INSTITUTE OF AERONAUTICAL ENGINEERING · The procedure of finding slope and deflection for simply...
Transcript of INSTITUTE OF AERONAUTICAL ENGINEERING · The procedure of finding slope and deflection for simply...
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INSTITUTE OF AERONAUTICAL ENGINEERING(AUTONOMOUS)
Dundigal – 500 043, Hyderabad
Regulation: R16 (AUTONOMOUS)
Course code: ACE004
STRENGTH OF MATERIALS - II
Prepared By
SURAJ BARAIK
Assistant Professor
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Unit 1
Deflection in Beams
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Unit 1 Deflection in Beams
• Topics Covered
Review of shear force and bending moment diagram
Bending stresses in beams
Shear stresses in beams
Deflection in beams
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Beam Deflection
Recall: THE ENGINEERING BEAM THEORY
y
M E
I R
y
x
NAA B
A’ B’
If deformation is small (i.e. slope is “flat”):
Moment-Curvature Equation
v (Deflection)
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A’
Alternatively: from Newton’s Curvature Equation
1 d2y
R dx2
1 d
R dx
xand
y(slope is “flat”)
y
x
R
y f (x)
1 d2y
R dx2
dx
dy2
1if
R
1
d2y
dx 2
1
dx
3
dy 22
R
B’
y
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From the Engineering Beam Theory:
M E
I R
1 M
R EI
d2y
dx2
d2y EI M
dx 2
Flexural
Stiffness
Bending
MomentCurvature
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Relationship
A BC
BA C
y
Deflection = y
Slope =dy
dx
Shearing force = EI
d2yBending moment = EI
dx2
d3y
dx3
Rate of loading = EId4y
dx4
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Methods to find slope and deflection
Double integration method
Moment area method
Macaulay’s method
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Since,d2y
2
1
dx EI M Curvature
1 dy
dx EI M dx C
1Slope
1
EI y 1 M dx dx C dx C
2Deflection
Where C1 and C2 are found using the boundary conditions.
Curvature Slope Deflection
R dy
dx
y
Double integration method
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Double integration method
Slope Deflection
B
L
L/2A C
L/2
yc
Slope =dy
dx
A B
2WL
16EI
Deflection = yc
WL3
48EI
Slope DeflectionA C
yc
L
x w/Unit length
B Slope =dy
dx
A B
2WL
24EI
Deflection = yc
5 WL3
384 EI
Simple supportedW
Uniform distributed load
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Macaulay’s method The procedure of finding slope and deflection for
simply supported beam with an eccentric load is very
laborious.
Macaulay’s method helps to simplify the calculations
to find the deflection of beams subjected to point
loads.
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9 - 11
Moment-Area Theorems
D
C xC
xD M d
EIdx
xC
xD MD C EI
dx
• Consider a beam subjected to arbitrary loading,
d d2y M
dx dx2 EI
• First Moment-Area Theorem:
area under BM diagram betweenC and D.
dx
R d
CD Rd dx
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9 - 12
Moment-Area Theorems• Tangents to the elastic curve at P and P’
intercept a segment of length dt on the vertical
through C.
dt xd xM
EIdx
xC xC
xD xDM 1 1tC D x
EI dx
EI xMdx
EI Ax
A= total area of BM diagram between C & D
x = Distance of CG of BM diagram from C
• Second Moment-Area Theorem:The tangential deviation of C with respect to D is equal to the first moment with respect to
a vertical axis through C of the area under the BM diagram between C and D.
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Moment Area Method
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15
An Exercise- Moment of Inertia – Comparison
Load
2 x 8 beam
Maximum distance of 1 inch to
the centroid
I1
I2 > I1 , orientation 2 deflects less
1
Maximum distance of
4 inch to the centroid I2
Load 2
2 x 8 beam
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UNIT 2
DEFLECTION BY ENERGY METHODS
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Elastic DeflectionCastigliano’s Method
If deflection is not covered by simple cases in Table 5.1 (p186)
Complementary
Energy U’
Stored Elastic Energy
U
U U' ∆ Q 2dU dU' ∆ dQ
∆ dU dQ
Incremental:
Deflection:
When a body is elastically deflected by any combination of loads, thedeflection at any point and in any direction is equal to thepartial derivative of strain energy (computed with all loads acting)with respect to a load located at that point and actingin that direction
Castiglino’s Theorem: ∆ U Q
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Elastic Deflection
Castigliano’sMethodTable 5.3 (p193): Energy and Deflection Equations
Example: Axial TensionStored Elastic Energy:
Case 1 from Table 5.1:
gives: For varying E
and A:
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Elastic DeflectionCastigliano’s Method
(1) Obtain expression for all components of energyTable 5.3
(2) Take partial derivative to obtain deflection
∆ U QCastiglino’s Theorem:
Energy and Deflection Equations
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Elastic Deflection: Castigliano’s Method
Table 5.3
1. Energy: here it has two components:
first compute Energy, then Partial Derivative to get deflection
Here 2 types of loading: Bending and Shear magnitude @ x:
2. Partial Derivatives for deflection:(23=8)*3*4 = 96
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Table 5.3
Elastic Deflection: Castigliano’s Method
TWO METHODS
Differentiate after Integral Differentiate under Integral
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Elastic Deflection: Castigliano’s Method
m
m
Transverse shear contributes only <5% to deflection
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Elastic Deflection: Castigliano’s Method
Use of “Dummy Load” Q=0
•90° bend cantilever beam
•shear neglected
•Shear neglected => only 4 energy components:
1) BENDING portion a_b: Mab=Py2) BENDING portion b_c: Mbc=Qx +Ph
3) TENSION portion a_b: Q4) COMPRESSION portion b_c: P
(Tension and Compression mostly
negligible if torsion and bending
are present)
:
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Elastic Deflection: Castigliano’s Method
•Eccentrically Load Column
•No Buckling
Redundant Support
500kg x 9.8m/s2
=4900 N
Guy wire
•Now Deflection known (=0) •Find necessary Tension Force F
•Hence partial derivative of total elastic energy with respect to F
must be zero
•Omit zero derivatives
- all energy terms above a
- compression term below a
•Only bending term is left: M= (4900 N)(1.2m)-Fy = 5880 Nm - Fy
finite value=! 0
F=2940 N
(Nm)2 m Nm3 m3
(Nm)2 Nm
Nm3 m3
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U P/2e
EnergyMethod
External Work
When a force F undergoes adisplacement dx in the same directionas the force, the work done is
dUe F dx
If the total displacement is x the work become
xUe F dx
0
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The work of a moment is defined by the product of the magnitude
of the moment M and the angle d then if the total
angle of
e
U M/2
Ue M d
0
rotation is the work become:
dUe M d
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Maxwell–Betti Reciprocal theorem
Consider a simply supported beam of span L as
shown. Let this beam be loaded by two systems of
forces P1and P2 separately as shown in the figure. Let
u21 be the deflection below the load point P2 when only
load P1 is acting.
Similarly let u12 be the deflection below load P1 , when
only load P2 is acting on the beam.
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The reciprocal theorem states that the work done by forces
acting through displacement of the second system is the same
as the work done by the second system of forces acting through
the displacements of the first system. Hence, according to
reciprocal theorem,
P1 u12 P2 u21
Now, u12 and u21 can be calculated using Castiglinao’s first
theorem. Substituting the values of u12 and u21 in equation we
get,
48EI 48EI
5P L3 5PL3
P 2 P 1
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Hence it is proved. This is also valid even when the first system of forces is
P1, P2 ,...., Pn and the second system of forces is given by Q1, Q2 ,...., Qn . Let
u1 , u2 ,...., un be the displacements caused by the forces P1, P2 ,...., Pn only and
1, 2 ,...., n be the displacements due to system of forces Q1, Q2 ,...., Qn only
acting on the beam as shown in Fig
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UNIT III
STRESSES IN CYLINDERS AND
SPHERICAL SHELLS
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THINCYLINDER
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Introduction
Cylindrical and spherical vessels are used in the
engineering field to store and transport fluids. Such vessels
are tanks ,boilers , compressed air receivers , pipe lines etc.
these vessels, when empty, are subjected to
atmospheric pressure internally as well as externally and the
resultant pressure on the walls of the shell is nil.
but whenever a vessel is subjected to an intenal
pressure
(due to air , water , steam etc.) its walls are subjected to
tensile stresses.
