DE2-EA 2.1: M4DE Dr Connor Myant

9. Stress Analysis

Comments and corrections to connor.myant@imperial.ac.uk

Lecture resources may be found on Blackboard and at http://connormyant.com

1

Contents Analysis of Stress and Strain ................................................................................................................... 2

Stress States ............................................................................................................................................ 2

Stress Matrices ........................................................................................................................................ 4

Plane Stress ............................................................................................................................................. 5

Stresses transformations ........................................................................................................................ 6

Stress on an Inclined Plane ................................................................................................................. 6

Principal stresses and maximum in-plane shear stress .................................................................... 10

Special Cases of Plane Stress ................................................................................................................ 11

Strain transformation............................................................................................................................ 13

Transformation equations for plane strain ....................................................................................... 18

Principle Strains and maximum shear strains ................................................................................... 18

Theories of Failure ................................................................................................................................ 19

Maximum Shear Stress Theory or Tresca Yield Criterion ................................................................. 19

Maximum Distortion Energy Theory or Von Mises Criteria .............................................................. 20

Maximum Normal Stress Theory ...................................................................................................... 20

2

Analysis of Stress and Strain

Normal and shear stresses in beams, shafts and bars can be calculated from the basic formulas given

to you in DE1 EA1.1 Mechanics. However, in this course we will use stress elements to represent the

state of stress at a point in a body.

When working with stress elements, it is good to keep in mind that only one intrinsic state of stress

exists at a point in a stressed body, regardless of the orientation of the element being employed to

portray that state of stress. When we have two elements with different orientations at the same

point in the body, the stresses acting on the faces of the two elements are different, but they

represent the same state of stress – the stress at the point under consideration. This situation is

analogous to the representation of a force vector by its components; although the components are

different when a coordinate axes are rotated to a new position, the force itself is the same.

Note: stresses are not vectors! Although we represent stresses by arrows just as we would a force

vector. The arrows employed to represent stresses have a magnitude and direction, they are not

vectors because they do not combine according to the parallelogram law of addition. Instead,

stresses are more complex quantities than vectors, and in mathematics they a called tensors. Other

tensor quantities in mechanics are strains and moments of inertia.

We usually have information about the strengths of different materials, such as yield stress or

ultimate strength, but how should we use these to determine whether a structural element will fail

under given loading conditions? We determine the state of stress in ‘hot spots’ of the structure. This

means we have a bunch of stress components in these spots, but we have to reduce them into one

number to compare with the known failure strength. The following sections describe the stress

components at a given point (or let’s call it element) in the structure and tells you how you can

reduce the components into important numbers, e.g. principal values.

Stress States In order to analyse stresses, let us first consider an in infinitesimal 2D element in a material. On each

face of the element, there must be a maximum of four (net) forces acting, 𝑃, 𝑄, 𝑅 and 𝑆, one on each

plane of the element.

3

Figure 9.1: Forces on a 2D element.

Immediately we can resolve these forces into their 𝑥 and 𝑦 components, denoted 𝑃𝑥, 𝑃𝑦, 𝑄𝑥, 𝑄𝑦,

𝑅𝑥, 𝑅𝑦, 𝑆𝑥, and 𝑆𝑦.

Figure 9.2: Forces on a 2D element, resolved into components.

We can then convert these forces into stresses (stress = force / area) if we consider the element to

be of unit thickness and divide the force by the element lengths, 1 and 𝑑𝑥 or 𝑑𝑦. If we require that

the element is in equilibrium we find that the net moments and forces must be zero, otherwise the

element would be in motion.

4

Figure 9.3: Stresses on a 2D element.

Note: there are other sign and notation conventions in use in different textbooks.

Stress Matrices Stress can be written as a second rank tensor, or matrix. Hence in the 3D case it can be written as a

3x3 matrix 𝜎𝑖𝑗, with 𝑖, 𝑗 = 𝑥, 𝑦, 𝑧, where 𝑖 is the direction of the stress and 𝑗 is the direction of the

plane normal to the face of the volume element. So the normal stresses are 𝜎𝑖𝑖 (summation

convention) and the shear stresses are the 𝜎𝑖𝑗, 𝑖 ≠ 𝑗. We can visualise these components as acting

on the cube as follows.

