(4)Str.analysis

DEFLECTION OF BEAMS

Structures undergo deformation when subjected to loads. As a result of this deformation, deflection and rotation occur in structures. This deformation will disappear when the loads are removed provided the elastic limit of the material is not exceeded. Deformation in a structure can also occur due to change in temperature & settlement of supports.

Deflection in any structure should be less than specified limits for satisfactory performance. Hence computing deflections is an important aspect of analysis of structures.

There are various methods of computing deflections. Two popular methods arei) Moment area Method, and ii) Conjugate beam method

In both of these methods, the geometrical concept is used. These methods are ideal for statically determinate beams. The methods give a very quick solution when the beam is symmetrical.

Moment Area Method

This method is based on two theorems which are stated through an example. Consider a beam AB subjected to some arbitrary load as shown in Figure 1.

Let the flexural rigidity of the beam be EI. Due to the load, there would be bending moment and BMD would be as shown in Figure 2. The deflected shape of the beam which is the elastic curve is shown in Figure 3. Let C and D be two points arbitrarily chosen on the beam. On the elastic curve, tangents are drawn at deflected positions of C and D. The angles made by these tangents with respect to the horizontal are marked as and . These angles are nothing but slopes. The change is the angle between these two tangents is demoted as

. This change in the angel is equal to the area of the diagram between the two points

C and D. This is the area of the shaded portion in figure 2.

Hence = = Area of diagram between C and D

= Area BM 1 (a) EI

It is also expressed in the integration mode as

= 1 (b)

Equation 1 is the first moment area theorem which is stated as follows:

Statement of theorem I:

The change in slope between any two points on the elastic curve for a member

subjected to bending is equal to the area of diagram between those two points.

1

In figure 4, for the elastic curve a tangent is drawn at point C from which the vertical intercept to elastic curve at D is measured. This is demoted as KCD. This vertical intercept is given by

KCD = (Area BM X)CD 2 (a) EI

Where is the distance to the centroid of the shaded portion of diagram measured

from D. The above equation can be expressed in integration mode as

2

Fig. 1

Fig. 2

Fig. 3

Fig. 4

KCD = 2 (b)

Equation (2) is the second moment area theorem which is stated as follows.

Statement of theorem II :

The vertical intercept to the elastic curve measured from the tangent drawn to the

elastic curve at some other point is equal to the moment of diagram, moment being

taken about that point where vertical intercept is drawn.

Sign Convention:

While computing Bending moment at a section, if free body diagram of Left Hand Portion (LHP) is considered, clockwise moment is taken as positive. If free body diagram of Right Hand Portion (RHP) is considered, anticlockwise moment is taken as positive. While sketching the Bending Moment Diagram (BMD), Sagging moment is taken as positive and Hogging moment is taken as negative.

Proof of Moment Area Theorems:

Figure 5 shows the elastic curve for the elemental length dx of figure 2 to an enlarged scale. In this figure, R represents the radius of curvature. Then from equation of bending, with usual notations,

= (3)

From figure 5,

Rdθ = dx

Hence R =

Substituting this value of R in equation (3),

=

= E

dθ = dx

dθ is nothing but change in angle over the elemental length dx. Hence to compute change in angle from C to D,

θCD = = dx

Hence the proof.

3

Figure 6 shows the elastic curve from C to D. Change in slope from 1 to 2 is dθ. Distance of elemental length from D is x.

dΔ = xdθ = x dx

Therefore, Δ from C to D = xdx

4

dθFig. 5

Fig. 5Fig. 6

KCD

dθ

d

dθ

Problem 1 : Compute deflections and slopes at C,D and E. Also compute slopes at A and B.

5

To Compute Reactions:

+

6

VA = W ; VB =W

Bending Moment Calculations:

Section (1) – (1) (LHP, 0 to L/3)+ Mx-x = WxAt x = 0; BM at A = 0 x = ; BM @ C =

Section (2) – (2) (LHP, to )

+ Mx-x = Wx – W(x - )

At x = , BM @ C =

=

At x = , BM @ D =

=

=

Section (3) – (3) RHP (0 to )+ Mx-x = Wx

At x = 0; BM @ B = 0

At x = , BM @ D =

This beam is symmetrical. Hence the BMD & elastic curve is also symmetrical. In such a case, maximum deflection occurs at mid span, marked as δE. Thus, the tangent drawn at E will be parallel to the beam line and θE is zero.

Also, δc = δD, θA = θB and θC = θD

To compute θC

From first theorem,θCE = Area of BMD between E&C

EI

θC~ θE =

=

θE being zero, θC = WL 2 ( ) 18EI

To compute θΔ

From First theorem,

7

θΔE = Area of BMD between A&EEI

θA~ θE =

=

θE being zero, θA = ( )

θB = ( )

To compute δE

From 2nd theorem

KEA =

KEA =

=

=

=

From figure, KEA is equal to δE.

Therefore δE =

To compute θC

From 2nd theorem

KEC =

8

=

=

=

δc = δE - KEC

=

=

=

=

9

Problem 2. For the cantilever beam shows in figure, compute deflection and slope at the free end.

Consider a section x-x at a distance x from the free end. The FBD of RHP is taken into account.

(RHP +) BM @ X-X = MX-X = -10 (x) (x/2) = -5x2

At x = 0; BM @ B = 0At x = 4m; BM @ A = -5(16) = -80 kNm

The BMD is sketched as shown in figure. Note that it is Hogging Bending Moment. The elastic curve is sketched as shown in figure.

10

To compute θB

For the cantilever beam, at the fixed support, there will be no rotation and hence in this case θA = 0. This implies that the tangent drawn to the elastic curve at A will be the same as the beam line.

From I theorem,

θAB = θA ~ θB =

=

=

=

θA being zero,

θB = ( )

To compute δB

From II theorem

KAB =

=

=

=

From the elastic curve,

KAB = δB =

Problem 3: Find deflection and slope at the free end for the beam shown in figure by using moment area theorems. Take EI = 40000 KNm-2

11

Calculations of Bending Moment:

Region AC: Taking RHP + Moment at section = -6x2/2

= -3x2

At x = 0, BM @ A = 0 x = 4m; BM @ C = -3(16) = - 48kNm

Region CB: (x = 4 to x = 8)

Taking RHP +, moment @ section = -24 (x-2) = -24x+48;

12

At x = 4m; BM @ C = -24(4) + 48 = -48kNm; x = 8 m BM @ B = -144 kNm;

To compute θB:

First moment area theorem is used. For the elastic curve shown in figure. We know that θA = 0.

θAB = θA ~ θB =

=

=

= -0.0112 Radians

= 0.0112 Radians ( )

To compute δB

=

=

=

=

=

Problem 4: For the cantilever shown in figure, compute deflection and at the points where

they are loaded.

13

To compute θB :

θBA = θB ~ θA =

θB = ( )

θC =

= ( )

δB =

14

=

=

δC =

δC =

STRAIN ENERGY4.1 Introduction

Under action of gradually increasing external loads, the joints of a structure deflect and the member deform. The applied load produce work at the joints to which they are applied and this work is stored in the structure in the form of energy known as Strain Energy. If the material of structure is elastic, then gradual unloading of the structure relieves all the stresses and strain energy is recovered.

The slopes and deflections produced in a structure depend upon the strains developed as a result of external actions. Strains may be axial, shear, flexural or torsion. Therefore, ther is a relationship can be used to determine the slopes and deflections in a structure.4.2 Strain energy and complementary strain energy

When external loads are applied to a skeletal structure, the members develop internal force ‘F’ in the form of axial forces (‘P’), shear force (‘V’) , bending moment (M) and twisting moment (T). The internal for ‘F’ produce displacements ‘e’. While under goint these displacements, the internal force do internal work called as Strain Energy

Figure 1 shows the force displacement relationship in which F j is the internal force and ej is the corresponding displacement for the jth element or member of the structure.

