx , σy and τxy acting on the x and y planes of an element in plane stress (Fig. 7-

15a). As we will see, this information is sufficient to construct the circle. Then, with the circle drawn, we can

determine the stresses σx1 , σy1 and τx1y1 acting on an inclined element (Fig. 7-15b). We can also obtain the

principal stresses and maximum shear stresses from the circle.

(a) (b)


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With σx , σy and τxy known, the procedure for constructing Mohr’s circle is as follows:

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).

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.

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.

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).

6. Using point C as the center, draw Mohr’s circle through points A and B. The circle drawn in this manner has

radius R (Eq. 7-31b), as shown in the next paragraph.

Now that we have drawn the circle, we can verify by geometry that lines CA and CB are radii and have lengths

equal to R. We note that the abscissas of points C and A are (σx + σy)/2 and σx , respectively. The difference in

these abscissas is (σx - σy)/2, as dimensioned in the figure. Also, the ordinate to point A is τxy. Therefore, line

CA is the hypotenuse of a right triangle having one side of length (σx - σy)/2 and the other side of length τxy

Taking the square root of the sum of the squares of these two sides gives the radius R:

which is the same as Eq. (7-31b). By a similar procedure, we can show that the length of line CB is also equal

to the radius R of the circle.

Stresses on an Inclined Element:

Now we will consider the stresses σx1, σy1, and τx1y1 acting on the faces of a plane-stress element oriented at

an angle θ from the x axis (Fig. 7-15b). If the angle θ is known, these stresses can be determined from Mohr’s

circle. The procedure is as follows.

On the circle (Fig. 7-15c), we measure an angle 2θ counterclockwise from radius CA, because point A

corresponds to θ = 0 and is the reference point from which we measure angles. The angle 2θ locates point D

on the circle, which (as shown in the next paragraph) has coordinates σx1 and τx1y1. Therefore, point D

represents the stresses on the x1 face of the element of Fig. 7-15b. Consequently, this face of the element is

labeled “D” in Fig. 7-15b.

Note that an angle 2θ on Mohr’s circle corresponds to an angle θ on a stress element. For instance, point D on

the circle is at an angle 2θ from point A, but the x1 face of the element shown in Fig. 7-15b (the face labeled

“D”) is at an angle θ from the x face of the element shown in Fig. 7-15a (the face labeled “A”). Similarly, points

A and B are 180° apart on the circle, but the corresponding faces of the element (Fig. 7-15a) are 90° apart.

To show that the coordinates σx1 and τx1y1 of point D on the circle are indeed given by the stress-

transformation equations (Eqs. 7-4a and 7-4b), we again use the geometry of the circle. Let β be the angle

between the radial line CD and the sx1 axis. Then, from the geometry of the figure, we obtain the following

expressions for the coordinates of point D:

Noting that the angle between the radius CA and the horizontal axis is 2θ + β, we get


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Expanding the cosine and sine expressions gives

Multiplying the first of these equations by cos2θ and the second by sin2θ and then adding, we obtain

Also, multiplying Eq. (a) by sin2θ and Eq. (b) by cos2θ and then subtracting, we get

When these expressions for cosβ and sinβ are substituted into Eqs. (7-33a) and (7-33b), we obtain the

stress-transformation equations for σx1 and τx1y1 (Eqs. 7-4a and 7-4b). Thus, we have shown that point D on

Mohr’s circle, defined by the angle 2θ, represents the stress conditions on the x1 face of the stress element

defined by the angle θ (Fig. 7-15b).

Point D’ which is diametrically opposite point D on the circle, is located by an angle 2θ (measured from line

CA) that is 180° greater than the angle 2θ to point D. Therefore, point D’ on the circle represents the stresses

on a face of the stress element (Fig. 7-15b) at 90° from the face represented by point D. Thus, point D’ on the

circle gives the stresses σy1 and - τx1y1 on the y1 face of the stress element (the face labeled “D’ ” in Fig. 7-15b).

From this discussion we see how the stresses represented by points on Mohr’s circle are related to the

stresses acting on an element. The stresses on an inclined plane defined by the angle θ (Fig. 7-15b) are found

on the circle at the point where the angle from the reference point (point A) is 2θ. Thus, as we rotate the x1y1

axes counterclockwise through an angle θ (Fig. 7-15b), the point on Mohr’s circle corresponding to the x1 face

moves counterclockwise through an angle 2θ. Similarly, if we rotate the axes clockwise through an angle, the

point on the circle moves clockwise through an angle twice as large.

