SFD BMD

UNIT 4 SHEAR FORCES AND BENDING MOMENTS

Shear Forces and Bending Moments

Structure 4.1

I ntroduct ntroductiion Objectives

4.2

Beams 4.2.1 Typ Types of Sup Support 4.2.2 Typ Types of Beam 4.2.3 Typ Types of Loading ing

4.3

Shear Shear For Force ce and Bendi ending ng Mom Moment 4.3.1

Definitions

4.3.2

Sign Conventions

4.3.3

Relation elation between Loading, Loading, Shear Forc Force e and Bending Mome Moment

4.3.4

MaximumValue of Be Bending Mom M oment

4.4

Shear Shear For Force ce and Bendi ending ng Mom Moment ent Di Diagrams

4.5

Beams Subjecte Subjected to Couples Couples

4.6

Beams Subjected Subjected to Obli Oblique L oadi oading

4.7

Summary

4.8

A nswe nswers rs to to SAQs SAQs

4.1 INTRODUCTION In practice, practice, all all beams are subj subjected to some type of external loadings. oadings. As A s the beam beamis loaded, it deflects from its original position, which develops the moment and shear on the beam. Shear force (SF (SF) and bending nding moment (BM (B M) are very important portant in in desi designing gning any type type of beam because the beamis designed designed for for maxim aximum bending nding moment and also also the shear reinf reinforcem orcemen entt is is provide provi ded d for for maxim xi mum shear force. force. Strength Strength of the beam is design sign cri criterion terion for for bending bending moment and shear for force, ce, whereas whereas the the sti stifffness ness of the beamis the design criterion for deflection of the beam. In this unit, unit, you you wil will be introduced ntroduced to various various types types of beam beams, supports supports and and loa loadi ding ngs. s. The Thereafte fter, methods of drawing ing Shear For Force Diag Diagram (SFD) SFD) and Ben Bending ing Mom Moment Dia Di agram(BM (B M D) for cantileve cantil ever, sim simply ply supported supported be beamand and overhan overhangi ging ng beamfor dif different erent types of of loadi oadings have have been discus discusse sed.

Objectives A fter studyi studying ng this this uni unit, t, you shoul should be be able ble to identify the types of beam and types of support when it appears in building, structural system, etc., describe descri be the shear for force ce and bending bending moment and the relati relation on between shear force, orce, bendi bending ng moment and rate of of loading, oading, draw the shear force orce and and bendi bending ng mom momen entt diag diagram rams for for any types of beam subjecte subjected to any any type type of loadi l oading ng system, and show the positi position on of maxim aximum bending ding moment and position positi on of zero bending moment (point of contraflexure). 5

Forces and Stresses in Beams

4.2 BEAMS A beam beamis a structural member of suff sufficien cientt leng length compared pared to lateral dime dimensions nsions and supported so as to bein equil quilibrium bri um and and generally, nerall y, subjecte subjected to a systemof external external forces orces to to produce bendi bending ng of the member in in an axial xi al plan plane. e. Ti Ties, struts, shafts shafts and and beams are all one-dim one-dimensi ensional onal or or li l ine-elem ne-elements where the length is is much greater than the depth or width. width. The These are also known as rod-li rod-l ike or skeletal skeletal elements; and and have have diff dif ferent names depen depending ding upon upon the main ain action action they they are desi designed gned to res resiist. Thus, T hus, ties ties and struts resist resist uniaxi uniaxial al ten tensi sion on or compressi compression, shafts shafts resist resist torque and beams resist resist bending nding momen oments ts (and shear forces). The The clea lear horizo izontal dis disttance between the supports rts is called lled the clea lear span of the beam. The The horizo izontal dis disttance between the centers of th the end bearing ings is called lled the effec ffective ive span of the beam. I f the intensi nsity of the bearing aring reac reacti tion on is is not uni unifform, the effective effective span is the horizontal distance between the lines of action of the end reactions. Beam

Effective Span

Support

Support

Clear Span

Figure 4.1 : Clear and Effective Spans in a Beam

Whenever a horizontal horizontal beam beamis loade loaded, it it bends bends due due to the action action of loads. oads. The T he amount with wi th which which a beam bends, depends upon upon the amoun amountt and type of loads, oads, length of the beam beam, elas elasti tici city ty properti properties es of the beam material ri al and thetype of the bea beam. Beams may may be concrete, steel or even composite posite beam, havi having ng any any type type of secti sections ons such as angles, ngles, channels, els, I -section, ction, recta rectangle, squa square, hat secti section on etc.

4.2.1 Types of Support The The supports can be clas lassified ified int into follo follow wing ing three categories ies : (a)

A simpl simple e or free freesupp support ort or roll rol ler support,

(b)

Hinged or pinne pinned support, support, and

(c) (c)

A bui built-in t-i n or fixe fi xed d or enca encastre supp support. ort.

