STRUCTURAL ANALYSIS

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

below.

Cyclic Process

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 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. The images given below 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.

i. Real structure

ii. A physical model

iii. A mathematical model

Real structure

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 the figure given below.

A Physical model

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 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

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 below.

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. This image shows two dimensional and three dimensional structures used in

mathematical modeling.

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. These images show two dimensional and three dimensional structures used in

mathematical modeling.

Two Dimensional

Three Dimensional

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. In the figure given below shows the 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.

Typical Support Conditions

Mathematical modeling requires considering 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.

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.

∑Fx=0 ; ∑Fy=0 ; ∑Mo=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 are

∑Fx=0 ; ∑Fy=0 ; ∑Fz=0;

∑Mx=0 ; ∑My=0 ; ∑Mz=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.

General procedures for determining the degree of indeterminacy of two-dimensional structures

are given below:

NUK = Number of unknown forces

NEQ = Number of equations available

IND = Degree of indeterminacy

IND = NUK - NEQ

Indeterminacy of Planar Frames

For entire structure to be in equilibrium, each member and each joint must be in equilibrium

(See the Figure given below)

NEQ = 3NM+3NJ

NUK = 6NM+NR

IND = NUK – NEQ = (6NM+NR)-(3NM+3NJ)

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

Free body diagram of Members and Joints

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

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)

Examples

Determine Degrees of Statical indeterminacy and classify the structures

Degree of freedom or Kinematic Indeterminacy

Members 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.

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

Expression for degrees of freedom

1. 2D Frames: NDOF = 3NJ – NR NR ≥ 3

2. 3D Frames: NDOF = 6NJ – NR NR ≥ 6

3. 2D Trusses: NDOF = 2NJ – NR NR ≥ 3

4. 3D Trusses: NDOF = 3NJ – NR NR ≥ 6

Where, NDOF is the number of degrees of freedom

In 2D analysis of frames sometimes axial deformation is ignored.

Then NAC=No. of axial condition is deducted from NDOF.

Examples

1.2 Determine Degrees of Kinematic Indeterminacy of the structures given below

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 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.

Stress-Strain Graph

Geometric Nonlinearity

Exercise Problems

Determine Degrees of Kinematic indeterminacy

PLANE TRUSS

Introduction

A truss is a structure having number of straight members connected at ends in such a way that

they primarily carry avail factors. If all the members lie in one plane it is called a plane truss. If

the members are in different planes it is called space truss. In 20th century and early 20th

century place trusses were extensively used as economical means for a bridge construction. Due

to development of cable staid constructions, reinforced concrete construction, use of trusses for

bridge has declaimed. Trusses are widely used in pre fabricated truss joist (beam) and other roof

systems in which large span supporting systems are required square trusses are commonly used

in tower structures like transmission towers and also used in aerospace industry.

The basic form of truss in triangle formed by 3 members joined together at their common ends

forming 3 joints such a triangle is rigid and do not lose its shape due to application of loads at

joints. Addition of 2 members at 2 joints of the previous triangle and connected at for end forms

a stable system of 2 triangles. If the whole structure is built up like this then the truss is rigid. If

such a truss is supported suitably will be a stable structure for this the reactions of the trusses

should be non – parallel and non-concurrent. Such a truss is called simple truss. If a simple truss

resists the load without undergoing appreciable deformation in shape is called as a perfect

truss. Perfect truss with 3, 4 & 5 joints are shown in figure.

From the basic triangle having 3 joints 3 members is observed that to 1 joint 2 members are

required. If NM = No. of members, NJ = No. of joints then for a following relationship exists.

NM = 2NJ – 3 (1)

Equation (1) is only necessary condition and not a sufficient condition. In addition to equation (1)

the truss should retain its shape when load is applied at any joint in any direction.

When the number of members in a truss is less than that computed from equation (1) then the

truss is called deficient truss i.e., NM<2NJ-3. On the other hand if NM>2NJ-3 then it is called

indeterminate truss or Redundant truss. These 2 types are shown in figure.

Assumptions

Following assumptions are made in the analysis of plane trusses:

a. Ends of the members are plane connected

b. The loads lie in the plane of truss & are applied only at joints

c. Self weight of the members are neglected

d. The centroidal axis of the of the members meeting at a joint will intersect at a common

point

e. Cross- section of the member is uniform (prismatic)

f. Members are straight

In practice it is difficult to achieve hinge condition at joint. Due to this effect secondary stresses

are developed in the members. Magnitudes of secondary stresses are very small & it can be

ignored. Primary stresses are developed in the members are due to axial forces developed due to

application of loads at joint. Axial forces developed are either compression or tension as shown in

figure.

