Lecture Strength of Materials I: Chapter 4 - PhD. Tran Minh Tu

Chapter 4 - State of stress and strength hypothese. The following will be discussed in this chapter: State of stress at a point, plane stress, mohr’s circle, special cases of plane stress, stress – strain relations, strength hypotheses.

STRENGTH OF MATERIALS

TRAN MINH TU -University of Civil Engineering,

4.1 State of stress at a point 4.2 Plane Stress

4.3 Mohr’s Circle 4.4 Special cases of plane stress 4.5 Stress – Strain relations

4.6 Strength Hypotheses

4.1 State of stress at a point

K x

y z

n





• External loads applied to the body =>

The body is deformed =>The stress is

occurred

• At a point K on the arbitrary section, there

are 2 types of stress: normal stress s and

shearing stress t

• The state of stress at a point K is a set of

all stresses components acting on all

sections, which go through this point

• The most general state of stress at a point

may be represented by 6 components,

, ,

, ,

normal stresses shearing stresses (Note:

• Principal planes: no shear stress acts on

4.1 State of stress at a point

• Principal directions: the direction of the principal planes

• Principal stresses: the normal stress act on the principal plane

• There are three principal planes , which are perpendicular to each otherand go through a point

• Three principal stresses: s1, s2, s3 with: s1 ≥ s2 ≥ s3

• Types of state of stress:

- Simple state of stress: 2 of 3 principal

stresses equal to zeros

- Plane state of stress: 1 of 3 principal

stresses equal to zeros

- General state of stress: all 3 principal

stresses differ from zeros

• Plane Stress – the state of stress in which two

faces of the cubic element are free of stress.For the illustrated example, the state of stress

is defined by

0 ,

Sign Convention：

• Normal Stress: positive: tension; negative: compression

• Shear Stress: positive: the direction associated with its subscripts areplus-plus or minus-minus; negative: the directions are plus-minus orminus-plus

4.2 Plane Stress

y

4.2.1 Complementary shear stresses：

• The shear stresses with the same subscripts

in two orthogonal planes (e.g txy and tyx)

are equal

F   0



x y

-  > 0: counterclockwise from the x axis to u axis

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4.2 Plane Stress

4.2.3 Principal stresses are maximum and minimum stresses ：

By taking the derivative of su to  and setting it equal to zero:

xy u

4.2.5 The first invariant of plane stress

The sum of the normal stresses has the same value in each coordinatesystem

sx  sy  su  sv  const

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4.4 Special Cases of Plane Stress

4.4.1 Uniaxial tension

4.4.2 Pure shear

t

t

tt

4.4.3 Special plane stress

4.5 Stress – Strain relations

 x y z 

z y

x x

E

E E

E

s s

 s

3 General state of stress

- Assumption : The normal strain causes only the normal stress The Shear strain causes only the shear stress

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a Normal stress – normal strain relation

b Shear stress – shear strain relation

with E, , G are Young modulus, Poisson ratio,

shear modulus, which the relation among them:

4.5 Stress – Strain relations

Generalized Hooke’s law

1a a a

V 

) (

a ) (

a ) (

a

V1  1 1  1 2 1  2 3 1  3

3 2

2 1

1

2

1 2

1 2

2 2

4.6 Strength Hypothesis

- For a bar under tensile loading, one can conclude at which stressfailure will occur from the stress – strain diagram To prevent suchfailure, an allowable stress sallow is introduced and it is postulated thatthe stresses in the bar must not exceed sallow, i.e: s ≤ sallow

- In an arbitrary structural member, a spatial stress state is present and

it is necessary to determine the circumstances under which the loadcarrying capacity is lost and the material starts to fail

- There exists no experimental setup which can provide a generalanswer, hypothesis on the basic of specific experiments are used.These so-called strength hypothesis allow us to calculate according to

a specific rule, an equivalent stress se from the normal and shearstresses It is assumed that the stress se , when applied to the uniaxialcase of a bar, has the same effect regarding failure through plastic flow

or fracture as the given spatial stress state in the body underconsideration

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4.6 Strength Hypothesis

- Since the stress state in the body and in a tensile bar are then said to

be equivalent, the stress se is called equivalent stress Therefore, if astructural element shall not lose its load carrying capacity, theequivalent stress must not exceed the allowable stress:

1 Maximum – normal – stress hypothesis: It is assumed that thematerial starts to fail when the largest principal stress reaches a criticalvalue Strength condition:

2 Maximum – normal – strain hypothesis: This hypothesis is based

on the assumption that the failure occurs when the maximum normalstrain reaches a critical value Strength condition :

3 Maximum – shear – stress hypothesis: This hypothesis is based

on the assumption that the failure occurs when the maximum shearstress reaches a critical value Strength condition :

s s

3 1

] [

] [

y

L M

N

4.6 Strength Hypothese

Sample Problem 4.1

From the established sign convention, it is seen that

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Sample Problem 4.1

Sample Problem 4.2

The state of plane stress state at a

point is represented by the element in

Fig Determine the principal stresses

and the principal directions of this state

of plane stress With β =60o

With coordinates xy shown in fig., We have

 is the angle measured from the x axis

to the normal axis u of the inclined plane

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Sample Problem 4.2

t



ss

y x

Homework

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Homework

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ATTENTION !