15

The element of internal work or strain energy represented by the area the strip with

horizontal shading is expressed as:

Strain energy stored in the jth element represted by the are under force-displacement

curve computed as :

For m members in a structure, the total strain energy is

The area above the force-displacement curve is called Complementary Energy. For

jth element, the complementary strain energy is represented by the area of the strip with

vertical shading in Fig.1 and expressed as

Complementary strain energy of the entire structure is

Complementary strain energy of the entire structure is

16

Fj

Strain Energy(Ui)j

ejej ej+ej

Fj

Fj+Fj

Complementry SE(Ui)j

Fig.1 FORCE-DISPLACEMENT RELATIONSHIP

When the force-displacement relationship is linear, then strain energy and

complimentary energies are equal

4.3 Strain energy expressions

Expression for strain energy due to axial force, shear force and bending moment is

provided in this section

4.3.1 Strain energy due to Axial force

A straight bar of length ‘L’ , having uniform cross sectional area A and E is the Young’s

modulus of elasticity is subjected to gradually applied load P as shown in Fig. 2. The bar

deforms by dL due to average force 0+(P/2) = P/2. Substituting Fj = P/2 and dej = dl in

equation 2, the strain energy in a member due to axial force is expressed as

From Hooke’s Law, strain is expressed as

Hence

17

LA,E

dL

Fig.2

Substituting equation 9 in 8, strain energy can be expressed as

For uniform cross section strain energy expression in equation 10 can be modified as

If P, A or E are not constant along the length of the bar, then equation 10 is used instead of

10a.

4.3.1 Strain energy due to Shear force

A small element shown in Fig.3 of dimension dx and dy is subjected to shear force V x . Shear

stress condition is shown in Fig. 4. Shear strain in the element is expressed as

Where, Ar= Reduced cross sectional area and G= shear modulus

Shear deformation of element is expressed as

Substituting Fj = Vx/2, dej = dev in equation (2) strain energy is expressed as

18

dydx

dy

dx

Fig.3Fig.4

4.3.2 Strain energy due to Bending Moment

An element of length dx of a beam is subjected to uniform bending moment ‘M’. Application

of this moment causes a change in slope d is expressed as

Where , , Substituting Fj = Mx/2, dej= deM in equation (2), Strain energy due to

bending moment is expressed as

4.3 Theorem of minimum Potential Energy

Potential energy is the capacity to do work due to the position of body. A body of weight

‘W’ held at a height ‘h’ possess an energy ‘Wh’. Theorem of minimum potential energy

states that “ Of all the displacements which satisfy the boundary conditions of a

structural system, those corresponding to stable equilibrium configuration make the

total potential energy a relative minimum”. This theorem can be used to determine the

critical forces causing instability of the structure.

4.3 Law of Conservation of Energy

From physics this law is stated as “Energy is neither created nor destroyed”. For the purpose

of structural analysis, the law can be stated as “ If a structure and external loads acting on

it are isolated, such that it neither receive nor give out energy, then the total energy of

the system remain constant”. With reference to figure 2, internal energy is expressed as in

equation (9). External work done We = -0.5 P dL. From law of conservation of energy Ui+We

=0. From this it is clear that internal energy is equal to external work done.

4.3 Principle of Virtual Work:

Virtual work is the imaginary work done by the true forces moving through imaginary

displacements or vice versa. Real work is due to true forces moving through true

19

displacements. According to principle of virtual work “ The total virtual work done by a

system of forces during a virtual displacement is zero”.

Theorem of principle of virtual work can be stated as “If a body is in equilibrium under a

Virtual force system and remains in equilibrium while it is subjected to a small

deformation, the virtual work done by the external forces is equal to the virtual work

done by the internal stresses due to these forces”. Use of this theorem for computation of

displacement is explained by considering a simply supported bea AB, of span L, subjected to

concentrated load P at C, as shown in Fig.6a. To compute deflection at D, a virtual load P’ is

applied at D after removing P at C. Work done is zero a s the load is virtual. The load P is

then applied at C, causing deflection C at C and D at D, as shown in Fig. 6b. External work

done We by virtual load P’ is . If the virtual load P’ produces bending moment

M’, then the internal strain energy stored by M’ acting on the real deformation d in element

dx over the beam equation (14)

Where, M= bending moment due to real load P. From principle of conservation of energy

We=Wi

20

If P’=1 then

Similarly for deflection in axial loaded trusses it can be shown that

Where,

= Deflection in the direction of unit load

P’ = Force in the ith member of truss due to unit load

P = Force in the ith member of truss due to real external load

n = Number of truss members

L = length of ith truss members.

Use of virtual load P’ = 1 in virtual work theorem for computing displacement is called

Unit Load Method

4.4 Castigliano’s Theorems:

Castigliano published two theorems in 1879 to determine deflections in structures and

redundant in statically indeterminate structures. These theorems are stated as:

21

A BC D

a

xL

P

A BC D

a

xL

P P’C D

Fig.6a

Fig.6b

1st Theorem : “If a linearly elastic structure is subjected to a set of loads, the

partial derivatives of total strain energy with respect to the deflection at any point is

equal to the load applied at that point”

2nd Theorem: “If a linearly elastic structure is subjected to a set of loads, the

partial derivatives of total strain energy with respect to a load applied at any point is

equal to the deflection at that point”

The first theorem is useful in determining the forces at certain chosen coordinates. The

conditions of equilibrium of these chosen forces may then be used for the analysis of

statically determinate or indeterminate structures. Second theorem is useful in computing the

displacements in statically determinate or indeterminate structures.

4.5 Betti’s Law:

It states that If a structure is acted upon by two force systems I and II, in equilibrium

separately, the external virtual work done by a system of forces II during the

deformations caused by another system of forces I is equal to external work done by I

system during the deformations caused by the II system

A body subjected to two system of forces is shown in Fig 7. Wij represents work done by ith

system of force on displacements caused by jth system at the same point. Betti’s law can be

expressed as Wij = Wji, where Wji represents the work done by jth system on displacement

caused by ith system at the same point.

22

I II

Fig. 7

Numerical Examples

1. Derive an expression for strain energy due to bending of a cantilever beam of length

L, carrying uniformly distributed load ‘w’ and EI is constant

Solution:

Bending moment at section 1-1 is

Strain energy due to bending is

23

w

x

1

1

2. Compare the strain energies due to three types of internal forces in the rectangular

bent shown in Fig. having uniform cross section shown in the same Fig. Take E=2 x 105

MPa, G= 0.8 x 105 MPa, Ar= 2736 mm2

Solution:

Step 1: Properties

A=120 * 240 – 108 * 216 = 5472 mm2,

E= 2 * 105 MPa ; G= 0.8 * 105 MPa ; Ar = 2736 mm2

Step 2: Strain Energy due to Axial Forces

Member AB is subjected to an axial comprn.=-12 kN

Strain Energy due to axial load for the whole str. is

Step 3: Strain Energy due to Shear Forces

Shear force in AB = 0; Shear force in BC = 12 kN

Strain Energy due to Shear for the whole str. Is

Step 4: Strain Energy due to Bending Moment

Bending Moment in AB = -12 * 4 = -48 kN-m

Bending Moment in BC = -12 x

Strain Energy due to BM for the whole structure is

24

120 mm

240 mm12 mm

5m

4m

12kN

A

B C

Step 5: Comparison

Total Strain Energy = (Ui)p + (Ui)V+ (Ui)M

Total Strain Energy =328.94 +1315.78 +767.34 x 103

= 768.98 x 103 N-mm

Strain Energy due to axial force, shear force and bending moment are 0.043%, 0.17% &

99.78 % of the total strain energy.

3. Show that the flexural strain energy of a prismatic bar of length L bent into a complete

circle by means of end couples is

Solution:

Circumference = 2p R =L or

From bending theory

25

M M

L

R

4. Calculate the strain energy in a truss shown in Fig. if all members are of same cross-

sectional area equal to 0.01m2 and E=200GPa

Solution: To calculate strain energy of the truss, first the member forces due to external force

is required to be computed. Method of joint has been used here to compute member forces.

Member forces in the members AB, BC, BD, BE, CE and DE are only computed as the truss

is symmetrical about centre vertical axis.

Step1: Member Forces:

i) Joint A: From triangle ACB, the angle = tan-1(3/4)=36052’

The forces acting at the joint is shown in Fig. and the forces in members are computed

considering equilibrium condition at joint A

SFy=0; FABsin+30=0; FAB=-50kN (Compression)

SFx=0; FABcos + FAC=0; FAC=40kN (Tension)

26

FAC

FAB

RA= 30 kN

ii) Joint C: The forces acting at the joint is shown in Fig. and the forces in

members are computed considering equilibrium condition at joint C

SFy=0; FCB=0;

SFx=0; FCE - 40=0; FCE=40kN(Tension)

iii) Joint B: The forces acting at the joint is shown in Fig. and the forces in

members are computed considering equilibrium condition at joint B

SFy=0; -30+50 sin-FBEsin=0; FBE=0

SFx=0; 50 cos - FBD=0; FBD=-40kN (Compression)

iv) Joint D: The forces acting at the joint is shown in Fig. and the forces in

members are computed considering equilibrium condition at joint D

SFy=0; FDE=0; SFx=0; FDF + 40=0; FDF=-40kN (Comprn.)