A truss which satisfies all the assumptions discussed earlier is called an ideal truss. The members

of an ideal truss would be acted upon by only two forces along the longitudinal axis, one at each

end and each force representing the action on the member by the joint at that end. Hence the

members of the truss are subjected only to either tensile or compressive forces. The stresses

developed in the members of an ideal truss are called Primary stresses.

In reality the joints of the actual truss are bolted, riveted or welded. Hence moments and shears

will exist at the ends of the members. Another reason for moments and shears to exist in the

members is that the centroidal axes of all the members meeting at a joint may not be

concurrent. The members in such a truss will develop in addition to Primary stresses, stresses

due to moments and shears which are called as Secondary stresses.

Coplenar Concurrent Force System

If all the forces in a system lie in a single plane and their line of action passes through a single

point, then the system is called as coplanar concurrent force system. Since in the case of a plane

truss the centroidal axes of all the members meeting at a joint pass through the centre of the

joint, the forces at the joint forms a coplanar concurrent force system. Hence the analysis of a

plane truss needs an understanding of the principles of composition, resolution and equilibrium

of concurrent system of forces.

Composition of Forces

Parallelogram law of forces Triangle law of forces

Polygon Law of Forces

Analytical Method

Resolution of Forces

Analysis of a truss means determination of unknown forces in the various members of the truss

and unknown reaction at the supports of the truss when subjected to externally applied loads.

There are analytical methods & one graphical method available for analysis of plane trusses.

The analytical methods are:

1. Methods of Joints (MOJ)

2. Methods of Sections (MOS

In methods of joints free body diagram of forces at a joint is considered and equations of

equilibrium applicable for coplanar concurrent force system are used to compute the forces in

members. In method of sections the truss is cut into 2 parts & equilibrium equations are written

for free body of one part of the unit.

In MOS the forces are coplanar non-concurrent system. In Graphical method the forces polygons

are constructed graphically to some scale and then forces in each members is computed.

Method of Joints

The Methods of Joints consists of applying the 2 equilibrium equation SFx = 0 and SFy = 0 to

each joint at a time & proceeding around the structure until all members forces have been

determined.

Procedure:

1. Compute all support reactions using equations of equilibrium for the whole

structure. This step may be omitted for cantilever trusses.

2. Identify the joint having 2 members only to start with and write FBD of this joint. Forces

in these 2 members can be computed using equations of equilibrium.

3. Identify next joint where there are 2 remaining unknowns. Repeat the procedure till step

2, till the forces in all the members are computed

Method of Selections

In this method a section line as drawn passing through not mere than 3 members and a virtual

cut is made through the section line making the truss into 2 parts. Each part should be under

equilibrium. To make the FBD under equilibrium the effect of the removed part is replaced by the

member forces. The support reaction has to be computed before making any parallel cut through

the section line. The FBD of the separated truss portion have non-concurrent system of forces.

As there are 3 equations of equilibrium for non-concurrent force system, there should be only 3

unknown forces in each FBD.

The method of section is preferred under the following two conditions:

1. In a large truss in which forces is only a few members are required.

2. In the situation where the method of joints fails to start (or) proceed with analysis.

Procedures:

1. Compute reactions of the support using equations of equilibrium for the entire truss

considering it as a rigid body.

2. Depending on the members under consideration pass an imaginary section which may be

horizontal (or) vertical (or) inclined at any angle & then draw the FBD of one of the

portions of the unit passing through the imaginary section. While drawing the FBD,

replace the unit member of the removed part by tensile forces.

3. Determine the unknown member forces using equations of equilibrium i.e. ΣFx = 0, ΣFy

= 0 & ΣM = 0. While selecting the moment centre for ΣM=0 care should be taken such

that altleast 2 membr forces are concurrent at the moment centre.

4. Repeat the procedure from step 2 onwards by taking another section such that the

section passes through another 3 member forces.

