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The bending moment at a section tends to bend or
deflect the beam, and the material resists this
bending by developing internal resisting stresses.
The resistance offered by the internal stresses, to the bending, is called “ BENDING STRESSES”.
Assumptions made in the theory of simple bending:-
(1) The material of the beam is perfectly homogeneous (I.e. of the same kind throughout) and Isotropic ( i.e. Having same elastic properties in all the directions). (2) The beam material is stressed within the elastic limit and obeys Hooke’s law. (3) The transverse sections which were plane before the bending remain plane after the bending. (4) Each layer of the beam is free to expand or contract independently on the layer, above or below it. (5) The value of E (Young’s Modulus) is the same in tension & Compression.
Theory of Simple Bending:--
Consider a simply supported beam, subjected to a loading as shown.
BEFORE BENDING AFTER BENDING A’ C’ B’^ D’ R’^ S’ A C B D R S M M dx Consider a very small part of the beam in this the curvature of the beam can be taken as circular. All the layers of the beam, which were originally of the same length, do not remain of the same length after the application of the bending moment. Compressed Elongated Top Layer AC :- Subjected to compression finally A’C’ Bottom Layer BD :- Subjected to Tension finally B’D’ There is one layer (RS) in between which is neither compressed, nor stretched is known as Neutral Layer. The amount by which a layer is compressed or stretched is depend upon it’s distance from the middle layer. AFTER BENDING A’ C’ B’^ D’ R’^ S’ A C B^ D R^ S M M Compressed Elongated BEFORE BENDING A’ C’ B’^ D’ S’ R’ R P’^ Q’ A C B D R S P^ Q Y dY M M Let, M= Moment acting at a point = Angle subtended at center by the arc. R = Radius of the curvature. Consider a layer PQ at a distance y from the neutral layer which is compressed to P’Q’ after bending. So decrease in length of this layer PQ, l = PQ - P’Q’ Strain = l / original length e = [PQ - P’Q’] / PQ From the geometry of the fig. OP’Q’ and OR’S’ are similar. P’Q’ / R’S’ = (R- y) / R 1 – (P’Q’ / R’S’) = 1 – {(R- y) / R} (R’S’ - P’Q’) / R’S’ = y / R [PQ - P’Q’] / PQ = y / R (R’S’=PQ=RS=N.L.)
e = y / R
The strain at any layer is directly proportional to it’s distance from the neutral layer. A’ C’ B’ D’ R’^ S’ R P’^ Q’ y
o
A C B^ D R S P (^) y^ Q M M
So, we have to locate the neutral axis in such a way that, moment of entire area about the neutral axis becomes zero. We know that, Moment of any area about an axis passing through it’s centroid, is also zero. So neutral axis will always passes through the centroid of the section. Thus to locate the neutral axis first find out the centroid of the cross-section and then draw the line passing through the centroid and perpendicular to the of bending. This will be the neutral axis of the cross-section. plane of bending cross-section N A cross-section N A On one side of the NA : - Compression On other side of the NA : - Tension These two form a couple, whose moment will be equal to the external; moment applied. The moment of this couple, which resists the external bending moment, is known as the MOMENT OF RESISTANCE or MR. cross-section N A The algebraic sum of moment of all the forces about the neutral axis,
= { y^2 * (E / R) * a }
M = (E / R) * {y^2 *a}
In the above equation, {y^2 *a} , represents second moment
of area, or moment of inertia of the cross-section about the neutral axis. Moment of this force @ the neutral axis, = Force X Perpendicular Dist.
= (E / R) * a * ( y ) * y
= ( y 2 )(E / R) * a
M = (E / R) * I or M / I = E / R ……. Eq. II
We had derived the following expressions,
/ y = E / R ……. Eq. I
M / I = E / R……. Eq. II
So we can write,
M / I = / y = E / R
cross-section N A On one side of the NA : - Compression On other side of the NA : - Tension
M / I = / y = E / R = (E / R)* y
From above equation, the variation in bending stress with respect to ‘y’ is linear. Stress will be maximum when ‘y’ is maximum. Stress of opposite nature. max on outer fiber max on outer fiber
M / I = / y = E / R M = * (I / y)
Bending Stress ()will be maximum when ‘y’ is maximum.
(Either top or bottom).