Forces in all the members are shown in Fig.

27

FCE

FAC=40kN

FCB

FBD

30kN

FAB= 50kN FCB=0FBE

FDF

FBD=40 FDE

Step 2: Strain Energy

A= 0.01m2; E=2*105 N/mm2 = 2*108 kN/m2, AE = 2*106 kN

(Ui)p=15.83*10-3 kN-m

5. Determine the maximum slope and maximum deflection in a cantilever beam of span L

subjected to point load W at its free end by using strain energy method. EI is constant

Solution:

i) Maximum Deflection

BM at 1-1 Mx= -Wx

From 2nd theorem of Castigliaino

ii) Maximum Slope

28

L

x

1

1 W

A B

Maximum slope occurs at B, Virtual moment M’ is applied at B

Bending moment at 1-1 is Mx = -Wx – M’

From 2nd theorem of Castiglano

Substituting M’=0

6. Calculate max slope and max deflection of a simply supported beam carrying udl of

intensity w per unit length throughout its length by using Castigliano’s Theorem

29

L

x

1

1 W

A M’B

L

w

i) Maximum Slope:

Maximum slope occurs at support. A virtual moment M’ is applied at A.

Reactions:

BM at 1-1

Put M’=0

ii) Maximum Deflection:

Maximum Deflection occurs at mid span. A virtual downward load W’ is applied at

mid-span.

30

L

w

1

1

x

M’

RA RB

Reactions:

BM at 1-1

Put W’=0

31

L

w

1

1

x

M’

RA RB

L/2

CONJUGATE BEAM METHOD

This is another elegant method for computing deflections and slopes in beams. The principle of the method lies in calculating BM and SF in an imaginary beam called as Conjugate Beam which is loaded with M/EI diagram obtained for real beam. Conjugate Beam is nothing but an imaginary beam which is of the same span as the real beam carrying M/EI diagram of real beam as the load. The SF and BM at any section in the conjugate beam will represent the rotation and deflection at that section in the real beam. Following are the concepts to be used while preparing the Conjugate beam.

It is of the same span as the real beam. The support conditions of Conjugate beam are decided as follows:

Some examples of real and conjugate equivalents are shown.

32

Problem 1 : For the Cantilever beam shown in figure, compute deflection and rotation at (i) the free end (ii) under the load

Conjugate Beam:

By taking a section @ C´ and considering FBD of LHP,

BM @ C´=

Similarly by taking a section at A’ and considering FBD of LHP;

33

SF @ A’ =

BM @ A’ =

SF @ a section in Conjugate Beam gives rotation at the same section in Real Beam

BM @ a section in Conjugate Beam gives deflection at the same section in Real Beam

Therefore, Rotation @ C = ( )

Deflection @ C=

Rotation @ A = ( )

Deflection @ A =

Problem 2: For the beam shown in figure, compute deflections under the loaded points. Also compute the maximum deflection. Compute, also the slopes at supports.

34

Note that the given beam is symmetrical. Hence, all the diagrams for this beam should be symmetrical. Thus the reactions are equal & maximum deflection occurs at the mid span. The bending moment for the beam is as shown above. The conjugate beam is formed and it is shown above. For the conjugate beam:

=

=

To compute δC : A section at C’ is placed on conjugate beam. Then considering FBD of LHP;

35

+ BM @ C’=

=

δD = δC (Symmetry)

To compute δE:

A section @ E’ is placed on conjugate beam. Then considering FBD of LHP;

+ BM @ E’=

i.e δE =

θA = ( ) θB = ( )

36

Problem 3: Compute deflection and slope at the loaded point for the beam shown in figure. Given E = 210 Gpa and I = 120 x 106mm4. Also calculate slopes at A and B.

Note that the reactions are equal. The BMD is as shown above.

To Compute reactions in Conjugate Beam:

37

;

SF and BM at C’ is obtained by placing a section at C’ in the conjugate beam.

SF @ C’ =

+ BM @ C’ =

Given E = 210 x 109 N/m2

= 210 x 106 kN/m2

I = 120 x 106 mm4

= 120 x 106 (10-3 m)4

= 120 x 106 (10-12) = 120 x 10-6 m4; EI = 210 x 106 (120 x 10-6) = 25200 kNm-2

Rotation @ C = = 1.19 x 10-3 Radians ( )

Deflection @ C = = 0.0107 m

= 10.71 mm ( ) θA = 4.76 X 10-3 Radians

θB = 5.95 X 10-3 Radians:

38

Problem 4: Compute slopes at supports and deflections under loaded points for the beam shown in figure.

39

To compute reactions and BM in real beam:

+

BM at (1) – (1) = 66.67 xAt x = 0; BM at A = 0, At x = 3m, BM at C = 200 kNm

BM at (2) – (2) = 66.67 x – 50 (x-3) = 16.67 x + 150 At x = 3m; BM at C = 200 kNm, At x = 6m, BM at D = 250 kNm

BM at (3) – (3) is computed by taking FBD of RHP. Then BM at (3)-(3) = 83.33 x (x is measured from B)At x = 0, BM at B = 0, At x = 3m, BM at D = 250 kNm

To compute reactions in conjugate beam:

+

i.e

( ) ( )

To Compute δC :

40

A Section at C’ is chosen in the conjugate beam:

+ BM at C’ =

=

δC =

To compute δD:

Section at D’ is chosen and FBD of RHP is considered.

+ BM at D’ =

=

41

Problem 5: Compute to the slope and deflection at the free end for the beam shown in figure.

42

The Bending moment for the real beam is as shown in the figure. The conjugate beam also is as shown.

Section at A’ in the conjugate beam gives

SF @ A’ =

=

=

θA = ( )

BM @ A’ =

=

δA =

43

STRUCTURAL ANALYSIS-I (CV42)CHAPTER 1

1.1 Introduction

A structure can be defined as a body which can resist the applied loads without appreciable deformations. Civil engineering structures are created to serve some specific functions like human habitation, transportation, bridges, storage etc. in a safe and economical way. A structure is an assemblage of individual elements like pin ended elements (truss elements), beam element, column, shear wall slab cable or arch. Structural engineering is concerned with the planning, designing and the construction of structures. Structural analysis involves the determination of the forces and displacements of the structures or components of a structure. Design process involves the selection and or detailing of the components that make up the structural system. The cyclic process of analysis and design is illustrated in the flow chart given in Fig.1.1

1.2 Forms of Structures

Engineering structure is an assemblage of individual members. Assemblage of members forming a frame to support the forces acting is called framed structure. Assemblage of

44

No

Preliminary Design

Structural Analysis

Compute Stresses and deformations

Is stresses and Deformation within limits?

Final Design

Revise sections

Yes

Fig.1.1 Cyclic Process

continuous members like flat plates, curved members etc., are called continuous system. Buildings, bridges, transmission towers, space crafts, aircrafts etc., are idealized as framed structures. Shells, domes, plates, retaining walls, dams, cooling towers etc., are idealized as continuous systems.

A frame work is the skeleton of the complete structure. This frame work supports all intended loads safely and economically. Continuous system structures transfers loads through the in-plane or membrane action to the boundaries. Fig.1.2 and Fig.1.3 illustrates framed structure and continuous system.

Actual structure is generally converted to simple single line structures and this process is called idealization of structures. The idealization consists of identifying the members of structure as well known individual structural elements. This process requires considerable experience and judgment. Structural analyst may be required to idealize the structure as one or more of following

45

Fig.1.2 Framed Structure

i) Real structure ii) A physical model iii) A mathematical model

In a real structure the response of the structure is studied under the actual forces like gravity loads and lateral loads. The load test is performed using elaborate loading equipment. Strains and deformations of structural elements under loads are measured. This is very expensive and time consuming procedure, hence performed in only exceptional cases. Load testing carried out on a slab system is illustrated in Fig. 1.4

47

Fig.1.3 Continuous system

Physical models which are scaled down and made up of plastic, metal or other suitable materials are used to study the response of structure under loading. These models are tested in laboratories. This study requires special techniques and is expensive. This study is carried out under compelling circumstances. Examples includes laboratory testing of small

48

Fig.1.4 Load test on slab

scale building frames, shake table test of bridges and building, photo elastic testing of a dam model, wind tunnel testing of small scale models of high rise buildings, towers or chimneys. Fig. 1.5 shows testing of a slab model under uniformly distributed load.