A Simple or Free Support or Roller Support

I t is is a support support in in which which beamrests rests free freelly. A roll roller support support is is the the si simplest and gi gives only one reaction, either in x direction or y direction ( R or R y) norm normal to the the R x or support support and off offers no resi resistance stance agai gainst rotation rotation or late lateral movemen ovements. ts. The The reaction will be taken as acting at a point, though actually it will be distributed over some length.

Actual

Symbolic Ry

Ry

Symbolic (Line Element) VA

Figure 4.2 : Simple or Free Support or Roller Support

6

Hinged or Pinned Support

In this this case case,, the beamis hinge hinged or pinne pinned to the support. support. A pinned pinned or hinged hinged support support gives gi ves onl only two rea reactions, ctions, one against gainst verti vertical cal movement and another another or R y) but off offers no resistan resistance ce to the angul angular ar agai against nst horizontal horizontal movement (say R x or rotation rotation of the beam at the hinge. hinge. The The reaction reaction wil will pass through thecentre of the hinge or pin. Rx

A

HA

VA

Ry

Actual


Symbolic (Line Element)

Symbolic

H A – Horizontal Reaction at the Support ‘A’ V A – Vertical Reaction at the Support ‘A’

Figure 4.3 : Hinged or Pinned Support

A Built-in or Fixed or Encastre Support

It is is a support which which restrains restrains complete plete movement of the beam both in in positi position on as well as direction. The support gives all the three relevant reactions (say R x, R y and I n other other words, words, M z), i.e. the reactions in x and y directions and fixing moment M z. In the fixed support offers resistance against translation in both the directions and also against against the rotati rotation. on. A Rx

HA MA MZ

Ry

Actual

Symbolic

VA

Symbolic

M A – Fixing Moment at the Support ‘A’

Figure 4.4 : A Built-in or Fixed or Encastre Support

4.2.2 Types of Beam Depending epending upon the type and num number of supports, the bea beams are divi divide ded d into into two categories : (a)

Statical Staticallly determi determinate nate beam, and

(b)

Staticall Statically indet ndeterm erminate nate bea beam.

Statically Determinate Beam

A beam beamis said to be statical staticallly determi determina nate, when it it can be anal analyz yzed ed using using three static equilibrium equations, i.e. H x =0, V x =0 =0 and and M x =0, where H x =algebraic sum of horizontal forces at section x, V x =algebraic sum of vertical forces at section x, and

M x =al =alge gebrai braic c sum of momen omentt of all all the forces about the secti section on x. Examples are as follows : (a)

Cantilevers,

(b) (b)

Simply supported beams, and

(c) (c)

Overhan Overhanging ging beams. 7

It can also be stated, if the number of unknown reactions or stress components can be found by means of three equilibrium equations, then the beam is called statically determinate beam.


Statically Indeterminate Beam

When the number of unknown reactions or stress components exceed the number of static equilibrium equations available, i.e. H x =0, V x =0 and M x =0 then, the beamis said to be statically indeterminate. This means, the three equilibrium equations are not adequate to analyse the beam. Examples are as follows : (a)

Fixed beams,

(b)

Propped cantilevers, and

(c)

Continuous beams.

Now, we will discuss briefly each type of beam in subsequent paragraphs. Cantilever Beams

A beamfixed at one end and free at the other end is known as cantilever.

A

B

Span (l)

Figure 4.5 : Cantilever Beams

Simply Supported Beam

A beamsupported or resting freely on the walls or columns at its both ends is known as simply supported beam. The support may be hinged or roller support. B

A

Span (l)

Figure 4.6 : Simply Supported Beam

Overhanging Beam

A beamhaving its end portion (or portions) extended in the form of a cantilever beyond the support is known as overhanging beam. A beammay be overhanging on one side or on both sides. The support may be fixed, hinged or roller support. B

A

l

l1

A

B

l2

Figure 4.7 : Overhanging Beam

l1, l2, l3 are the lengths of overhanging portions.

8

Fixed Beam

l

l3

A beam rigidly fixed at its both ends or built-in walls is known as rigidly fixed beamor a built-in beam. A


B

Span (l)

Figure 4.8 : Fixed Beam

Propped Cantilever

If a cantilever beamis supported by a simple support at the free end or in between, is called propped cantilever. It may or may not be having overhanging portion. A

B

A

B

Span (l)

Span (l)

Propped Cantilever

l1

Propped Cantilever with Overhang Figure 4.9

l1 =length of overhanging portions.

