243-Wikarta-KULIAH v & VII BEAM Gaya Internal, Diagram Gaya Geser Dan Momen

7/23/2019 243-Wikarta-KULIAH v & VII BEAM Gaya Internal, Diagram Gaya Geser Dan Momen

1/22

KULIAH V & VII

MEKANIKA TEKNIK TI

BEAMGAYA INTERNAL, DIAGRAM

GAYA GESER DAN MOMEN

OLEH:

ALIEF WIKARTA, ST

JURUSAN TEKNIK MESIN

FTI ITS SURABAYA, 2007


2/22

Defii!i Be"#

Beam- structural member designed to support

loads applied at various points along its length.

Beam designis two-step process:

1) determine shearing forces and bending

moments produced by applied loads2) select cross-section best suited to resist

shearing forces and bending moments

Beam can be subjected to concentratedloads or

distributedloads or combination of both.


3/22

A$" i% '"(" I%e)"* +

traight two-force memberABis in

e!uilibrium under application of Fand

-F. Internal forcese!uivalent toFand -Fare

re!uired for e!uilibrium of free-bodies

ACand CB.

'"(" I%e)"* : "(" (" #ei-"% .e)!"#" .e).""i."i" !%)-%) !e/i" !%)-%) %e)!e.% #e"1i --/


4/22

Re"-!i $"1" Be"#

Beams are classified according to way in which theyare supported.

"eactions at beam supports are determinate if they

involve only three un#nowns. $therwise% they are

statically indeterminate.


5/22

'"(" 'e!e) 1" M#e $"1" Be"#

&ish to determine bending moment

and shearing force at any point in a

beam subjected to concentrated and

distributed loads.

'etermine reactions at supports by

treating whole beam as free-body.

(ut beam at Cand draw free-body

diagrams forACand CB. By

definition% positive sense for internal

force-couple systems are as shown.

rom e!uilibrium considerations%

determineM and VorMand V.


6/22

Di")"# '"(" 'e!e) 1" M#e $"1" Be"#

*ariation of shear and bending

moment along beam may beplotted.

'etermine reactions at

supports.

(ut beam at Cand consider

memberAC%22 PxMPV +=+=

(ut beam atEand consider

memberEB%

( ) 22 xLPMPV +== or a beam subjected to

concentrated loads% shear is

constant between loading points

and moment varies linearly.


7/22

3%/ S"* 4

'raw the shear and bending moment

diagrams for the beam and loading

shown.

$+,$/:

a#ing entire beam as a free-body%

calculate reactions atBandD.

ind e!uivalent internal force-couplesystems for free-bodies formed by

cutting beam on either side of load

application points.

0lot results.


8/22

3%/ S"* 4 5"6"."

0lot results.

/ote that shear is of constant value

between concentrated loads and

bending moment varies linearly.


9/22

3%/ S"* 2

Given: beam is supported

by a hinge at % a roller at B.

orce applied at (. oment

applied at '.

Find: 'raw the shear and

bending moment diagrams

Plan:

a) 'raw a B' of the beam.

b)(alculate support reactions.

c) ind e!uivalent internal force-couple systems for free-bodies formed

by cutting beam on either side of load application points.


10/22

3%/ S"* 2 5"6"."

3y4 5

y6 By7 855 4 5

5.1 6 By7 5.8 4 5

By4 5.9 #ip

3B4 5

- y25) 6 5.815) 7 ; 4 5

25y4 8 - ;

y4 5.1 #ip

y By


11/22

3%/ S"* 2 5"6"."

0otongan 1-1

31-14 5

5.1 7 * 4 5

* 4 5.1 #ip

< 4 5 ft *4 5.1 #ip

< 4 15 ft *(4 5.1 #ip

31-14 5

7 5.1


12/22

3%/ S"* 2 5"6"."

0otongan 2-2

32-24 5

5.1 7 5.8 7 * 4 5

* 4 5.1 7 5.8 #ip

* 4 - 5.9 #ip

< 4 15 ft *(4 - 5.9 #ip

< 4 19 ft *'

4 - 5.9 #ip

32-24 5

7 5.1


13/22

3%/ S"* 2 5"6"."

0otongan =-=

5.1 #ip 5.9 #ip 0.1 kip15 x < 20 ft

V

M

x

3=-=4 5

5.1 7 5.8 7 * 4 5

* 4 5.1 7 5.8 #ip

* 4 - 5.9 #ip

< 4 19 ft *'4 - 5.9 #ip

< 4 25 ft *'

4 - 5.9 #ip

3=-=4 5

7 5.1


14/22

3%/ S"* 2 5"6"."

'iagram >eser dan omen

31-14 5 4 5.1


15/22

3%/ S"* 8

'raw the shear and bending moment

diagrams for the beamAB. he

distributed load of ?255 /@m. e


16/22

3%/ S"* 8 5"6"."

$+,$/:

a#ing entire beam as a free-body% calculatereactions atAandB.

:5= AM

( ) ( ) ( ) ( ) ( ) 5m99.5/1A55m19.5/2185mA.5 =B

/18;2=B

:5= BM

( ) ( ) ( ) ( ) ( ) 5mA.5m29.5/1A55m89.5/2185 =+ A

/2=1A=

A

:5= xF 5=xB

/ote: he 1A55 / load atEmay be replaced by

a 1A55 / force and 1A5 /m. couple atD.


17/22

3%/ S"* 8 5"6"."

:51 =M ( ) 5?2552=1A 21

=++ Mxxx

2=8552=1A xxM =

:52 =M ( ) 519.521852=1A =++ Mxx

( )m/19A=2;

+=xM

rom CtoD:

= :5F 521852=1A = V

/19A=V

valuate e!uivalent internal force-couple systems

at sections cut within segmentsAC% CD% andDB.

rom to C:

= :5F 5?2552=1A = Vx

xV ?2552=1A=


18/22

3%/ S"* 8 5"6"."

:52 =M

( ) ( ) 5;9.51A551A519.521852=1A =+++ Mxxx

( ) m/18;21=1; = xM

valuate e!uivalent internal force-couple

systems at sections cut within segmentsAC%

CD% andDB.

romDtoB:

= :5F 51A5521852=1A = V

/18;2=V


19/22

3%/ S"* 8 5"6"."

0lot results.rom to C:

xV ?255=21A=

225=21A xxM =

rom CtoD:

/19A=V

( ) m/19A=2; += xM

romDtoB:/18;2=V

( ) m/18;21=1; = xM


20/22

3%/ S"* 9

'raw the shear and bending-moment diagrams

for the beam and loading shown.


21/22

3%/ S"* 9 5"6"."


22/22

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243-Wikarta-KULIAH v & VII BEAM Gaya Internal, Diagram Gaya Geser Dan Momen