A mathematical model is the development of mathematical equations. These equations describe the structure loads and connections. Equations are then solved using suitable algorithm. These solutions generally require electronic computers. The process of mathematical modeling is shown in block diagram given in Fig.1.6

A structure is generally idealized as either two dimensional structure (Plane frame) or as three dimensional structure (Space frame). The selection of idealization depends on the desire and experience of structural engineer. A two dimensional structure or a plane frame structure is that which has all members and forces are in one plane. Space frame or a three dimensional structure has members and forces in different planes. All structures in practice are three dimensional structures. However, analyst finds more convenient to analyze a plane structure rather than a space structure. Fig. 1.7 shows two dimensional and three dimensional structures used in mathematical modeling.

49

Fig.1.5 Testing of a Model

Fig.1.5 b. Three Dime

A mathematical modeling should also idealize the supports of the structure. Roller supports or simple supports, pinned supports or hinged supports and fixed supports are generally assumed type of supports in practice. Fig. 1.8 shows different types of supports. In a roller support the reaction is perpendicular to the surface of the roller. Two components of reaction are developed in hinged support and three reaction component, one moment and two forces parallel to horizontal and vertical axis are developed in fixed support.

50

Actual Structure

Idealize Structure

Idealize Loads

Development of Equations

Response of str.

Interpretation of Results

Fig.1.6 Block Diagram of Mathematical modeling

Fig.1.7 a. Two Dimensional Structures

Fig.1.7 b. Three Dimensional Structures

Mathematical modeling requires to consider the loads acting on structure. Determination of the loads acting on the structure is often difficult task. Minimum loading guidance exists in codes and standards. Bureau of Indian standards, Indian road congress and Indian railways have published loading standards for building, for roads and for railway bridges respectively. Loads are generally modeled as concentrated point loads, line loads or surface loads. Loads are divided into two groups viz., dead loads and live loads. Dead loads are the weight of structural members, where as live loads are the forces that are not fixed. Snow loads, Wind loads, Occupancy loads, Moving vehicular loads, Earth quake loads, Hydrostatic pressure, earth pressure , temperature and fabrication errors are the live loads. All the live loads may not act on the structure simultaneously. Judgment of analyst on this matter is essential to avoid high loads.

51

Fig.1.8 Typical Support Conditionsa) Fixed support, b) Hinged support, c) simple support

1.3 Conditions of Equilibrium and Static Indeterminacy

A body is said to be under static equilibrium, when it continues to be under rest after application of loads. During motion, the equilibrium condition is called dynamic equilibrium. In two dimensional system, a body is in equilibrium when it satisfies following equation.SFx=0 ; SFy=0 ; SMo=0 ---1.1

To use the equation 1.1, the force components along x and y axes are considered. In three dimensional system equilibrium equations of equilibrium areSFx=0 ; SFy=0 ; SFz=0;

SMx=0 ; SMy=0 ; SMz=0; ----1.2

To use the equations of equilibrium (1.1 or 1.2), a free body diagram of the structure as a whole or of any part of the structure is drawn. Known forces and unknown reactions with assumed direction is shown on the sketch while drawing free body diagram. Unknown forces are computed using either equation 1.1 or 1.2

Before analyzing a structure, the analyst must ascertain whether the reactions can be computed using equations of equilibrium alone. If all unknown reactions can be uniquely determined from the simultaneous solution of the equations of static equilibrium, the reactions of the structure are referred to as statically determinate. If they cannot be determined using equations of equilibrium alone then such structures are called statically indeterminate structures. If the number of unknown reactions are less than the number of equations of equilibrium then the structure is statically unstable.

The degree of indeterminacy is always defined as the difference between the number of unknown forces and the number of equilibrium equations available to solve for the unknowns. These extra forces are called redundants. Indeterminacy with respect external forces and reactions are called externally indeterminate and that with respect to internal forces are called internally indeterminate.

A general procedure for determining the degree of indeterminacy of two-dimensional structures are given below:

NUK= Number of unknown forcesNEQ= Number of equations availableIND= Degree of indeterminacyIND= NUK - NEQ

52

Indeterminacy of Planar Frames

For entire structure to be in equilibrium, each member and each joint must be in equilibrium (Fig. 1.9)

NEQ = 3NM+3NJ

NUK= 6NM+NR

IND= NUK – NEQ = (6NM+NR)-(3NM+3NJ)

IND= 3NM+NR-3NJ ----- 1.3

Degree of Indeterminacy is reduced due to introduction of internal hinge

NC= Number of additional conditions

NEQ = 3NM+3NJ+NC

NUK= 6NM+NR

IND= NUK-NEQ = 3NM+NR-3NJ-NC ------------1.3a

Indeterminacy of Planar Trusses

Members carry only axial forces

NEQ = 2NJ

NUK= NM+NR

IND= NUK – NEQ

IND= NM+NR-2NJ ----- 1.4

53

Three independent reaction components

Two independent reaction components

Fig. 1.9 Free body diagram of Members and Joints

Indeterminacy of 3D FRAMES

A member or a joint has to satisfy 6 equations of equilibrium

NEQ = 6NM + 6NJ-NC

NUK= 12NM+NR

IND= NUK – NEQ

IND= 6NM+NR-6NJ-NC ----- 1.5

Indeterminacy of 3D Trusses

A joint has to satisfy 3 equations of equilibrium

NEQ = 3NJ

NUK= NM+NR

IND= NUK – NEQ

IND= NM+NR-3NJ ----- 1.6

Stable Structure:

Another condition that leads to a singular set of equations arises when the body or structure is improperly restrained against motion. In some instances, there may be an adequate number of support constraints, but their arrangement may be such that they cannot resist motion due to applied load. Such situation leads to instability of structure. A structure may be considered as externally stable and internally stable.

Externally Stable:

Supports prevents large displacements

No. of reactions ≥ No. of equations

Internally Stable:

Geometry of the structure does not change appreciably

For a 2D truss NM ≥ 2Nj -3 (NR ≥ 3)

For a 3D truss NM ≥ 3Nj -6 (NR ≥ 3)

54

f)

Examples:

1.1 Determine Degrees of Statical indeterminacy and classify the structures

55

a)

b)

c)

d)

e)

f)

NM=2; NJ=3; NR =4; NC=0IND=3NM+NR-3NJ-NCIND=3 x 2 + 4 – 3 x 3 -0 = 1INDETERMINATE

NM=3; NJ=4; NR =5; NC=2IND=3NM+NR-3NJ-NCIND=3 x 3 + 5 – 3 x 4 -2 = 0DETERMINATE



NM=1; NJ=2; NR =6; NC=2IND=3NM+NR-3NJ-NCIND=3 x 1 + 6 – 3 x 2 -2 = 1 INDETERMINATE


56

R2

R1

R5

R4

g)


Each support has 6 reactions

NM=18; NJ=15; NR =18; NC=0IND=6NM+NR-6NJ-NCIND=6 x 18 + 18 – 6 x 15 = 36 INDETERMINATE

Each support has 3 reactions


h)

i)

Truss NM=2; NJ=3; NR =4; IND=NM+NR-2NJIND= 2 + 4 – 2 x 3 = 0 DETERMINATE

j)

1.4 Degree of freedom or Kinematic IndeterminacyMembers of structure deform due to external loads. The minimum number of parameters required to uniquely describe the deformed shape of structure is called “Degree of Freedom”. Displacements and rotations at various points in structure are the parameters considered in describing the deformed shape of a structure. In framed structure the deformation at joints is first computed and then shape of deformed structure. Deformation at intermediate points on the structure is expressed in terms of end deformations. At supports the deformations corresponding to a reaction is zero. For example hinged support of a two dimensional system permits only rotation and translation along x and y directions are zero. Degree of freedom of a structure is expressed as a number equal to number of free displacements at all joints. For a two dimensional structure each rigid joint has three displacements as shown in Fig. 1.10

In case of three dimensional structure each rigid joint has six displacement.

