Shear Stresses in Beams PowerPoint Slides (1)

8/10/2019 Shear Stresses in Beams PowerPoint Slides (1)

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EN2701 Mechanics of Solids

Shear Stresses in Beams


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School of Engineering, RGU

Slide 2 of 60


Introduction

The Shear Formula

Shear Stresses in Rectangular Beams !or"ed E#am$le 1

Shear Stresses in !ide%Flange &I' Beams

(imitations in the )se of the Shear Formula !or"ed E#am$le 2

Shear Stresses in *ircular Beams

Shear Stresses in Built%)$ Beams !or"ed E#am$le +


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Slide of 60

Shear Stresses in Beams % Intro Beams generall, su$$ort -oth shear and moment loadings.

Normal or -ending stresses/ / associated ith -endingmoments are found from the -ending euation

!hen -eam is su-3ected to non%uniform -ending in e4er, case e#ce$t hen -ending moment/ M/ constant along

length of uniform -eam

-oth -ending moments/ M/ and shear forces/ 5/ act on *S6

remem-er shear force and -ending moment diagrams from le4el 1

Shear force/ 5/ is result of trans4erse shear stress distri-utionthat acts o4er -eams cross%section.

Relationshi$ needed to allo distri-ution of shear stresses/ /associated ith shear force to -e determined.

R

E

,I

M=

=


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Slide ! of 60

Shear Stresses in Beams % Intro *om$lementar, $ro$ert, of shear.

Results in associated longitudinal shear stresses act alonglongitudinal $lanes of -eam.

Element of material from interior of -eam ill -e su-3ected to-oth trans4erse and longitudinal shear stress.

"


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Slide # of 60

E#istence of longitudinal shear stresses illustrated -,considering -eam made u$ of to -oards.

If to$ and -ottom surfaces of -oards are smooth and not-onded together then a$$lication of a load ill cause -oards toslide relati4e to each other.

Each -oard ill -e in com$ression a-o4e its neutral a#is and intension -elo its neutral a#is.

Shear Stresses in Beams % Intro

(oer longitudinal fi-res ofu$$er -oard ill sliderelati4e to u$$erlongitudinal fi-res of loer

-oard.


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Slide 6 of 60

If -oards are -onded together to ma"e a solid -eam/longitudinal shear stresses e#ist hich $re4ent relati4e sliding.

6lso longitudinal shear stresses at free surfaces on to$ and-ottom of -eam ill not e#ist.

8ue to com$lementar, nature/ trans4erse shear at free surfaceill also -e 9ero.

Shear Stresses in Beams % Intro


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Slide 7 of 60

The Shear Formula Formula for shear stress is de4elo$ed indirectl, using

-ending euation/

relationshi$ -eteen -ending and shear &5 : dM;d#'/

consideration of longitudinal shear stress.

*onsider an element of length d# cut from a -eam.

FB8 shos shear forces 5 and -ending moments M and

&M < dM' acting on element.

$d $

"

"

MM % d M

d $


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Slide & of 60

The Shear Formula Then consider to$ segment of

element that has -een

sectioned at a distance of ,from the neutral a#is.

Segment has a *S6 of 6 anda idth of t at section.

t' () (


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Slide * of 60

The Shear Formula No consider hori9ontal

stresses acting on segment.

(inearl, 4ar,ing normalstresses and due to Mand &M < dM'

Shear stress acting on-ottom surface.

*onsidering force euili-riumof segment

0+d$t,d)d)--)-)

=

M M % d M

d $

(

%%%%=1>

shear stress acts o4er an area t d#assuming is constant through thic"ness t


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Slide 10 of 60

From -ending euation

and

Su-stituting into euation =1>

E#$and first term

The Shear Formula

IM,= ( )

I,dMM +=-

0+-- =+

d#td6,IM

d6,I

dMM66

0+---

=+ d#td6,I

Md6,

I

dMd6,

I

M666

0+,-

= d#td6,IdM

6


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Slide 11 of 60

Therefore

Rearranging to o-tain

Sim$lified further

: first moment of area 6 a-out neutral a#is : ?

@ence

The Shear Formula+,

-d#td6,

I

dM6

=

= -1

6d6,

d#

dM

It

d#

dM5=

-) d)'

It

5?

=E./ation non as

the Shear orm/la


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Slide 12 of 60

8eri4ation considered shear stresses acting on -eams

longitudinal $lane. Because of com$lementar, nature of shear also used to find

trans4erse shear stress on -eams *S6.