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Thin cylindricalshell.
When t/d <= d/10 to d/15, it is called thin
cylindrical shell. t = thickness of the
shell
d =internal diameter of shell.
in thin cylindrical shells hoops stress and longitudinal stresses
are constant over the
Thickness and radial stresses are negligible.
When t/d > d/10 to d/15, it is called THICK cylindrical shell.
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Stresses in thin cylindrical shells :
whenever, a thin cylindrical shell is subjected to
an internal pressure (p). Its Walls are subjected
to two types of tensile stresses.
(a) Hoop stress (circumferential stress)
(b) Longitudinal stress.
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Consider a thin cylindrical shell subjected to
an internal Pressure as shown in fig.𝜎𝑐= circumferential stress in the shell
material. p =internal pressure
d =internal
diameter of shell
t =thickness of the
shell.
Total pressure,
p = Pressure * Area
=p.d.l
Resisting area = A = 2.t.l𝜎𝑐= P/A = p.d.l/2.t.l
𝝈𝒄= p.d/2t.
Hoop stress (circumferential stress) : 𝝈𝒄
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(b) Longitudinal stress (𝜎𝑙 ):
Total pressurep = Pressure *
Area4
p = p𝜋
𝑑2
Resistingarea,A = 𝜋𝑑𝑡
𝜎𝑙= P/A =p4
𝜋𝑑2
𝜋𝑑𝑡
𝝈𝒍=pd/4t
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Change in dimensions of a thin cylindrical shell due to internal pressure:
1휀=
𝛿𝑑
𝜎
𝑐 𝜎𝑙= −
=
𝑝𝑑
−
𝑝𝑑𝑑 𝐸 𝑚𝐸 2𝑡𝐸
4𝑡𝑚𝐸 𝟏𝜺𝟏 = 𝒑𝒅/𝟐𝒕𝑬(𝟏−
𝟐𝒎)
Longitudinal strain,
2휀=
=𝛿𝑙
𝜎
𝑙
𝑙
𝐸− 𝑐
𝑚𝐸 4𝑡𝐸
𝜎 = 𝑝𝑑
−
𝑝𝑑2𝑡𝑚𝐸
𝒑𝒅 𝟏𝟏
𝜺𝟐 =𝟐𝒕𝑬 𝟐
−𝒎
Let, 𝛿𝑑=change in dia. Of
shell
𝜎𝑙=change in length
of shell
Circumferential strain,
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Change in volume of a thin cylindrical shell due to internal
pressure:
Volume of
shell,
V = 𝜋 ∗ 𝑑2
∗𝑙
4
Finalvolume,
4V+𝛿𝑣 = 𝜋 𝑑 +𝛿𝑑
2
∗𝑙 +𝛿𝑙
Change in
volume,
𝛿𝑣=
𝑉 + 𝛿𝑣
−𝑉= [𝜋
𝜋
44
22
𝑑 + 𝛿𝑑 ∗ (𝑙 + 𝛿𝑙)] − ∗ 𝑑 ∗ 𝑙
22
𝜋
44
2= [𝜋 (𝑑 + 2𝛿𝑑 ∗ 𝑑 + 𝛿𝑑 ) ∗ (𝑙 + 𝛿𝑙)]
− ∗ 𝑑∗ 𝑙
=𝜋 4
𝑑2𝑙 + 2𝛿𝑑 ∗ 𝑑 ∗ 𝑙 + 𝛿𝑑2 ∗ 𝑙 + 𝑑2 ∗ 𝛿𝑙 + 2𝛿𝑑 ∗ 𝑑 ∗ 𝛿𝑙+ 𝛿𝑑2 ∗𝛿𝑙
𝜋
4
2− 𝑑∗ 𝑙
=𝜋 4
𝑑2𝛿𝑙 + 2𝛿𝑑 ∗ 𝑑∗ 𝑙
𝛿𝑣=𝜋𝑉
4
4𝑑2𝛿𝑙 + 2𝛿𝑑 ∗ 𝑑 ∗ 𝑙 /
𝜋 ∗ 𝑑2 ∗ 𝑙
=𝛿𝑙 + 2
𝛿𝑑
12
𝑑
𝑙
(휀 = 𝛿𝑑
, 휀 =𝛿
𝑙)
𝑙 𝑑
=휀2 + 2휀1
𝜹𝒗 =V(𝝐𝟐 +𝟐𝜺𝟏)
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Thin spherical shellsConsider a thin spherical shell subjected to internal pressure p
as shown in fig.
p =internal pressure
d =internal
diameter of
shell t
=thickness of
the shell
𝜎 = stress in the shell material
Total forceP =
𝜋∗ 𝑑2
∗𝑝4
Resistingsection
=𝜋𝑑𝑡
Stress in the shell𝜎 = 𝑇𝑜𝑡𝑎𝑙 force/resisting
section
4=
𝜋𝑑2𝑝/𝜋
𝑑𝑡 𝑝𝑑
𝜎 =4
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Chenge in diameter and volume of a thin spherical shell due to
internal pressureConsider a thin spherical shell subjected to
internal pressure.
p =internal pressure
d =internal diameter of shell
t =thickness of the shell
𝜎 = stress in the shellmaterial
We know that for thin spherical shell
𝜎 = 𝑝𝑑
4𝑡
Strain in anydirection
𝐸
𝑚𝐸
𝜖=𝜎
−=
𝜎 𝑝𝑑
−
𝑝𝑑
4𝑡𝐸 4𝑡𝑚𝐸 𝟏
𝒎𝜺 = 𝒑𝒅/𝟒𝒕𝑬(𝟏−)
We know that,
strain,∈=
𝛿𝑑𝑑
𝑑
𝛿𝑑
=
𝟏
𝒎𝒑𝒅/𝟒𝒕𝑬(𝟏−)
……(
1)
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CLASSIFICATIONS OFSHELLS
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• Shell is a type of building enclosures.
• Shells belong to the family of arches . They can be defined as curved or angled structures capable of transmitting loads in more than two directions to supports.
• A shell with one curved surface is known as a vault (single curvature ).
• A shell with doubly curved surface is known as a dome (double curvature).
SHELLS
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Classification of shells
• There are many different ways to classify shell structures buttwo ways are common:
1. The material which the shell is made of: like reinforced concrete, plywood or steel, because each one has different properties that can determine the shape of the building and therefore, these characteristics have to be considered in the design.
2. The shell thickness: shells can be thick or thin.
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Thin Concrete ShellsThe thin concrete shell structures are a lightweight construction composed of a relatively thin shell made of reinforced concrete, usually without the use of internal supports giving an open unobstructed interior. The shells are most commonly domes and flat plates, but may also take the form of ellipsoids or cylindrical sections, or some combination thereof. Most concrete shell structures are commercial and sports buildings or storage facilities.
There are two important factors in the development of the thin concrete shell structures:
• The first factor is the shape which was was developed along the history ofthese constructions. Some shapes were resistant and can be erected easily.However, the designer’s incessant desire for more ambitious structures didnot stop and new shapes were designed.
• The second factor to be considered in the thin concrete shell structures is the thickness, which is usually less than 10 centimeters. For example, the thickness of the Hayden planetarium was 7.6 centimeters.
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Types of Thin Concrete Shells
1. Barrels shellsThe cylindrical thin shells, also called barrels, should not be confused with the vaults even with the huge similarity in the shape of both structures, because each of these structures has a different structural behavior as well as different requirements in the minimum thickness and the shape.