Figure 9.4. Three-dimensional view of an element in plane stress.

5

So we write the general 3D stress matrix as;

𝜎𝑖𝑗 = (

𝜎𝑥𝑥 𝜏𝑥𝑦 𝜏𝑥𝑧

𝜏𝑦𝑥 𝜎𝑦𝑦 𝜏𝑦𝑧

𝜏𝑧𝑥 𝜏𝑧𝑦 𝜎𝑧𝑧

) = (

𝜎𝑥 𝜏𝑥𝑦 𝜏𝑥𝑧

𝜏𝑦𝑥 𝜎𝑦 𝜏𝑦𝑧

𝜏𝑧𝑥 𝜏𝑧𝑦 𝜎𝑧

)

Plane Stress The stress conditions you encountered earlier in DE1 EA1.1 when analysing bars in tension and

compression, shafts in torsion, and beams in bending are examples of the stress state – Plane Stress.

To explain plane stress, we will consider the three dimensional stress element shown in Figure 9.5.

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. When the element 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. If there are no external forces acting on the element, the 𝑧 face will be free of stress.

A normal stress, 𝜎, has a subscript that identifies the face on which the stress acts. Example: 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 opposite faces.

The sign convention for normal stresses is – tension is positive, compression is negative.

A shear stress, 𝜏, has two subscripts – the first denotes the face on which the stress acts, and the

second gives direct on that face. Example: the stress 𝜏𝑥𝑦 acts on the 𝑥 face in the direction of the 𝑦

axis, and the stress 𝜏𝑦𝑥 acts on the 𝑦 face in the direction of the 𝑥 axis.

The sign convention for shear stresses is a little more complicated compared to normal stresses. A

shear stress is positive when it acts on a positive face of an element in a positive direction of an axis,

and it is negative when it acts on a positive face in a negative direction. Similarly, on a negative face

of the element, a shear stress is positive when it acts in a negative direction of an axis. Hence the

stresses 𝜏𝑥𝑦 and 𝜏𝑦𝑥 shown in Figure 9.5 are all positive. This sign convention for shear stress is easy

to remember if we state the following: a shear stress is positive when the direction associated with

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

or minus-plus.

The sign convention for shear stresses is consistent with the equilibrium of the element, because we

know that shear stresses on the 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 and downward on the negative face.

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

direction such that both stresses point toward or away from the point of intersection of the faces.

This tells us the element is in equilibrium (not moving or rotating), and therefore; 𝜏𝑥𝑦 = 𝜏𝑦𝑥.

In many instances when one plane is stress free we usually sketch only a two dimensional view of

the element, as shown in Figure 9.5. Although a 2D sketch is adequate for showing all the stresses

6

acting on the element, we must remember that the element is a solid body with thickness

perpendicular to the plane of the sketch.

Figure 9.5. Elements in plane stress: a) 3D view of an element oriented to xyz axes, b) two

dimensional view of the same element.

Stresses transformations Stress on an Inclined Plane

If we have a stress element with known 𝜎𝑥, 𝜎𝑦, 𝜏𝑥𝑦 (as shown in Figure 9.5). If we want to know the

stresses acting on an inclined section within that stress element, we can do this by considering a new

stress element that is located at the same point in the material as the original element. Let’s denote

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

are rotated counter clockwise through an angle 𝜃 with respect to the 𝑥𝑦 axes, as shown in Figure

9.6.

7

Figure 9.6. Two dimensional view of an element oriented to the 𝑥1𝑦1𝑧1 axes.

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

the same subscript designations and sign conventions as before. The previous conclusions regarding

equilibrium still hold so that; 𝜏𝑥1𝑦1 = 𝜏𝑦1𝑥1.

Because the element is still in equilibrium we can express the stresses acting on the inclined 𝑥1𝑦1

element in terms of the stresses acting on the 𝑥𝑦 element. To do this form a wedge-shaped stress

element (Figure 9.7) having an inclined face that is the same as the 𝑥1 face (of the inclined element

shown in Figure 9.6). The other two side faces of the wedge are parallel to the 𝑥 and 𝑦 axes.