Principal Stresses:

The determination of principal stresses is probably the most important application of Mohr’s circle. Note that

as we move around Mohr’s circle (Fig. 7-15c), we encounter point P1 where the normal stress reaches its

algebraically largest value and the shear stress is zero. Hence, point P1 represents a principal stress and a

principal plane. The abscissa σ1 of point P1 gives the algebraically larger principal stress and its angle 2θp1

from the reference point A (where θ = 0) gives the orientation of the principal plane. The other principal

plane, associated with the algebraically smallest normal stress, is represented by point P2, diametrically

opposite point P1. From the geometry of the circle, we see that the algebraically larger principal stress is

which, upon substitution of the expression for R (Eq. 7-31b), agrees with the earlier equation for this stress

(Eq. 7-14). In a similar manner, we can verify the expression for the algebraically smaller principal stress σ2

The principal angle θp1 between the x axis (Fig. 7-15a) and the plane of the algebraically larger principal

stress is one-half the angle 2θp1 which is the angle on Mohr’s circle between radii CA and CP1. The cosine and

sine of the angle 2θp1 can be obtained by inspection from the circle:

These equations agree with Eqs. (7-18a) and (7-18b), and so once again we see that the geometry of the circle

matches the equations derived earlier. On the circle, the angle 2θp2 to the other principal point (point P2) is

180° larger than 2θp1; hence, θp2 = θp1 + 90° , as expected.

Maximum Shear Stresses:

Points S1 and S2 representing the planes of maximum positive and maximum negative shear stresses,

respectively, are located at the bottom and top of Mohr’s circle (Fig. 7-15c). These points are at angles 2θ =

90° from points P1 and P2 which agrees with the fact that the planes of maximum shear stress are oriented at

45° to the principal planes. The maximum shear stresses are numerically equal to the radius R of the circle


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(compare Eq. 7-31b for R with Eq. 7-25 for τmax). Also, the normal stresses on the planes of maximum shear

stress are equal to the abscissa of point C, which is the average normal stress σaver

Alternative Sign Convention for Shear Stresses:

An alternative sign convention for shear stresses is sometimes used when constructing Mohr’s circle. In this

convention, the direction of shear stress acting on an element of the material is indicated by the sense of the

rotation that it tends to produce (Figs. 7-16a and b). If the shear stress τ tends to rotate the stress element

clockwise, it is called a clockwise shear stress, and if it tends to rotate it counterclockwise, it is called a

counterclockwise stress. Then, when constructing Mohr’s circle, clockwise shear stresses are plotted upward

and counterclockwise shear stresses are plotted downward (Fig. 7-16c).

It is important to realize that the alternative sign convention produces a circle that is identical to the

circle already described (Fig. 7-15c). The reason is that a positive shear stress τx1y1 is also a

counterclockwise shear stress, and both are plotted downward. Also, a negative shear stress τx1y1 is a

clockwise shear stress, and both are plotted upward.

Thus, the alternative sign convention merely provides a different point of view. Instead of thinking of the

vertical axis as having negative shear stresses plotted upward and positive shear stresses plotted downward

(which is a bit awkward), we can think of the vertical axis as having clockwise shear stresses plotted upward

and counterclockwise shear stresses plotted downward (Fig. 7-16c).

FIG. 7-16 Alternative sign convention for shear stresses: (a) clockwise shear stress, (b) counterclockwise

shear stress, and (c) axes for Mohr’s circle. (Note that clockwise shear stresses are plotted upward and

counterclockwise shear stresses are plotted downward.)

General Comments about the Circle:

From the preceding discussions in this section, it is apparent that we can find the stresses acting on any

inclined plane, as well as the principal stresses and maximum shear stresses, from Mohr’s circle. However,

only rotations of axes in the xy plane (that is, rotations about the z axis) are considered, and therefore all

stresses on Mohr’s circle are in-plane stresses.

For convenience, the circle of Fig. 7-15 was drawn with σx , σy and τxy as positive stresses, but the same

procedures may be followed if one or more of the stresses is negative. If one of the normal stresses is

negative, part or the entire circle will be located to the left of the origin.

Point A in Fig. 7-15c, representing the stresses on the plane θ = 0, may be situated anywhere around the

circle. However, the angle 2θ is always measured counterclockwise from the radius CA, regardless of where

point A is located.

Besides using Mohr’s circle to obtain the stresses on inclined planes when the stresses on the x and y planes

are known, we can also use the circle in the opposite manner. If we know the stresses σx1 , σy1 and τx1y1 acting

on an inclined element oriented at a known angle θ, we can easily construct the circle and determine the

stresses σx , σy and τxy for the angle θ = 0. The procedure is to locate points D and D’ from the known stresses


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and then draw the circle using line DD’ as a diameter. By measuring the angle 2θ in a negative sense from

radius CD, we can locate point A, corresponding to the x face of the element. Then we can locate point B by

constructing a diameter from A. Finally, we can determine the coordinates of points A and B and thereby

obtain the stresses acting on the element for which θ = 0.

If desired, we can construct Mohr’s circle to scale and measure values of stress from the drawing. However, it

is usually preferable to obtain the stresses by numerical calculations, either directly from the various

equations or by using trigonometry and the geometry of the circle.

Mohr’s circle makes it possible to visualize the relationships between stresses acting on planes at various

angles, and it also serves as a simple memory device for calculating stresses. Although many graphical

techniques are no longer used in engineering work, Mohr’s circle remains valuable because it provides a

simple and clear picture of an otherwise complicated analysis. Mohr’s circle is also applicable to the

transformations for plain strain and moments of inertia of plane areas, because these quantities follow the

same transformation laws as do stresses.

plane stress gere

Engineering

stress element

positive face

positive x face

stress acts

positive y face

negative face

stress x acts

negative x face