Moment of Resistance M = max * I / ymax = max * Z
Where Z = I / ymax is known as section modulus. cross-section N A Stress of opposite nature. max on extreme fiber
Section Modulus:
max on extreme fiber
1. Rectangle :-
Z = I / ymax Zxx = Ixx / ymax Zxx = (bd^3 /12) / ( d/2) Zxx = bd^2 / Similarly, Zyy = db^2 /
2. Hollow Rectangle:-
Zxx = Ixx / ymax Zxx = (BD^3 /12- bd^3 /12) / (D/2) =(1/6D) * (BD^3 - bd^3 ) Zyy = (DB^3 /12- db^3 /12) / (B/2) =(1/6B) * (DB^3 - db^3 ) b d d/ B D/2 D b d
3. Circle:--
Ixx= IYY = d^4 Zxx = Ixx / ymax Zxx ={ }/ (d/2) 64 d^4 Zxx = 32 d^3
4. Hollow Circle:--
= (D
Ixx= IYY^4 - d^4 )
Zxx = Ixx / ymax Zxx = { }/ (D/2) 64 (D^4 -d^4 ) Zxx = 32D (D^4 -d^4 ) d d/ D D/ d
4m b d 20 kN/m M = Max. Moment in the beam= wL^2 / = (204^2 )/8 = 40 kN.m = 4010^6 N.mm (^) max = 8 N/mm^2 Z = (1/6)* b * d^2 = (1/6)(d/1.5)(d)^2 = d^3 /9 mm^3 Mr = max* Z = 8 * (d^3 /9) 4010^6 = (8/9)d^3 d/b =1. b = d/1. d = 355.7 mm 360 mm b = 240 mm Example :- 4 A simply supported beam of 4m span is loaded with a UDL of 20 kN/m over entire span. If the maximum stress allowed is 8 N/mm^2. Find the width (b) and depth (d) of the beam. Take ratio of depth to width = 1.
Example :- 5 A beam with the cross section as
shown in figure, is simply supported over a span of 4m.
Determine the Uniformly Distributed Load “w ” kN/m that
the beam can carry, if the allowable stress is not to
exceed 100 N/mm^2.
4m
w kN/m
100mm 25 250mm 25 200mm 25mm
Solution:-
Assume that the beam can carry a load w kN/m.
Max B.M. at centre = wL^2 /8 kN.m
M.R = wL^2 /8 = M
4m w kN/m 100mm 25 250mm 25 200mm 25 Y = (A 1 y 1 + A 2 y 2 + A 3 y 3 )/(A 1 +A 2 + A 3 ) =125 mm Y max = 175 mm IXX = Ix 1 x 1 +A 1 h 12 + Ix 2 x 2 + A 2 h 22 + Ix 3 x 3 +A 3 h 32 = 1.53 * 10^8 mm^4 125 175 125 175 4m w kN/m 100mm 25 250mm 25 200mm 25mm (^) max = 100 N/mm^2 y (^) max = 175 mm IXX = 1.53 * 10^8 mm^4 Mr = max* (Ixx /y (^) max) wL^2 /8 = 100 (1.5310^8 /175) W = 43.67 kN/m
M
I
y
E
R
Example :- 6 A beam with the cross section as
shown in figure, is simply supported over a span of 3m.
Determine the Uniformly Distributed Load “w ” kN/m that
the beam can carry, if the allowable stress is not to
exceed 120 N/mm^2.
3m
w kN/m
75mm 8 200mm 8 mm
Solution:-
Assume that the beam can carry a load w kN/m.
Max B.M. at centre = wL^2 /8 kN.m
M.R = wL^2 /8 = M
3m
w kN/m
Y = (A 1 y 1 + A 2 y 2 )/(A 1 +A 2 ) =124.09 mm
Y max = 124.09 mm
IXX = Ix 1 x 1 +A 1 h 12 + Ix 2 x 2 + A 2 h 22
= 9.036 * 10^6 mm^4
75mm 8 200mm 8 mm
3m
w kN/m
max = 120 N/mm^2
y max = 124.09 mm
IXX = 9.036 * 10^6 mm^4
Mr = (/y max)* Ixx
(120/124.09) * 9.036 * 10^6 = wL^2 /
W = 7.77 kN/m
M
I
y
E
R
75mm 8 200mm 8 mm
Example :- 7 A beam with the cross section as
shown in figure, is simply supported over a span of 5m.
It carry a Uniformly Distributed Load 8 kN/m. find the
magnitude & nature of stresses induced in extreme
fibres.