Continuous Beam

A beam which is provided with more than two supports is called a continuous beam. It may be noted that a continuous beam may or may not be an overhanging beam. B

A

l1

C

l3

l2

D

l4

Figure 4.10 : Continuous Beam

l1 =length of overhanging portion, and l2 =span AB; l3 =span BC ; l4 =span CD.

4.2.3 Types of Loading A beammay be subjected to the following types of loads. (a)

Concentrated or point load,

(b)

Uniformly distributed load, and

(c)

Uniformly varying load.

Concentrated or Point Load

A load acting at a point on a beamis known as concentrated or a point load. I n practice, it is not possible to apply a load at a point, i.e. at a mathematical point as it will be distributed over a small area. But this very small area, as compared to the length of the beam, is negligible. W

W

9 A

B


Figure 4.11 : Concentrated or Point Load

A beammay be subjected to one or more point loads. Uniformly Distributed Load

A load which is spread over a beamin such a manner that each unit length is loaded to the same extent, is known as uniformly distributed load (briefly written as u.d.l.). Sometimes, the intensity of load may not be uniform and varying from point to point. For all calculation purposes, the uniformly distributed load is assumed to act at the centre of gravity of load. A

A

B

B

Figure 4.12 : Uniformly Distributed Load

Uniformly Varying Load

A load which is spread over a beamin such a manner that its extent varies uniformly on each unit length is known as uniformly varying load. Sometimes, the load is zero at one end and increases uniformly to the other end. Such a load is known as triangular load.

B

A

A

B

Figure 4.13 : Uniformly Varying Load

A beammay carry anyone of the above load systems or a combination of two or more systems at a time. Loads may be static or dynamic. Static loads are those which are applied gradually and do not change their magnitude, direction or point of application with time. Dynamic loads are those, which vary in time with speed.

4.3 SHEAR FORCE AND BENDING MOMENT 4.3.1 Definitions Shear Force : General

The static equilibrium of a space can be ensured if the algebraic sum of all the forces acting on the particle in the x, y, z directions are separately zero [Figure 4.14(a)]. Mathematically, F x =0, F y =0, and F z =0. where, F x =algebraic sum of all the forces acting on the particle in the x direction =F x F x , 1

2

F y =algebraic sum of all the forces acting on the particle in the y direction =F y F y , and 1

10

2

F z =algebraic sum of all the forces acting on the particle in the z direction = F z F z . 1


2

y axis

Fy1 Fz2

Fx2

x axis Fx1

Fz1 Fy2 z axis

Figure 4.14(a) : Equilibrium of a Particle under a System of Forces

These forces are general forces and their positive or negative sign will depend on, whether the forces are directed away from the origin or towards the origin. Let us consider a section A-A parallel to y-z plane in the body (Figure 4.14(b)). y-axis A Fy1

Fy2

Fz2

Fx3

Fx1 x-axis Fx2

Fz1 Fy3 A

z-axis

Figure 4.14(b) : Equilibrium of a Body under System of Forces y-axis A

A

Fy

Fx

x-axis

Fz

z-axis

A

A

Figure 4.14(c) : A Section A-A Parallel to y-z Plane Cut and Forces Exposed at A-A

11


The right hand portion is removed and the forces exposed on section A- A in the left hand portion of the body are shown in Figure 4.14(c). The three resultant forces in the x, y, z directions, denoted as F x, F y and F z do not create the sametype of stresses on the section. A s the area of the section A- A is normal to x axis, F x will create axial stresses or normal stress x on the section. However, the remaining two forces F y and F z act tangential or along the plane and create shear stresses xy and xz. Using the common nomenclature of beams and bending, these forcesF y and F z are known asshear forces, whereas F x is simply an axial force. F x =resultant of all the forces acting on the left portion of the body in the x direction. F y =resultant of all the forces acting on the left portion of the body in the y direction. F z =resultant of all the forces acting on the left portion of the body in the z direction. P C B

A

RA=P P C

RA=P S

C

S

Figure 4.14(d) : Development of Shear Force at Section C

Now let us consider a beam (for example, cantilever beam) subjected to point load P. Consider a section at C . At this section there is a possibility of failure by shear as shown in Figure 4.14(d). If such a failure occurs at section C , the cantilever is liable to be sheared off into two parts. For the equilibrium of the cantilever, the fixed support at A will provide a vertical reaction vertically upwards, of magnitude R A =P. It is clear that the force acting normal to the centre line of the member on each part equals S =P. The force acting on the right part on the section C is downward. The resultant force acting on the left part is upward. The resultant force acting on anyone on the parts normal to the axis of the member is calledshear force at the section C . Thus, the shear force (briefly written as SF) at the cross section of a beam may be defined as the unbalanced vertical force to the right or left of the section.

