sm lesson important learning should 2.pdf

Supports

1. ARS202 Unt 2 Bending of 5eams (CompatibilityModej- Microsoft PowerPoint
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Reactions at supports of beamns
A beam is a structural member used to support loads applied at
variouspointsalongits length. Beams are generally long, straight
and prismatic (i.e. of the same cross-sectional area throughout the
length of the beam). II
Types of Supports
Beams are supported on roller, hinged or fixed supportsas shown
n Fig.
Simple Support:
If one end of the beam rests in a fixed support,the supportis
known as simple support.The reactionof the simple supportis
always perpendicularto the surfaceof support.The beam is free
to slide and rotate at the simple support.
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Roller Support:
Here one end of the beam is supported on a roller.The only
reactionof the roller supportis normal to the surfaceon which
the roller rolls withoutfriction. Fig. in which four possible
situationsare illustrated.Examplesof roller supportsare wheels of
a motorcycle, or a handcart, or an over-head crane, or of a car, etc.
lerelte
Hinged Support:
At the hinged support,the beam does not move either along or
normaltoits axis. The beam, however,may rotate at the hinged
support.
The total supportreactionis R and its horizontal and
verticalcomponents are H and V, respectively.Since the beam is
free to rotate at the hinged support, no resistingmoment will
exist. The hinged support behaves like the hinges provide to doors
**** ***
and windows.
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Fixed Support:
At the fixed support,the beam is ot free to rotate or slide along
the length of the beam or in the direction normal to the beam.
Therefore, there are threereactioncomponents, viz.,vertical
reactioncomponent (V), horizontal reactioncomponent (H) and
the moment (M),
as shown in Fig. Fixed support
is also known as
built-insupport.
. .
eam
Racton
Support oued
Reacton-V
TReacton
() Sanple support
idined ufece
Roleron horuoriasure
Roor ouppo
nd ho
VARApeveyeo ta (V
V
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Sign
conventions for Shear force and Bending
moment
) Shear force. Fig. 6.9 shows a simply supported beam A8, carryinga load of 1000 N
at its middle point. The reactions at the supportswill be equal to 600 N. Hence R = Rp
-500 N.
Now imaginethe beam to be dividodinto two portionsby the sectionX-X. The resultant
of the lond and ractionto the left of X.X is 500 N vertically upwards.(Notein this case, there
is no load to the left of X-X). And the resultant of the load and
reactionto the
right of
X-X is
(1000-
500 f
= 600 J N) 500N downwards. The resultant force
acting on any
one
of the pArts
normal
to
the axis
of the
beam
is called the shear force at the sectionX-X. Here the shear force
at the sectionX-X is 500 N.
**
The shear
force at a section will be consideredpositivewhen the resultantof the foreces
to the left to the section is upwards,
or to the right
of the section is downwards.Similarly
the
shear force at a sectionwill be
considered negative
if the resultant of the forces to the lef> of
the sectionis downwards,
or to
the right
of
the aetion is upwards. Here the
reaultant force to
the left of the sectionis upwardsand hence the shear force will be poeitive.
*
1000 N
| 1000 N
500 N
500 N
Fig. 6.9
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SHEAR FORCE AND BENDINGMOMENT DIAGRAMSFOR A CANTILEVER
wrTH A POINT LOAD AT TIIE FREE END
Fig. 6.14 showa a cantilever AB o> length L fxed at A and free at B and earryinga point
load W at the free end B
bose ine Baoe ine
***
BM. dagram
Fg 6.14
F,
Shear force
at X,
and
M, Bendingmomentat X.
Let
Takea soction X at a distancez from the tree end. Considerthe right portionof the
Bection.
The shear foree al this sectionts equal to the resultant force acting on the right portion
at
the givensoctlon.
But
the resultant foroe neting
on
the right portion
at
the pecetion
X
is W
and acting in the downwarddirection. But a force on the right portion acting downwards is
considered positive. Hence shear force at X is positive.
+W
The shear force
will bo
constantat
all
soections of the cantilover between
A and B as
there ie no otherload betweenA and B. The sbear force diagramis shown in Fiz. 6.14 (6).
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Sign conventions for Shear force and Bendingmoment
() Bendingmoment.The bendingmomentat a sectionis considered positiveif the
bendingmomentat that sectionis such that it tends to bend the beam ta a curvature having
concavity
at
the
top
us
shown
in
Fig.