5m
8 kN/m
200mm 10 200mm 10 200mm 10mm
Max B.M. at centre = wL^2 /8 kN.m
M.R = wL^2 /8 = (8 * 5^2 ) / 8
=25 kN.m
Solution:-
Y = (A 1 y 1 + A 2 y 2 + A 3 y 3 )/(A 1 +A 2 + A 3 ) =134.17 mm
Y top = 185.83 mm (Compression)
Y bottom = 134.17 mm (Tension)
IXX = Ix 1 x 1 +A 1 h 12 + Ix 2 x 2 + A 2 h 22 + Ix 3 x 3 +A 3 h 32
= 9.06 * 10^7 mm^4
100mm 10 300mm 10 200mm 10mm (^) ten = 75 N/mm^2 Y (^) tensile = 134.17 mm (Ten) (^) comp = 125 N/mm^2 Y (^) comp = 185.83 mm (Comp.) IXX = 9.06 * 10^7 mm^4 Mr (Comp.) = comp* (Ixx / ycomp) = 125 * ( 9.0610^7 /185.83) = 60.94 kN.m Mr (Ten.) = Ten (Ixx / yTen) = 75 * (9.0610^7 /134.17) = 50.64 kN.m Max. Moment of resistance = 50.64 kN/m 100mm 10 300mm 10 200mm
Ten Comp.
Composite Beams (Flitched Beams)
A beam made up of two or more materials rigidly
connected together and behaving like a single member
is known as a composite beam or flitched beam.
The strain at the common surfaces will be same for both
materials. Also the total moment of resistance will be
equal to the sum of the moments of individual sections.
Steel Plate Wooden Beam
y
N A
t
d
Wooden Beam
b
Strain in Steel at distance y from N.A.,
s = Stress/E = fs/Es
Strain in Wood at distance y from N.A.,
w = Stress/E = fw/Ew
But the strain at common surface is same;
s = w
fs/Es = fw/Ew
Let; Es = Young’s Modulus of Steel Ew = Young’s Modulus of Wood Is = Moment of Inertia of Steel Iw = Moment of Inertia of Wood Ms= Moment of Resistance of Steel Mw= Moment of Resistance of Wood fs = Stress in Steel fw = Stress in Wood
y
N A
t
d
Steel Plate Wooden Beam
b
strain at common surface is same;
s = w
fs/Es = fw/Ew
fs = fw (Es/Ew)
fs = m. fw
Where, m = Es/Ew is known as Modular Ratio between steel & timber.
Total Moment of resistance;
M = Ms + Mw
= fs Zs + fw Zw = (1/6)t d^2 * fs + (1/6) b d^2 *fw
= (1/6) td^2 * m fw + (1/6) bd^2 * fw
= (1/6) fw (m t + b) d^2 ……. Equivalent wooden section
= (1/6) fs (b/m +t) d^2 …….. Equivalent Steel section
Example: 9 A flitched beam consists of
wooden beam 100 mm wide and 200 mm deep is
strengthened by a steel plate 200 mm wide and 15 mm
thick at one side of the beam section. Find the moment
of resistance of the section, if the allowable stress in
wood = 7.5 N/mm^2. Take ES=20 Ew.
Solution:
m = ES / Et=
Width b=100 mm
Depth, d = 200 mm
Thick ness of steel, t = 15 mm
fw = allowable stress for wood = 7.5 N/mm^2
fs=allowable stress for steel
Steel stress at the level of top most fibre of wood.
fs =m x fw = 20 x 7.5 =150 N/mm^2
Let, Mw = Moment of resistance of wood
Ms = Moment of resistance of Steel
Mw = fw * Zw
= 7.5 * (1/6) 100 (200)^2
= 5000 N.m
Ms = fs * Zs
= 150 * (1/6) 15 (200)^2
= 15000 N.m
Total MR = Mw + Ms = 5000 + 15000 = 20000N.m
Wood
Alternate method Equivalent Section:-
y
N
15 200 Steel Plate Wooden Beam b=
y 1
A
t
200 Steel Plate Steel Beam b/m =100/20 = Equivalent width of steel = b/m =100/20 =5mm M = fs * Zs = 150 *(15 + 5) *200^2 / 6 = 20 kN.m Equivalent
fs =165. Steel 130 fs =
t=110/ = 5.5 mm M = fs * Zs = 165.45 {(1/12)(110260^3 - 104.5 220^3 )} / = 87035780 N-mm = 87.035 kNm
Alternate method
Equivalent Section:-