Problem01

BE = 3m; CD = 6m

No. of joints (j) = 5

No. of members (m) = 7

No. of restraints(r) = 3

m = 2j - r

7 = 2*5 – 3

Hence truss is a “perfect truss”

In order to determine the forces in various members of the truss there is no need to find the

reactions from the supports apriori since free body diagram of joint “A” contains only two

unknowns to be determined. Applying the two equations of equilibrium viz. åFX=0 and åFY=0 at

the joint “A” we get, Let us arbitrarily assume that the nature forces in the members AB and AE

to be tensile. The free body diagram of joint “A” is; + _FY=0

Negative sign for FAB implies that the assumed sense of force in member is wrong, and hence

needs to be reversed.

Let us find the reactions at supports. For this the entire truss has to be considered as rigid and

equilibrium of the entire truss involves three equations of equilibrium viz. åFX=0, åFY=0 and

åMC=0.

Since the sign of RDH is negative

The assumed direction of RDH has to be reversed.

and the assumed direction is correct

To determine forces in other members of the truss, let us consider joint “C” since there are only

two unknowns, i.e. FCB and FCD. Free body diagram of joint “C” is

Hence force in member CD is 420kN (Tension)

Next let consider equilibrium of joint “D”

Hence force in member DB is 200kN (Tension) and in member DE is 400kN

(Tension) Considering equilibrium of joint “E”

Problem02

A.Determine the degree of indeterminacy of the truss ?

J = 11, m = 18, r = 4, m = 2j- r, 18 = 22 - 4

Hence the truss is determinate. Degree of indeterminacy is ZERO

B.Determine the degree of indeterminacy of the truss ?

J = 11, m = 18, r = 4, m = 2j- r, 18 = 22 - 4

Hence the truss has one member more than a perfect truss. Degree of indeterminacy is ONE

Method of Sections

Steps in the method of sections

1. Determine the reactions from the supports of the truss.

2. Divide the truss into two portions by passing a section through the members in which the

forces has to be determined.

3. Each part of the truss has to be in equilibrium under the action of loads, reactions and

forces in the members that are cut by the section.

4. The forces acting on each part of the truss constitutes a system of coplanar non-

concurrent forces. Hence for equilibrium of the truss three equations of equilibrium has

be considered viz. åFX=0, åFY=0 and åM=0.

5. Since we have only three equations of equilibrium, the section should cut only three

members.

Numerical Example

Step 1: Find Reactions.

åMA=0 gives

Step 1: Find forces in members DB, DC and EC.

To find forces in the members DB, DC and EC, consider equilibrium of the part of the truss shown

Hence force in the member BD is 30kN (Compression)

Problem03

Determine the degree of indeterminacy of the truss ?

J = 11, m = 18, r = 4, m = 2j- r, 18 = 22 - 4

Hence the truss is determinate. Degree of indeterminacy is ZERO

DEFLECTION OF BEAMS

Introduction

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 are:

i. 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.

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.

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

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

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,

From figure 5,

Substituting this value of R in equation (3),

d is nothing but change in angle over the elemental length dx. Hence to compute change in

angle from C to D,

Hence the proof.

above figure shows the elastic curve from C to D.Change in slope from 1 to 2 is d .Distance of

elemental length from D is x.

Problem01

To Compute Reactions:

Bending Moment Calculations:

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

To compute θΔ

To compute δE

Problem02

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 = 0

At 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.

To compute θB

For the cantilever beam, at the fixed support, there will be no rotation and hence in this case

A = 0. This implies that the tangent drawn to the elastic curve at A will be the same as the beam

line. From I theorem,

To compute δB

From II theorem

Problem03

Find deflection and slope at the free end for the beam shown in figure by using moment area

theorems.Take EI = 40000 KNm-2

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;

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 A = 0.

To compute δB

Problem04

To compute θB

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.

1. It is of the same span as the real beam.

2. The support conditions of Conjugate beam are decided as follows:

Some examples of real and conjugate equivalents are shown.

Problem01

For the Cantilever beam shown in figure, compute deflection and rotation at (i) the free end (ii)

under the load

Conjugate Beam:

Problem02

For the beam shown in figure, compute deflections under the loaded points. Also compute the

maximum deflection. Compute, also the slopes at supports.

ote 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:

Problem03

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.

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.

To Compute reactions in Conjugate Beam:

Problem04

To compute reactions and BM in real beam:

BM at (1) – (1) = 66.67 x

At 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 and BM in real beam:

To Compute δC

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

To compute δB

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

Problem05

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

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

shown.

