u1l-3
TRANSCRIPT
-
8/12/2019 u1L-3
1/27
PH0101 UNIT 1 LECTURE 3 1
PH0101 UNIT 1 LECTURE 3
Bending of Beams
Bending moment of a BeamUniform Bending (Theory and Experiment)
Worked Problem
-
8/12/2019 u1L-3
2/27
PH0101 UNIT 1 LECTURE 3 2
Bending of Beams
A beam is a rod or bar of uniform crosssection(circular or rectangular) whose length is verymuch greater than its thickness as shown inFigure
A beam is considered to be made up of a largenumber of thin plane layers called surfacesplaced one above the other.
AB
MN
DC
-
8/12/2019 u1L-3
3/27
PH0101 UNIT 1 LECTURE 3 3
Consider a beam to bebent into an arc of acircle by the applicationof an external couple
as shown Figure Taking the longitudinal
section ABCD of thebent beam ,the layers
in the upper half areelongated while thosein the lower half arecompressed.
-
8/12/2019 u1L-3
4/27
PH0101 UNIT 1 LECTURE 3 4
In the middle there is a layer (MN) which is notelongated or compressed due to bending of the
beam.
This layer is called theneutral surfaceand the
line (MN) at which the neutral layer intersects the
plane of bending is called the neutral axis.
It is obvious that the length of the filamentincreases or decreases in proportion to its distanceaway from the neutral axis MN.
-
8/12/2019 u1L-3
5/27
PH0101 UNIT 1 LECTURE 3 5
The layers below MN are compressed and thoseabove MN are elongated and there will be such
pairs of layers one above MN and one below
MN experiencing same forces of elongation and
compression due to bending and each pairforms a couple.
The resultant of the moments of all these
internal couples are called the in ternal bend ing
momentand in the equilibrium condition, this is
equal to the external bending moment.
-
8/12/2019 u1L-3
6/27
PH0101 UNIT 1 LECTURE 3 6
Bending Moment of a Beam
Consider the section PBCP, the extended filaments
lying above the neutral axis MN are in state oftension and exert an inward pull on the filamentadjacent to them towards the fixed end of the beam.
M
N
C
B
P
D
AP
Load
-
8/12/2019 u1L-3
7/27
PH0101 UNIT 1 LECTURE 3 7
In the same way the shortened filaments lying below theneutral axis MNare in a state of compression and exertan outward push on the filaments adjacent to themtowards the loaded end of the beam.
As a result tensile and compressive stresses develop inthe upper and lower halves of the beam respectively andform a couple which opposes to bending of the beam.
The moment of this couple is called the moment of the
resistance.
When the beam is in equilibrium position the bendingmoment and restoring moment or moment of resistanceshould be equal.
-
8/12/2019 u1L-3
8/27
PH0101 UNIT 1 LECTURE 3 8
To find an expression for
the moment of therestoring couple considera fiber DCat a distance rfrom the neutral axis MNas shown in Figure
Let the radius ofcurvature be Rand bethe angle subtended by itat the centre of curvature.In unstrained position ofthe beam, the length ofthe fiber DC= MN = R.
In the strained positionthe length of the fibre,
C
A
M
B
D
N
DC= (R + r)
-
8/12/2019 u1L-3
9/27
PH0101 UNIT 1 LECTURE 3 9
Strain in the fiber DC, =
or
lengthOriginal
lengththeinChange
R
r
R
RrR
Strain
)(
-
8/12/2019 u1L-3
10/27
PH0101 UNIT 1 LECTURE 3 10
i.e., strain is proportional to the distance from the
neutral axis.
Let the area of the fiber beAand its neutral axis
be at a distance r from neutral axis of the beam
and the strain produced be r/R. We haveStress = Yx Strain = Y r / R
Hence, force on the areaA
F = Y(r/R) xA
Therefore the moment of this force about MN
= Y(r/R) xAx r= Y Ar2/ R
-
8/12/2019 u1L-3
11/27
PH0101 UNIT 1 LECTURE 3 11
As the moment of the forces acting on both the upper andlower halves of the section are in the same direction, thetotal moment of the forces acting on the filaments due tostraining
gIR
YarR
Y
R
raY
22
where Igis the geometrical moment of inertia and
is equal to AK2
, A being the total area of thesection and Kbeing the radius of gyration of the
beam
-
8/12/2019 u1L-3
12/27
PH0101 UNIT 1 LECTURE 3 12
gIRY
gIR
Y
:. moment of the forces
In equilibrium, bending moment of the beam is equal and
opposite to the moment of bending couple due to the loadon one end.