Figure 9.7. Wedge-shaped stress element in plane stress.

8

Now consider the free body diagram in Figure 9.8 with the forces acting on the faces (force = stress

x area). If we donate the area of the inclined face 𝐴0, then the normal and shear forces acting on

that face are 𝜎𝑥1𝐴0 and 𝜏𝑥1𝑦1𝐴0. The area of the bottom face (negative 𝑦) is 𝐴0𝑠𝑖𝑛 𝜃, and the area

of the left hand side face (negative 𝑥) is 𝐴0𝑐𝑜𝑠 𝜃.

Figure 9.8. Free body diagram of the forces acting on the wedge-shaped stress element.

Resolving forces parallel to 𝜎𝑥1, we find;

𝜎𝑥1𝐴0 = 𝜎𝑥(𝐴0𝑐𝑜𝑠 𝜃 𝑐𝑜𝑠 𝜃) + 𝜎𝑦(𝐴0𝑠𝑖𝑛 𝜃 𝑠𝑖𝑛 𝜃) + 2𝜏𝑥𝑦(𝐴0𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜃)

Which gives

𝜎𝑥1 = 𝜎𝑥𝑐𝑜𝑠2 𝜃 + 𝜎𝑦𝑠𝑖𝑛2 𝜃 + 2𝜏𝑥𝑦𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜃

Resolving forces parallel to 𝜏𝑥1𝑦1 gives;

𝜏𝑥1𝑦1𝐴0 = −𝜎𝑥(𝐴0𝑐𝑜𝑠 𝜃 𝑠𝑖𝑛 𝜃) + 𝜎𝑦(𝐴0𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜃) + 𝜏𝑥𝑦(𝐴0𝑐𝑜𝑠 𝜃 𝑐𝑜𝑠 𝜃)

− 𝜏𝑥𝑦(𝐴0𝑠𝑖𝑛 𝜃 𝑠𝑖𝑛 𝜃)

So,

9

𝜏𝑥1𝑦1 = (𝜎𝑦 − 𝜎𝑥)𝑐𝑜𝑠 𝜃 𝑠𝑖𝑛 𝜃 + 𝜏𝑥𝑦(𝑐𝑜𝑠2 𝜃 − 𝑠𝑖𝑛2 𝜃)

We now have equations giving the normal and shear stresses acting on the 𝑥1 plane in terms of the

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

We can then use the identities cos 2𝜃 = 𝑐𝑜𝑠2 𝜃 − 𝑠𝑖𝑛2 𝜃 and 𝑠𝑖𝑛 2𝜃 = 2𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠 𝜃 to write

these expressions for 𝜎𝑥1 and 𝜏𝑥1𝑦1 as

𝜎𝑥1 = (𝜎𝑥+𝜎𝑦)

2+

(𝜎𝑥−𝜎𝑦)

2𝑐𝑜𝑠 2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛 2𝜃 (9.1)

𝜏𝑥1𝑦1 = −(𝜎𝑥 − 𝜎𝑦)

2𝑠𝑖𝑛 2𝜃 + 𝜏𝑥𝑦𝑐𝑜𝑠 2𝜃 (9.2)

If the normal stress acting in the 𝑦1 direction is needed, it can be obtained by simply substituting

(𝜃 = 𝜃 + 90) for 𝜃 into equation (9.1);

𝜎𝑦1 = (𝜎𝑥+𝜎𝑦)

2−

(𝜎𝑥−𝜎𝑦)

2𝑐𝑜𝑠 2𝜃 − 𝜏𝑥𝑦 𝑠𝑖𝑛 2𝜃 (9.3)

If 𝜎𝑦1 is calculated as a positive quantity, this indicates that it acts in the positive 𝑦1direction.

These are known as the transformation equations for place stress because they transform the stress

components from one set of axes to another. Since the transformation equation were derived solely

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

or nonlinear, elastic or inelastic.

By plotting Equations. (9.1) and (9.2), we obtain Figure 8.9. We can observe the manner in which

the normal and shear stresses vary. The graph is plotted for the particular case of 𝜎𝑥 = 1, 𝜎𝑦 =

0.2𝜎𝑥, and 𝜏𝑥𝑦 = 0.8𝜎𝑥. We see from the plot that the stresses vary continuously as the orientation

of the element is changed.

