stresses & strains in axial loadings_2012

7/27/2019 Stresses & Strains in Axial loadings_2012

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1. Stresses, Strains and Axial Loadings

Prepared b y: Pro f. Nabil El-Tayeb 1

Chapters Outline

Stresses

Normal Stresses in Axially Loaded Bar

Shearing Stresses

Bearing (Bore or compressive) Stresses

Stress Analysis & Design

Stress in two-force MembersStress in Oblique Planes

Factor of Safety &Allowable stress

1a. Stresses


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To provide a way or means ofanalyzing

and designing various machines parts and

load bearing structures.

Specifically, to determine stresses and

deformations for a given elementstructure

Objectives:


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Stresses (ax

ial / normal load)

N/m2=PaA

P

1kPa=103 Pa, 1 MPa = 106 Pa, 1GPa=109 Pa

The actual distribution of stresses in any

given section is statically indeterminate. It isusually assumed that uniformly distributed

over the cross section and its concentrated

value acts at the centroid of the section.

Tensile stress: normal force pulls or

stretches the areaA

Compressive stress: normal forcepushes or compresses areaA

N/mm2=MPa


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StressesTension Stresses: Cross-sectional area of a tension member

Condition:

The force has a line of action that passes

through the centroid of the cross section.

Centroid


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Normal Stresses in Axially Loaded Bar

Maximum average normal stress IF the force P and x-sec area A were constantalong

the longitudinal axis of the bar, then normal stress

= P/A is also constant

If the bar is subjected to several external loads along its

axis, change in x-sectional area may occur. Thus, it isimportant to find the maximum average normal stress,

So we need to find the critical x-sec or the locationwhereratio P/A is a maximum.


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NORMAL STRESS IN AXIALLY LOADED BAR

Maximum average normal stress

Draw an axial or normal force diagram ND

ND

Sign convention:Pis positive (+) if it causes tension in the member

Pis negative () if it causes compression

Identify the maximum average normal stress from the plot


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EXAMPLE 1

Bar width = 35 mm, thickness = 10 mm

Determine the max. average normal stress in the

bar when subjected to the loadings as shown below. 10

35

A

P

Different loads and constant x-sec,result in different stresses

Different x- sections and constant

load result in different stresses


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EXAMPLE 1 (Cont.)

Find the Internal loading, section method

Normal force diagram

By inspection,

largest loading isonBC portion,

wherePBC= 30 kN

N-D


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EXAMPLE 1 (Cont.)

Maximum Average normal stress

BC=

PBC

A

30(103) N

(0.035 m)(0.010 m)= = 85.7 MPa

10

35


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Shearing Stress (transverse / tangent load)

Applied transverse

forces (P and P`) tothe member AB

produce internal

forces at section C

which are called

shearingforces.

The corresponding average shear

stress is,

Shear stress distribution varies

from zero at the member surfaces

to maximum values that may be

much larger than the average

value.

Ash

Pave

1 S S i d A i l L di


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Shearing Stress (Cont.)

Average shear stress over each section is:

avg = average shear stress at section,

assumed to be same at each

point on the sectionAsh = sheared area

of section

P

A shavg =

This loading case is known

as direct or simple shear

1 St St i d A i l L di


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A

F

A

Pave

Single Shear

A

F

A

P

2ave

Double Shear

Single & Double Shear Stresses



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Cross-sectional area of a connecter

subjected to shear

Assumption:

If bolt is loose or clamping force of bolt is unknown,

assume frictional force between plates to be

negligible.



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Single shear (Single Shear Connection)

Steel and wood joints shown below are

examples ofsingle-shear connections, also

known as lap joints.

Since we assume members are thin, there are

no moments caused byF

WoodSteel



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Double shear (Double Shear Connection)

Examples of double-shear connections are shownbelow, often called double lap joints.

For equilibrium, x-sectional area of bolt and

bonding surface between two members subjectedto double shear force, V=F/2



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Although actual shear-stress distribution along roddifficult to determine, we assume it is uniform.

Thus useA = V/ allow to calculate l, provided dand

allow

is known.

Required area to resist shear caused by axial load

1 Stresses Strains and Axial Loadings


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Bearing Stresses Bolts, rivets, and pins create

stresses on the points of contact

or bearing sur facesof themembers they connect.

dt

P

A

Pb

Corresponding average force

intensity is called the bearing

stress,

WStress on bolt:1-Single Shear

2-Bearing stress

Stress on plate:1-Tensile stress

2-Bearing stress

Self readingExample



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The structure is

designed to support a30 kN load. The

structure consists of a

boom and rod joined by

pins at the junctions andsupports.

