Esforcos_Combinados

8/2/2019 Esforcos_Combinados

1/50

ME HANI F MATERIALThird

Edition

Beer Johnston DeWolf

Transformao de Tenses e Deformaes

DIMENSIONAMENTO DE VEIOS CIRCULARES

2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 1


2/50


Edition


Unsymmetric Bending

=

===

dA

dAc

ydAF mxx

0or

0

neutral axis passes through centroid

Wish to determine the conditions underwhich the neutral axis of a cross section

dAc

yyMM mz

==

of arbitrary shape coincides with theaxis of the couple as shown. defines stress distribution

inertiaofmomentIIc zm

===Mor

dAcy

zdAzM mxy

===0

The resultant force and moment

from the distribution of

couple vector must be directed along

inertiaofproductIdAyz yz ===0or

must satisfy

coupleappliedMMMF zyx ==== 0


a principal centroidal axis


3/50


Edition


Unsymmetric Bending

Superposition is applied to determine stresses inthe most general case of unsymmetric bending.

Resolve the couple vector into components alongthe principle centroidal axes.

sincos MMMMz ==

Superpose the component stress distributions

M zM y

x

z yI I

=

Along the neutral axis,cos sin

0 yzxz y z y

z

M z M y M zM y

I I I I

Iy

= = + = +

= =

yz I



4/50


Edition


Example 4.08

SOLUTION:

components along the principle

centroidal axes and calculate the . sincos MMMM yz ==

component stress distributions.yz

M zM y = +

A 1600 lb-in couple is applied to arectangular wooden beam in a plane

formin an an le of 30 de . with the

z yI I

Determine the angle of the neutralvertical. Determine (a) the maximumstress in the beam, (b) the angle that the

.

tantany

z

I

I

z

y==


plane.


5/50


Edition


Example 4.08

Resolve the couple vector into components and calculatethe corresponding maximum stresses.

( )

( ) inlb80030sininlb1600

inlb138630cosinlb1600

43

==

==

y

z

M

M

( )( ) in9844.0in5.1in5.3

n.n.n.

43121

12

==

==

y

z

I

( )( )psi6.452

in359.5

in75.1inlb1386

41=

==

z

z

z

I

yM

( )( )

psi5.609

in75.0inlb800

alongoccurstoduestressnsilelargest teThe

42 =

==

y

z

zM

ADM

.

The largest tensile stress due to the combined loadingoccurs atA.


5.6096.45221max +=+= psi1062max =


6/50


Edition


Example 4.08

Determine the angle of the neutral axis.

30tan

in9844.0

in359.5tantan

4

4==

y

z

I

I

143.3=

o4.72=



7/50


Edition


General Case of Eccentric Axial Loading

Consider a straight member subject to equaland opposite eccentric forces.

The eccentric force is equivalent to the system

of a centric force and two couples.

PbMPaM

P

zy ==

= forcecentric

By the principle of superposition, thecombined stress distribution is

yz

zx

IIA+=

,be found from

Pz

MM yz =


AII yz


8/50


Edition


Combined Loads - Introduction

Plane Stress - state of stress in which two faces ofthe cubic element are free of stress. For theillustrated example, the state of stress is defined by

.0,, andxy === zyzxzyx

to forces acting in the midplane of the plate.

of a structural element or machine component, i.e.,at any point of the surface not subjected to an


ex erna orce.


9/50


Edition


Transformation of Plane Stress

Consider the conditions for equilibrium of aprismatic element with faces perpendicular to

( ) ( )

cossinsinsin

sincoscoscos0

AA

AAAF xyxxx

==

, , .

( ) ( )

( ) ( )

sinsincossin

coscossincos0

AA

AAAF

xyy

xyxyxy

+

+==

The equations may be rewritten to yield

2sin2cos22

yxyx

xyyxyx

x

+=

+

+=

2cos2sin2

22

xyyx

yx +

=



10/50


11/50


Edition


Maximum Shearing Stress

Maximum shearing stress occurs for avex =

22

max xyyx

R

+

==

and90byseparatedanglestwodefines:Note

22tan

o

xy

ys

=

45byfromoffset o

yx

p

+==

2



12/50


Edition


Example 7.01

SOLUTION:

Find the element orientation for the principalstresses from

xyp

=

22tan

Determine the principal stresses from2

For the state of plane stress shown,determine (a) the principal panes,

minmax, 22 xyyxyx

+

=

Calculate the maximum shearin stress with(b) the principal stresses, (c) themaximum shearing stress and the

corres ondin normal stress.