10

Figure 9.9. Variation of normal and shear stress on the inclined plane with angle 𝜃.

Principal stresses and maximum in-plane shear stress The principal stresses represent the maximum and minimum normal stress at a point.

To determine the maximum and minimum normal stress, we differentiate and solve Equation (9.1)

with respect to 𝜃 and set the result to zero.

𝑑𝜎𝑥1

𝑑𝜃= −

(𝜎𝑥 + 𝜎𝑦)

2(2𝑠𝑖𝑛 2𝜃) + 2𝜏𝑥𝑦 𝑐𝑜𝑠 2𝜃 = 0

𝑡𝑎𝑛 2𝜃𝑝 =𝜏𝑥𝑦

(𝜎𝑥−𝜎𝑦)/2 (9.4)

The solution to this has two roots 𝜃𝑝1, and 𝜃𝑝2; which are 90° apart.

Solving equation (8.4) into (8.1), we get;

𝜎1,2 = (𝜎𝑥+𝜎𝑦)

2± √(

𝜎𝑥−𝜎𝑦

2)

2+ 𝜏𝑥𝑦

2 (9.5)

Depending upon the sign chosen, this result gives the maximum or minimum in-plane normal stress

acting at a point, where 𝜎1 ≥ 𝜎2. This particular set of values are called the in-plane principle

stresses, and the corresponding planes on which they act are called the principal planes of stress.

When the stress state is represented by the principal stresses, no shear stress will act on the element

(𝜏𝑥𝑝𝑦𝑝 = 0).

To determine the maximum shear stress on an elements faces, we differentiate and solve Equation

(9.1) with respect to 𝜃 and set the result to zero.

11

𝑡𝑎𝑛 2𝜃𝑠 =−(𝜎𝑥−𝜎𝑦)/2

𝜏𝑥𝑦 (9.6)

The solution to this has two roots 𝜃𝑠1, and 𝜃𝑠2; which are 90° apart. Thus, the roots 𝜃𝑠 and 𝜃𝑝 are

45° apart.

The maximum in-plane shear stress is defined as;

𝜏𝑚𝑎𝑥𝐼𝑃 = √(𝜎𝑥−𝜎𝑦

2)

2+ 𝜏𝑥𝑦

2 =𝜎1−𝜎2

2 (9.7)

Substituting the values of sin 2𝜃𝑠 and cos 2𝜃𝑠 into Equation (9.1) we that there is also a normal

stress acting on the planes of maximum in-plane shear stress;

𝜎𝑎𝑣𝑔 =(𝜎𝑥+𝜎𝑦)

2 (9.8)

Special Cases of Plane Stress The general case of plane stress reduces to simpler cases of stress under special conditions.

Example: Uniaxial Stress - This is the situation for a simple tensile test, Figure 9.10.

Figure 9.10: Element in uniaxial tension.

So the stress matrix, for this arrangement of the axes, is given by;

𝜎𝑖𝑗 = (𝜎 0 00 0 00 0 0

)

12

Here all the stresses acting on the 𝑥𝑦 element are zero except for the normal stress 𝜎𝑥, then the

element is in uniaxial stress. Figure 3.9 shows an element in uniaxial tension; this is true for 𝜎 > 0,

however if 𝜎 < 0 we would have a compressive test, uniaxial compression. The corresponding

transformation equations are obtained by simply setting 𝜎𝑦 = 0 and 𝜏𝑥𝑦 = 0;

𝜎𝑥1 =𝜎𝑥

2(1 + 𝑐𝑜𝑠 2𝜃)

𝜏𝑥1𝑦1 =−𝜎𝑥

2(𝑠𝑖𝑛 2𝜃)

Example: Pure Shear - Here 𝜎𝑥 = 𝜎𝑦 = 0, as follows.

𝜎𝑥1 = 𝜏𝑥𝑦𝑠𝑖𝑛 2𝜃

𝜏𝑥1𝑦1 = 𝜏𝑥𝑦𝑐𝑜𝑠 2𝜃

Figure 9.11: Pure Shear applied to a unit cube.