:. Bending moment of the beam =
The quantity YIg (=YAK2) is called the flexural rigidity of the
beam. Flexural r igid i ty is defined as the bending moment
required to produce a unit radius of curvature.
-
8/12/2019 u1L-3
13/27
PH0101 UNIT 1 LECTURE 3 13
Uniform Bending The beam is loaded
uniformly on its both ends,
the bent beam forms an
arc of a circle. An elevation
in the beam is produced.
This bending is calleduni form bending.
Consider a beam (or bar)
AB arranged horizontally
on two knife edges C
and D symmetrically so
that AC = BD = a ,as
shown in Figure
-
8/12/2019 u1L-3
14/27
PH0101 UNIT 1 LECTURE 3 14
The beam is loaded with equal weights W and Wat the ends A and B.
The reactions on the knife edges at C and D areequal to W and W acting vertical upwards.
The external bending moment on the part AF ofthe beam is
= W x AFW x CF = W (AFCF)= W x AC = W x a
-
8/12/2019 u1L-3
15/27
PH0101 UNIT 1 LECTURE 3 15
R
YIWa
g
Internal bending moment =where
Y - Youngsmodulus of the material of the bar
Ig - Geometrical moment of inertia of the cross-
section of beamR - Radius of curvature of the bar at F
In the equilibrium position,
external bending moment = internal bending
moment
R
YIg
-
8/12/2019 u1L-3
16/27
PH0101 UNIT 1 LECTURE 3 16
Since for a given value of W, the values of a, Y and
Igare constants, R is constant so that the beam isbent uniformly into an arc of a circle of radius R.
CD = l and yis the elevation of the midpoint E of the
beam so that y= EF
D
F
C
O
y
R
l/2 E
-
8/12/2019 u1L-3
17/27
PH0101 UNIT 1 LECTURE 3 17
EF (2REF) = (CE)2
y(2Ry) =
y2R = (y2is negligible)
y=
2
l
R8
2
l
4
2
l
-
8/12/2019 u1L-3
18/27
PH0101 UNIT 1 LECTURE 3 18
or
Wa =
or
g2 I8 Yy
2
8R1
ly
yIalWY
g8
2
-
8/12/2019 u1L-3
19/27
PH0101 UNIT 1 LECTURE 3 19
If the beam is of rectangular cross-section,
, where b is the breath and d is the
thickness of beam. If M is the mass, the
corresponding weight W = Mg
12
3bd
Ig
-
8/12/2019 u1L-3
20/27
PH0101 UNIT 1 LECTURE 3 20
Hence
ybd
aMglY
3
2
2
3
from which Y the Youngs modulus of thematerial of the bar is determined.
-
8/12/2019 u1L-3
21/27
PH0101 UNIT 1 LECTURE 3 21
Exper iment
A rectangular beam (or bar) AB of uniformsection
is supported horizontally on two knife edges A
and B as shown in Figure
-
8/12/2019 u1L-3
22/27
PH0101 UNIT 1 LECTURE 3 22
Two weight hangers of equal masses aresuspended from the ends of the beam.
A pin is arranged vertically at the mid-pointof the beam.
A microscope is focused on the tip of the pin.
Initial reading of the microscope in thevertical scale is noted.
-
8/12/2019 u1L-3
23/27
PH0101 UNIT 1 LECTURE 3 23
Equal weights are added to both hangerssimultaneously and the reading of the
microscope in the vertical scale in noted.
The experiment is repeated for decreasing
order of magnitude of the equal masses.
The observations are then tabulated and the
mean elevation (y) at the mid point of the bar
is determined.
-
8/12/2019 u1L-3
24/27
PH0101 UNIT 1 LECTURE 3 24
Load in Kg
Microscope reading elevation Meanelevation(y)
for
a load of M.
Load
increasing
Load
decreasingMean
WW+50 gms
W+100 gms
W+150 gms
W+200 gms
W+250 gms
-
8/12/2019 u1L-3
25/27
PH0101 UNIT 1 LECTURE 3 25
The length of the bar between the knifeedges lis measured.
The distance of one of the weight hangersfrom the nearest knife edge p is measured.
The breadth (b) and thickness (d) of the barare measured using vernier calipers andscrew gauge.
-
8/12/2019 u1L-3
26/27
PH0101 UNIT 1 LECTURE 3 26
The youngs modulus of the material of the
beam is determined by the relation
Nm-2
ybd
lagMY
3
2
2
3
-
8/12/2019 u1L-3
27/27
PH0101 UNIT 1 LECTURE 3 27
THANK YOU
THANK YOU