From Static analysis, the

internal forces in AB and BC

members were:

Rod (+50kN)

Boom (-40kN)

Example 2: Statics Analysis of a Structure

FAB= 40 kN (compression) FBC= 50 kN (tension)

B

C

A



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Example 2: Stress Analysis

Conclusion: the strength of member BCisadequate.

MPa165all

From the material properties for steel, theallowable stress is

Can the structure safely support the

30kN load? ? ? ? ?

MPa159m10314

N105026-

3

A

PBC

Internal force at any section through BC

memberis 50 kN with a force intensity or

stress of

dBC= 20 mm

From a Statics analysis

FAB= 40 kN (compression)

FBC= 50 kN (tension)B

C

A



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Example 2: Design(selection of materials, dimensions, safety)

Design of new structures requires selection of

appropriate materials and component

dimensions to meet performance requirements.

Selection of material is based on Cost, Weight,

Availability, etc., the choice is made to construct

the rod from aluminum (sal l= 100 MPa). What

is an appropriate choice for the rod diameter?

( mm2.25m1052.2m10500444

m10500Pa10100

N1050

226

2

266

3

Ad

dA

PA

A

P

all

all

An aluminum rod 26 mm or more in diameter

is adequate.

Strength ??

Stress ??

Allowable stress??



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Prepared b y: Pro f. Nabil El-Tayeb21

Example 3: Shearing & bearing Stresses in a Pin

The x-sec. area for pins atA, and Care

equal, Single shear at C and double shear

at A .

262

2 m104912

mm25

rA

MPa102m10491

N105026

3

,

A

PaveC

The pin at A subjected to a double

shear with a total force equal to 40kN

MPa7.40m104912

kN40

226,

A

P

aveA

Pin at C, Single

shear & bearing

Pin at A, Double

shear & bearingSt.

The pin at C subjected to a single

shear with a total force equal to 50kN



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Example 3: (Cont.) Pin Bearing Stresses

To determine the bearing stress at pin A,

we have t = 30 mm and d = 25 mm,

projected area

( ( MPa3.53

mm25mm30kN40

td

Pb

( ( MPa0.32

mm25mm50

kN40

td

Pb

Find the bearing stresses in the pin A and in

thebracket

as well

To determine the bearing stress at A in the

bracket, we have t= 2(25 mm) = 50 mmand d= 25 mm,



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Stresses in Two-Force Members

The next slides will show that either

axial or transverse forces may produceboth normal and shear stresses with

respect to a plane other than one cut

perpendicular to the member axis.

Axial forces on a two-force

member result in only normal

stresses on a plane cutperpendicular to the member axis.

Transverse forces on bolts

and pins result in only shearstresses on the planeperpendicular to bolt or pin axis.



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Pass a section through the member forming

an angleq

with the normal plane.

Stresses on Oblique Planes

qq qq sincos PVPF

ResolvePinto components normal andtangential to the oblique section,

Find the oblique area Aq in terms of

reference area

From equilibrium conditions, the

distributed forces (stresses) on the plane

must be equivalent to the forceP.

qq cos/cos AAAA

Reference

area

Oblique

area



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Prepared b y: Pro f. Nabil El-Tayeb

Stress on an Oblique Plane (Cont.)

qq

q

q

qq

q

q

q

q

q

cossin

cos

sin

coscos

cos 2

A

P

A

P

A

V

A

P

A

P

A

F

The average normal ()and shear () stresses on theoblique plane (Aq) are:

25

PP q

b Normal plane

Ad

Aq

a

q

ba

Oblique plane

F=Pcos q P

V= P sin q A = Aq cos q , Aq=A/cosq

q = F/Aq & q= V/ Aq

Aq

A q q



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At q= 45o , maximum shear stressoccurs. (oblique plane).

Maximum Stresses on an Oblique Plane

qqqqq

cossincos2

A

P

A

P

At q=0, maximum normal stressoccurs (reference plane is perpendicular

to the member axis, no oblique plane).

0max qq andA

P

A

P

A

P

2

45cos2 q

2245cos45sinmax

q

A

P

A

P



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Example 4:

Depth and thickness = 40 mm

Determine average normal stress and average

shear stress acting along (a) section planes a-a,and (b) section plane b-b.

=max & =0 max & 0



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Example 4 (Cont.):

At sec a-a: Internal loading

Based on free-body diagram, Resultantloading of axial force, P = 800 N and Averagenormal stress,

=P

A800 N

(0.04 m)(0.04 m)= 500 kPa= No shear stress on

section, since shearforce at section is zero.