• Expression for degrees of freedom1. 2D Frames: NDOF = 3NJ – NR NR ³32. 3D Frames: NDOF = 6NJ – NR NR ³63. 2D Trusses: NDOF= 2NJ – NR NR ³34. 3D Trusses: NDOF = 3NJ – NR NR ³6

Where, NDOF is the number of degrees of freedom

In 2D analysis of frames some times axial deformation is ignored. Then NAC=No. of axial condition is deducted from NDOF

57

Truss NM=11; NJ=6; NR =4; IND=NM+NR-2NJIND= 11 + 4– 2 x 6 = 3 INDETERMINATE

Truss NM=14; NJ=9; NR =4; IND=NM+NR-2NJIND= 14+ 4 – 2 x 9 = 0 DETERMINATE

k)

l)

Fig 1.10

Examples:

1.2 Determine Degrees of Kinermatic Indeterminacy of the structures given below

58

ExtensibleNJ=2; NR =3; NDOF=3NJ-NRNDOF=3 x 2 – 3= 3 (1, 2, 2)

InextensibleNJ=2; NR =3; NAC=1NDOF=3NJ-NR-NACNDOF=3 x 2 – 3-1= 2 (1, 2)

ExtensibleNJ=4; NR =5; NDOF=3NJ-NRNDOF=3 x 4 – 5= 7 (1, 21, 23 3 ,y2,e1,e2)

InextensibleNJ=4; NR =5; NAC=2NDOF=3NJ-NR-NACNDOF=3 x 4 – 5-2= 5(1, 21, 23 , 3 y2)

a)

b)

c)

59

ExtensibleNJ=4; NR =6; NDOF=3NJ-NRNDOF=3 x 4 – 6= 6 (2, 3 , , e1,e2, e3)

InextensibleNJ=4; NR =6; NAC=3NDOF=3NJ-NR-NACNDOF=3 x 4 – 6-3= 3(2, 3, )

d)

e)

f)

60

ExtensibleNJ=4; NR =5; NDOF=3NJ-NRNDOF=3 x 4 – 5+1= 8 (21, 23, 3 , 2, 3, e1,e2, e3)

InextensibleNJ=4; NR =5; NAC=3NDOF=3NJ-NR-NACNDOF=3 x 4 – 5-3= 4 (21, 23, 3 , 2=3=)

ExtensibleNJ=5; NR =6; NDOF=3NJ-NRNDOF=3 x 5 – 6 = 9 (2, 3, 4 , x2, x3, y3, x4 e1, e4)

InextensibleNJ=5; NR =6; NAC=3NDOF=3NJ-NR-NACNDOF=3 x 5 – 6 - 3= 6 (2, 3, 4 , x2= x4=1, x3, y3)

g)

1.5 Linear and Non Linear Structures

Structural frameworks are commonly made of wood, concrete or steel. Each of them has different material properties that must be accounted for in the analysis and design. The modulus of elasticity E of each material must be known for any displacement computation. Typical stress-strain curve for these materials is shown in Fig.1.11. The structure in which the stresses developed is within the elastic limit, and then the structure is called Linear Structure. If the stress developed is in the plastic region, then the

61

NJ=6; NR =3; NDOF=2NJ-NRNDOF=2 x 6 – 3 = 9

A Truss

NJ=6; NR =4; NDOF=2NJ-NRNDOF=2 x 6 – 4 = 8

A Truss

structure is said to Non-Linear Structure. In addition to material nonlinearities, some structures may behave in a nonlinear fashion due to change in the shape of the overall structure. This requires that the structure displace an amount significant enough to affect the equilibrium relations for the structure. When this occurs the structure is said to be Geometrically nonlinear. Cable structures are susceptible to this type of nonlinearity. A cantilever structure shown in Fig. 1.2 has geometrical nonlinearity

62

Fig.1.11 Stress-Strain Graph

63

Fig.1.12 Geometric Nonlinearity

Exercise Problems

P1.1 Determine Degrees of indeterminacy and classify the structures

P1.2 Determine Degrees of Kinematic indeterminacy

Reference Books

64

1. C.S.Reddy, Basic Structural Analysis, Tata Mc Graw Hill, 2nd Edition, New Delhi, 1996.

2. Ashok.K.Jain, Elementary Structural Analysis, Newchand and brothers, Roorkee, India, 1990

3. L.S.Negi and R.S.Jangid, Structural Analysis, Tata Mc Graw Hill, 2nd Edition, New Delhi, 1997.

4. G.S.Pandit, S.P.Gupta and R.Gupta, Theory of Structures, Vol-1, Tata Mc Graw Hill, 2nd Edition, New Delhi, 1999.

5. Jeffrey P. Laible, Structural Analysis, Holt-Saunders International Editions, 1985

THREE HINGED ARCHES

An arch is a curved beam in which horizontal movement at the support is wholly or partially prevented. Hence there will be horizontal thrust induced at the supports. The shape of an arch doesn’t change with loading and therefore some bending may occur.

Types of Arches On the basis of material used arches may be classified into and steel arches,

reinforced concrete arches, masonry arches etc.,

On the basis of structural behavior arches are classified as :

Three hinged arches:- Hinged at the supports and the crown.

65

Types of Arch

Material used Structural behavior

Loaded areaLoaded area

Hinged at the support

Hinged at the crown

Springing

Rise

Span

Two hinged arches:- Hinged only at the support.

Fixed arches:- The supports are fixed.

A 3-hinged arch is a statically determinate structure. A 2-hinged arch is an indeterminate structure of degree of indeterminancy equal to 1. A fixed arch is a statically indeterminate structure. The degree of indeterminancy is 3.

Depending upon the type of space between the loaded area and the rib arches can be classified as open arch or closed arch (solid arch).

66

Hinges at the support

Rib of the archRise

Span

Filled Spandrel Open Arch

Braced Spandrel Solid Arch

Comparison between an arch and a beam

-H x y

Owing to its geometrical shape and proper supports, an arch supports loading with less bending moment than a corresponding straight beam. However in case of arches there will be horizontal reactions and axial thrust.

Analysis of 3-hinged arches It is the process of determining external reactions at the support and internal quantities

such as normal thrust, shear and bending moment at any section in the arch.

Procedure to find reactions at the supports Step 1. Sketch the arch with the loads and reactions at the support.Apply equilibrium conditions namely Apply the condition that BM about the hinge at the crown is zero (Moment of all the forces either to the left or to the right of the crown). Solve for unknown quantities.

P1: A 3-hinged arch has a span of 30m and a rise of 10m. The arch carries UDL of 0.6 kN/m on the left half of the span. It also carries 2 concentrated loads of 1.6 kN and 1 kN at 5 m and 10 m from the ‘rt’ end. Determine the reactions at the support. (sketch not given).

67

WE

CE

BE

HB = HAE

HA = H

y RISE = h

SPANba

VA = VB =

VA =

VB =

H1 H1

Wa b

x

------ (1)

To find vertical reaction.

------ (2)

68

B HB = 4.275

0.6 kN/mC

5 m 5 m

h = 10m

L = 30mVA = 7.35 VB = 4.25

HB = 4.275 A

1 kN 1.6 kN

To find horizontal reaction.

OR

To find total reaction

69

RA

A

A HA = 4.275 kN

VA = 7.35 kN RB

A HB = 4.275 kN

VB = 4.25 kN

2) A 3-hinged parabolic arch of span 50m and rise 15m carries a load of 10kN at quarter span as shown in figure. Calculate total reaction at the hinges.


------ (1)

To find Horizontal reaction

To find total reaction.

70

B HB

10 kNC

15 m

50 mVA VB

HA A

12.5 m

RA

A

A HA = 4.17

VA = 7.5RB

HB = 4.17

VB = 4.25

3) Determine the reaction components at supports A and B for 3-hinged arch shown in fig.

To find Horizontal reaction

------ (1)


------ (2)

71

B HB

10 kN/m C

2.5 m

10 mVA

VB

HA A

2 .4 m

180 kN

8 m 6 m

------ (3)

------ (4)

Adding 2 and 3

72

Bending moment diagram for a 3-hinged arch

We know that for an arch, bending moment at any point is equal to beam BM-Hy (Refer comparison between arch and beam). Hy is called H-Moment. It varies with respect to Y. Therefore the shape of BM due to Hy should be the shape of the arch. Therefore to draw the BMD for an arch, draw the BMD for the beam over that superimpose the H-moment diagram as shown in fig.

73

B

W C

h

a b

A

L

L

Wab

+

–

+–

Shape – same as archBEAM B.M.

