mechanics of materials chp10

Stress transformation is about:

- Changing the orientation of the axes used to describe stress so that you

can get the stress components in other directions, in particular, the

orientation in which maximum stresses occur. - We will only be considering Plane Stress which is the common assumption used in engineering practice. Plane stress assumes all stresses

can be analyzed in a single plane. Therefore, we need only consider three

stress components and not all six. . .

Chapter 10 - Stress Transformation

2

Stress Transformation Procedure is pretty simple actually. You just have to cut off the sample element along

the inclined direction you want to know the stresses and then do a free body diagram

of the remaining wedge – use equilibrium and solve for the forces along the inclined

plane

Equals...

3

General Equations of Plane Stress Transformations

Positive sign convention

4

Principle Stresses and Maximum In-Plane Shear Stress

The MAXIMUM and MINIMUM normal stress and MAXIMUM shear stress by

differentiating the previous equations with respect to theta and setting the result to

zero . .

IN-PLANE PRINCIPLE STRESSES

Which solves to give. . .

5

IN-PLANE PRINCIPLE STRESSES CONTINUED

Shear stress is ZERO on the principal planes, ie

6

MAXIMUM IN-PLANE SHEAR STRESSES

Similarly, by differentiating the equation for shear stress with respect to theta and

setting the result to zero we get . .

This equation has two roots (solutions) for theta that are 45 degrees apart –

which means that

The planes for maximum shear stress can be determined by orienting an element 45 degrees from the position of an element than defines the planes of principal stress. The solution is:

There is a normal stress on the planes of maximum shear stress and it is found via

substitution to be equal to:

7

Recall from the past the following:

The standard equation for a circle is

(x - h)2 + (y - k)2 = r2

where h and k are the x- and y coordinates of the center of the circle and r is the

radius.

8

Otto Mohr rewrote the equations for normal and shear stress as

Notice the form of these

stress equations looks like

that of a circle?

(x - h)2 + (y - k)2 = r2

9

The coordinate system requires sigma positive to the right and tau positive

downwards

Procedure for analysis

The following steps are required to draw and use Mohr’s circle

10

11

12

13

14

Problem 9-1: The state of stress at a point in a member is

shown on the element. Determine the stress components acting

on the inclined plane AB. Solve the problem using the stress-

transformation equations.

15

Solution X’

y’

16

Problem 9-2: The state of stress at a point in a member is

shown on the element. Determine the stress components acting

on the inclined plane AB. Solve the problem using Mohr’s circle.

17

Solution

ksiR xy

yx0623.88

2

35

2

2

2

2

2

ksiyx

average 42

35

2

875.82

2

35

8tan

2

tan 11

yx

xy

18

P