To compute reactions and BM in real beam:

Section at A’ in the conjugate beam gives

STRAIN ENERGY METHOD

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, there is a

relationship can be used to determine the slopes and deflections in a structure.

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 gone these displacements, the internal

force do internal work called as Strain Energy.

Below figure shows the force displacement relationship in which Fj is the internal force and ej is

the corresponding displacement for the jth element or member of the structure.

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

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

energies are equal

Strain energy due to axial force

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

this section

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 above figure. 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

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.

Strain energy due to shear force

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

Strain energy due to Bending Moment

A small element shown in figure of dimension dx and dy is subjected to shear force Vx . Shear

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

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.

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 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 DC at C and DD 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)

If P’=1 then

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

Where,

d = 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.

Castigliano’s Theorems

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:

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.

Betti’s Law

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.

Example01

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

Example02

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

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

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.

Example03

Show that the flexural strain energy of a prismatic bar of length L bent into a complete circle by

means of end couples is

Example04

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 q = 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)

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 -FBEsinq=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.

Example05

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

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

Example06

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

Maximum Slope

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

Maximum Deflection:

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

Example07

A beam is simply supported on a span of 5m. A point load of 31kN is acting on the beam at

3.75m from right end calculate the deflection under load by strain energy method. E= 200 GPa,

I= 13*10-6m4

Solution:

Example08

A beam is simply supported on a span of 5 m. A point load of 10kN is acting on the beam at 2 m

from right end. Calculate the deflection under load and slope at left hand support by strain

energy method. Flexural rigidity for AC= 2EI and for CB = EI. Take E= 200 GPa, I= 25*10-6 m4

Solution:

Region Flexural Rigidity Origin Limit Mx

AC 2EI A 0 to 3 0.4Wx 0.4x

CB EI A 3 to 5 0.4Wx-W(x-3) 0.4x-(x-3)

Reactions and Moments:

Region Flexural Rigidity Origin Limit Mx

AC 2EI A 0 to 3

CB EI A 3 to 5

Example09

A cantilever having varying cross-section as shown in Fig., carries a concentrated load at free

end. Calculate deflection at free end.

Solution:

Example10

Compute horizontal and vertical deflection at joint C of the truss shown in Figure by

Castigliano’s. Take AE= 10,000 kN for all the members.

Let the horizontal and vertical forces at C be denoted as W and P' respectively. Moment equation

and partial derivative used for both horizontal and vertical deflection is tabulated in the table.

Region Flexural Rigidity Origin Limit Mx

AC 2EI A 0 to 3 -Wx -x 0

CB EI A 3 to 5 -3W-P’x 3 -x

E=200GPa; I= 80*106 mm4

EI=1.6 *104 kN-m2

Procedure for Computation of Displacement in Trusses

Step 1: Apply Virtual load at the point under consideration in the direction of required

displacement

Step 2: Compute Forces (P) in all the members due to external load and also due to Virtual force

Step 3: Differentiate P wrt virtual force

Step 4: Write deformation equation using Castigliano’s Second theorem as

Step 5: Equate Virtual load to zero before performing summation

Example11

A Compute horizontal and vertical deflection at joint C of the truss shown in Figure by

Castigliano’s. Take AE= 10,000 kN for all the members.

Solution:

q=tan-1(3/4) =36052

H and W are the horizontal and vertical Virtual forces acting at C as shown in Fig.

Member L (m)

Forces in member

P (kN)

AB 5 -12.5- 0.833W+0.625H -0.83 0.625

AD 4 10+0.67W+0.5H 0.67 0.50

CD 3 15 0 0

BD 4 10+0.67W+0.5H 0.67 0.5

BC 5 -12.5- 0.833W-0.625H -0.83 -0.625

Example12

Determine horizontal and vertical displacement of the joint D of truss shown in Fig., using the

Castigliano’s Theorem. The area of cross section A = 500 mm 2E=200GPa for all members.

