INSTITUTE OF AERONAUTICAL ENGINEERING · The procedure of finding slope and deflection for simply supported beam with an eccentric load is very ... DEFLECTION BY ENERGY METHODS. Elastic

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

Unit 1

Deflection in Beams

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

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)

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

From the Engineering Beam Theory:

M E

I R

1 M

R EI

d2y

dx2

d2y EI M

dx 2

Flexural

Stiffness

Bending

MomentCurvature

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

Methods to find slope and deflection

Double integration method

Moment area method

Macaulay’s method

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



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

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.

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

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.

Moment Area Method

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

UNIT 2

DEFLECTION BY ENERGY METHODS

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

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:

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

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

Table 5.3


TWO METHODS

Differentiate after Integral Differentiate under Integral


m

m

Transverse shear contributes only <5% to deflection


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)

:


•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

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

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

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.

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

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

UNIT III

STRESSES IN CYLINDERS AND

SPHERICAL SHELLS

THINCYLINDER

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.

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.

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.

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

(b) Longitudinal stress (𝜎𝑙 ):

Total pressurep = Pressure *

Area4

p = p𝜋

𝑑2

Resistingarea,A = 𝜋𝑑𝑡

𝜎𝑙= P/A =p4

𝜋𝑑2

𝜋𝑑𝑡

𝝈𝒍=pd/4t

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,

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)

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

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)

CLASSIFICATIONS OFSHELLS

• 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

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.

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.

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.

• 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.


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.

Folded platetypes

Folded Plates system

Folded-Plate Hut in Osaka

Folded Plates Library


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

Types of shells4. Various Double Curvature


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.


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.

• 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.

Unit IV

Indeterminate Beams:

PROPPED CANTILEVER AND

FIXED BEAMS

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.

89

Indeterminatebeams

• Staticallydeterminatebeams:

– Cantileverbeams

– Simplesupportedbeams

– Overhangingbeams

• Staticallyindeterminatebeams:

– Proppedcantileverbeams

– Fixedbeams

– Continuousbeams

• Propped cantileverBeams:

Indeterminatebeams

Degree of staticindeterminacy=

N0. of unknown reactions – static equations=3-2=1

90

• 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



91

Indeterminatebeams



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.



92

•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

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

• (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

• 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

--

FixedBeams

𝑀𝐵

𝑀𝐴

𝑉𝐴𝑉𝐵

𝑊1

𝑊2

𝑉𝑎 𝑉𝑏


𝑀𝐴

𝑀𝐵𝑉 𝑉

(iiia) Fixedbeam

(iia) Effect of endcouples𝑊1 𝑊2

𝑀𝑥

𝑀𝑥 ′

+

--

(iib) FixedB.M.D.


(ib) FreeB.M.D.

97

• 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

• 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,

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

FixedBeams

𝑀𝐵

𝑀𝐴

𝑉𝐴𝑉𝐵

𝑊1

𝑊2

𝑉𝑎 𝑉𝑏


𝑀𝐴

𝑀𝐵𝑉 𝑉

(iiia) Fixedbeam

(iia) Effect of endcouples𝑊1 𝑊2

𝑀𝑥

𝑀𝑥′

+

--

(iib) FixedB.M.D.


(ib) FreeB.M.D.

101

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

• 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

Fixed beamproblems

• Find the fixed end moments of a fixed beam subjected to a

eccentric point load.

W

a

A B

b

𝑙

104

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)

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

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

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

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


𝑙1

109

𝑙2


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.

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

• 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)


• 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


𝑑𝑥−𝑦

0

𝑙1𝑙1 𝑙1


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


114

• Therefore,


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

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


𝟎 =𝟔𝒂𝟏𝒙𝟏

𝒍𝟏+

𝟔𝒂𝟐𝒙𝟐

𝒍𝟐−𝑴𝑨𝒍𝟏 +𝟐𝑴𝑩(𝒍𝟏 + 𝒍𝟐) +𝑴𝑪𝒍𝟐

𝑴𝑨𝒍𝟏 +𝟐𝑴𝑩(𝒍𝟏 + 𝒍𝟐) + 𝑴𝑪𝒍𝟐 =𝟔𝒂 𝒙𝟏 𝟏

𝒍𝟏+𝟔𝒂 𝒙𝟐 𝟐

𝒍𝟐

117

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.

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

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

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

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

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

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.

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

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

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

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

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

Problems

A

B

20kN/m20×𝑥

6

𝑥 X

6m

X𝑑𝑥

-+

46.18kNm

-

6 3

Resultant BMD

130

24 kNm

36kNm

ResultantBMD:

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

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

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

UNIT 5

• INDETERMINATE BEAMS: CONTINUOUS BEAMS

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


A C

𝑙1 𝑙2𝑀𝐴

𝑀𝐶

B𝑀𝐵

𝛿

135

1

• 𝐸𝐼𝑥2𝑑 𝑦


𝑑2𝑦𝑙1 𝑙1

• Integrating from 𝑥= 0 𝑡𝑜𝑥= 𝑙1, we get,

𝐸𝐼 𝑥𝑑𝑥2 = 𝑀𝑥𝑥 𝑑𝑥 − 𝑀𝑥

′𝑥 𝑑𝑥

𝑙1

0 0

0


𝑑𝑥−𝑦

136

0

𝑙1𝑙1 𝑙1


0 0

−−−(1)


• 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



𝑑𝑥−𝑦

0

𝑙1𝑙1 𝑙1


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


138

• 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


∴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)

• 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 =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

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

• 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

• 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

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

• ∴ 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

• 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

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

• 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

Problems

+

5.24tm

-

5.39tm

-

8tm 8tm+ 4tm

+

BMD

2t/m

4m 4m 4m

A 4 t/m B C D

0.25cm

149

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.

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

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

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

BA

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

BA

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

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

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

• 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

• 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

• 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

INSTITUTE OF AERONAUTICAL ENGINEERING · The procedure of finding slope and deflection for simply supported beam with an eccentric load is very ... DEFLECTION BY ENERGY METHODS. Elastic

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