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Example 4 (Cont.):

At sec b-b: Average Normal stress at the

oblique Area

N= 800 sin 60 V= 800 N cos 60

kPammxA

N

oave 37560sin/04.004.0

60sin800

q

60o



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30

Example 4 (Cont.):

At sec b-b: Average shear stress

Stress distribution shown below

kPammxA

V

oave217

60sin/04.004.0

60cosN800

q

Identify thestress stateat point aa



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31

Factor of Safety & Allowable Stress

When designing a structural member or

mechanical element, the stress in it must berestricted to safe level

Choose an allowable load that is less than theload the member can fully support

Factor of safety (F.S.) is determined by:

F.S. =

Ffail

Fallow stressallowablestressultimate

safetyofFactor

all

u

FS

FS



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St esses, St a s a d a oad gs


32


If load applied is linearly related to stress

developed within member, then F.S. can also

be expressed as:

F.S. =

fail

allow F.S. =

fail

allow

F.S. must be chosen greater than 1, to avoid potential

failure

Specific values will depend on types of material used

and its intended purpose



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, g


33


To determine dimensions (area) of section

subjected to a normal force:

A =Vallow

To determine dimensions ofsheared areasubjected to a shear force:

allow

PA



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, g


34

uncertainty in material properties

uncertainty of loadings uncertainty of analyses

number of loading cycles

types of failure

maintenance requirements and deteriorationeffects

importance of member to integrity of whole

structure

risk to life and property influence on machine function

Factor of safety considerations:

Next: Strains and Axial Loadings



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, g


1b. Strains and AxialLoadings

35

1. Stresses, Strains and Axial LoadingsNormal Strain


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, g


strainnormal

stress

L

A

P

L

A

P

A

P

2

2

LL

A

P

2

2

36

Normal Strain



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, g


ELASTIC DEFORMATION OF AN AXIALLY

LOADED MEMBER

Sign convention

Sign Forces Displacement

Positive (+) Tension Elongation

Negative () Compression Contraction

37



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g


Deformations Under Axial Loading

AEP

EE

From Hookes Law:

From the

definition of

strain: L

Equating and

solving for the

deformation,AE

PL

With variations in loading (P), cross-section (A) or material properties (E):

i ii

ii

EA

LP

38



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g


Different loads and different X-sectional areas If a bar has different x-sec area, differentEor subjected to

different axial forces, then the equation below can beapplied to each segment of the bar and addedalgebraically to get the total deformation

# ELASTIC DEFORMATION OF AN AXIALLY LOADED

MEMBER

=PL

AE

t= AB BC CD 39

1. Stresses, Strains and Axial Loadings# Example 3.5


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# Example 3.5

Determine the

deformation of the steel

rod shown under the

given loads.

SOLUTION:

Divide the rod into components at the load application points B, C, & D.

Apply a free-body analysis on each component to determine the

internal forces.

Evaluate the deflection of each component () and find the total value (t).40

1. Stresses, Strains and Axial Loadings# Example 3.5 (Cont.)


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Divide the rod intothree components:

# Example 3.5 (Cont.)

Comment41

1. Stresses, Strains and Axial Loadings# Example 3.6


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Composite A-36 steel bar shown made from two segments

AB and BD.

# Example 3.6

AreaAAB

= 600 mm2 and

ABD = 1200 mm2.

Determine the vertical

displacement of end A (A) anddisplacement ofB relative to C(A/B)

.

i ii

ii

EA

LP

AB

BC

CD

42



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Solution:

Due to external loadings, internal axial forces in

regionsAB,BCand CD are different.

Find the Internal force

in each portion by using

section method andFy=0 for verticalforces.

Plot the variation of P.

Plot the variation of


43



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Est= 210(103) kN/m2 (Given).

Vertical displacement ofArelative to fixed supportD is:

[+75 kN](1 m)(106)

[600 mm2 (210)(103) kN/m2]

[+35 kN](0.75 m)(106)

[1200 mm2 (210)(103) kN/m2]+

= +0.61 mm[45 kN](0.5 m)(106)

[1200 mm2 (210)(103) kN/m2]+


A =PL

AE =

Comment 44



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A is +ve, so the bar elongates andso displacement at A is upward.Find the displacement between B

and C,

B-C=PBCLBC

ABCE=

= +0.104 mm[+35 kN](0.75 m)(106)

[1200 mm2 (210)(103) kN/m2]

Here,B moves away from C, since segment

elongates

# a p e 3 6 (Co t )

45



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AB

BC

CD

0.125

0.0292

-0.0375

Stress, , kN/m

Length of the bar, m

Deflection, , mm

0.595

0.104

-0.0893


75

35

-45

Int. Load, P, kN


Section i=(Pi/Ai) i=(PiLi/AiEi)i=(iLi/Ei)

p ( )

stresses & strains in axial loadings_2012

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