mathematical proof yourself if you like!). As the world is three-dimensional, you can imagine there to be a third direction: the intermediate principal stress. This stress, denoted with
2, is perpendicular to both
1 and
3 (see Figure 5.8). We always define
3
2
1. In Figure 5.8 the principal stresses are given on the axes of the coordinate system.5.3.2 Stress on any plane
If you know the three principal stresses and their directions, you can calculate the stress on any other plane you want. Triangle a1a2a3. with area A, is marked gray in Figure 5.8. The orientation of
the triangle is denoted by its normal vector: a vector perpendicular to the surface starting in the origin O. The normal touches the surface of the triangle in point h and makes an angle
3 with the surface of σ1 and σ2. The angles
1 and
2 are defined in the same manner – they have beenomitted in the figure for clarity. We don’t need to know the magnitude of all three angles: only two are necessary to calculate the third because it is known that (sin
1)2 + (sin
2)2 + (sin
3)2 = 1.(You can prove this by expressing the distance between h and the surface of σ1 and σ2 in Oh and
sin
3. If you do that for all three directions and if you think of Pythagoras’ theorem in threedimensions, you’re almost there.)
We want to determine the stress on triangle a1a2a3. The stress on a plane can be expressed as a vector with components in the three principal directions. The value of each component can be determined in the following way: we need to know what the ‘effect’ of triangle a1a2a3 is in that
direction. If you look at the triangle from the direction of σ3, for example, the triangle you see has
the size of triangle Oa1a2. Look at the figure and try to visualise this. Exercise 5-4**: Stress on triangle a1a2a3
a. Angle
3 is also present at point a3. Draw this in the figure.Hint: draw the large triangle O e3 a3 and draw the small one (Oe3h) in it.
b. Now show that Oe3 = sin
3
a3e3 and therefore that surface(a1a2O) =sin
3
A.Now that we know how to calculate the component of the stress on the surface in the direction of σ3, we can fill in the whole stress vector. Stress is defined as force per unit area, so the component
in the stress vector is
3
sin
3 (to see this: you just found that the area in the direction of
3 was A
sin
3 so you multiply this by the stress in this direction -
3 – and divide it by the area of the whole triangle – A). You can calculate the components in the other two directions in the sameFigure 5.8: Three components of stress on a surface
way. The whole stress vector, written in standard vector notation (x component, y component, z component), in the direction of triangle a1a2a3 becomes:
(
1
sin
1,
2
sin
2,
3
sin
3)5.3.3 Normal stress and shear stress
The stress on a plane depends on the principal stresses (
1,
2,
3) and on the orientation of the surface (given by its normal). It is seldom the case that the stress is exactly perpendicular to the plane. For each stress vector, we can however determine to what extent it is perpendicular to the plane, and to what extent it is parallel. For a given plane, we can decompose the stress vector in exactly those components: the normal stress perpendicular to the plane and the shear stress parallel to it (see Figure 5.9).Exercise 5-5**: Shear stress
Prove the following: If triangle a1a2a3 is parallel to one of the principal stresses, it follows that the
shear stress on a1a2a3 is zero.
Using some heavier mathematical techniques (a branch of mathematics called Linear Algebra) we can express the stress components in terms of σ and θ. This is rather cumbersome, so we will just discuss the two-dimensional case, where only σ1 and σ2 are involved. In reality, such a
simplification proves to be very useful.
5.3.4 The 2D case
Consider the following two-dimensional figure (Figure 5.10).
In Figure 5.10 line A is the 2D section of a 3D surface like we used in Figure 5.10: Two-dimensional view of . Because this is a 2D case, we only need one angle θ. Here, it is the angle between the normal vector on line A and the direction of σ1. The stress on A is indicated by the vector σA,
Figure 5.9: Deformation as a result of different types of stress.
Figure 5.10: Two-dimensional view of Figure 5.8
Exercise 5-6**: Stress vectors
a. Show for the 2D case that
A = (
1
cos
,
2
sin
). Use vector notation: (x component, ycomponent). In other words, prove that the x component is equal to
1
cos
, and the ycomponent equal to
2
sin
This means that you have to decompose
A in an x and a ycomponent. Hint: remember that stress vector = force vector / area. And the force vector is force: length. After that you can substitute using the trigonometric functions: cos
= ...b. How can you be sure that the
1 component of vector σA is smaller than the σ2 component?(the quantities of both components are not given numerically, but you can see that θ in the figure is clearly smaller than 45
In the optional exercise 5.1 (see end of this chapter) you can derive yourself how τ and σn are
derived from σ1, σ2 and θ. This derivation results in the following equations:
1 2 1
-
2cos(2 )
2
2
n
1 2sin(2 )
2
where Θ is the angle between the normal stress (perpendicular to the plane) and the direction of σ1.
5.3.5 Mohr’s circle
These equations look very complicated and difficult to work with. That is why the German civil engineer Christian Otto Mohr devised a method to visualise stress situations in a material.
To see how this works, think of a circle of radius r. You may know that the coordinates in every point on a circle are defined by x = r cos (α) and y = r sin (α). (Check this!)
If we move the whole circle in the positive x direction by a distance p, the y value for every point in the circle stays the same, but every x coordinate changes by an amount p, resulting in:
x = p + r cos (α) and y = r sin (α), see Figure 5.11 (right hand).
x
y
(p+r cos α, r sin α)α
p
r
r
y
α
(r cos α, r sin α )X
Figure 5.11: Coordinates in a circle.
Now compare the equations for a circle to the equations we have for the normal and shear stress: circle stresses X coordinate p + r cos (α) Y coordinate r sin (α) If p is equal to
2
2 1
, r to2
2 1
and α = 2θ then the equations are actually the same! This means that we have to take σn for the x coordinate and τ for the y coordinate, and if we fill
everything in, we end up with the following Figure 5.12: The Mohr circle cirkel.) – we call this Mohr’s Circle:
If you know the principal stresses σ1 and σ2 in a situation, you can draw the Mohr diagram by
putting them on the σn-axis and drawing a circle through both of them, with its centre exactly in
between the maximum and minimum principal stress. Using this diagram, you can very easily determine the normal and shear stress for any plane. You just need to known the angle between the plane and the principal stresses. So the only thing you have to do is determine the angle and just read the stress values from the circle.
You have to be aware of a few pitfalls here: the angle that you measure in the diagram is twice as large as θ, so be certain to always fill in 2θ! Also remember that this diagram is only valid for the two-dimensional case!