22

max 2 xy

yx

+

=

2yx

+

=



13/50


Edition


Example 7.01

SOLUTION:

Find the element orientation for the principalstresses from

( )=

+

== 333.11050

4022

2tanxy

p

= 1.233,1.532 p

= 6.116,6.26

Determine the principal stresses fromMPa10

MPa40MPa50

=

+=+=

x

xyx

2minmax, 22

+

+= xy

yxyx

+=

MPa70max =


min


14/50


Edition


Example 7.01

Calculate the maximum shearing stress with2

22

2max 2

+

= xy

yx

MPa50max =

MPa10

MPa40MPa50

=

+=+=

x

xyx

45=ps

= 6.71,4.18s

1050 + yx

The corresponding normal stress is

22=== ave

MPa20=



15/50


Edition




16/50


Edition




17/50


Edition




18/50


Edition




19/50


Edition




20/50


Edition



ME HANI F MATERIALTE


21/50


Edition


Sample Problem 7.1

SOLUTION:

-system at the center of the transverse

section passing throughH. Evaluate the normal and shearing stresses

atH.

Determine the principal planes andcalculate the principal stresses.

s ng e or zonta orce omagnitude is applied to end D of lever

ABD. Determine (a) the normal andshearing stresses on an element at pointHhaving sides parallel to thex andyaxes, b the rinci al lanes and


principal stresses at the pointH.



22/50


Edition


Sample Problem 7.1

SOLUTION:

Determine an equivalent force-couplesystem at the center of the transversesection passing throughH.

=

( )( )

( )( ) inkip5.1in10lb150

inkip7.2in18lb150

==

==

xM

T

Evaluate the normal and shearing stressesatH.

( )441

in6.0inki7.2

in6.0

n.np.

+=+=

Tc

I

cy

( )421 in6.0

+=+=J

xy

===




23/50


Edition


Sample Problem 7.1

Determine the principal planes andcalculate the principal stresses.

( )=

=

= 8.1

84.80

96.7222tan

yx

xyp

= ,.p

= 5.59,5.30p

22

minmax, 22+

+= xy

yxyx

( )22

96.72

84.802

84.80+

+=

ksi68.4

ksi52.13

min

max

=

+=




24/50


Edition


Mohrs Circle for Plane Stress

With the physical significance of Mohrs circlefor plane stress established, it may be applied

.values are estimated graphically or calculated.

or a nown s a e o p ane s ressplot the pointsXand Yand construct thecircle centered at C.

xyy ,,

2

2

22 xyyxyx

ave R +

=+=

The principal stresses are obtained atA andB.

ave R=

minmax,

yx

xyp

=

22tan


the same as CXto CA.



25/50


Edition


Mohrs Circle for Plane Stress

Mohrs circle for centric axial loading:

0, === xyyxA

P

A

Pxyyx 2

===

Mohrs circle for torsional loading:


J

xyyx === 0 0=== xyyxJ

ME HANI F MATERIALTh

Ed


26/50

ME HANI F MATERIALhird

dition


Sample Problem 8.2

SOLUTION:

Determine reactions atA andD.

Determine maximum shear andbending moment from shear andbending moment diagrams.

uniformly distributed load and aconcentrated load. Knowing that for

Calculate required section modulusand select appropriate beam section.

t e gra e o stee to use all= sand all= 14.5 ksi, select the wide-

flange beam which should be used. Find maximum shearing stress.

n max mum norma stress.



Ed


27/50


dition


Sample Problem 8.2

SOLUTION: Determine reactions atA andD.

kips410

kips590

==

==

AD

DA

RM

RM

Determine maximum shear and bendingmoment from shear and bending moment

.

ki s43

kips2.12withinkip4.239max=

==

V

VM

Calculate required section modulus

max

.

in7.119ksi24

inkip24 3maxmin =

==

all

MS


sectionbeam62select W21


Ed


28/50


dition


Sample Problem 8.2

Find maximum shearing stress.Assuming uniform shearing stress in web,

ksi14.5ksi12.5

in8.40

kips432

maxmax


29/50


dition


Design of a Transmission Shaft

If power is transferred to and from theshaft b ears or s rocket wheels, theshaft is subjected to transverse loadingas well as shear loading.

Normal stresses due to transverse loadsmay be large and should be included in

e erm na on o max mum s ear ng

stress.

Shearing stresses due to transverseloads are usually small and

may be neglected.



Ed

B J h t D W lf


30/50

ME HANI F MATERIALirdition


Design of a Transmission Shaft

At any section,

MMMMc

zm +==222where

J

Tcm =

Maximum shearing stress,

222

max

2section,-crossannularorcircularafor22

JI

JIm

m

=

+

=+

=

22max TM

c+=

Shaft section requirement,

TM 22

+


all

c

min

=

ME HANI F MATERIALThi

Edi



31/50

ME HANI F MATERIALirdition


Sample Problem 8.3

SOLUTION:

Determine the gear torques andcorresponding tangential forces.