Solutions:

AE= 500*2*105=1*108 N

AE=1*105

q=tan-1(3/4)=36052’

Member L (m)

Forces in member

P (kN)

AB 4 -33.33- 0.667W’+0.5H’ -0.67 0.5

BD 3 25+0.5W’-0.375H' 0.5 -0.375

DC 4 33.33+ 0.667W’- 0.5H' 0.67 -0.5

BC 5 -41.67-0.83W’+0.625H' -0.83 0.625

AD 5 41.67+0.83W’+0.625H' 0.83 0.625

-1/(1*105)

or

Unit Load Method

If the Unit load produces BM as M’

Strain energy stored by M’ acting on the real deformation d in element dx is expressed as

Procedure For Computation Displacement

Step 1: Apply Unit load at the point under consideration

Step 2: Consider the structure with real load and unit load separately

Step 3: Write BM equation M for real load and M’ for unit load

Step 4: Substitute M and M’ in equation 16 and 17

Step 5: Integrate equation to get the required deformation

Example13

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 unit load method. EI is constant.

Maximum Deflection

Maximum Slope

Maximum slope occurs at B, Unit moment is applied at B

Example14

Calculate max slope and max deflection of a simply supported beam carrying udl of intensity w

per unit length throughout its length by using Unit Load method

Maximum Slope

Maximum slope occurs at support. A unit moment is applied at A.

Maximum Deflection

Maximum Deflection occurs at mid span. A unit load is applied at mid-span

Example15

A beam is simply supported on a span of 5m. A point load of 31kN is acting on the beam at

3.75m from right end. Calculate the deflection under load by unit load method. E= 200 GPa, I=

13*10-6 m4

Solution:

E=200GPa=2*108 kN/m2; I= 13*10-6 m4

EI=2600 kN-m2

Deflection at C:

A unit downward load is applied at C.

Reactions:

BM for Region AC :

BM for Region CB :

Using Unit Load Method

Example16

A beam is simply supported on a span of 5m. A point load of 10kN is acting on the beam at 2m

from right end. Calculate the deflection under load and slope at left hand support by Unit Load

method. Flexural rigidity for AC= 2EI and for CB = EI. Take E= 200 GPa, I= 25*10-6 m4

Solution

E=200GPa=2*108 kN/m2; I= 25*10-6 m4

EI=5000 kN-m2

Deflection at C:

Maximum Deflection occurs at mid span. A Unit downward load is applied at mid-span.

Reactions and Moments:

Region Flexural Rigidity Origin Limit Mx M'

AC 2EI A 0 to 3 4x 0.4x

CB EI A 3 to 5 4x-10(x-3) 0.4x-(x-3)

Using Unit Load method

Maximum Slope

Maximum slope occurs at support. A unit moment is applied at A.

Reactions and Moments:

Region Flexural Rigidity Origin Limit Mx M'

AC 2EI A 0 to 3 4*x 1-0.2x

CB EI A 3 to 5 4x-10(x-3) 1-0.2x

Using Unit Load Method

Substituting EI= 5000 kN-m2

Example17

Determine slope and deflection at free end of cantilever beam shown in Fig. using unit load

method Take E= 200 GPa, I= 3*108 mm4

Solution:

E=200GPa=2*108 kN/m2; I= 3*108 mm4

EI=6*104 kN-m2

Region Flexural

Rigidity Origin Limit Mx M1' M2'

BC EI B 0 to 3 -50x -x -1

CA 2EI B 3 to 6 -50x -x -1

Deflection at B:

A Unit downward load is applied at B. Using concept of unit load method

Slope at B

A Unit moment is applied at B. Using concept of unit load method

Example18

Calculate the horizontal deflection of joint B and support D of the frame shown in Fig. EI is

constant

Solution:

Reaction due to P: VA=-P; HA=-P; RD=P

Reaction due to unit load at B: VA=-1 ;HA=-1 ; RD=1

Reaction due to unit load at D: VA=0; HA=-1 ; RD=0

Region Flexural

Rigidity Origin Limit Mx M1' M2'

AB EI A 0 to L Px x x

BC EI C 0 to L Px x x

DC EI D 0 to L 0 0 x

Deflection at B

Deflection at D

Example19

Calculate the vertical deflection of point C in the loaded electrical pole shown in Fig. EI is

constant

Solution :

For the semicircular portion BC, arc length ds is used instead of dx and ds= R d

Region Origin Limit Mx M'

BC A 0 to p -10(R-R cos q) -(R-R cos q)

AB A 0 to 4 -20x -10 * 2 -2

Deflection at C:

Procedure for Analysis of Trusses

Step1: Apply Unit load at the point under consideration in the direction of required displacement

Step2: Compute Forces (P) in all the members due to external load and also due to Unit load

Step3: Write deformation equation using Unit load theorem

the values of P, P’ and product of P and P’ are tabulated

Example20

Determine the vertical and horizontal deflection of joint E of the truss shown in Fig. Take AE=

3.6 * 105 kN

P1' and P2'are the forces in members due to unit loads along vertical and horizontal direction

applied at E.