L

Wab H – Moment diagram

Hh

Hh

Normal thrust and radial shear in an arch

Total force acting along the normal is called normal thrust and total force acting along the radial direction is called radial shear. For the case shown in fig normal thrust

= + HA Cos + VA Cos (90 - )= HA Cos + VA Sin

(Treat the force as +ve if it is acting towards the arch and -ve if it is away from the arch).Radial shear = + HA Sin -VA Sin (90 - )

= HA Sin + VA Cos

(Treat force up the radial direction +ve and down the radial direction as -ve).

Note: 1) To determine normal trust and tangential shear at any point cut the arch into 2 parts. Consider any 1 part. Determine net horizontal and vertical force on to the section. Using these forces calculate normal thrust and tangential shear.

2. Parabolic arch: If the shape of the arch is parabolic then it is called parabolic arch.

If A is the origin then the equation of the parabola is given by y = cx [L – x] where C is a constant.

We have at X = y = h

h = C

74

A

Radial

Normal

VERTICAL

NORMAL

HORIZONTAL

VA

HA

M

AHA

VA RADIAL

Normal Radial

90 -

B

C

h

A

Horizontal

y

x2

Normal

Equation of parabola is

is given by the following equation.

1. A UDL of 4kN/m covers left half span of 3-hinged parabolic arch of span 36m and central rise 8m. Determine the horizontal thrust also find (i) BM (ii) Shear force (iii) Normal thrust (iv) Radial shear at the loaded quarter point. Sketch BMD.

------ (1)

------ (2)

75

4 kN/mC

h = 8 m

36 mVA VB

HA A

B HB

BM at M =

- 40.5 x 6 + 54 x 9

- 4 x 9 x 4.5

= 81 kN.m

Shear force at M = + 54 – 4 x 9 = 18 kN (only vertical forces)

tan

=

Normal thrust = N = + 40.5 Cos 23.96 + 18 Cos 66.04 = 44.32 kN

S = 40.5 Sin 23.96 – 18 Sin 66.04S = - 0.0019 0

76

VERTICAL

NORMAL

HORIZONTAL

66.04

40.5 kN = 23.96

M

A

40.5 kN RADIAL

4 kN/m

54 kN9 m

y = 618 kN

54 x 13.5 – 4 x 13.5

A symmetrical 3-hinged parabolic arch has a span of 20m. It carries UDL of intensity 10 kNm over the entire span and 2 point loads of 40 kN each at 2m and 5m from left support. Compute the reactions. Also find BM, radial shear and normal thrust at a section 4m from left end take central rise as 4m.

77

13.5A B

36 m 54 kN 18 kN

18 – x

x = 13.5 18 kN 18 kN

324 kN/m

Beam BM

364.5 kN/m kN/m

+

4 kN/m

54 kN

+–

364.5 kNm324 kNm

A C B13.5 m

------ (1)

------ (2)

78

C

4 m

20 m

40 kN

M

2 m 3m

10 kN/m40 kN

BM at M = - 160 x 2.56 + 166 x 4 – 40 x 2 - (10 x 4)2 = + 94.4 kNm

tan

=

Normal thrust = N = + 160 Cos 25.64

+ 86 Cos 64.36 = 181.46 kN

S = 160 Sin 25.64 - 86 x Sin 64.36S = - 8.29 kN

79

VERTICAL

NORMAL

HORIZONTAL

64.35

160 kN = 25.64

M

REDIAL

10 kN/m

160 kN4 m

y = 2.56

86 kN

40 kN

2 m166 kN

Segmental arch

A segmental arch is a part of circular curve. For such arches y = is not

applicable since the equation is applicable only for parabolic arches. Similarly equation for will be different.

To develop necessary equations for 3-hinged segmental arch

Relationship between R, L and h:From

80

B

C

h(90-)x

A

O

RR

Origin

EM

90

x

y

L/2 D L/2

1) A 3-hinged segmental arch has a span of 50m and a rise of 8m. A 100 kN load is acting at a point 15 m from the right support. Find horizontal thrust at the supports (ii) BM, Normal thrust and radial shear at a section 15 m from the left support.

Reactions:

------ (1)

------ (2)

81

B HB 93.75 kN

C

Origin

25 mVA = 30 kN VB = 70 kN

HA = 93.75 kN A

100 kN

15 mZ

25 m

15 m

NORMAL30 kN 76.97

93.75 kN

= 13.43

93.75 kNREDIAL

30 kN15 m

y

Z

* In case of segmental arch A is not origin

B.Mz = 30 x 15 – 93.75 x 6.822 = - 189.562 kNm

N = 93.75 Cos 13.43 + 30 Cos 76.57 N = 98.15 kNS = 93.75 Sin 13.43 – 30 Sin 76.57 = - 7.41 kN

CABLES AND SUSPENSION BRIDGES

Cables are used to support loads over long spans such as suspension bridges, roof of large open buildings etc. The only force in the cable is direct tension. Since the cables are flexible they carry zero B.M.

Analysis of cables Analysis of cable involves determination of reactions at the support and tension over

different parts of the cable.

To determine the reactions at the support and tension equilibrium conditions are used. In addition to that BM about any point of the cable can be equated to zero.

P1: Determine the reactions components and tension in different parts of the cable shown in figure. Also find the sag at D and E.

82

------ (1)

------ (2)

(About the point where sag is given)

Point A:

83

A 20 m 20 m 20 m 20 m B1

T110 m yD = 14.55

yE = 11.82

HA = 220 kN HB = 110 kN

VA = 55 kN VB = 65 kNT4

T2

T3

30 kN40 kN

50 kN

C

DE

2

= 26.56

55 kN

110 kN

T1

tan

To find YE:- We have

To Find YD:-

84

Point D:

tan

tan

85

12.81 =

40 kN

110 kN

3 = 7.77

T2 T3

Part B:

tan

Note: It can be observed that more the inclination or the slope more is the tension. Near the supports slope will be more and hence tension will also be more.

2) A Chord Supported at its ends 40 m apart carries loads of 200 kN, 100 kN and 120 kN at distances 10 m, 20 m and 30 m from the left end. If the point on the chord where 100 kN load is supported is 13 m below the level of the end supports. Determine

(a) Reactions at the support.(b) Tension in different part.(c) Length of the chord.

------ (1)

86

110 kN

65 kN

30.18

T4

A 10 m 10 m 10 m 10 m E

T1 yB 13 myD

HA = 200 HB = 200

VA = 230 VE = 190 T4

T2 T3

200 kN100 kN

120 kN

B

C

D

1

2

3

3

------ (2)

(Always take BM about the point for which sag is given)

87

Point A:

Point B:

tan 2 =

2 = 8.53Point C:

88

49°

230 kN

200 kN

T1

2

200 kN

204.85 kN

T2

49°

tan

89

100 kN

202.22 kNT3

8.53 19.3

Point E:

tan

Length of the Chord =

Cos

Cos

Cos

Cos

Total length of Chord = 49.76m

3) Determine reactions at supports and tension indifferent parts of the cable shown in figure.

90

200 kN

190 kN

4

T4

------ (1)

------ (2)

------ (3)

------ (4)

(3) – (4) gives

Point A:

91

HA A 25 m 25 m 25 m 25 m 1

3

2

T4T3

T2

T1

VA

VB

40 kN

50 kN

30 kN

C

D

E

8.757.5 m determined

10 m (given)

8.95

26.25

15 m

2.5

92

1 = 19.29°

89.38 kN

212.42 kN

T1

30 kN

225.05 kN

T2

19.292 = 5.71

Point C:

4) A cable is used to support loads 40 kN and 60 kN across a span of 45 m as shown in figure. The length of the cable is 46.5 m. Determine tension in various segments.

------ (1)

------ (2)

------ (3)

93

A 15 m 15 m 15 m D

T1

yB = 4.4 m yC = 5.025

HA = 159.2 kN HD

VA = 46.67 = 0 VD = 53.33T2

T3

40 kN

60 kN

B

D

We have = 46.5m

We have

Point A:

94

Point C:

95

1

46.67 kN

159.2 kN

T1

60 kN

T2 T3

2 3

General Equation of a cable or Differential Equation.

General shape of a cable depends on nature of loading, location of loads, type of supports etc. The equilibrium of a part of a cable shall be considered to obtain equation for cable when the cable is subjected to all over UDL.

Let us consider the equilibrium of a small length ‘ds’ of the cable shown in figure. Let the cable be subjected to UDL of intensity W over horizontal span figure shows tension and horizontal, vertical reactions in the part of the cable we have.