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• On one hand, the structural behavior of the vault is based on connected parallel arches, which transmit the same effort to the supports . Therefore, the materials used in these structures have to be able to resists compressions (e.g. stone) and the thickness is usually higher. Furthermore, the shape of the vaults must be as similar as possible to the arch in order to achieve the optimum structural behavior.
• On the other hand, the structural behavior of the barrels shell is that it carries load longitudinally as a beam and transversally as an arch. and therefore, the materials have to resist both compression and tension stresses. This factor takes advantage of the bars of the reinforced concrete, because these elements can be placed where tension forces are needed and therefore, the span to thickness Ratios can be increased. Furthermore, the shape has fewer requirements than the vaults and therefore, new curves like the ellipse or the parabola can be used improving the aesthetic quality of the structure.
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Types of Thin Concrete Shells
2. Folded plate
A thin-walled building structure of the shell type.
Advantages of Folded Plate Roofs over Shell Roofs are:
(a) Movable form work can be employed.
(b) Form work required is relatively simpler.
(c) Design involves simpler calculations.
Disadvantages of Folded Plate Roofs over Shell Roofs are:
(a) Folded plate consumes more material than shells.
(b) Form work may be removed after 7 days whereas in case of shells it can be little earlier.
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Folded platetypes
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Folded Plates system
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Folded-Plate Hut in Osaka
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Folded Plates Library
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Types of Thin Concrete Shells
3. Hyperbolic Paraboloid (Hypar)
A Hypar is a surface curved in two
directions that can be designed as a shell
or warped lattice.
A hypar is triangular, rectangular or rhomboidal in plan, with corners raised to the elevation desired for use and/or appearance. The edges of Hypars are typically restrained by stiff hollow beams that collect & transfer roof loads to the foundations.
Rhomboid
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Types of shells4. Various Double Curvature
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Types of Thin Concrete Shells
5. DomeA rounded roof, with a circular base, shaped like an arch in all directions.. First used in much of the Middle East and North Africa whence it spread to other parts of the Islamic world, because of its distinctive form the dome has, like the minaret, become a symbol of Islamic architecture.
Dome has double curvature and the resulting structure is much stiffer and stronger than a single curved surface, such as a barrel shell.
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Types of Thin Concrete Shells
6. Translation ShellsA translation shell is a dome set on four arches. The shape is different from a spherical dome and is generated by a vertical circle moving on another circle. All vertical slices have the same radius. It is easier to form than a spherical dome.
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• Advantages of Concrete Shells:The curved shapes often used for concrete shells are naturally strong structures.Shell allowing wide areas to be spanned without the use of internal supports, giving an open, unobstructed interior.The use of concrete as a building material reduces both materials cost and the construction cost.As concrete is relatively inexpensive and easily cast into compound curves.
• Disadvantages of Concrete ShellsSince concrete is porous material, concrete domes often have issues with sealing. If not treated, rainwater can seep through the roof and leak into the interior of the building. On the other hand, the seamless construction of concrete domes prevents air from escaping, and can lead to buildup of condensation on the inside of the shell. Shingling or sealants are common solutions to the problem of exterior moisture, and ventilation can address condensation.
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Unit IV
Indeterminate Beams:
PROPPED CANTILEVER AND
FIXED BEAMS
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Contents:
• Concept of Analysis -Propped cantilever and fixed
beams-fixed end moments and reactions
• Theorem of three moments – analysis of
continuous beams – shear force and bending
momentdiagrams.
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89
Indeterminatebeams
• Staticallydeterminatebeams:
– Cantileverbeams
– Simplesupportedbeams
– Overhangingbeams
• Staticallyindeterminatebeams:
– Proppedcantileverbeams
– Fixedbeams
– Continuousbeams
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• Propped cantileverBeams:
Indeterminatebeams
Degree of staticindeterminacy=
N0. of unknown reactions – static equations=3-2=1
90
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• Fixedbeam:
A fixed beam is a beam whose end supports are such that the end slopes
remain zero (or unaltered) and is also called a built-in or encasterbeam.
Indeterminatebeams
Degree of staticindeterminacy=
N0. of unknown reactions – static equations=4-2=2
91
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Indeterminatebeams
Degree of staticindeterminacy=
N0. of unknown reactions – static equations=5-2=3
Continuous beam: Continuous beams are very common in the structural design.
For the analysis, theorem of three moments isuseful.
A beam with more than 2 supports provided is known as continuous beam.
Degree of staticindeterminacy=
N0. of unknown reactions – static equations=3-2=1
92
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•B.M. diagram for a fixed beam :
Figure shows a fixed beam AB carrying
an external load system. Let VA and VB
be the vertical reactions at the
supports A andB.
Let MA and MB be the fixed end
Moments.
FixedBeams
𝑀𝐵
𝑀𝐴
𝑉𝐴𝑉𝐵
𝑊1
𝑊2
𝑊1
𝑊2
93
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vb
The beam may be analyzed in the followingstages.
(i) Let us first consider the beam as Simply supported.
Let va and vb be the vertical reactions at the supports A and B. Figure (ib)
shows the bending moment diagram for this condition. At any section the
bending moment Mx is a saggingmoment.
𝑊1 𝑊2
va
FixedBeams
(ib) FreeB.M.D.
94
(ia) Freely supportedcondition
𝑀𝑥
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• (ii) Now let us consider the effect of end couples MA and MB alone.
Let v be the reaction ateach
end due to this condition.
Suppose 𝑀𝐵 >𝑀𝐴.
Then 𝑉 =𝑀𝐵−𝑀𝐴.
𝐿
If 𝑀𝐵 > 𝑀𝐴 the reaction Vis
upwards at B and downwards atA.
Fig (iib). Shows the bending moment
diagram for thiscondition.
At any section the bending moment Mx’ is hoggingmoment.
FixedBeams
𝑀𝐴
𝑀𝐵 vv
(iia) Effect of endcouples
(iib) FixedB.M.D.
𝑀𝑥
′
95
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• Now the final bending moment
diagram can be drawnby
combining the above two B.M.
diagrams as shown in Fig. (iiib)
Now the final reaction VA =va-v
and VB =vb+v
The actual bending moment at any
𝑑2𝑦section X, distance 𝑥 from the end A is givenby,
𝐸𝐼𝑑𝑥2 = 𝑀𝑥 −𝑀𝑥 ′
FixedBeams
𝑀𝐴
𝑉𝐴
𝑀𝐵
𝑉𝐵
𝑊1 𝑊2
(iiib) ResultantB.M.D.
+
96
--
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FixedBeams
𝑀𝐵
𝑀𝐴
𝑉𝐴𝑉𝐵
𝑊1
𝑊2
𝑉𝑎 𝑉𝑏
(ia) Freely supportedcondition
𝑀𝐴
𝑀𝐵𝑉 𝑉
(iiia) Fixedbeam
(iia) Effect of endcouples𝑊1 𝑊2
𝑀𝑥
𝑀𝑥 ′
+
--
(iib) FixedB.M.D.
(iiib) ResultantB.M.D.
(ib) FreeB.M.D.
97
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• Integrating, weget,
• 𝐸𝐼𝑑𝑦
𝑑𝑥
𝑙0
= 𝑥𝑀 𝑑𝑥− 𝑥′𝑀 𝑑
𝑥
𝑙
0
𝑙
0
• But at x=0, 𝑑𝑦
=0𝑑𝑥𝑑𝑦
and at 𝑥= 𝑙, 𝑑𝑥
= 0
𝑙Further0 𝑀𝑥𝑑𝑥 = area of the Free BMD=𝑎
𝑥
𝑙
𝑀 ′ 𝑑𝑥 = area of the fixed B. M. D = 𝑎′
0
Substituting in the above equation, weget,
0 = 𝑎 −𝑎′
∴𝑎 =𝑎′
𝑑2𝑦𝐸𝐼𝑑𝑥2 = 𝑀𝑥 −𝑀𝑥 ′
98
FixedBeams
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• Integrating weget,
•2
𝑑𝑥2𝐸𝐼𝑥 = 𝑥𝑀 𝑥𝑑𝑥 − 𝑥
′𝑀 𝑥𝑑𝑥
𝑙
0
𝑙
0
𝑙 𝑑 𝑦
0
𝑑𝑥 0• ∴𝐸𝐼 𝑥 𝑑𝑦
−𝑦 𝑙=a𝑥-a’𝑥′
• Where 𝑥= distance of the centroid of the free B.M.D. from A. and 𝑥′=
distance of the centroid of the fixed B.M.D. fromA.