6.10 (a). Similarly
the bendingmomé
(B.M.)at a section
is
considored nogative
if the bending
momentat
that
sectionis such that it tends to bend the
beam
to a curvature
having convexity
at thhe top
as shown in Fig.
6.10 (6). The positiveB.M. is
often called saggingmomentand negative B.M. as
hoggingmoment
1000 N
Considerthe simply supported beam AB,
Carrying
a load of 1000 N at its middle point.
toactions A
and
Rg
are equal and are having
magnitude 500 N as shown in Fig 6.11. Imagine
the bcam to be dividedinto two portionsby
the
RectionX-Xx.Let the sectionX-X ia at a distunceof
R500N
Re S00 N
Im from A. Pig
6.11
.
Convoxity
Concavity
Concavily
Comvodty
(a) PosiiveB.M. (6) NagativeB.M.
** **** Fig. 6.10
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Sign
conventions for Shear force and Bending
moment
The momentsof all the forces (ie, losd and reaction) to the left of XX at the section
xX is R
x 1
500
x 1
= 500 Nm (clockwise). Also the momentsof all the foreea üe, oad and
roactions to
tho
right
of
Kx at the sectionXXis R, x3 (antielockwise)-
1000x 1 (clockwine)
500 x 3 Nm-1000x 1 Nm = 1500-1000=500 Nm (anti-clockwise).
Hence the tandency
of
the bending
momentat X-X is to bend the beam so as to produo
concavity
at the top as shown in Fig.
6.12.
Clockwise Anticlockwiso
AnticlockwiseClockwa
Fig
6.12 Fig
6.13
The bendingmomentat a section
is
the
algebraie
sum of
the momentsof forces and
reactions acting on one side of the section. Henco bending moment at the
sectionX-X
is 500 Nn.
The
bendingmomentwill be considered positive
when the
moment of the forces and
reactionon
the left
portion
is
elockwise,
and
on
the right portionanti-cloekwise. In Fig. 6.12,
the bendingmomentat the sectionX-X is positive.
Similarly the bandingmomentwill be
considered negativewhen the moment of
the
forces and reactions on the left portionis anti-clockwise, and on the right portionclockwise as
shown in FRig
6.18. In Fig. 6.13, the bending moment at the sectionX-X is
negative
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affice Thema

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Important points for drawingShearforceand Bending
moment diagrams
The following are the important points for drawingshearforceand bendingmoment
diagrams
1. Considerthe left or the right portionof the section.
2.
Add
the
forees(including reaction) normal
to
the beam on
one of
the portion.If right
portionof the
section
is choson,a
force on the right portionactingdownwards is positivewhile
:****
a force acting upwardsis negative.
Ir the lef portionof the
sotion
is chosen,a
loroe on the let portionactingupwardeis
***
positivewhilea force acting downwards is negative.
3. The positivevaluesof shear force and bendingmomentsare plottedabovethe base
line, and negativevaluesbelowthe base line.
4. The
shear force diagramwill inereaseor decreasesuddenlyi.e.,bya verticalstraight
line at a sectionwherethereis a verticalpointload.
5. The shear force betweenany
two verticalloadswill be constantand hencethe shear
force diagrambetweentwo verticalloadswill be horizontal.
6. The
bendingmorment
at the two supportsof a simply supported beam and
at the free
end of a cantilever will be zero.
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9. nit 2
Bending
of Beams [compatibiny ode
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SHEAR
FORCE AND BENDINGMOMENT DIAGRAMS FOR A CANTILEVER
WITH A PoINT LOAD AT THE FRER END
Pig. 6.14 showa a eantilever
AB of length L fxed atA and free at B and carrying
a point
load Wat the free end B.
(a)
dingam
Base
Base ine
Wxx
BM dagram
******
14
F
Shear force
at X,
and
M,
-
Bondingmomentat A.
Let
Take a soctionX at a distancefrom the hee end. Conaider the nght portionof the
Bection
The shear
foree
at
this section
ts
equalto the reraltant force acting on
the
rieht portion
at the glven section.But
the resultant forco ncting
on
the right portion
at
the section
X
is W
and acting in the downwarddirection. Buta forceon the right portionactingdownwards is
conusidered positive. Hence shear force at X is positive.
The shear toros will be constantat all sectionsol the eantilever betweenA and B as
there ie no other load betweenA and B. The shear force dingTamis shown in Pu.