6s deri4ation used -ending euation shear formula onl, 4alid if

material -eha4es in a linear%elastic manner/ material is homogeneous and isotro$ic/

material has same Aoungs modulus in tension and com$ression.

The Shear Formula


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Slide 1 of 60

The Shear Formula

: shear stress at distance , from neutral a#is.

6ssumed constant across idth/ t/ of -eam.5 : internal resultant shear force/ determined from sectioning

-eam and considering euili-rium.

I : 2nd moment of area of entire*S6 a-out neutral a#is.

t : idth of -eam/ at $osition here to -e determined.? : 1st moment of area of to$ &or -ottom' $ortion of *S6/defined from section here t is measured.

? calculated using

6 : area of to$ &or -ottom' $ortion of *S6/ defined from sectionhere t is measured

: distance to centroid of 6 measured from neutral a#is.

It

5?=

-)-'3=

-'


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Slide 1! of 60

Shear Stresses in Rectangular Beams *onsider a -eam ith rectangular

cross%section.

idth : - and height : h.

8istri-ution of shear stressthroughout cross%section found -,determining shear stress at

ar-itrar, distance , from neutrala#is.

"

4

h

'


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Slide 1# of 60

Shear Stresses in Rectangular Beams First ste$ is to calculate ?

1st moment of area of 6

a-out neutral a#is.

) (

' 'h2

h2

4

-)'3=

= '

2

h4-)

''2h

21' +

=

+

= '

2

h4''

2

h

2

13

4'!

h

2

13 2

2

=


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Slide 16 of 60

Shear Stresses in Rectangular Beams Second ste$/ calculate I.

For rectangular -eam of idth

- and height h/ I/ a-outneutral a#is is

6$$l,ing shear formula

) (

' 'h2

h2

4

12

-hI=

--h

-,h

5

It

5?

12

!2

1

22

==

= 2

2

'

!

h

4h

"6


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Slide 17 of 60

Shear Stresses in Rectangular Beams

Shos shear stress distri-utiono4er cross section isparabolic.

Intensit, 4aries from 0 at to$

and -ottom here , : h;2.

To ma#imum at neutral a#ishere , : 0.

= 2

2

'

!

h

4h

"6

h2

h2


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Slide 1& of 60

Shear Stresses in Rectangular Beams To calculate maximum4alue

Remem-er area of total crosssection/ 6 : -h

Ma# shear stress is 1. timesa4erage 4alue calculated fromsim$le formula : 5;6

h2

h2

= 0Ch

-h5D

2

+ma#

-h2

5+ma# =

6

5

2

+ma# =


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Slide 1* of 60

!or"ed E#am$le 1 Beam shon made of ood.

Su-3ected to resultant internal

4ertical shear force/ 5 : +0"N.

8imensions in mm.

&a'8etermine shear stress in-eam at $oint .

&-' *alculate ma#imum shearstress in the -eam.

"

5# 0

2 0

1 #! 0


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Slide 20 of 60

!or"ed E#am$le 1 % Solution

&a' First calculate section

$ro$erties I and ?.

Remem-er I is for wholesection.

12

-hI

+

=

( )12

0B.00C.0I

+=

CF m107.C1DI =

! 0

2 #

2 #

N )


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Slide 21 of 60


? is first moment of area of

shadedarea.

Shaded area is area a-o4e .

is distance from neutral a#isto centroid of area.

) (

'' 7 1 #

! 0

2 #

2 #

2 0

5

N )

26 m10&000200!0-) ==

( ) m01#001000#0' =+=

D

10G0001B.0H6,?

==+D m1012? =

,


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Slide 22 of 60


!idth of the section &t' at is

C0 mm. 6$$l,ing shear formula

) (

'' 7 1 #

! 0

2 #

2 #

2 0

5

N )

tI

5? =

MaD.21 =

0C.0107.C1D101210+0

D+

=

Ea10D.21 D =


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Slide 2 of 60


&-' Shear stress is ma#imum at

neutral a#is.*onsider area of -eam a-o4eneutral a#is ) (

' 7 1 2 #

! 0

2 #

2 #

N )

2 m10102#00!0-) ==

m012#0'=

+101012.0H6,? ==

+D m10.12? =


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Slide 2! of 60


I and t are as -efore.