FixedBeams
𝑑2𝑦
𝑎 = 𝑎′
∴Area of the free B.M.D. =Area of the fixedB.M.D.
Again consider the relation,
𝐸𝐼𝑑𝑥2 = 𝑀𝑥 −𝑀𝑥 ′
𝑑2𝑦
𝑑𝑥2 𝑥 𝑥
99
′𝐸𝐼𝑥 = 𝑀 𝑥− 𝑀 𝑥
𝑀𝑢𝑙𝑡𝑦𝑖𝑛𝑔𝑏𝑦 𝑥𝑤𝑒𝑔𝑒𝑡,
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𝑑𝑥• Further at x=0, y=0 and 𝑑𝑦 =0
or
100
• and at x=l, y=0 and 𝑑𝑦
=0.𝑑𝑥
• Substituting in the above relation, wehave
0 = 𝑎𝑥-𝑎′𝑥′
𝑎𝑥=𝑎′𝑥′
𝑥= 𝑥
∴The distance of the centroid of the free B.M.D . From A= The distance
of the centroid of the fixed B.M.D. fromA.
∴𝑎 = 𝑎′
𝑥= 𝑥
FixedBeams
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FixedBeams
𝑀𝐵
𝑀𝐴
𝑉𝐴𝑉𝐵
𝑊1
𝑊2
𝑉𝑎 𝑉𝑏
(ia) Freely supportedcondition
𝑀𝐴
𝑀𝐵𝑉 𝑉
(iiia) Fixedbeam
(iia) Effect of endcouples𝑊1 𝑊2
𝑀𝑥
𝑀𝑥′
+
--
(iib) FixedB.M.D.
(iiib) ResultantB.M.D.
(ib) FreeB.M.D.
101
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Fixed beamproblems
• Find the fixed end moments of a fixed beam subjected to a
point load at thecenter.
W
l/2
A
102
B
l/2
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• 𝐴′ =𝐴
𝑊 𝑙𝑀 =
8= 𝑀𝐴 =𝑀𝐵
𝑙/2
Fixed beamproblems
WA B
𝑊 𝑙
4
FreeBMD
𝑙/2
MM
+
-
𝑀 ×𝑙= 1
×𝑙×𝑊𝑙
2 4
+
- -
FixedBMD
𝑊 𝑙
4
ResultantBMD
103
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Fixed beamproblems
• Find the fixed end moments of a fixed beam subjected to a
eccentric point load.
W
a
A B
b
𝑙
104
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Fixed beamproblems• 𝐴′ =𝐴
19
• 𝑥′ =𝑥
𝑀𝐴 +2𝑀𝐵𝑀𝐴 +𝑀𝐵
×3
=𝑙 𝑙+𝑎
3
𝐵𝑀 =𝑀 ×𝑎
𝑙−𝑎𝐴
𝑎𝑀𝐵 =
𝑏−−−(2)
𝑎W
B
𝑙
𝑏
+
𝑀𝐴 +𝑀𝐵
1×𝑙=
2 ×𝑙×
A
𝑊𝑎𝑏
2 𝑙
𝑙𝑊𝑎𝑏
FixedBMD
𝑀𝐴 𝑀
𝐵
Free-BMD
-
𝐴
𝐵
𝑙𝑀 +𝑀 =
𝑊𝑎𝑏−−−−(1)
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Fixed beamproblems
𝑎
𝑎W
A B𝑏
𝐴
𝐵
𝑙𝑀 +𝑀 =
𝑊𝑎𝑏−−−−(1)
+
𝑊 𝑎𝑏2𝑙2 𝑊 𝑏
𝑎
2
𝑙2
𝑙
𝑊 𝑎𝑏𝑙
ResultantBMD
- -
𝑀𝐵 = 𝑀𝐴 ×𝑏
−−−(2)
By substituting (2) in (1),
𝑀𝐴 =𝑊 𝑎 𝑏
2
𝑙2From(2),
𝑀𝐵 =𝑊 𝑏𝑎
2
𝑙2
106
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Clapeyron’s theorem of threemoments
• As shown in above Figure, AB and BC are any two successive spans of
a continuous beam subjected to an external loading.
• If the extreme ends A and C fixed supports, the support moments 𝑀𝐴,
𝑀𝐵and 𝑀𝐶 at the supports A, B and C aregiven by therelation,
𝑀 𝐴 𝑙1 +2𝑀 𝐵 𝑙1 + 𝑙2 +𝑀 𝐶 𝑙2 = 6𝑎1𝑥1 +
6𝑎2𝑥2
𝑙1 𝑙2
A C
𝑙1 𝑙2𝑀𝐴
𝑀𝐶
B𝑀𝐵
107
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Clapeyron’s theorem of three moments(contd…)
𝑀 𝐴 𝑙1 +2𝑀 𝐵 𝑙1 +𝑙2 +𝑀 𝐶 𝑙2 = 6𝑎1𝑥1 +
6𝑎2𝑥2
𝑙1 𝑙2• Where,
• 𝑎1 =area of the free B.M. diagram for the spanAB.
• 𝑎2 =area of the free B.M. diagram for the spanBC.
• 𝑥1= Centroidal distance of free B.M.D on AB fromA.
• 𝑥2= Centroidal distance of free B.M.D on BC fromC.
108
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𝑑𝑥 𝑀𝑥
+
𝑥1
𝑥
+
𝑥2
A B C
(a)
(b)
+ve +ve𝑀𝐴
𝑀𝐶 (d)
𝑑𝑥𝑥
𝑀
𝑥’
𝑥2′𝑥1
′
𝑀𝐴
𝑀𝐶
(c)++
FreeB.M.D
𝑀𝐵
FixedB.M.D
𝑀𝐵
-ve
Clapeyron’s theorem of three moments(contd…)
𝑙1
109
𝑙2
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Clapeyron’s theorem of three moments(contd…)
𝑑𝑥 𝑀𝑥
+
𝑥1
𝑥
+
𝑥2
A B C
𝑑𝑥𝑥
𝑀
𝑥’
𝑥2′𝑥1
′
𝑀𝐴
110
𝑀𝐶
--
FreeB.M.D
𝑀𝐵
FixedB.M.D
(a)The givenbeam
(b) FreeB.M.D.
(c) Fixed B.M.D.
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Clapeyron’s theorem of three moments(contd…)• Consider the spanAB:
• Let at any section in AB distant 𝑥 from A the free and fixed bending
moments be 𝑀𝑥 and 𝑀𝑥′ respectively.
• Hence the net bending moment at the section is given by
𝑑2𝑦𝐸𝐼𝑑𝑥2 = 𝑀𝑥 −𝑀
𝑥
′
• Multiplying by 𝑥, weget
𝐸𝐼𝑥𝑑2𝑦
𝑑𝑥2 𝑥 𝑥
111
′= 𝑀 𝑥−𝑀 𝑥
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• 𝐸𝐼𝑥2𝑑 𝑦
𝑑𝑥2 𝑥 𝑥′= 𝑀 𝑥−𝑀 𝑥
𝑑2𝑦𝑙1 𝑙1
• Integrating from 𝑥= 0 𝑡𝑜𝑥= 𝑙1, we get,
𝐸𝐼 𝑥𝑑𝑥2 = 𝑀𝑥𝑥 𝑑𝑥 − 𝑀𝑥
′𝑥 𝑑𝑥
𝑙1
0 0
0
𝑑𝑦𝐸𝐼 𝑥.