6.14 (6).
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10. BendingMomentDiagram
The bending
momentat the sectionX is givenby
M,-W*x
CBending moment will be negativeas
for the right portionof
the section,the momentof
Wat Xis eloekwise. Also the bendingof cantilever will take placein such a mannerthat
convexity will be at the top of the beam).
From equation(i), it is clear that B.M. at any sectionis proportional
to the distanceof
the sectionfrom the free end.
Atx
= 0 ie., at B, B.M.
= 0
Atr= Lie., at A, B.M. -WxL
Hence B.M. followsthe straight line law. The B.M. diagramis shown in Fig. 6.14 (c). At
point A, take AC
= WxL in the downwarddirection Join pointB to C.
The shear forco and bendingmomentdiagramsfor severalconcentrated loadsacting
on
a cantilever, will be drawn in the similar
manner
V.Nithya

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SFD and BMD for a cantilever beam with a uniformly distributed
load
w Per unit kength
shows a cantilever of length
L fixed at A and carry1ng
a
unitormy
aistriouea
Fig.
load of w per
unit length
over the entire length
of the cantilever.
Take a sectionX at a distanceofx from the free end B.
F,
Shear force at X,
and
M, Bending
momentatX
Here
we have consideredthe rightportiorof the section. The shear force at the section
X
will
be equal to the resultant force acting
on theright portion
of the section.But the result
ant force on the right portion
= to x Length of right portion
=
wx
This resultant force is acting downwards. But the resultant force on the right portion
acting downwards is considered positive. Henceshearforceat X is
positive
Let
20
The above equation
shewsthat the shear force Sollows a straightline law.
At B,x = 0 and hence P=0
At A,
x
= Land hence P-
vl
The shear force diagramis shown in Mg
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SFD and BMD for a cantilever beam with a uniformly distributed
load
w Por unit kngt
a)
Fi hows a cantilever of length
L fixed at A and carrying
a unitormiy
austrnouzea
loadof w
per
unit length
over the entire length
of the cantilever.
Take a sectionXat a distanceofz from the free end B.
F,
Shear force
at X,
and
M, Bendingmomentat X.
Here we have considered the rightportiorof the section.The shear force at the section
X
will
be equal to the resultant force acting on the right portion
of the seetion.But the result
ant force on the right portion
= to x Length of right portion
= wx.
This reultantforceis acting downwarda. But the resultant fonce
on the right portion
actingdowwardsis considered positive. Hence shear force at X is positive.
Let
* *
The above equationshowsthat the shear foroe followsa straightline law.
At B,x = 0 and hence P,=0
AtA,x =L and hence P w.L
The shearforcediagramis shown in Fig.
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w Por unit length
()
(6) wL
A 8.F diagram
Base line
8ase line
rgzz
in
B.M. diagram
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14. Unit 2 Bendingof Beanms [Compatibility Modej- Microsoft PowerPoint
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(
2-
Base ine
wwwwwwu TTTTTTtmnn
SF. 6agromn
usuuuuuuun
BM.dagram Base in
3
* **..
* ***
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******
Fig. 6.24 showsa beam AB of length
L simply supported
at the ends A and B and
carry
ing a pointload W at its middle point C.
The reactionsat the support will be equal toas theloadis acting at the middle point
of the
beam. Hence RR
Take a sectionX at a distance x from the end A botween A and C.
F
Shear
force at
X,
-Bendingmomentat X.
Iet
and
Here we havo cohsidered the let portion
of
the section.
The
shear
foree
at X will
be
equal
to the resultant force acting
on the loft portionof tho section.But the resultant force on
the left portioniactingupwarda.
But sccording
to the sigm convention, the vesultant
fore
on the lef portionáctingupwardsis considered postive.Henceshoarfore at Xis positiveand
its magnitude is
-
Hence the shear foree betweenA andCis constantand equal to+
..4.t
55
** **
-.*TT
m***
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** ********
Now considor any
sectionbetweenC and B at distancez from end A. The resultant force
on the loft portion will be
-
This force will also remain conatantbetweenCand B. Hence shear force betwoonCand B
W
is qual
to
-
At the sectionC the shear forodVhanges
from Lo-
2
The sbear force diagramis shown in Fig. 6.24 (6).
