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SEV320 Week4Tute
MemberActions
(InternalLoadings)
Memberactions:
N=Normal/axial
V=Shear
M=Moment
SignConventions:
Note:TechnicalimagesCopyright2012PearsonEducation.
+(elongates)
+(dishes)
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SEV320 Week4Tute
Methodofsections:
Usesequilibriumequationsandimaginarysectionthroughmemberto
solvefor
specific
actions
at
various
locations.
Solveforreactionsatsupports(andinternalpins)first.
Make
cut
through
member.
Drawfreebodydiagramofsegmentwithleastloads.
Showmemberactionsintheirpositive directions.
Solveequilibriumequationtodeterminememberactions.
Typicallysolvingformomentfirst(mostdirectsolution).
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SEV320 Week4Tute
ShearandMomentFunctions:
Describetheshearandmomentalongmemberwithanequation.
Axialaction(N)isoftensmallinbeams,thussimplifytoN=0.
Notethatanappliedcompressiveforcewouldchangethis
sobecareful!
Shear
(V)
and
Moment
(M)
change
with
distance
(x)
along
beam.
Useequilibriumequationstowritefunctionswhichallowusto
determinemagnitudeofMandVwithchangeinx.
Vis
obtained
from: 0
Misobtainedfrom: 0
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SEV320 Week4Tute
Shear&MomentDiagrams:
Summarise ShearandMoment
alongentire
beam
in
diagram
form.
SFD=ShearForceDiagram
BMD=BendingMomentDiagram
Tools(equations):
SlopeofSheardiagram=Intensityofload
Or:ChangeinShearwithdistance=Load
SlopeofMomentdiagram=Shear
Or:ChangeinMomentwithdistance=Shear
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SEV320 Week4Tute
SFDs&BMDs(continued):
Therefore:
WhenV=0,slopeofBMD=0
AreaundertheSFD=M
Dont
forget
that: Momentisalwayszeroatapin.
Alsohelpful:
Point
loads
induce
instant change
in
member
shear
(vertical
line
on
SFD)
UDLsinducegradualchangeinmembershear(ie.Vchangeswithx)
Appliedmomentsinduceinstant
changeinmoment(verticallineonBMD)
SFD
BMD
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DrawtheSFDandBMDfortheframe.
First findreactionsatsupports:
FX 0: 0=16x0.6 A X
A X =9.6
kN (
)
MA 0: 0=0.6x162/2+0.8x202/2 20xDY
DY =11.84kN ( )
FY 0: A Y =0.8x20 DY
A Y =4.16kN ( )
Next DrawSFD:
Remember shearactsperpendicular tomember
SimplesttostartatDonmemberCD:
DX =0,thusV=0
Noperp load
on
member,
so
no
change
in
V
ThensaymemberAB:
Ax =9.6kN,thusV=9.6kN inmemberABatA.
UDLmeansgradualslopeonSFDV=9.60.6y
So,aty=16m,V=0kN.
Wk 4TuteQuestion
AXA Y
D Y
0kN
0kN
0kN
11.8kN
(=DY)
4.2kN
(=A Y)
9.6kN SFD
SlopeofSFD=changeinload
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DrawtheSFDandBMDfortheframe.
Lastly DrawBMD:
Tips:
Draw
M
=0kNm points
first
(pins)
AtJointsA&D
Thenuse:M=AreaunderSFD
WorkupmemberABfromAtoB.
NotefixedbeamendconnectionatBandC
MomentatBtransfersfromABtoBC.
NotelocationwhereShear=0kN
SlopeofBMDmustbezeroatthislocation
Wk 4TuteQuestion
0kNm
87.6kNm
0kNm76.8
kNm
BMD
Shear=zero,
soBMDslope=0.
SFD
76.8
kNm
11.8kN
4.2
kN
9.6kN