6$$l,ing shear formula

) (

' 7 1 2 #

! 0

2 #

2 #

N ) a10.22 Dma# =

M5a#22ma$ =

0C.0107.C1D

10.1210+0

tI

5?

D+

ma#

==

Ma# shear stress can also -e calculated from

5a10#220#00!0

100

2

)

"

2

6

ma$ =

==


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Slide 2# of 60

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Slide 26 of 60

Shear Stresses in !ide Flange &I' Beams

!ide%flange -eam &or I%-eam'consists of to &ide'

Jflanges and a Je-.

!hen su-3ected to shear force5/ shear stresses de4elo$edthroughout cross%section.

f l a n g e s

e 4

8istri-ution of stresses much more com$licated than inrectangular -eam.

ossi-le to determine stresses using same techniues as forrectangular -eams

h d l & '


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Slide 27 of 60


6s ith rectangular -eam/ shear stress 4aries $ara-olicall, o4erde$th.

For flanges thic"ness/ t/ in shear formula is idth of flange.

For e-/ thic"ness/ t/ is thic"ness of e-.

Shear stresses in flanges small com$ared to those in e-.

5ariation of shear stress o4er de$th of e- is small.

8 a r a 4 o l a

f l a n g e

e 4

f l a n g e

Sh S i id l & '


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Slide 2& of 60


Most of 4ertical shearforce is carried -, e-.

Ma#imum shear stressoften a$$ro#imated -,di4iding shear force 5 -,area of e-

8 a r a 4 o l a

a 9 gf l a n g e

e 4

f l a n g e

e4a9g

)

"=

(i it ti i ) f Sh F l


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Slide 2* of 60

(imitations in )se of Shear Formula

Kne assum$tion used in deri4ationof shear formula is shear stress is

uniforml, distri-uted o4er idth/ t. 6ccurac, tested -, com$aring

results ith more rigorousmathematical anal,sis -ased ontheor, of elasticit,.

If -eams cross%section isrectangular/ shear stressdistri-ution calculated from theor,is shon

Ma#imum 4alue/ ma#/ occurs at

edges of cross%section. Magnitude de$ends on ratio -;h

&idth;de$th'.

(m a $m a $

4

hN )



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Slide 0 of 60


For sections ha4ing a -;h ratio of0./ ma#onl, a-out +L greater

than calculated from shearformula.

For flat sections ith -;h : 2/ma#is C0L greater than ma#.

Error e4en greater as -;h ratioincreases.

For flanges of ide%flange -eamsa realistic 4alue of shear stress isnot e4en a$$ro#imated.

( m a $ m a $

4 7 0 # h

hN )

( m a $ m a $

4 7 2 h

h

N )



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Shear Stresses in BeamsEN2701 Mechanics of Solids

School of Engineering, RGUSlide 1 of 60


!ill not gi4e accurate results for shearstress at flange;e- 3unction of I%

-eam. Inner regions of flanges are free

-oundaries and shear stresses must -e9ero.

)sing shear formula a non%9ero 4alue is

found. (imitations for flanges of I%-eams not

im$ortant in engineering $ractice.

Kften/ engineers onl, calculatema#imum shear stress.

This occurs at neutral a#is/ here -;hratio is 4er, small.

*alculated result 4er, close to actualma#imum shear stress.

! " d E l 2


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!or"ed E#am$le 2 Steel ide%flange -eam

has dimensions shon.

Su-3ected to shearforce 5 : G0 "N.

lot shear stressdistri-ution acting o4er-eams cross%sectionalarea.

0 0 m m

2 0 0 m m

2 0 m m

2 0 m m

1 # m m

! " d E l 2 S l ti


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School of Engineering, RGUSlide of 60

!or"ed E#am$le 2 % Solution Shear stress

distri-ution ill -e

$ara-olic and ha4eform shon.

8ue to s,mmetr,/onl, shear stresses at$oints B/ B and *

needed. Second moment of

area/ I/ ill -e samefor each $osition.

B

:

B ( B ( B

:

! " d E l 2 S l ti


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School of Engineering, RGUSlide ! of 60

!or"ed E#am$le 2 % Solution For flanged cross section I is

found in one of to a,s

1. *alculate I for large rectangle

0 0 m m

2 ! 0 m m

( )12 2C.0+.0I+

=

!o "ed E am$le 2 Solution


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School of Engineering, RGUSlide # of 60

!or"ed E#am$le 2 % Solution For flanged cross section I is

found in one of to a,s

1. *alculate I for large rectangleand su-tract I for 2 smallrectangles.