So the stress matrix is given by;

𝜎𝑖𝑗 = (0 𝜏 0𝜏 0 00 0 0

)

This appears simple because of the choice of axes to coincide with the symmetry of the stress state.

It would appear more complex for arbitrary axes. Also notice that by convention the system of axes

is right handed and so 𝜏𝑦𝑥 is in the anticlockwise direction. If the direction of the arrows on the shear

stresses in the diagram were reversed, then the shear stress in the stress tensor would be negative.

13

Example: Biaxial Stress - The element is subjected to normal stresses in both the 𝑥 and 𝑦 planes but

without any shear stresses. The transformation equations are simply obtained by dropping the 𝜏𝑥𝑦

component;

𝜎𝑥1 = ½ (𝜎𝑥 + 𝜎𝑦) + ½ (𝜎𝑥 − 𝜎𝑦) 𝑐𝑜𝑠 2𝜃

𝜏𝑥1𝑦1 = −½ (𝜎𝑥 − 𝜎𝑦)𝑠𝑖𝑛 2𝜃

Figure 9.12: Biaxial stress applied to a unit cube.

Here 𝜎𝑥1 = 𝜎𝑥2 = 𝜎𝑥3 = 𝜎 where 𝜎 < 0. Then the stress matrix is written as;

𝜎𝑖𝑗 = (𝜎 0 00 𝜎 00 0 0

)

Whereas a uniaxial stress is not the same for all axis orientations, and some rotations may give shear

stresses, a purely hydrostatic stress state is invariant to axis rotation and so it is an isotropic tensor.

Strain transformation Strain is a quantity similar to stress and can be shown with matrix convention as well. The

transformation of strain at a point is similar to the transformation of stress. You were introduced to

the idea in DE1 EA1.1 that the general state of strain at a point in a body is represented by a

combination of three components of normal strain (𝜖𝑥,𝑦,𝑧) and three components of shear strain

(𝛾𝑥,𝑦,𝑧). These tend to deform each face of an element of material, and like stress the normal and shear

strain components at a point will vary according to the orientation of the element.

14

Like above we will confine our attention to plain strain; a plain strained element is subjected to two

components of normal strain and one component of shear strain (Figure 9.14).

Figure 9.14. The deformations of an element caused by normal strain in the 𝑥 plane, normal strain in

the 𝑦 plane and shear strain in the 𝑥𝑦 plane.

It should not be inferred that plane strain and plane stress occur simultaneously. In general, an

element in plane stress will undergo strain in the 𝑧 direction; hence, it is not in plane strain. Also, an

element in plane strain usually will have stresses 𝜎𝑧 acting on it because of the requirement that 𝜀𝑧 =

0; therefore, it is not in plane stress. Thus, under ordinary conditions plane stress and plane strain do

not occur simultaneously.

To determine the normal strain 𝜀𝑥1 in the 𝑥1 direction, we consider a small element of material

selected so that the 𝑥1 axis is along a diagonal of the 𝑧 face of the element and the 𝑥 and 𝑦 axes are

along the side of the element (Figure 9.15).

15

Figure 9.15. Deformation of an element in plain strain due to a) normal strain 𝜖𝑥, b) normal strain 𝜖𝑦,

and c) shear strain 𝛾𝑥𝑦,

Consider first the strain 𝜀𝑥 in the 𝑥 direction. This strain produces an elongation in the 𝑥 direction

equal to 𝜀𝑥𝑑𝑥, where 𝑑𝑥 is the length of the corresponding side of the element. As a result of this

elongation, the diagonal of the element increases in length by an amount 𝜀𝑥𝑑𝑥𝑐𝑜𝑠 𝜃 (Figure 9.15a).

Next, consider the strain 𝜀𝑦 in the y direction. This strain produces an elongation in the 𝑦 direction

equal to 𝜀𝑦𝑑𝑦, where 𝑑𝑦 is the length of the side of the element parallel to the 𝑦 axis. As a result of

this elongation, the diagonal of the element increases in length by an amount 𝜀𝑦𝑑𝑦𝑠𝑖𝑛 𝜃 (Figure

9.15b).

