Mechanics of Materials – MAE 243 (Section 002)

Spring 2008

Aaron Kessman, substitute for Dr. K.A. Sierros

CHAPTER 7

ANALYSIS OFSTRESS

Overview

• Introduction – haven’t we been analyzing stress the whole time???

• 7.1-2 Plane stress – how uniaxial normal stress creates a shear component

• Problem solving example

• 7.3 Principal stresses and max shear stress – will the material break under loading?

• Problem solving example

Introduction

• Up till now, you’ve been dealing mostly with the big picture: uniaxial loading, torsion, and some combined loading in 2-D and 3-D.

• The end result has been to solve for the stress/moment at a given location on some loaded object – either explicitly or by a shear/moment diagram.

• Now we’ll take a microscopic look at the combined stresses and the effects of those loadings on the fabric of the material that is being loaded.

Introduction – stresses at a point

• When a body is loaded by normal and shear stresses, we can consider any point in that body as a stress element.

• The stress element can be depicted by a little square (in 2-D – or more correctly a cube in 3-D) with the stresses acting upon it. We’ll just ignore 3-D for the meantime…

*https://ecourses.ou.edu

Plane Stress – components and conventions

• And that’s what we mean by plane stress: the 2-D representation of combined stresses on the four faces of a stress element

• Two normal stress components, x, y

• One shear stress component xy

– Which btw, xy = yx

Elements in plane stress, note sign conventions:

(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 x1y1 axes -

rotated by some angle from original

For now we’ll deal with plane stress, the 2-D biaxial stress

projection of the 3-D cube

Plane Stress – How do we look at stresses in rotation?

• If you were to rotate that little square stress element some angle , what would happen?

• Well, stresses aren’t vectors, so they can’t be resolved the same (easy) way.

• We have to account for:– Magnitude

– Direction

– AND the orientation of the area upon which the force component acts

Stress Transformation - equations

• The stress transformation is a way to describe the effect of combined loading on a stress element at any orientation.

• From geometry and equilibrium conditions (F = 0 and M = 0),

)2cos()2sin(2

)2sin()2cos(22

1

x

1

1

xyyx

yx

xyyxyx

)º90()2sin()2cos(22 11 xy

xyyxyx

Stress Transformation - Ramifications

• Given stresses at one angle we can calculate stresses at any arbitrary angle• Even a uniaxial loading (x) will create both perpendicular (y) and shear (xy)

loadings upon rotation

• Why this is important:If any of the transformed stresses at angle

exceed the material’s yield stress, the material will fail in this direction, even if it was loaded by lower stresses.

• Sometimes the way this works out is failure by shear, which is not obvious.Materials are often weaker in shear.

*https://ecourses.ou.edu

)2sin()2cos(221x

xyyxyx

)2cos()2sin(

211

xy

yxyx

Stress Transformations – Example 7.2-11

)2cos()2sin(2

)2sin()2cos(22

1

x

1

1

xyyx

yx

xyyxyx

Approach:

1. Determine x, y, xy,

2. Plug

3. Chug)2sin()2cos(

221y

xyyxyx

Principal Stresses and Maximum Shear Stress

• If material failure is what we ultimately care about, then we really want to know what are the – maximum and minimum normal stresses– maximum shear stress– orientation () at which these occur

• These are called the principal stresses (1, 2) and maximum shear stress (xy).

• The equations for these can be found from the stress transformation equations by differentiation ( ) and some algebraic manipulation.

• This is really just a more general look at the material in the previous section.

0d

d

1, 2, xy, and - equations

22

21,2 2

2/)(

)2tan(

2

xyyx

yx

xyp

yxavg

yx

p = planes of principal stresses

p = p1, p2, 90º apart

no shear stress acts on the principal planes

2

2/)(

)2tan(

21max

22

2max

xy

xy

yxs

yx

IP

s = planes of max shear stress

s = s1, s2, 90º apart, 45º offset p

maxIP = max in-plane shear stress

Summary

• Principal stresses represent the max and min normal stresses at the point.

• At the orientation at which principal stresses act, there is no acting shear stress.

• At the orientation at which maximum in-plane shear stress acts, the average normal stress acts in both normal directions (x, y)

• The element acted upon by the maximum in-plane shear stress is oriented 45º from the element acted upon by the principal stresses

*https://ecourses.ou.edu

Principal Stresses and Max. Shear Stress - Example 7.3-18

Approach:

• Determine x, xy

• Find y(x, xy, 0)

• Find numerical rangey cannot be = 0 because at some angles the combined effect will raise xy above 0.

22

2max xyyx

IP