Member L(M) P (kN) P1' P2'

AB 4.24 -28.3 -0.47 0.00 1.57 0.00

AF 3 40.0 0.33 3.33 1.10 3.33

BC 3 -60.0 -0.67 0.00 3.36 0.00

BF 4.24 28.3 0.47 0.00 1.57 0.00

BF 3 00.0 00.0 0.00 0.00 0.00

CD 4.24 -84.9 -0.94 0.00 9.4 0.00

CE 3 60.0 0.67 0.00 3.35 0.00

DE 3 60.0 0.67 0.00 3.35 0.00

EF 3 40.0 0.33 1.00 1.10 3.33

24.8 6.66

Example21

Determine the vertical point D in the truss shown in Fig. Take A= 1500 mm2 for members AD

and DE, A=1000 mm2 for other members

Solution:

P1'and P2'are the forces in members due to unit loads along vertical and horizontal direction

applied at E.

Member L(mm) Area mm2 P(kN) P1'

AB 4000 1000 45.0 0.00 0.00

BC 5000 1000 75.0 0.00 0.00

CD 3000 1000 -45.0 0.00 0.00

DE 4000 1500 -105 -1.00 280

DB 4000 1000 -60.0 0.00 0.00

AD 5856 1000 84.84 1.414 452.55

ARCHES AND CABLES

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 :

Hinged at the supports and the crown.

Three hinged arches

Two hinged arches

Fixed arches

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).

Comparison between an arch and a beam

Analysis of 3-hinged arches

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.

Procedure to find reactions at the supports

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.

Step 1. Sketch the arch with the loads and reactions at the support.

Step 2. Apply equilibrium conditions namely. Fx= 0,Fy= 0 and M = 0

Step 3. 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.

Problem01

A 3-hinged arch has a span of 30 m and a rise of 10 m. 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.

Fx= 0

HA- HB= 0

HA= HB

To find vertical reaction.

FY= 0

VA+ VB= 0.6x15+1+1.6

= 11.6

MA= 0

- VBx30+1+1.6x25+1x20+(0.6x15)7.5 = 0

VB= 4.25 kN

VA= 4.25 kN = 11.6

AA= 7.35 kN

To find horizontal reaction.

MC= 0

-1x5-1.6x10+4.25x10+4.25x15-HBx10= 0

HB= 4.25 kN

HA= 4.25 kN

Or

MC= 0

7.375x15-HAx10-(0.6x15)7.5

HA= 4.275kN

HB= 4.275kN

To find total reaction

Problem02

A 3-hinged parabolic arch of span 50 m and rise 15 m carries a load of 10kN at quarter span as

shown in figure. Calculate total reaction at the hinges.

Fx= 0

HA= HB

To find vertical reaction.

FY= 0

VA+ VB=10

MA= 0

- VBx50+10x12.5 = 0

VB= 2.5 kN

VA= 7.5 kN

To find horizontal reaction.

MC= 0

HB= 4.25 kN

To find total reaction

Problem03

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

Fx= 0

HA- HB = 0

HA = HB ------ (1)

To find vertical reaction.

FY= 0

VA+ VB = 180+10x10

VA+ VB = 280 ------ (2)

MA= 0

- VBx24+HBx2.4+180x18+10x10x5 = 0

2.4HB- 24VB= -3740

HB-10VB= -1588.33 ------ (3)

MC= 0

-180x8-VBx14-14-HBx4.9=0

HBx 4.9-VB14 = -1440

-HB+ 2.857VB = +293.87 ------ (4)

Adding (2) and (3)

-10VB+2.87VB= -1558.33+293.87

VB= 177 kN

VA= 103 kN

HB-10x177 = -1558.33

HB = 211.67 kN = HA

Bending moment diagram for a 3-hinged arch

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).

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

constant.

Problem01

A UDL of 4kN/m covers left half span of 3-hinged parabolic arch of span 36 m and central rise 8

m. Determine the horizontal thrust also find:

(i) BM (ii) Shear force

(iii) Normal thrust (iv) Radial shear at the loaded quarter point. Sketch BMD.