Finally, consider the shear strain 𝛾𝑥𝑦 in the 𝑥𝑦 plane. This strain produces a distortion of the element

such that the angle at the lower left corner of the element decreases by an amount equal to the shear

strain. Consequently, the upper face of the element moves to the right (with respect to the lower face)

by the amount 𝛾𝑦𝑧𝑑𝑦. The deformation results in an increase in the length of the diagonal equal to

𝛾𝑥𝑦𝑑𝑦𝑐𝑜𝑠 𝜃 (Figure 9.15c).

The total increase Δ𝑑 in the length of the diagonal is the sum of the preceding three expressions; thus,

Δ𝑑 = 𝜀𝑥𝑑𝑥𝑐𝑜𝑠 𝜃 + 𝜀𝑦𝑑𝑥𝑠𝑖𝑛 𝜃 + 𝛾𝑥𝑦𝑑𝑦𝑐𝑜𝑠 𝜃

The normal strain 𝜀𝑥1in the 𝑥1 direction is equal to this increase in length divided by the initial length

𝑑𝑠 of the diagonal;

16

𝜀𝑥1 =Δ𝑑

𝑑𝑠= 𝜀𝑥

𝑑𝑥

𝑑𝑠𝑐𝑜𝑠 𝜃 + 𝜀𝑦

𝑑𝑥

𝑑𝑠𝑠𝑖𝑛 𝜃 + 𝛾𝑥𝑦

𝑑𝑦

𝑑𝑠𝑐𝑜𝑠 𝜃

Observing that 𝑑𝑥 𝑑𝑠⁄ = cos 𝜃 and 𝑑𝑦 𝑑𝑠⁄ = sin 𝜃, we obtain the following equation for normal

strain;

𝜀𝑥1 = 𝜀𝑥𝑐𝑜𝑠2 𝜃 + 𝜀𝑦𝑠𝑖𝑛2 𝜃 + 𝛾𝑥𝑦 sin 𝜃 𝑐𝑜𝑠 𝜃 (9.9)

Thus we have obtained an expression for the normal strain in the 𝑥1 direction in terms of the strains

𝜀𝑥, 𝜀𝑦, and 𝛾𝑥𝑦 associated with the 𝑥𝑦 axes.

This strain is equal to the decrease in angle between line in the material that were initially along the

𝑥1 and 𝑦1 axes. Consider Fig. 8.16, which shows both the 𝑥𝑦 and 𝑥1𝑦1 axes, with the angle 𝜃 between

them.

Figure 9.16. Shear strain 𝛾𝑥1𝑦1 associated with the 𝑥1𝑦1 axes.

Let line 𝑂𝑎 represent a line in the material that was initially along the 𝑥1 axis. The deformations caused

by the strains 𝜀𝑥, 𝜀𝑦, and 𝛾𝑥𝑦 cause line 𝑂𝑎 to rotate through a counter-clockwise angle 𝛼from the 𝑥1

axis to the position shown in Figure 9.16. Similarly, line 𝑂𝑏 was originally along he 𝑦1 axis, but because

of the deformations it rotates through a clockwise angle 𝛽. The shear strain 𝛾𝑥1𝑦1 is the decrease in

angle between the two lines that originally were at right angles; therefore;

𝛾𝑥1𝑦1 = 𝛼 + 𝛽

Thus, in order to find the shear strain 𝛾𝑥1𝑦1, we must determine the angles 𝛼 and 𝛽.

The angle 𝛼1 is equal to the distance 𝜀𝑥𝑑𝑥𝑠𝑖𝑛 𝜃 divided by the length 𝑑𝑠 of the diagonal;

17

𝛼1 = 𝜀𝑥

𝑑𝑥

𝑑𝑠sin 𝜃

Similarly, the strain 𝜀𝑦 produces a counter-clockwise rotation of the diagonal through an angle 𝛼2;

𝛼2 = 𝜀𝑦

𝑑𝑦

𝑑𝑠cos 𝜃

Finally, the strain 𝛾𝑥𝑦 produces a clockwise rotation through an angle 𝛼3;