H = T Cos V = T Sin

V = H tan

Let us consider the equilibrium of the part of the cable shown in fig. We have.

(V + dv) – V – W x dx = 0 dv = wdx

96

C

D

W kN/m

VD = v + dv T + d

HD = H

V = VC = TSin T

C

H = HC = TCos ds

+ 2

dy

dx

To derive equations for cable profile and tension in the cable when it is supported at the same level and subjected to horizontal UDL.

Let us consider a cable of span L and max sag H subjected to UDL of intensity ‘W’ as

shown in fig. From general equation we have

Integrating with respect to x we have

We have at X = 0

Substituting 0 = 0 + C1

C1 = 0

Integrating with respect to x we get

We have at X=0 Y=0

=0

At X= Y = h

Substituting

We have

If A is considered as the origin then

97

W

yLA B

h

x

To fin the tension at any point on the cable: Tension at any point T = H Sec = H

= H

T = H

To derive an expression for cable profile when it is subjected to horizontal UDL and supports are at different levels:General equation for cable profile is

Let us consider each part separately we have

At x = -L1, Y = b

------ (1)

At X = L2, Y= a + b

------ (2)

Add (1) on either side

98

w

a

L1

B

b

L2y

x

L1

O

Substituting in (2)

Tension at any point is given by T = H Sec T = HT = H

= H

T = H

Y =

To derive an expression for length of the parabolic cable profile when the supports are at the same level

We have ds =

Ds = dx

When the supports are at the same level we have

(taking origin at the center)

99

dx

dy

ds

ds = dx

Total length of the cable is given by

LC =

= 2

= 2

LC = 2 (expanding by binomial theorem)

LC = 2

LC =

To derive an expression for length of the cable profile when the supports are at different levels

LC = Length of cableBetween A & C + Length of cable between C&B

LC = Length of cable of span 2L1 + length of cable of span 2L2

But LC = L +

LC =

1) A cable suspends across a gap of 250m and carries UDL of 5kN/m horizontally calculate the maximum tensions if the maximum sag is 1/25

th of the span. Also calculate the sag at 50m from left end.

100

aL1

B

b

L2

c

A

We have

T = H

Take origin as left end

Tension is maximum at the supports i.e., at X = 0 from A

We have [x measured from A]

2) Determine the length of the cable and max tension developed if the cable supports a load of 2kN/m on a horizontal span of 300m. The maximum sag is 25m

101

5 kN/m

250 mA B

h = = 10 m

2 kN/m

300 mA B

25

[X from A]

Tension is maximum at the supports i.e., at x = 0

Length of cable LC =

3) Determine the maximum span for a mild steel cable between supports at the same level if the central dip. is 1/10th of the span and permissible stress in steel is 150 N/mm2. Steel weighs 78.6 kN/m3. Assume the cable to hang in a parabola.

Here the weight of the cable itself is acting as UDL on the cable. We have SP weight

=

Weight = Specific Weight x Volume = Specific Weight x Area x Length

= Specific Weight x Area

We have

102

L

A

Tension is maximum at X = 0

Bridges supported by cables

103

Suspension Bridge

Anchor cable

Anchoring of cables There are 2 methods by which suspension cable can be anchored:

1) Continuous cable or pulley type anchoring.2) Non- Continuous cable or saddle type anchoring.

Continuous cable or pulley type anchoringIn this method suspension cable itself passes over roller or guide pulley on the top of

the tower or abutment and then anchored. The tension remains same in the suspension cable and anchor cable at the supports.

104

Fan Type Cable stayed Bridge

Harp type Cable stayed Bridge

Pulley Anchor cable

Suspension cable

Abutment

A TA

TA

is the inclination of the suspension cable with the horizontal. Net horizontal force on tower HT = TA Cos ~ TA Cos

Where ht is the height of the tower.

2) Saddle type anchoring or Non-continuous cable In this method of anchoring suspension cable are attached to saddles mounted on

rollers on the top of the tower as result in suspension cable and anchor cable will be differed. However horizontal components of tension will be equal.

105

ht

Tower

B TATA

Anchor cable

Suspension cable

Abutment

B TATa

1) A cable of span 150m and dip 15m carries a load of 6 kN/m on horizontal span. Find the maximum tension for the cable at the supports. Find the forces transmitted to the supported pier if (a) Cable is passed over smooth rollers or pulleys over the pier. (b) Cable is clamped to saddle with smooth rollers resting on the top of the pier.

For each of the above case anchor cable is 30 to horizontal. If the supporting pier is 20m tall. Determine the maximum BM on the pier.

Case 1: Cable over smooth pulley

106

6 kN/m

150 mA B

15 mVB

HBHA

VA

HT = 1211.66 Cos 21.8 1211.66 Cos 300

= 75.73kNVT = 1211.66 Sin 21.8 + 1211.66 Sin 300 VT = 1055.77kNM = HT X ht = 75.73 X 20 = 1513.4kNm

Case 2 : Cable clamped to saddle

Here TA Cos 1211.66 Cos 21.8 = Ta Cos 30Ta 1299.05kN

In this case HT = 0M =

VT = 1099.5 kN

3) A Suspension cable is suspended from 2 pier A and B 200m apart, B being 5m below A the cable carries UDL of 20kN/m and its lowest point is 10m below B. The ends of the cable are attached to saddles on rollers at the top of the piers and backstays anchor cables. Backstays may be assumed to be straight and inclined at 600 to vertical. Determine maximum tension in the cable, tension in backstay and thrust on each pier.

Let C be the origin

107

Tower

300 = TA = 1211.66 kNTA

= 21.8

300 TA = 1211.66 kNTa

21.8

y5 m

C

200 m

10 m

20 kN/m

VA

= 2202.05

AA

= 8081.6 kN

L1 L2

= 110.12 m = 89.89 m

x

We have

But

---- (1)

---- (2)

---- (3)

108

300 TA = 8376.26 kNTa

15.24600

----

- (4)

Since VA> VB tension at A is maximum

109

ANALYSIS OF CONTINUOUS BEAMS (using 3-moment equation)

Stability of structure

If the equilibrium and geometry of structure is maintained under the action of forces

than the structure is said to be stable.

External stability of the structure is provided by the reaction at the supports. Internal

stability is provided by proper design and geometry of the member of the structure.

Statically determinate and indeterminate structures

A structure whose reactions at the support can be determined using available

condition of equilibrium is called statically determinate otherwise it is called statically

indeterminate.

Ex:

No. of unknowns = 6

No. of eq . Condition = 3

Therefore statically indeterminate

110

W

A B

HA HB

VA VB

MA MB

End moments

FIXED BEAM

W W

A

RA RB

AC

RC

Degree of indeterminacy =6 – 3 = 3

No. of unknowns = 3

No. of equilibrium Conditions = 2

Therefore Statically indeterminate

Degree of indeterminacy = 1

Advantages of Fixed Ends or Fixed Supports

1. Slope at the ends is zero.

2. Fixed beams are stiffer, stronger and more stable than SSB.

3. In case of fixed beams, fixed end moments will reduce the BM in each section.

4. The maximum defection is reduced.

Bending Moment Diagram for Fixed Beam

Draw free BMD

Draw fixed end moment diagram, superimpose one above the other.

Ex:

Continuous beams

111

W

4

WL2

L

2

L

+

+

MM

Beams placed on more than 2 supports are called continuous beams. Continuous

beams are used when the span of the beam is very large, deflection under each rigid support

will be equal zero.

BMD for Continuous beams

BMD for continuous beams can be obtained by superimposing the fixed end moments

diagram over the free bending moment diagram.

Three - moment Equation for continuous beams OR CLAYPREON’S

THREE MOMENT EQUATION

Ex:

112

FREE B.M.

1x

a1 a2

8

2WL

L2L2

A B C

NN

4

WL

2x

The above equation is called generalized 3-moments Equation.

MA, MB and MC are support moments E1, E2 Young’s modulus of Elasticity of 2 spans.

I1, I2 M O I of 2 spans,

a1, a2 Areas of free B.M.D.

Distance of free B.M.D. from the end supports, or outer supports.

(A and C)

A, B and C are sinking or settlements of support from their initial position.

Normally Young’s modulus of Elasticity will be same through out than the equation

reducers to

If the supports are rigid then A = B = C = 0

If the section is uniform through out

113

Note: 1.

If the end supports or simple supports then MA = MC = 0

2.

MC = - WL3

If three is overhang portion then support moment near the overhang can be computed

directly.