𝑑𝑥−𝑦
112
0
𝑙1𝑙1 𝑙1
= 𝑀𝑥𝑥 𝑑𝑥 − 𝑀𝑥′𝑥𝑑𝑥
0 0
−−−(1)
Clapeyron’s theorem of three moments(contd…)
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• But it may be suchthat
At 𝑥 =0, deflection 𝑦 =0
𝑑𝑥• At 𝑥= 𝑙1, 𝑦 = 0; 𝑎𝑛𝑑 𝑠𝑙𝑜𝑝𝑒𝑎𝑡𝐵 𝑓𝑜𝑟𝐴𝐵,
𝑑𝑦=𝜃𝐵𝐴
• 𝑥𝑙1
0𝑀 𝑥𝑑𝑥=𝑎 𝑥1 1 =Momentof thefreeB.M.D.onABaboutA.
•𝑙1
0 𝑥 1′
1′
′
𝑀 𝑥𝑑𝑥=𝑎 𝑥 =Momentof thefixedB.M.D.onABaboutA.
Clapeyron’s theorem of three moments(contd…)A B C
𝑙1 𝑙2𝑥
113
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𝑑𝑦𝐸𝐼 𝑥.
𝑑𝑥−𝑦
0
𝑙1𝑙1 𝑙1
= 𝑀𝑥𝑥 𝑑𝑥 − 𝑀𝑥′𝑥𝑑𝑥
0 0
−−(1)
• Therefore the equation (1) is simplifiedas,
1𝐸𝐼 𝑙1𝜃𝐵𝐴−0 =𝑎1𝑥1 −𝑎′𝑥1′.
1But 𝑎′ =area of the fixed B.M.D. on AB=𝑀 𝐴 +
𝑀 𝐵2𝑙1
𝑥1′ =Centroidof thefixedB.M.D.fromA=
𝑀 𝐴 +2𝑀 𝐵 𝑙1
𝑀 𝐴 +𝑀 𝐵 3
Clapeyron’s theorem of three moments(contd…)
114
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• Therefore,
Clapeyron’s theorem of three moments(contd…)
𝟏𝒂′ 𝒙𝟏
′ =(𝑴𝑨 +𝑴𝑩 ) 𝑴 𝑨 + 𝟐𝑴𝑩 𝒍𝟏×
𝒍𝟏
𝟐 𝑴 𝑨 +𝑴 𝑩
𝟑
= (𝑴𝑨 + 𝟐𝑴𝑩 )𝟏
𝒍𝟐
𝟔
𝑬𝑰𝒍𝟏𝜽𝑩𝑨 = 𝒂𝟏𝒙𝟏 − (𝑴𝑨 + 𝟐𝑴𝑩 ) 𝟏𝒍𝟐
𝟔
𝒍𝟏𝟔𝑬𝑰𝜽𝑩𝑨 =
𝟔𝒂𝟏𝒙𝟏 −(𝑴𝑨 +𝟐𝑴𝑩)𝒍𝟏 − − −−(𝟐)
Similarly by considering the span BC and taking C as origin it can be
shownthat,
𝟔𝑬𝑰𝜽𝑩𝑪 =𝟔𝒂𝟐 𝒙𝟐 − (𝑴𝑪 + 𝟐𝑴𝑩)𝒍𝟐 − − −−(𝟑)𝒍𝟐
𝜃𝐵𝐶= slope for span CB atB
115
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Clapeyron’s theorem of three moments(contd…)• But𝜃𝐵𝐴 = −𝜃𝐵𝐶 as the direction of 𝑥 from A for the span AB,
and from C for the span CB are in oppositedirection.
• And hence,𝜃𝐵𝐴+ 𝜃𝐵𝐶 =0
𝟔𝑬𝑰𝜽𝑩𝑨 =𝟔𝒂𝟏𝒙𝟏 −(𝑴𝑨 +𝟐𝑴𝑩)𝒍𝟏 − − −−(𝟐)𝒍𝟏
𝒍𝟐𝟔𝑬𝑰𝜽𝑩𝑪 =
𝟔𝒂𝟐𝒙 𝟐 − (𝑴𝑪 + 𝟐𝑴𝑩)𝒍𝟐 − − −−(𝟑)
• Adding equations (2) and (3), weget
𝑬𝑰𝜽𝑩𝑨 +𝟔𝑬𝑰𝜽𝑩𝑪 =𝟔𝒂𝟏𝒙𝟏 +
𝟔𝒂𝟐𝒙𝟐 −(𝑴𝑨 +𝟐𝑴𝑩)𝒍𝟏 −(𝑴𝑪 + 𝟐𝑴𝑩)𝒍𝟐𝒍𝟏 𝒍𝟐
𝟔𝑬𝑰(𝜽𝑩𝑨 + 𝜽𝑩𝑪) =𝒍𝟏
𝟏 𝟏 𝟐 𝟐𝟔𝒂 𝒙 𝟔𝒂 𝒙
𝒍𝟐𝑨 𝟏 𝑩 𝟏 𝟐 𝑪𝟐
116
+ − 𝑴 𝒍 +𝟐𝑴 (𝒍 + 𝒍) + 𝑴 𝒍
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Clapeyron’s theorem of three moments(contd…)
𝟎 =𝟔𝒂𝟏𝒙𝟏
𝒍𝟏+
𝟔𝒂𝟐𝒙𝟐
𝒍𝟐−𝑴𝑨𝒍𝟏 +𝟐𝑴𝑩(𝒍𝟏 + 𝒍𝟐) +𝑴𝑪𝒍𝟐
𝑴𝑨𝒍𝟏 +𝟐𝑴𝑩(𝒍𝟏 + 𝒍𝟐) + 𝑴𝑪𝒍𝟐 =𝟔𝒂 𝒙𝟏 𝟏
𝒍𝟏+𝟔𝒂 𝒙𝟐 𝟐
𝒍𝟐
117
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118
Problems• A continuous beam of three equal span is simply supported over two
supports. It is loaded with auniformly distributed load of w/unit length,
over the two adjacentspansonly.Using the theoremof three moments,
find the support moments and sketch the bending moment diagram.
AssumeEI constant.
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Problems• Solution:
• The theorem of three
moments equation for two spansis,
𝑀 𝐴 (𝑙1) +2𝑀 𝐵 𝑙1 +𝑙2 +𝑀 𝐶 𝑙2 = 6𝑎1𝑥1 +
6𝑎2𝑥2
𝑙1 𝑙2Apply the theorem of three moment equation for spans AB and BC is,
𝑀 𝐴 𝑙 +2𝑀 𝐵 𝑙+𝑙 +𝑀 𝐶 𝑙
= 6𝑎1𝑥1 +
6𝑎2𝑥2
𝑙1 𝑙2
𝑙
A
w/ unitlengthB C D
𝑙
𝑤 𝑙2
8
𝑙
𝑤 𝑙2
8
FreeB.M.D.
119
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Problems
1
• Solution:
• 𝑎 = 2 ×𝑙×𝑤𝑙2
3 8
1
12= 𝑤 𝑙3
• 𝑥1 =𝑙
2
2• 𝑎2 =
3 ×𝑙×
𝑤 𝑙2
8
• 𝑥2 =𝑙
2
𝑙
6×1𝑤𝑙3×
𝑙6×
1 𝑤𝑙3×
𝑙
= 12 2 + 12
2 𝑙
𝐵 𝐶 2• 4𝑀 +𝑀 =
𝑤𝑙2
−−−− −(1)
𝑙
A
w/ unitlengthB C D
𝑙
𝑤 𝑙2
8
𝑙
𝑤 𝑙2
8
FreeB.M.D.