𝛼3 = 𝛾𝑥𝑦

𝑑𝑦

𝑑𝑠sin 𝜃

Therefore;

𝛼 = −𝛼1 + 𝛼2 − 𝛼3 = −𝜀𝑥

𝑑𝑥

𝑑𝑠sin 𝜃 + 𝜀𝑦

𝑑𝑦

𝑑𝑠cos 𝜃 − 𝛾𝑥𝑦

𝑑𝑦

𝑑𝑠sin 𝜃

Observing that 𝑑𝑥 𝑑𝑠⁄ = cos 𝜃 and 𝑑𝑦 𝑑𝑠⁄ = sin 𝜃, we obtain;

𝛼 = −(𝜀𝑥 − 𝜀𝑦) sin 𝜃 cos 𝜃 + 𝛾𝑥𝑦 sin2 𝜃

The rotation of line 𝑂𝑏, which initially was at 90° to line 𝑂𝑎, can be found by substituting 𝜃 + 90 for

𝜃 in the expression for 𝛼. The resulting expression is counterclockwise when positive, hence it is equal

to the negative of angle 𝛽. Thus;

𝛽 = (𝜀𝑥 − 𝜀𝑦) sin(𝜃 + 90) cos(𝜃 + 90) + 𝛾𝑥𝑦 sin2(𝜃 + 90)

𝛽 = −(𝜀𝑥 − 𝜀𝑦) sin 𝜃 cos 𝜃 + 𝛾𝑥𝑦 𝑐𝑜𝑠2 𝜃

Adding 𝛼 and 𝛽 gives the shear strain 𝛾𝑥1𝑦1;

𝛾𝑥1𝑦1 = −2(𝜀𝑥 − 𝜀𝑦) sin 𝜃 cos 𝜃 + 𝛾𝑥𝑦(𝑐𝑜𝑠2 𝜃 − 𝑠𝑖𝑛2 𝜃)

18

We have now obtained an expression for the shear strain 𝛾𝑥1𝑦1 associated with 𝑥1𝑦1 axis in terms of

the strains 𝜀𝑥, 𝜀𝑦, and 𝛾𝑥𝑦 associated with the 𝑥𝑦 axes.

Transformation equations for plane strain The equations for plane strain can be expressed in terms of the angle 2𝜃 by using the following

trigonometric identities;

𝑐𝑜𝑠2 𝜃 =1

2(1 + cos 2𝜃)

𝑠𝑖𝑛2 𝜃 =1

2(1 − cos 2𝜃)

sin 𝜃 cos 𝜃 =1

2sin 2𝜃

Thus;

𝜀𝑥1 =𝜀𝑥+𝜀𝑦

2+

𝜀𝑥−𝜀𝑦

2𝑐𝑜𝑠 2𝜃 +

𝛾𝑥𝑦

2sin 2𝜃 (9.10)

𝜀𝑦1 =𝜀𝑥+𝜀𝑦

2−

𝜀𝑥−𝜀𝑦

2𝑐𝑜𝑠 2𝜃 −

𝛾𝑥𝑦

2sin 2𝜃 (9.11)

And

𝛾𝑥1𝑦1

2= −

𝜀𝑥−𝜀𝑦

2sin 2𝜃 +

𝛾𝑥𝑦

2cos 2𝜃 (9.12)

These equations are the counterparts of Equations (9.1) and (9.2) for plane stress.

Principle Strains and maximum shear strains Principal strains exist on perpendicular planes with the principle angles 𝜃𝑝 calculated from the

following equation;

tan 2𝜃𝑝 =𝛾𝑥𝑦

𝜀𝑥−𝜀𝑦 (9.13)

The principal strains can be calculated from the equation;

19

𝜀1,2 =𝜀𝑥+𝜀𝑦

2± √(

𝜀𝑥−𝜀𝑦

2)

2+ (

𝛾𝑥𝑦

2)

2 (9.14)

The maximum shear strains in the 𝑥𝑦 plane are associated with the axes at 45° to the directions of

the principal strains;

tan 2𝜃𝑠 = −𝜀𝑥−𝜀𝑦

𝛾𝑥𝑦 (9.15)