Fx= 0

HA- HB= 0

HA= HB

FY= 0

VA+ VB= 14x18

VA+ VB= 72

MA= 0

- VBx36+4x18x9 = 0

VB= 18 kN

VA= 54 kN

MC= 0

VBx18-HBx8=0

HB=40.5 kN

HA=40.5 kN

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

= 44.32 kN

S = 40.5 Sin 23.96 – 18 Sin 66.04

S = - 0.0019 » 0

54 x 13.5 – 4 x 13.5

= 364.5 kN

Problem02

A symmetrical 3-hinged parabolic arch has a span of 20 m. It carries UDL of intensity 10 kNm

over the entire span and 2 point loads of 40 kN each at 2 m and 5 m 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 4 m.

Fx= 0

HA- HB=0

HA= HB

FY= 0

VA+ VB- 40 - 40 -10 x 20 = 0

VA+ VB= 280

MA= 0

+ 40 x 2 + 40 x 5 + (10x20)10-VB x 20 = 0

VB= 114 kN

VA= 166 kN

MC= 0

-(10x10) 5 - HB x 4 + 114 x 10 = 0

HB=160 kN

HA=160 kN

Normal thrust = N = + 160 Cos 25.64

+ 86 Cos 64.36

= 181.46 kN

S = 160 Sin 25.64

- 86 x Sin 64.36

S = - 8.29 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 f will be different

To develop necessary equations for 3-hinged segmental arch

Problem01

A 3-hinged segmental arch has a span of 50 m and a rise of 8 m. A 100 kN load is acting at a

point 15 m from the right support.

(i) Find horizontal thrust at the supports

(ii) BM, Normal thrust and radial shear at a section 15 m from the left support.

Fx= 0

HA= HB

FY= 0

VA+ VB= 100 kN

MA= 0

- VBx 50 + 100 x 35 = 0

VB= 70 kN

VA= 30 kN

MC= 0

VBx25-HAx8=0

HB=40.5 kN

HA=93.75 kN = HB

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 kN

S = 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.

Problem01

Determine the reactions components and tension in different parts of the cable shown in figure.

Also find the sag at D and E

Fx= 0

-HA+ HB= 0

HA = HB

FY= 0

VA+ VB= 120 kN

MA= 0

30 x 20 + 40 x 40 + 50 x 60 – VB x 80 = 0

VB= 65 kN

VA= 55 kN

MC= 0

-HA x 10 + 55 x 20 = 0

HA=110 kN = H

Point A

Fx= 0

T1Cos26.56-110=0

T1=123 kN

tan

To find YE:-

We have B.ME

-110 x YE+ 65 x 20 = 0

YE = 11.82m

To find YD:-

B.MD= 0

50 x 20 + 65 x 40 – HB x YD = 0

YD = 11.45m

Point D

Fx= 0

T1Cos7.77-T2cos 12.81=0

T3=T20.984

FY= 0

T1Sin 7.77 + T2 Sin 12.81 – 40 = 0

T20.984Sin 7.77 + T2 Sin 12.81 – 40 = 0

T2 = 112.75 kN

T3 = 110.95 k

Point B

Fx= 0

110-T4Cos 30.58 = 0

T4= 127.77

tan

4= 30.58

Problem02

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

Fx= 0

-HA+ HE= 0

HA= HE

FY= 0

VA+ VE= 420 kN

MA= 0

200 x 10 + 100 x 20 + 120 x 30 – VE x 40 = 0

VE = 190 kN

VA= 230 kN

MC= 0

-HA x 13 - 200 x 10 + 230 x 20 = 0

HA=200 kN = HE

Point A

B.MB= 0

230 x 10 – 200 x YB = 0

YB = 11.5 m

Fx= 0

T1Cos49 - 200 = 0

T1=304.85 kN

Point B

Fx= 0

T1Cos8.53-304.8 = 0

Cos 49 = 0

T2= 202.22 kN

tan

3 = 19.3

Point E

Fx= 0

200 – T4Cos 43.53 =0

tan

3 = 43.53

The Total length of Chord = 49.76 m

Problem03

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

Fx= 0

-HA+ HB= 0

HA= HB -----------

(1)

FY= 0

VA+ VB = 120 kN -----------

(2)