0
• 𝑀 𝐴 𝑙 +2𝑀 𝐵 𝑙+𝑙 +𝑀 𝐶 𝑙
120
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Problems• Solution:
• The theorem of three
moments equation for two spansis,
𝑀 𝐴 (𝑙1) +2𝑀 𝐵 𝑙1 +𝑙2 +𝑀 𝐶 𝑙2 = 6𝑎1𝑥1 +
6𝑎2𝑥2
𝑙1 𝑙2Apply the theorem of three moment equation for spans BC and CD is,
𝑀 𝐵 𝑙 +2𝑀 𝐶 𝑙+𝑙 +𝑀 𝐷 𝑙
𝑙1=
6𝑎1𝑥1 +0
𝑙
A
w/ unitlengthB C D
𝑙
𝑤 𝑙2
8
𝑙
𝑤 𝑙2
8
FreeB.M.D.
121
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Problems
1
• Solution:
• 𝑎 = 2 ×𝑙×𝑤𝑙2
3 8
1
12= 𝑤 𝑙3
• 𝑥1 =𝑙
2
6×1 𝑤𝑙3×
𝑙
𝑙= 12 2+0
𝐵 𝐶 4• 𝑀 +4𝑀 =
𝑤𝑙2
−−−−−(2)
𝑙
A
w/ unitlengthB C D
𝑙
𝑤 𝑙2
8
𝑙
𝑤 𝑙2
8
FreeB.M.D.
0
• 𝑀 𝐵 𝑙 +2𝑀 𝐶 𝑙+𝑙 +𝑀 𝐷 𝑙
122
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Problems• 4𝑀𝐵 + 16𝑀𝐶 =w𝑙2 −−−− −(2) ×4
4𝑀𝐵 + 𝑀𝐶 =𝑤 𝑙2
2−−−−−(1)
𝑤 𝑙2𝑀 𝐶 =
30
15𝑀𝐶 =𝑤 𝑙2
2−−−−− 2 ×4 −(1)
𝑤 𝑙2Substitute𝑀𝐶 =
30in equation(2),
𝑀 𝐵 +4×30
=𝑤 𝑙2 𝑤 𝑙2
4
𝑀𝐵 =7𝑤 𝑙2
60
123
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124
Fixed beamProblems
• A fixed beam AB of span 6 m carries uniformly varying load of intensity
zero at A and 20 kN/m at B. Find the fixed end moments and draw the
B.M. and S.F. diagrams for the beam.
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Problems
Consider any section XX distant 𝑥 from the end A, the intensityof
loading at XX= =𝑤 𝑥 20𝑥
𝐿 6
Hence the load acting for an elemental distance 𝑑𝑥 =20𝑥
6𝑑𝑥
Due to this elemental load the fixed moments are as follows:
𝑑𝑀𝑎 =Wa𝑏2
𝐿2(Formula is derived from firstprinciples)
= 620𝑥𝑑𝑥 ×𝑥× 6 − 𝑥 2
62=
20𝑥2 6 − 𝑥 2𝑑𝑥
63
A B
20kN/m20×𝑥
6
𝑥 X 6m
X𝑑𝑥
125
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Problems
and
𝑑𝑀𝑏 =W𝑏𝑎2
𝐿2Formula is derived from basicprinciples
= 6
20𝑥𝑑𝑥× 6−𝑥×𝑥2
62=
20𝑥3 6−𝑥 𝑑𝑥
63
𝑀𝐴 =𝑑𝑀𝑎
Taking fixing moment atA,𝑙
0216
6
0
= 20
𝑥2 6 − 𝑥 2𝑑𝑥
A B
20kN/m20×𝑥
6
𝑥 X
6m
X𝑑𝑥
126
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Problems
𝐴216
𝑀 =20
𝑥2 36 + 𝑥2 − 12𝑥 𝑑𝑥
6
0
20 36𝑥3
=216 3
𝑥5
+ 5
−12𝑥4
40
6
20 36 ×63 65
+ 5
−12 ×64
4=
216 3
∴𝑀𝐴 = 24kNm
127
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Problems
𝐵𝑀 =𝑑𝑀 𝐵
𝑙
063
= 20𝑥3 6 − 𝑥 𝑑𝑥
6
0
20=
216
𝑥4 𝑥5
4 ×6 −
50
6
20 64 ×6=
216 4
65
−5
∴𝑀𝐵 = 36kNm
128
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Problems
A
B
20kN/m20×𝑥
6
𝑥 X
6m
X𝑑𝑥FreeBMD:
𝑀𝑚𝑎𝑥 =𝑤 𝑙2
=20 ×62
9 3 9 3= 46.18 kNm (Cubic paraboliccurve)
Will occurat6 3 m from left endA.
46.18kNm
+
6 3
FreeBMD
FixedBMD
129
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Problems
A
B
20kN/m20×𝑥
6
𝑥 X
6m
X𝑑𝑥
-+
46.18kNm
-
6 3
Resultant BMD
130
24 kNm
36kNm
ResultantBMD:
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ProblemsCalculation of supportreactions:
A
B
20kN/m20×𝑥
6
𝑥 X
6m
X𝑑𝑥
𝑅𝐵
𝑅𝐴
24 36𝑀𝐴 =0
𝑅1 2
𝐵 ×6 + 24 = 36 + 2
×6 ×20 ×3
×6
𝐵6
𝑅 = 252
= 42 kN
1𝑅𝐴 + 42 =
2 ×6 × 20
𝑅𝐴 = 60 − 42 = 18kN
131
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ProblemsSFD:
A
B
20kN/m20×𝑥
6
𝑥 X
6m
X𝑑𝑥
42kN18kN
24 36
S.F. @ A =+18kN
S.F. @ B =-42kN
SFD between A and B is
aparabola.
S.F. @ XX=01 𝑥
18 − 2
×𝑥×20 ×6
= 0
18=10𝑥2
6𝑥= 3.29𝑚
Parabola
3.29m
132
18kN
42kNSF
D
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ProblemsResultantBMD
& SFD:
A
B
20kN/m20×𝑥
6
𝑥 X 6m
X𝑑𝑥
--
+
46.18kNm
Resultant BMD24 kNm
36 kNm
3.29m
133
6 3 =3.46m
18kN
42kN
SF
D
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UNIT 5
• INDETERMINATE BEAMS: CONTINUOUS BEAMS
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Continuous beam with supports at different levels
• Consider the continuous beam shown in above Figure. Let thesupport B be 𝛿1 below A and below C.
• Consider the spanAB:
• Let at any section in AB distant 𝑥 from A the free and fixed bending moments be 𝑀𝑥 and 𝑀𝑥
′ respectively.
• Hence the net bending moment at the section is given by
𝑑2𝑦𝐸𝐼𝑑𝑥2 = 𝑀𝑥 −𝑀
𝑥
′
• Multiplying by 𝑥, weget
𝐸𝐼𝑥𝑑2𝑦
𝑑𝑥2 𝑥 𝑥′= 𝑀 𝑥−𝑀 𝑥
A C
𝑙1 𝑙2𝑀𝐴
𝑀𝐶
B𝑀𝐵
𝛿
135
1
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• 𝐸𝐼𝑥2𝑑 𝑦
𝑑𝑥2 𝑥 𝑥′= 𝑀 𝑥−𝑀 𝑥
𝑑2𝑦𝑙1 𝑙1
• Integrating from 𝑥= 0 𝑡𝑜𝑥= 𝑙1, we get,
𝐸𝐼 𝑥𝑑𝑥2 = 𝑀𝑥𝑥 𝑑𝑥 − 𝑀𝑥
′𝑥 𝑑𝑥
𝑙1
0 0
0
𝑑𝑦𝐸𝐼 𝑥.
