BEAMS-SHEAR AND MOMENT (17).ppt

8/9/2019 BEAMS-SHEAR AND MOMENT (17).ppt

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BEAMS

SHEAR AND MOMENT


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Beam Shear

♦ Shear and Moment

Diagrams

♦ Vertical shear:

tendency or one !arto a "eam to mo#e

#ertically $ith res!ect

to an ad%acent !art


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&

Beam Shear

♦ Magnitude (V) ' s(m o #ertical orces oneither side o the section ) can "e determined at any section along the

length o the "eam♦*!$ard orces +reactions, ' !ositi#e

♦Do$n$ard orces +loads, ' negati#e

♦Vertical Shear ' reactions ) loads

+to the let o the section,


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-

Beam Shear

♦.hy/

♦ necessary to 0no$ the maim(m #al(e othe shear ) necessary to locate $here the shear changes

rom !ositi#e to negati#e $here the shear !asses thro(gh 3ero

♦*se o shear diagrams gi#e a gra!hicalre!resentation o #ertical shear thro(gho(tthe length o a "eam


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4

Beam Shear )

♦ Sim!le "eam

) S!an ' 25 eet

) 2 concentrated loads

♦ 6onstr(ct shear

diagram


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7

Beam Shear ) Eam!le 1

1, Determine the reactions

Solving equation (3):


Figure 6.7a =>

( ) ( )

( ) ( )

( ) ( )∑∑∑

×−×+×==↵+

+−−==↑+

=↑+

,825+,8171255+,879555+5&

1255955552

51

2

::

1

2

::

1

R M

R R F

F

lblb

lblb

y

( )↑==

+=

−

−−

:

:

::

2

::::

2

&75;&25

255;7<

255;1=555;-9825

lb

!t

!t lb

!t lb !t lb

R

R

( )↑=−+=

:

1

:::

1

9-5;4

&75;&255;1555;9

lb

lblblb

R

R


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Beam Shear ) Eam!le 1 +!g 7-,

♦ Determine the shear at #ario(s

!oints along the "eam

:

,19+

:

,9+

:

,1+

&75;&255;1555;9-95;4

175;2555;9-95;4

-95;45-95;4

lb

lb

lb

V

V

V

−=−−=

−=−=

+=−=

=

=

=


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♦ 6oncl(sions ) ma #ertical shear ' 4;9-5 l"

ma #ertical shear occ(rs at greater

reaction and e>(als the greater reaction

+or sim!le s!ans,

) shear changes sign (nder 9;555 l" load

$here ma "ending occ(rs


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♦ Sim!le "eam

) S!an ' 25 eet

) 1 concentrated load

) 1 (niormly distr load

♦ 6onstr(ct shear diagram;

designate maim(m shear;

locate $here shear !asses

thro(gh 3ero


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♦ Determine the reactions



( ) ( )

( ) ( )

( ) ( )∑∑∑

×−×+×==↵+

+−×−==↑+

=↑+

,82-+,8177555+,?87,+812555;1@+5&

555;7,812555;1+52

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2

1

2

1

R M

R R F

F

lb !t lb

lb !t lb

y

( )↑==

+=

−

−−

:

:

::

2

::::

2

555;<2-

555;179

555;=7555;


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Shear and Moment Diagrams


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( )[ ]

( )[ ] :,2-+

:

,17+

:

,17+

:

,12+

:,2+

:

,1+

555;


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♦ 6oncl(sions

) ma #ertical shear ' 11;555 l"

at let reaction

) shear !asses thro(gh 3ero at some

!oint "et$een the let end and theend o the distri"(ted load

= eact location rom R 1

) at t"i# lo$ation% V = &

!eet

V

11

,555;1+555;115

=×−==


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Beam Shear ) Eam!le &

♦ Sim!le "eam $itho#erhanging ends ) S!an ' &2 eet

) & concentrated loads

) 1 (niormly distr loadacting o#er the entire "eam

♦ 6onstr(ct shear diagram;designate maim(m shear;

locate $here shear !assesthro(gh 3ero


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( ) ( )

( ) ( )

( ) ( ) ( )( )[ ]

( ) ( ) ( )( )[ ] ,825+-2&2,8&2455+,829555;2+,87555;12+,8-555;-+5-

,825+92

&2,8&2455+,82-555;-+,81-555;12+,89555;2+5&

,8&2455+555;-555;12555;252

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1

2

2

1

2

1

R M

R M

R R F

F

!t lblblblb

!t lblblblb

!t lblblblb

y

+−×−×−×−×==↵+

−−×+×+×+×−==↵+

+×−−−+−==↑+

=↑+

∑

∑∑∑

( )↑==

+++−=

−

−−−−

:

:

::

2

::::::::

2

955;19

25

555;&


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Determine the reactions


Solving equation ('):


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( )( )

( )

( )

( )( )