Find reactions atA andB.

Identify critical shaft section from

Solid shaft rotates at 480 rpm andtransmits 30 kW from the motor to

torque an en ng moment agrams.

Calculate minimum allowable shaft

gears G andH; 20 kW is taken off atgear G and 10 kW at gearH. Knowing

ameter.

all ,smallest permissible diameter for theshaft.


ME HANI F MATERIALThi

Edi



32/50

ME HANI F MATERIALrdition


Sample Problem 8.3

SOLUTION: Determine the gear torques and corresponding

tangential forces.

mN597Hz802

kW30

2 ===EP

T

kN73.3m0.16

mN597=

==

E

EE

r

TF

( ) kN63.6mN398Hz802 kW20 === CC FT

( )kN49.2mN199

Hz802=== DD FT

.

kN90.2kN80.2

kN22.6kN932.0

==

== zy

BB

AA


ME HANI F MATERIALThir

Edit Beer Johnston DeWolf


33/50

ME HANI F MATERIALrdtion


Sample Problem 8.3

Identify critical shaft section from torque andbending moment diagrams.

mN1357

5973731160 222max

22

=

++=

+ TM



34/50

ME HANI F MATERIALThir

Edit Beer Johnston DeWolf


35/50

ME HANI F MATERIALrdtion


Stresses Under Combined Loadings

Wish to determine stresses in slenderstructural members sub ected toarbitrary loadings.

Pass section throu h oints of interest.Determine force-couple system atcentroid of section required to maintain

.

System of internal forces consist of

couple vectors.

e erm ne s ress s r u on yapplying the superposition principle.



Editi Beer Johnston DeWolf


36/50

dion


Axial force and in-plane couple vectorscontribute to normal stress distributionin the section.

couple contribute to shearing stressdistribution in the section.



Editi Beer Johnston DeWolf


37/50

don


Normal and shearing stresses are used todetermine rinci al stresses, maximumshearing stress and orientation of principalplanes.

Analysis is valid only to extent that

principle and Saint-Venants principle aremet.



Editio Beer Johnston DeWolf


38/50

don

Sample Problem 8.5

SOLUTION:

Determine internal forces in SectionEFG.

Evaluate normal stress atH.

Calculate principal stresses and

.

Three forces are applied to a shortsteel post as shown. Determine the

.Determine principal planes.

maximum shearing stress at pointH.





39/50

don

Sample Problem 8.5

SOLUTION: Determine internal forces in SectionEFG.

( )( ) ( )( )m200.0kN75m130.0kN50

kN75kN50kN30

=

===

x

zx

M

VPV

( )( ) mkN3m100.0kN300

mkN5.8

===

=

zy MM

Note: Section properties,

( )( ) 463121 m1015.9m140.0m040.0

m.m.m.

==

==

xI

( )( ) 463121 m10747.0m040.0m140.0 ==zI





40/50

on

Sample Problem 8.5

Evaluate normal stress atH.

= xzbMaMP

( )( )m020.0mkN3kN504623-

+=

xz IIA

( )( )

m1015.9

m025.0mkN5.8

m.m.

46

( ) MPa66.0MPa2.233.8093.8=+=

Evaluate shearin stress atH.

( )( )[ ]( )m0475.0m045.0m040.0

36

11

=

==

yAQ

( )( )m040.0m1015.9

m105.85kN75

.

46

36

==

tI

QVzyz

2002 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 40MPa52.17=




41/50

on

Sample Problem 8.5

Calculate principal stresses and maximumshearing stress.

Determ ne pr nc pa p anes.

=+== MPa4.3752.170.3322

max R

===

=+=+=

MPa4.74.370.33

MPa4.704.370.33

min

max

ROC

ROC

=

=== 96.2720.33

52.17

2tan pp CD

CY

.

= MPa4.37max

=

=

MPa4.7

a.

min

max


= .p



S l P bl 9


42/50

on

Sample Problem 9



Edition


S l P bl 10


43/50

n

Sample Problem 10



Edition


Sample Problem 11


44/50

n

Sample Problem 11



Edition


Sample Problem 11


45/50

n

Sample Problem 11



46/50


Edition


Introd ction


47/50

Introduction



Edition


Introduction


48/50

Introduction



Edition


Introduction


49/50

Introduction



Edition


Introduction


50/50

Introduction


Esforcos_Combinados

Documents

Transcript of Esforcos_Combinados