I : 1.D # 10%DmC

0 0 m m

2 0 0 m m

2 0 m m

2 0 m m

1 # m m

1 ! 2 # m m 1 ! 2 # m m

( )12 2.01C2B.02

+

( )12 2C.0+.0I+

=

!or"ed E#am$le 2 Solution


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2. *alculate I for all + rectangles

and use $arallel a#is theorem.I for flanges must -e mo4edso that neutral a#es coincide.

I : 1.D # 10%DmC

0 0 m m

2 0 0 m m

2 0 m m

2 0 m m

1 # m m

( ) ( )( )

+

+

= 2

110020012

02002

12

2001#0I

++=

2++

6"12

-h

212

-h

I



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!or"ed E#am$le 2 % Solution Stress at $oint B

tB: 0.+ m

6 : area of flange

02.0+.011.0H6,? HB ==

+.010D.1

10DD.010G0

tI

5?D

++

HB

==

0 0 m m

1 0 0 m m

2 0 m m

' 7 1 1 0 m m ) (

N )++

HB m10DD.0? =

Ma1+.1HB =



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School of Engineering, RGUSlide & of 60

!or"ed E#am$le 2 % Solution Stress at $oint B

tB: 0.01 m

?B: ?B

01.010D.1

10DD.010G0

tI

5?D

++

B

==

0 0 m m

1 0 0 m m

2 0 m m

' 7 1 1 0 m m ) (

N )

MaD.22B =



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School of Engineering, RGUSlide * of 60

!or"ed E#am$le 2 % Solution Stress at $oint *

t*: 0.01 m

6 : area of section a-o4eneutral a#is.

? for this area is sim$l,sum of ?s for 2 rectanglesthat ma"e u$ sha$e.

?*: ?1< ?2

0 0 m m

1 0 0 m m

2 0 m m

1 # m m' 1

' 2

) ( 1

) (2

N )

2211* 6,6,? +=

?*: &0.11 # 0.+ # 0.02' < &0.0 # 0.01 # 0.1'

?*: 0.7+ # 10%+m+

MEatI

5?* 22#

01#01061##

107#010&06

=

==



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School of Engineering, RGUSlide !0 of 60


B: 1.1+ Ma

B : 22.D Ma

* : 2.2 Ma

B

:

B ( 1 1 2 2 6

2 # 2

T t i l ? ti


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Shear Stresses in *ircular Beams This t,$e of -eam im$ortant in

the transmission of $oer.

E.g. -ending and shearing loadsare induced in shafts -, forces atgears/ -earings and $ulle,s.

*onsider -eam ith solid circular

cross%section su-3ected to shearload 5.

6ccording to shear formula ashear force/ 5/ causes a shearstress/ / in same direction as 5.

"



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School of Engineering, RGUSlide ! of 60

Shear Stresses in *ircular Beams 6t $oint this stress can -e

resol4ed into to com$onents

normal &n' and tangential &t' tosurface. Kutside surface of shaft is free

surface and so nmust -e 9ero. Indicates shear stress at $oint

must -e tangential to surface andnot in direction of shear force.

6t neutral a#is shear stress ill -ein direction of shear force

Shear formula can -e used.

6lso $osition of ma#imum shearstress.

n t

5



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School of Engineering, RGUSlide !! of 60


For a semi%circle

and

For circular cross%section

Therefore

tI

5?=ma$

&

d-)

2=

=

d2'

12

d

d2

&

d

-)'3

2

=

==

6!

!dI =

!;

;

16;

;

6!;

12;2!

ma$6

5

d

5

dd

d5=

=

=

6

5

!ma$ =

m a $

N )

) (

'

Shear Stresses in Built%)$ Beams


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School of Engineering, RGUSlide !# of 60

Shear Stresses in Built%)$ Beams Fa-ricated from to or more $ieces of material 3oined together

to form single/ solid -eam.

*onstructed in a great 4ariet, of sha$es to meet s$ecial needsor to $ro4ide larger cross%sections that are ordinaril, a4aila-le.

!ooden box beamconstructed of to $lan"s/

as flanges/ connected -,$l,ood e-s.

ieces 3oined together ithnails/ scres or glue.

f l a n g e1 e 4

Sh St i B ilt ) B


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Shear Stresses in Built%)$ Beams lued/ laminated -eam "non

as glulam beam.