3.

If the end supports are fixed assume an extended span of zero length and apply 3-

moment equation.

114

A B C

NN

A B C

NN

D

L1 L2 L3

AB

CA1 D

Zero Span

Zero Span

Note: i)

In this case centroid lies as shown in figure

ii)

115

a bW

L

Wab

a b

3

a3

b

a b

Wa Wa

2

Lx

L

Problems

1. For the continuous beam shown in fig. draw BMD and SFD. Assume

uniform cross section.

(Take care of ordinates)

116

AB

C

20 kN 10 kN/N

RA = 6.25 RC = 11.25RB = 32.5

1.5 m 1.5 m 3 m

1.5 m1.5 m

FREE BMD11.25 kNm

+

–

–++

END MOMENT DIAGRAM

15 kN11.25 kNm

11.25 kNm

O O

15 kNm

11.25 kNm

To Calculate Bending Moment

Span AB

=

= 15 kNm

Span BC

=

= 11.25 kNm

a1 = 22.5

a2 = 22.5

MA = MC = 0

Applying 3 – moment equation.

- 11.25 = 3RC - 45

RC = 11.25 kN

- 11.25 = 3RA - 30

RA = 6.25 kN

SFy = 0

117

6RA + RB + RC = 20 + 30

6.25 + RB + 11.25 = 50

RB = 32.5 kN

118

AB

C

20 kN 10 kN/N

RA = 6.25 kN RC = 11.25 kNRB = 32.5 kN

1.5 m 1.5 m 3 m

+ +

––

6.25 6.25

13.75 13.75 11.25 kN

18.75 kN

2. Draw BMD and SFD for the continuous beam shown in Fig.

119

AB

D

4 kN/N

RA = 18.98 kN RD = 18.63RC = 48.58

8 m 10 m 6 m

kNm48

FREE BMD

32.19

+

–

END MOMENT DIAGRAMO O

6 kN/N 8 kN/N

C

RB = 49.81

kNm50 kNm36

40.16

+

48 kN 50 kNm36 kNm40.16

32.19++

++

– – –

20.79 kNm29.37 kNm

29.02 18.63

18.98


Span AB

= = 48 kNm

Span BC

=

= 50 kNm

Span CD

=

= 36 kNm

a1 = 256

333.33

MA = MD = 0

Applying 3 – moment equation between A, B and C or for spans AB and BC.

----- (1)

Applying 3 – moment equation between A, B and D.

120

----- (2)

Solving (1) and (2)

(1) and (2)

- 2.923 MC = - 94.09

MC = - 32.18 kNm

MB = 40.16 kNm

MC = RD x 6 – (8 x 6)3

- 32.2 = 6RD - 144

RD = 18.63 kN

MB – RA x 8 – (6 x 8) 4

- 40.16 = 8RA – 192

RA = 18.98 kN

MC = 18.98 x 18 – (6 x 8) (14) + RB x 10 – (4 x 10) 5

RB = 49.81 kN

SFy = 0

RA + RB + RC + RD = 48 + 40 + 48

RC = 48.58 kN

121

a1 = 333.33

a2 = 144

3. Draw SFD and BMD for the continuous beam shown in Fig.

122

A B

RA = 10.73 kN RD = 18.63

2 m 5 m

FREE BMD–

C

RB = 23.7225 kN

kNm16

+

16 kNm19.62 kNm 30 kNm

–+

16 kN

kNm30

19.62 kNm

++

––

12.4525

11.27 SFD3.548 kN

0.73

3.548

12 kN


Span AB

= = 16 kNm

Span BC

= = 30 kNm

a1 = 48

a2 = 120

MA = 0 MC = 0

Applying 3 – moment equation between A, B and C.

MB = - 16 x 3 + RC x 8

- 19.62 = - 48 + 8RC

RC = 3.5475 kN

123

MB = RA x 6 – 12 x 2

- 19.62 = 6RA – 24

RA = 0.73 kN

RA + RB + RC = 28

RB = 23.7225 kN

124

4. Draw BMD and SFD for the continuous beam shown in figure clearly indicate all salient

points.

Applying 3 – moment equation between A', A and C.

-----

(1)

Applying 3 – moment equation between A, C and E.

125

A

C

E

16 kN 20 kN/N

RA = 46.235 RE = 6.33RC = 83.435

2 m 2 m 2 m 4 m

kNm33.536

4x2x40

1.5 m1.5 m

56 kNm

56 kNm

–

–+

+

34.98 kNm

O

40 kN

B D

42.0453.33 kNm

++

––

6.235

9.765 kN6.33

46.235

6.33

49.765 kN

33.67 33.67

A

Zero

a2 = 149.33

But ME = 0

MA x 4 + 2MC (4 + 6) + 0

4MA + 20MC = - 980.79 ------ (2)

x by 2

8MA + 40MC = - 1961.58 ------ (3)

(3) – (1)

MC = RA x 4 – 16 x 2 – 20 x 4 x 2 – 34.98

- 42.04 = 4RA – 32 – 160 – 34.98

RA = 46.235 kN

SFy = 0

RA + RC + RE = 16 + 40 + (20 x 4)

RC = 83.435 kN

MC = - 40 x 2 + RE x 6

- 42.04 = - 80 + 6RE

126

A C

16 kN 20 kN/N

48 kN 48 kN

56 kNm

B

O O

MB = 48 x 2 – (20 x 2)

= 96 - 40 = 56 kNm

a1 = 149.33

= 1690

RE = 6.33 kN

127

5. Analyse the continuous beam shown in figure and draw BMD and SFD.

MB = RA x 4 – 60 x 3 -30.25

= - 18.24 x 4RA – 180 - 30.25

RA = 48 kN

MB = - 20 x 4 x 2 + RC x 4 - 30 x 5

RC = 72.94 kN

SFy = 0

RA + RB + RC = 30 + 60 + 20 x 4

RB = 49.06 kN

There is overhang portion CD

MC = - 30 x 1

128

+

AC

E

60 kN 20 kN//m

RA = 48kN RC = 72.94RB = 49.06

1 m 3 m 4 m 1 m

kNm454

3x1x60

40 kNm

40 kNm

–

+

18.24

B D

++

––

48 kN

12 kN

48

42.94

37.06 30 kN

30 kN

Zero3I 4I

CD 4I

–

+30.25

45

–

30 kNm

+30 kN

O

A'

+

hxbxa2

12

a2 = 90

= 2.333 m

MC = - 30 kNm

Applying 3 – moment equation between A',A and B.

----- (1)

Applying 3 – moment equation between A, B and C.

1.333 MA + 4.667MB + MC

= - 75.15 – 80.00

MA + 3.509MB + 94.099 ----- (2)

Saving (1) and (2)

MB = - 18.24 kNm

MA = - 30.25 kNm

6. Draw SFD and BMD for the beam shown in figure.

129

= 90

= 1.67m

a2 = 106.67

SFy = 0

RB + RC = 0

SMB = 0

120 – RC x 7 = 0

RC = 17.14 kN

RB = 17.14 kN

BM at C = 0

BM at B = 0

BM at E Just to left - 17.14 x 4 = - 68.56

BM at E Just to right - 17.14 x 3 = + 51.46 kNm

Applying 3 – moment equation between A, A and B.

130

AC

6 kN/m

10 m 4 m 3 m

BC'

120 kNm

ZERO ZERO

A'

75

51.46 67.85

68.56 kNm

OO

51.46kNm75

8

10x6 2

+

-

+

–

– +

68.56

14.285.9 kNm

120 kNE

B C

RB = 17.14 RC = 17.14

4 m 3 m

a2 = 500

a1 = 500

= - 439.80

2MA x 10 + MB x 10

20MA + 10MB = - 1500

MA + 0.5MB = - 75 ------ (1)

Applying 3 – moment equation for A, B and C.

MA x 10 + 2MB (10 + 7) + MC x 7

10MA + 34MB + 7MC = - 1123.02 ------ (2)

MA + 3.4MB + 0.7MC = - 112.303 ------ (2)

131

Applying 3 – moment equation between B, C and C.

7MB x 14MC

7MB x 14MC = 17.39 ------ (3)

Using (1), (2) and (3)

MA +0.5MB = -75

(1) – (2)

2.9MB + 0.7MC = –37.303 x 20 ------ (4)

58MB + 14MC = –746.06

7MA + 14MC = –17.39

51MB = –728.67

132

= –

= 20.29

(4)Str.analysis

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