( ) ,&2+

,29+

,29+

,22+

,22+

,9+

,9+

555;-8&2455555;12555;2955;19255;14

555;7829455555;12555;2955;19255;14955;12829455555;12555;2255;14

955;=822455555;12555;2255;14

255;2822455555;2255;14

255;=89455555;2255;14

555;789455555;25

lb !t lblblblblb

lb !t lblblblblb

lb !t lblblblb

lb !t lblblblb

lb !t lblblb

lb !t lblblb

lb !t lblb

V

V V

V

V

V

V

=×−−−+=

=×−−−+=

−=×−−−=

−=×−−−=

=×−−=

=×−−=

−=×−−=

−=

+=

−=

+=

−=

+=

−=


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♦ 6oncl(sions

) ma #ertical shear ' 12;955

l"

disregard C notations

) shear !asses thro(gh 3ero at

three !oints

R 1; R 2; and (nder the 12;555l"

load


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Bending Moment


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Bending Moment ) Eam!le 1

♦ Sim!le "eam ) s!an ' 25 eet

) 2 concentrated loads

) shear diagram romearlier

♦ 6onstr(ct moment

diagram


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1, 6om!(te moments at

critical locations

) (nder 9;555 l" load

1;255 l" load

( )

( ) ,817+

,87+

--5;1&,815555;9+,8179-5;4+

5-5;&45,879-5;4+

!t lblblb

!t lblb

M

M

−=

−=

=×−×=↵+

=−×=↵+


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2&


♦ Sim!le "eam ) S!an ' 25 eet

) 1 concentrated load

) 1 (niormly distr oad ) Shear diagram

♦ 6onstr(ct moment

diagram


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1, 6om!(te moments at

critical locations

) .hen ' 11 t and (nder

7;555 l" load

( ) ( )( )[ ]( ) ( ) ( )( )[ ] :::,817+

:::

,811+

555;47-2

1212555;1,817555;11+

455;752

1111555;1,811555;11+

!t lb !t lblb

!t lb !t lblb

M

M

−=

−=

=+×−×=↵+

=×−×=↵+


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Negati#e Bending Moment

♦Fre#io(sly; sim!le "eams s("%ected to !ositi#e

"ending moments only

) moment diagrams on one side o the "ase line

conca#e (!$ard +com!ression on to!,

♦O#erhanging ends create negati#e moments conca#e do$n$ard +com!ression on "ottom,


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Negati#e Bending Moment

♦ delected sha!e has

inlection !oint

) "ending moment ' 5

♦ See eam!le


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2<

Negati#e Bending Moment C

Eam!le ) Sim!le "eam $ith

o#erhanging end on right

side

S!an ' 25G

O#erhang ' 7G

*niormly distri"(ted load

acting o#er entire s!an

) 6onstr(ct the shear and

moment diagram

) ig(re 712


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Eam!le2, Determine the shear at

#ario(s !oints along

the "eam and dra$

the shear diagram

,25+

,25+

,15+

,1+

755;&,75525+1-5;15-75;4

4-5;7,75525+-75;4

4-5,75515+-75;4

975;-,7551+-75;4

lb

lb

lb

lb

V

V

V

V

=×−+=

−=×−=

−=×−=

=×−=

+=

−=

=

=


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&5


Eam!le&, Determine $here the

shear is at a

maim(m and $here

it crosses 3ero ) ma shear occ(rs at the right

reaction ' 7;4-5 l"

!eet

V

1:=

,755+-75;45

=

×−==


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&1


Eam!le-, Determine the

moments that the

critical shear !oints

o(nd in ste! &, anddra$ the moment

diagram

( )( )[ ]( )( )[ ] :::,25+

:::

,1:=+

955;15

2

25825755,825-75;4+

9-&;2-2

1:=81:=755,81:=-75;4+

!t lb !t lblb

!t lb !t lblb

M

M

−=

−=

−=×−×=

=×−×=


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&2


Eam!le-, ind the location o the inlection

!oint +3ero moment, and ma "ending moment

♦ since cannot '5; then $e (se '192G

♦ Ma "ending moment '2-;9-& l"Ct

( )( )[ ]

!eet !eet

M

M

M !t lb

2:19I5755

-75;44-75

,&55+2

,5,+&55+-,-75;4+-75;4

-75;4&55&55-75;45

2755-75;45

2755,-75;4+5

2

22

2

=−

±−=

−−−±−

=

+−=−==

−==

×−×==


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&&

R(les o Th(m"Re#ie$

♦ shear is de!endent on the loads and reactions

) $hen a reaction occ(rsI the shear J%(m!sK (! "y the

amo(nt o the reaction

) $hen a load occ(rsI the shear J%(m!sK do$n "y theamo(nt o the load

♦ !oint loads create straight lines on shear diagrams

♦ (niormly distri"(ted loads create slo!ing lines o

shear diagrams


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R(les o Th(m"Re#ie$

♦ moment is de!endent (!on the shear diagram

) the area (nder the shear diagram ' change in the

moment +ie Ashear diagram ' LM,

♦ straight lines on shear diagrams create slo!inglines on moment diagrams

♦ slo!ing lines on shear diagrams create c(r#es on

moment diagrams

♦ !ositi#e shear ' increasing slo!e

♦ negati#e shear ' decreasing slo!e


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Ty!ical oadings

♦ n "eam design; only

need to 0no$:

) reactions

) ma shear ) ma "ending moment

) ma delection

BEAMS-SHEAR AND MOMENT (17).ppt

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