Boards glued together to formmuch larger -eam than could-e cut from solid $iece oftim-er.

!elded steel plate girder/fa-ricated from three steel$lates elded together.

!ide flange -eamstrengthened -, ri4etingchannel section to each flange.



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8esign must ensure that entire cross%section acts as a single/solid unit.

*alculations in4ol4e to $hases

1. -eam anal,sed as if it as solid cross%section/ ta"ing intoaccount -oth -ending and shear stressesO

2. connecting elements &nails/ -olts/ glue/ elds' anal,sed toensure the, are strong enough and suita-l, $ositioned.


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School of Engineering, RGUSlide !* of 60


: shear flo/ measured as force $er unit length along -eam.

5 : internal resultant shear force/ determined from sectioning-eam and euili-rium.

I : 2nd moment of area of entirecross%sectional a-out neutral

a#is. ? : 1stmoment of area of to$ &or -ottom' $ortion of cross%

sectional area/ defined from section here shear flo to -ecalculated.

I

5? =

Shear Stresses in Built )$ Beams


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School of Engineering, RGUSlide #0 of 60

Shear Stresses in Built%)$ Beams 6$$lication of shear flo formula follos same $rocedure as for

shear stress formula.

5er, im$ortant to identif, correct 4alue for ? hendetermining shear flo at $articular 3oint in cross%section.

Reuired 4alue of ? calculated from shaded sections.

Note shear flo ill -e resisted -,

single ro of fasteners in &a' and &-'

to ros of fasteners in &c' and &d'

three ros of fasteners in &e'

, a + , 4 + , c + , e +, d +

!or"ed E#am$le +


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!or"ed E#am$le + !ood -o# -eam % 2 -oards as flanges and 2 e-s of $l,ood.

Each -oard : C0 mm # 1G0 mm. l,ood : 1 mm thic".

Total height of -eam : 2G0 mm. l,ood fastened to flanges -, scres ha4ing alloa-le load in

shear of 1100 N $er scre.

If shear force acting on cross%section is 10. "N/ determinema#imum $ermissi-le longitudinal s$acing/ s/ of scres.

! 0

! 0

2 & 0

1 # 1 & 0 1 #s s s

!or"ed E#am$le + Solution


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!or"ed E#am$le + - SolutionShear force transmitted -eteen one of flanges and to e-s

found from shear flo formula

? for to$ flange &shaded'

6 : 0.1G # 0.0C : 7.2 # 10%+m2

? : 0.12 # 7.2 # 10%+

? : GDC # 10%Dm+

m12.0, =

( ) ( )12

201&0

12

2&0210

12

==

-hI

!610226! mI =

I

5? =

! 0

1 ! 0

1 # 1 & 0 1 #

'

) (

!or"ed E#am$le + - Solution


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School of Engineering, RGUSlide # of 60

!or"ed E#am$le + Solution

: +C.+ "N;m

Shear flo $er metre of length that must -e carried -, scres.

If scres are s$aced a distance Js a$art load ca$acit, of scresis

F : load carried -, one scre

2 -ecause there are to lines of scres

6

6

10226!

10&6!10#10

==I

5?

s

2

!or"ed E#am$le + - Solution


54/61


School of Engineering, RGUSlide #! of 60

!or"ed E#am$le + SolutionEuating load ca$acit, of scres to shear flo

Rearranging

s : DC.1 mm

Therefore lines of scres must -e no more than DC.1 mm a$art.

.s

2

=

m06!1010!

11002

.

2s

=

==

Tutorial ?uestions


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School of Engineering, RGUSlide ## of 61

Tutorial ?uestions

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Summar,


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Summar,

Summar, !or"ed E#am$le C


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Summar, !or"ed E#am$le C&com-ined loading'

To forces :1G "N and F:1 "N are a$$lied to the shaft ith a radius ofR:20 mm as shon. 8etermine the ma#imum normal and shear stressesde4elo$ed in the shaft.

!or"ed E#am$le C % Solutions


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School of Engineering, RGUSlide #& of 61

!or"ed E#am$le C Solutions&com-ined loading'

To forces :1G "N and F:1 "N are a$$lied to the shaft ith a radius ofR:20 mm as shon. 8etermine the ma#imum normal and shear stressesde4elo$ed in the shaft.



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School of Engineering, RGUSlide #* of 61



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Shear Stresses in Beams PowerPoint Slides (1)