𝛾𝑚𝑎𝑥

2= √(

𝜀𝑥−𝜀𝑦

2)

2+ (

𝛾𝑥𝑦

2)

2 (9.16)

The minimum shear strain has the same magnitude but is negative. In the directions of maximum

shear strain, the normal strains are;

𝜀𝑎𝑣𝑔 =𝜀𝑥+𝜀𝑦

2 (9.17)

Theories of Failure

Maximum Shear Stress Theory or Tresca Yield Criterion For many ductile materials, such as metals, the most common cause of yielding is slipping. This

occurs along the contact planes of randomly ordered crystals that make up the material. Slipping is

due to the shear stress within the material reaching the point of yield, 𝜎 = 𝜎𝑌. Consider now an

element of material taken from a tension specimen (Tutorial sheet 8 Q5), using equation 9.1 we can

see that the maximum shear stress will occur at an angle 𝜃 = 45, and that 𝜏𝑚𝑎𝑥 = 𝜎𝑌/2.

Note: 𝜎𝑌 – not to be confuse this with 𝜎𝑦, the normal stress on the 𝑦 face!

Design Engineering Example: As design engineers you may be tasked with producing a product

or component from a specific material. Therefore, it is important to place a limit on the state of

stress that defines the material’s failure. You will have heard of this before; for instance a ductile

material will fail by the initiation of yielding, whilst a brittle material fails by fracture. These modes

of failure are easily defined when the material is subjected to a uniaxial state of stress (e.g.

tension), however, this becomes more difficult if it is subjected to biaxial or triaxial stress.

No single theory of failure can be applied to a specific material at all times, because a material

may behave in either ductile or brittle manner depending on the temperature, rate of loading,

chemical environment, or the way the material is shaped or formed. When using any theory of

failure we must first calculate the normal and shear stress components at the points where they

are largest in the member. Once the state of stress is established the principle stresses at these

point are then determined, since each of the following theories is based on knowing the principal

stress.

20

Using this principle that ductile materials fail by shear, Henri Tresca proposed the maximum shear

stress theory. The theory states that yielding of a ductile material begins when the absolute

maximum shear stress in the material reaches the yield stress for the same material under uni-axial

tension. So to avoid failure we must keep the absolute maximum shear stress in the material to be

less than or equal to 𝜎𝑌/2, where 𝜎𝑌 is determined from a simple tension test.

Maximum Distortion Energy Theory or Von Mises Criteria A material that is deformed by an external force tends to store energy internally throughout its

volume. The energy per unit volume of material is called the strain energy density. This strain energy

density can be considered as the sum of two parts, with respect to the initial principle stresses

𝜎𝑥, 𝜎𝑦, 𝜎𝑧,;

1. The energy needed to cause a volume change of the element with no change to its shape.

This is a result of the average principal stresses, 𝜎𝑎𝑣𝑔 = (𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧)/3. Also known as

the hydrostatic stress.

2. The energy needed to distort the element, which is caused by the remaining portion of the

principle stresses; (𝜎𝑥 − 𝜎𝑎𝑣𝑔), (𝜎𝑦 − 𝜎𝑎𝑣𝑔), and (𝜎𝑧 − 𝜎𝑎𝑣𝑔). Also known as the deviatoric

stress.

Yielding in a ductile material occurs when the distortion energy per unit volume of the material

equals or exceeds the distortion energy per unit volume of the same material when it is subjected

to yielding in a simple tension test.

Maximum Normal Stress Theory Brittle materials tend to fail suddenly by fracture (with no apparent yielding). In a tension test, the

fracture occurs when the normal stress reaches the ultimate stress, 𝜎𝑢𝑙𝑡. Similarly in a torsion test

brittle fracture occurs due to a maximum tensile stress, with the plane of fracture for an element at

45° to the shear direction. In other words the element is under tension 45° from the axis of torsion.

It has been shown that the tensile stress needed to fracture a specimen during a torsion test is

approximately the same as that needed to fracture it in simple tension.

Therefore, the maximum normal stress theory states that a brittle material will fail when the

maximum principal stress in the material reaches a limiting value that is equal to the ultimate normal

stress the material can sustain when it is subjected to simple tension.