MA= 0

30 x 25 + 50 x 50 + 40 x 75 –VB x 100 – HB x 15 = 0

15 HB + 100 VB = 6250

HB + 6.67 VB = 416.67

VA+ VB= 120 kN -----------

(3)

VA= 55 kN

B.MD= 0

- HB x 2.5 +VB x 50 – 40 x 25 =0

2.5 HB – 50 VB = - 1000

HB – 20 VB = -400 -----------

(4)

(3) - (4) gives

26.67VB = 416.67 + 400

VB = 30.62 kN

HA = 212.42 kN = HB

VA = 89.38 kN

Point A

Fx= 0

T1Cos 19.29 - 212.42 = 0

T1 = 225.05

Point C

Problem04

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.

Fx= 0

-HA+ HD= 0

HA= HD ----------

- (1)

FY= 0

VA+ VD= 100 kN ----------

-(2)

MA= 0

40 x 15 + 60 x 30 –VD x 45 = 0

VD = 53.33 kN

VA= 46.67 kN ---------

-- (3)

B.MB= 0

46.67 x 15 – HA x YB = 0

HAYB = 700.05

MC= 0

46.67 x 30 – 40 x 15 – HA x YC = 0

HAYC = 800.10

YB = 0.875 YC

We have

1 = 160.25

Point A

Fx= 0

T1Cos 16.350

-159.2 kN = 0

T1 = 165.90 kN

Point C

Fx= 0

T3Cos 18.52- T2Cos 2.385 = 0

T3 = 1.053 T2

FY= 0

T3 Sin 18.52 + T2 Sin 2.385 – 60 =0

1.053 T2 Sin 18.52 + T2Sin 2.385 – 60 = 0

0.376T2 = 60

T2 = 159.44 kN

T3 = 167.89 kN

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.

FY= 0

(V+ dv) – V – W x dx = 0

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

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

------- (1)

At X = L2, Y= a + b

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

same level

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

Problem01

A cable suspends across a gap of 250 m and carries UDL of 5kN/m horizontally calculate the

maximum tensions if the maximum sag is 1/25th of the span. Also calculate the sag at 50 m

from left end.

Problem02

Determine the length of the cable and max tension developed if the cable supports a load of

2kN/m on a horizontal span of 300 m. The maximum sag is 25 m

Problem03

Determine the maximum span for a mild steel cable between supports at the same level if the

central dip. is 1/10

th 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

Bridges supported by cables

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.

In 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.

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.

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.

Problem01

A cable of span 150 m and dip 15 m 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

HT = 1211.66 Cos 21.8

~ 1211.66 Cos 300

= 75.73kN

VT = 1211.66 Sin 21.8 + 1211.66 Sin 300

VT = 1055.77kN

M = 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 30

Ta 1299.05kN

In this case HT = 0

M = f

VT = 1099.5 kN

Problem02

A Suspension cable is suspended from 2 pier A and B 200 m apart, B being 5m below A the cable

carries UDL of 20kN/m and its lowest point is 10 m 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

MC= 0

VB=1797.95 kN

HB=8081.6 kN

HA=8081.6 kN

VA=2202.05 kN

Since VA> VB tension at A is maximum

MC= 0

VB=1797.95 kN

HB=8081.6 kN

HA=8081.6 kN

VA=2202.05 kN

Since VA> VB tension at A is maximum

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.

No. of unknowns = 6

No. of eq . Condition = 3

Therefore statically indeterminate

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.

Continuous Beams

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.

Three moment equation or Claypreon’s Three Moment Equation

Three - moment Equation for continuous beams OR CLAYPREON’S THREE MOMENT EQUATION.

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

Note :

1)

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

2)

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

Problem01

For the continuous beam shown in fig. draw BMD and SFD. Assume uniform cross section.

(Take care of coordinates)

To Calculate Bending Moment

Problem02

Draw BMD and SFD for the continuous beam shown in Fig

To Calculate Bending Moment

Problem03

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

To Calculate Bending Moment

Problem04

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

FY = 0

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

RC = 83.435 kN

MB = 48 x 2 – (20 x 2)

= 96 - 40

= 56 kNm

MC = - 40 x 2 + RE x 6

- 42.04 = - 80 + 6RE

RE = 6.33 kN

Problem05

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

Problem06

Draw SFD and BMD for the beam shown in figure