𝑑𝑥−𝑦
136
0
𝑙1𝑙1 𝑙1
= 𝑀𝑥𝑥 𝑑𝑥 − 𝑀𝑥′𝑥𝑑𝑥
0 0
−−−(1)
Continuous beam with supports at different levels
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• But it may be suchthat
At 𝑥 =0, deflection 𝑦 =0
𝑑𝑥• At 𝑥= 𝑙1, 𝑦 = −𝛿1; 𝑎𝑛𝑑 𝑠𝑙𝑜𝑝𝑒𝑎𝑡𝐵 𝑓𝑜𝑟𝐴𝐵,
𝑑𝑦=𝜃𝐵𝐴
• 𝑥𝑙1
0𝑀 𝑥𝑑𝑥=𝑎 𝑥1 1 =Momentof thefreeB.M.D.onABaboutA.
•𝑙1
0 𝑥 1′
1′
′
𝑀 𝑥𝑑𝑥=𝑎 𝑥 =Momentof thefixedB.M.D.onABaboutA.
A C
𝑙1 𝑙2𝑀𝐴
𝑀𝐶
B𝑀𝐵
𝛿1𝑥
137
Continuous beam with supports at different levels
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𝑑𝑦𝐸𝐼 𝑥.
𝑑𝑥−𝑦
0
𝑙1𝑙1 𝑙1
= 𝑀𝑥𝑥 𝑑𝑥 − 𝑀𝑥′𝑥𝑑𝑥
0 0
−−(1)
• Therefore the equation (1) is simplifiedas,𝐸𝐼 𝑙1𝜃𝐵𝐴− −𝛿1 =𝑎1𝑥1 −𝑎′𝑥1
′.1
1But 𝑎′ =area of the fixed B.M.D. on AB=𝑀 𝐴 +
𝑀 𝐵2𝑙1
𝑥1′ =Centroidof thefixedB.M.D.fromA=
𝑀 𝐴 +2𝑀 𝐵 𝑙1
𝑀 𝐴 +𝑀 𝐵 3
Continuous beam with supports at different levels
138
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• Therefore,
𝟏𝒂′ 𝒙𝟏
′ =(𝑴𝑨 +𝑴𝑩 ) 𝑴 𝑨 + 𝟐𝑴𝑩 𝒍𝟏×
𝒍𝟏
𝟐 𝑴 𝑨 +𝑴 𝑩
𝟑
= (𝑴𝑨 + 𝟐𝑴𝑩 )𝟏
𝒍𝟐
𝟔
Similarly by considering the span BC and taking C as origin it can be shown
that,
𝜃𝐵𝐶= slope for span CB at B
Continuous beam with supports at different levels
∴𝐸𝐼(𝑙𝜃1 𝐵𝐴+ 1𝛿 ) =𝑎 𝑥 − 𝑀 +2𝑀1 1 𝐴𝐵
𝑙12
6
𝐵𝐴
6𝐸𝐼𝜃 =6𝑎 𝑥1 1
−6𝐸𝐼𝛿 1
𝑙1 𝑙11− 𝑀𝐴 +2𝑀𝐵 𝑙 −−−(2)
6𝐸𝐼𝜃𝐵𝐶 =6𝑎 𝑥 6𝐸𝐼𝛿2 2 2
𝑙2 𝑙2𝐶 𝐵 2
139
− − 𝑀 +2𝑀 𝑙 −−−(3)
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• But𝜃𝐵𝐴 = −𝜃𝐵𝐶 as the direction of 𝑥 from A for the span AB,
and from C for the span CB are in oppositedirection.
• And hence,𝜃𝐵𝐴+ 𝜃𝐵𝐶 =0
Continuous beam with supports at different levels
6𝐸𝐼𝜃𝐵𝐴 =𝑙1
6𝑎 𝑥 6𝐸𝐼𝛿1 1 1
𝑙1𝐴 𝐵 1− − 𝑀 +2𝑀 𝑙 −−−(2)
6𝐸𝐼𝜃𝐵𝐶 =6𝑎 𝑥2 2 6𝐸𝐼𝛿 2
𝑙2 𝑙2− − 𝑀𝐶 +2𝑀𝐵
2𝑙 −−−(3)
Adding equations (2) and (3), weget
6𝐸𝐼(𝜃𝐵𝐴+𝜃𝐵𝐶)
= 6𝑎1𝑥1 +
6𝑎2𝑥2 − 6𝐸𝐼𝛿1 −
6𝐸𝐼𝛿2
𝑙1 𝑙2 𝑙1 𝑙2− 𝑀𝐴𝑙1 +2𝑀𝐵 𝑙1 +𝑙2 + 𝑀𝑐𝑙2
140
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Continuous beam with supports at different levels6𝐸𝐼(𝜃𝐵𝐴+𝜃𝐵𝐶)
= 6𝑎1𝑥1 +
6𝑎2𝑥2 − 6𝐸𝐼𝛿1 −
6𝐸𝐼𝛿2
𝑙1 𝑙2 𝑙1 𝑙2− 𝑀𝐴𝑙1 +2𝑀𝐵 𝑙1 +𝑙2 + 𝑀𝑐𝑙2
0=6𝑎 𝑥1 1
𝑙1+
𝑙2−
6𝑎 𝑥 6𝐸𝐼𝛿2 2 1
𝑙1−
6𝐸𝐼𝛿 2
𝑙2− 𝑀𝐴𝑙1 +2𝑀𝐵 𝑙1 +𝑙2 + 𝑀 𝑙𝑐2
𝑀𝐴𝑙1 +2𝑀𝐵 𝑙1 +𝑙2 + 𝑀 𝑙𝑐2=
6𝑎1𝑥1 + 6𝑎2𝑥2 − 6𝐸𝐼
𝑙1 𝑙2
𝛿1 +𝛿2
𝑙1 𝑙2
141
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• The following Figure shows a continuous beam carrying an external
loading. If the support B sinks by 0.25 cm below the level of the other
supports find support moments. Take I for section= 15000 cm4and
E=2x103 t/cm2.
Problems
2t/m
4m
142
4m 4m
A 4 t/m B C D
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• The theorem of three moments for two spans AB and BC is as follows,
• Consider the spans AB andBC,
• 𝑀𝐴 =0
• 𝛿1 = +0.25cm
• 𝛿2 = +0.25cm
Problems
𝑀𝐴𝑙1 +2𝑀𝐵 𝑙1 +𝑙2 + 𝑀𝑐𝑙2 = 6𝑎1𝑥1 +
6𝑎2𝑥2 −6𝐸𝐼𝛿1 +
𝛿2
𝑙1 𝑙2 𝑙1 𝑙2
2t/m
4m 4m
A 4 t/m B C D
0.25cm
4m
143
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2• 𝑎1 =
3 ×4 ×8
2
• 6𝑎1𝑥1 = 6×
3×4×8×2
=64𝑙1 4
6𝑎 𝑥
𝑙2
6×2×4×8×2
4• 2 2 = 3 =64
Problems
6𝐸𝐼=6 ×2 ×103 ×15000
1002
= 18000 t𝑚2
8tm 8tm4tm
FreeBMD
2t/m
4m 4m 4m
A 4 t/m B C D
0.25cm
144
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• ∴ 0+2𝑀𝐵 4+4 + 4𝑀𝐶 = 64 + 64 − 18000
• ∴ 16𝑀𝐵+4𝑀𝐶= 128 −22.5
• 16𝑀𝐵+4𝑀𝐶=105.5
• 4𝑀𝐵+𝑀𝐶 = 26.375 −−−−(1)
Problems
8 tm 8 tm
𝑀 𝑙 +2𝑀 𝑙 +𝑙 + 𝑀 𝑙𝐴 1 𝐵 1 2 𝑐2 𝑙1 𝑙2=
6𝑎1𝑥1 + 6𝑎2𝑥2 − 6𝐸𝐼
𝛿1 +𝛿2
0.25𝑙10.25𝑙2
400 +
400
4tm
2t/m
4m 4m 4m
A 4 t/m B C D
0.25cm
145
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• Now consider the spans BC andCD,
• 𝑀𝑑=0,
• 𝛿1 = −0.25 cm
• 𝛿2 =0
Problems
4tm
2t/m
4m8tm
4m
A 4 t/m B C D0.25cm 4 m
8tm
146
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𝑀𝐵 ×4+2𝑀𝐶 4+4 + 0 = 64 + 32 − 18000
• ∴ 4𝑀𝐵+16𝑀𝐶= 96 +11.25
• 4𝑀𝐵+16𝑀𝐶= 107.25 −−−−(2)
Problems
8 tm 8 tm
𝑀𝐴𝑙1 +2𝑀𝐵 𝑙1 +𝑙2 + 𝑀 𝑙𝑐2 𝑙1 𝑙2=
6𝑎1𝑥1 + 6𝑎2𝑥2 − 6𝐸𝐼
𝛿1 +𝛿2
−0.𝑙1 𝑙2
25+
0
400 400
4tm
2t/m
4m 4m 4m
A 4 t/m B C D
0.25cm
147
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• 4𝑀𝐵+𝑀𝐶 = 26.375 −−−−(1)
• 4𝑀𝐵+16𝑀𝐶 = 107.25 −−−−(2)
• Solving (1) and (2), weget,
• 𝑀𝐵 =5.24tm hogging .