BendingMomentDiagram
i)
The bending
momentat any
sectionbetwoenAand Cat a distanceofx from theend A,
is given by
W
M =RA* or M *
CB.M.will be
positive
as for the
left portionof
the
section,
the
moment
of all
foreos at
X
is
clockwise. Moreover, the bending
of beam takes place
in such a manner that concavity
is at
the top of the beam).
*
W
ALA, 0 hence M x0 -
0
L
At C,-hence Mc
3
***
*
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From equation(i), it is clearthat B.M. variesaccording to straightline law between
WxL
A and C. B.M. is zero at A and it increases to at C.
Cii) The bendingmomentat any sectionbetweenC and B at a distancex from the endA,
is given by
M,R-W )a-* W
AsC c M
At B,x
= L hanoe
M -2*=0.
WL
Honce bendingmomentat C is and it decreases to zero at B. Now the B.M.diagram
can be completed as shown in Fig. 6.24 (c).
Note. The bendingmomentis maximumat the middle point C, wbere the ehear forco changesics
sign.
V.Nithya
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Pig 6.25 ahowsa beam AB of length L simply supported at the enda A and B and
carryinga point load W at C at a distanceof'a from the end A.
b-
SE dagram
uuuu .
tallliuuuiuutiuuuútt
Bdegran
6.25
Let RA=
Reactionat the supportA, and
R
Reaction
at the supportB.
irst
calculatethe reactions,by laking momentsaboutA or about B.
Taking momentsof the forcos on tho beam about A, we got
*
RxL Wxa
R
:ar******
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Font
and RAW-R, W-
L
w1-w().W
Considera sectionX at a distancex trorntho end
Agetween
A and C.
(: L-a ö)
Tho shear torce F, at the sectionis given by,
W.b
F RA
..i)
(The shoar foreo will be positive
as the resultant forceon the lefe portionof the sectionis
acting upwards).
W.b
The shear force between A and C is constantand oqual
to
Now consider
any sectionbetwoonC
and B at a distancefrom the ond A. The resule
ant force on the
left portion
will be
Ra-
W
w-"{--
( L-b=a)
or
The shear force betweenC and B is conatantand equal to
-
,At
the section C, the
hear force changesfromto-"The
ehearforcediagramis showa ia Fiu.
6.26 6).
.
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SHEAR FoRCE AND
BENDINGMOMENTDIAGRAMsFOR A SIMPLY
SUPPORTED BEAM CARRYTNG A UNIPORMI.Y DISTRIBUTED LOAD
Fig. 6.27 shows a beam AR of length
L
simply supported
at
the ends A and B and
carrying a uniformly distributed load of w per
unit
length over the entire length. The reac
tions
at the supportswill be equal and their magnitudo will be half the total load on the
entire length
wU length
Base ino
b)
8F dagram
8M, diagran
Baco ine
.
****
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21. Unit 2 Bendingof Beams [Compatibility Model -Microsoft PowerPoint
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R
Reactionat A,
and
R,-Reactionat B
RA R-
Let
Considerany
sectionX at a distance
z from the left end A. Theshear forco at the section
(ie.,P)
is given by,
..i) +RA- w.I=* *
From equation(i), it is clear that the ehear forco varies according
to straight
line law.
The valuosof shear force at different points
are :
At A,
= 0 hence
AL B, x
=
L henee Fa= -w.L--
AtC,x
hence Fe=+
0
The shearforcediagram
is drawnas shown in Fig.
6.27 (6).
The bendingmomontat the sectionX at a diatance
x from left end A is given by, . .
M,+Ra x-w. x.
w.x
R
41
Fron equation(ii),it is clear that B.M. varies according
to parabolic
law.
**** .
- **** -**
** *****
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22. Unit 2 Bending of Beams [Compatibility Model -MicrosoftPowerPoint
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SHEAR FORCE AND BENDINGMOMBNT DIAGRAMSFOB OVER-HANGING
BEAMS
*"6***
If the end portionof a beam is
extended beyond the support,such
beam
is known
as
overhanging beam. In ease of ovarhanging beams, the B.M. is positivebetweenthe two sup
ports, wheroasthe B.M.
is
neçativo
for
the over-hanging portion.Henceat some point,
the
B.M. is zero after changingits sign from positive
to negativeor vice-verea. That point is known
as the point of contraflexure or point of inflexion.
Polnt of Contraflexure. It is the point where the B.M. is zero after changing
its sign from positive to negative or viceversa.
49
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