• 𝑀𝐶 =5.39tm hogging .
148
Problems
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Problems
+
5.24tm
-
5.39tm
-
8tm 8tm+ 4tm
+
BMD
2t/m
4m 4m 4m
A 4 t/m B C D
0.25cm
149
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BA
V
Fixed beam with ends at different levels (Effect ofsinking of supports)V𝑀
𝐴 𝑀𝐵
𝛿
150
𝑀𝐴 is negative (hogging) and 𝑀𝐵 is positive (sagging).Numerically
𝑀𝐴 and 𝑀𝐵 areequal.
Let V be the reaction at eachsupport.
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BA
V
Fixed beam with ends at different levels (Effect of sinking of supports)V
𝑀𝐴 𝑀
𝐵
𝛿
𝑑4𝑦
Consider any section distance 𝑥 from the endA.
Since the rate of loading is zero, we have, with the usualnotations
𝐸𝐼𝑑𝑥4 =0
Integrating, weget,𝑑3𝑦
Shear force =𝐸𝐼𝑑𝑥3 =𝐶1
Where 𝐶1is a constant
At 𝑥 =0, 𝑆. 𝐹.= +V
∴𝐶1 =V
𝑥
151
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BA
V
Fixed beam with ends at different levels(Effect of sinking of supports)V
𝑀𝐴 𝑀
𝐵
𝛿
𝑑2𝑦B.M. at any section =𝐸𝐼
𝑑𝑥2 = 𝑉𝑥 +𝐶1
At 𝑥 = 0, 𝐵. 𝑀.= −𝑀𝐴
∴ 𝐶2 =−𝑀𝐴
𝑑2𝑦∴𝐸𝐼
𝑑𝑥2 = 𝑉𝑥 −𝑀𝐴
Integratingagain,
𝑑𝑥 2𝐸𝐼𝑑𝑦 =
𝑉 𝑥2 − 𝑀𝐴𝑥 + 𝐶3 (Slopeequation)𝑑𝑦
But at 𝑥 = 0,𝑑𝑥
=0 ∴𝐶3 =0
𝑥
152
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BA
Fixed beam with ends at different levels (Effect of sinking of supports)V
𝑀𝐴 𝑀
𝐵
𝛿
V
Integratingagain,3
𝐸𝐼𝑦= − 𝐴𝑉𝑥 𝑀 𝑥2
6 2 4+𝐶 ------ (Deflectionequation)
But at 𝑥= 0, 𝑦 = 0
∴𝐶4 =0
At 𝑥= 𝑙, 𝑦 = −𝛿
𝑉𝑙3−𝐸𝐼𝛿=
6−𝑀 𝐴 𝑙2
2−−−−−−−−(i)
𝑙𝑥
153
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BA
V
Fixed beam with ends at different levels(Effect of sinking of supports)V
𝑀𝐴 𝑀
𝐵
𝛿
𝑑𝑦But we also know that at B, 𝑥= 𝑙𝑎𝑛𝑑
𝑑𝑥=0
𝑑𝑥 2And substitute inslopeEq. 𝐸𝐼𝑑𝑦 =
𝑉 𝑥2 −𝑀𝐴𝑥
𝑉𝑙2∴ 0 =
2−𝑀𝐴𝑙
∴𝑉 = 2𝑀𝐴 −−−−−−− − ii
𝑙
Substituting in deflection Eq.(i) i.e., −𝐸𝐼𝛿= 𝑉𝑙3
− 𝑀𝐴𝑙2
;we have,
2𝑀𝐴
𝑙3−𝐸𝐼𝛿= × −
6 2
𝑀𝐴𝑙2
𝑙𝑥
2 154
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BA
V
Fixed beam with ends at different levels(Effect of sinking of supports)V
𝑀𝐴 𝑀
𝐵
𝛿
𝐸𝐼𝛿 = 𝐴𝑀 𝑙2
6
∴𝑀
𝐴 =6𝐸𝐼𝛿
𝑙2
𝑑2𝑦
Hence the law for the bending moment at any section distant x
from A is givenby,
𝑀 = 𝐸𝐼𝑑𝑥2 = V𝑥−𝑀𝐴
𝑙∴𝑀 =
2𝑀𝐴 𝑥− 6𝐸𝐼𝛿
𝑙2
𝑙
155
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BA
V
Fixed beam with ends at different levels (Effect ofsinking of supports)V𝑀
𝐴 𝑀𝐵
𝛿
ButforB.M.atB,putx= l,
∴𝑀
𝐵 𝑙=
2𝑀𝐴 ×𝑙− 6𝐸𝐼𝛿
= 12𝐸𝐼𝛿
− 6𝐸𝐼𝛿
=6𝐸𝐼𝛿
𝑙2 𝑙2 𝑙2 𝑙2
Hence when the ends of a fixed beam are at different levels,
The fixing moment at each end =6𝐸𝐼𝛿
𝑙2numerically.
At the higher end this moment is a hogging moment and at the
lower end this moment is a saggingmoment.
𝑙
156
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Problems
A B3m
157
2m
• A fixed beam of span 5 metres carries a concentrated load of 20 t at 3
meters from the left end. If the right end sinks by 1 cm, find the fixing
moments at the supports. For the beam section take I=30,000 cm4 and
E=2x103 t/cm2. Find also the reaction at the supports.
20t
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• The right end sinks by 1 cm, find the fixing moments at the supports.
Problems
A B3m 2m
• A fixed beam of span 5 metres carries a concentrated load of 20 t at 3
meters from the left end.
20t
1 cm
158
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• 𝑀𝐴 = −Wa𝑏2−6𝐸𝐼𝛿
𝑙2
• =−
𝑙2
20×3×22
52+
6×2×103×30,000×1
52×1002tm
• = − 9.6+0.48 tm=-10.08 tm(hogging)
• 𝑀𝐵=−W𝑏𝑎2
+6𝐸𝐼𝛿
𝑙2 𝑙2
•52
= − 20×2×32 +6×2×103×30,000×1
52×1002tm
• = −14.4+0.48 tm= -13.92 tm(hogging)
Problems
A B3m 2m
20t
1cm𝑀𝐴
𝑀𝐵
159
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• Reaction atA:• 𝑀𝐵=0,
• 𝑉𝐴×5 + 13.92 −10.08− 20×2 =0
• ∴𝑉𝐴= 7.232t
• Reaction atB:
• ∴𝑉𝐵= 20 − 7.232 = 12.768t.
ProblemsA B
3m 2m
20t
10.08
13.92𝑉𝐴
160
𝑉𝐵