Bending Moment : General

For a rigid or deformable body, the rotational equilibrium can be ensured if the algebraic sumof moments of all the forces acting on the body in the x, y, z directions are separately zero (Figure 4.14(e)). Mathematically, M x =0, M y =0, and M z =0.

where M x, M y, M z are the moments of forces taken about x, y, z axes. 12


y-axis A

My1

My1 x-axis Mx1

Mx2 Mz2

Mz1

Fy3 A

z-axis

Figure 4.14(e) : Equilibrium of a Body under a System of Moments

These moments are general moments and their positive or negative sign will depend on, whether the moments are counter-clockwise or clockwise. M x =Algebraic sum of moments of all the forces about x axis = M x M x 1

2

M y =Algebraic sumof moments of all the forces about y axis = M y M y 1

2

M z =Algebraic sumof moments of all the forces about z axis = M z M z 1

2

Now, consider a section A- A parallel to y-z plane in the body (Figure 4.14(f)). The right hand portion is removed and the moments exposed on section in the left hand portion of the body are shown in Figure 4.14(f). The three moments, M x, M y, M z are shown to be positive for the sake of convenience. A little thought will show that all of the moments do not create same type of stresses on the section.  M x, acting on section x creates a twisting action on the section. M x renamed as M xx is a torque and produces torsional shear stresses on the section. The remaining moments M y and M z acting on section x, renamed as M xy, and M xz produce a bending of the body around y or z axis and create linearly varying tensile and compressive stresses on the section. These moments are known as bending moments on a section A- A. Thus, we see that same moments can be twisting or bending moments, depending on the orientation of the section considered. y-axis A

A My

Mx

x-axis

Mz

z-axis A

A

Figure 4.14(f) : A Section A-A Parallel to y-z Plane Cut and Moments Exposed at A-A

13


M x =Algebraic sum of the moments of all the forces acting on the left portion of the body, taken about x-axis. M y =Algebraic sum of the moments of all the forces acting on the left portion of the body, taken about y-axis. M z =Algebraic sumof the moments of all the forces acting on the left portion of the body, taken about z-axis.

Let us now study another effect of load applied on the cantilever. The cantilever is liable to bend due to the load on it. The cantilever has a tendency to rotate in clockwise direction about A (Figure 4.14(g)). Hence, the fixed support of A has to offer a resistance against this rotation. P D A

B C

MA =Pl1

l1

RA =P

l l2

Figure 4.14(g)

Moment of the applied load P about  A is equal to Pl1 (clockwise). For the equilibrium of the cantilever, the fixed support at A will provide a reacting or resisting anticlockwise moment of Pl1. Now consider, for instance, the section D. Suppose the part DB was free to rotate about D. Obviously the load on the part DB would cause the part DB to rotate in a clockwise order about D. Moment of any force is calculated as the product of the force and the perpendicular distance between the line of action of force and the section (or point) about which moment is required. Considering the part DB, taking moments about D, we find that there is a clockwise moment of P (l1 l2) about D. Hence, for the equilibrium of the part DB it is necessary that the part DA of the cantilever should provide a reacting or resisting anticlockwise of P (l1 l2) about D. Taking moments about D, considering the part DA, we have the following moments about D. (a)

RA DA =P

(b)

Couple =P

(clockwise)

l2

(anticlockwise)

l1

Net moment about D =Pl1 Pl2 =P (l1 l2) (anticlockwise). P

D

B

P (l1 l2) C A

D

MA =Pl1

P (l1 l2)

D RA =P

P (l1 l2)

Figure 4.14(h)

14

P (l1 l2)

Hence, for equilibrium of the part DA, the part DB should provide a clockwise moment of P (l1 l2). Hence, at the section D, the part DB provides a clockwise moment of P (l1 l2) and the part DA provides an anticlockwise moment of P (l1 l2). This moment is known as bending moment at D.


Thus, the bending moment (briefly written as BM ) at the cross section of a beam may be defined as the algebraic sum of the moment of the forces, to the right or left of the section.

While calculating the shear force or bending moment at a section, the end reactions must also be considered along with other external loads.

4.3.2 Sign Conventions Sign Convention for Shear Force (SF)

Since the shear force is the unbalanced vertical force, therefore it tends to slide one portion of the beam, upward or downward, with respect to the other as shown in Figure 4.15(a). Right Side

Left Side Right Side

Left Side

Section

Section

Positive Shear Force

Negative Shear Force

Figure 4.15(a) : Sign Convention for Shear Force (SF)

It is said to be positive, at a section, when the left hand portion tends to slide upwards with respect to the right hand portion. Or, in other words, all the upward forces to the left of the section will cause positive shear. It is said to be negative, at a section, when the left hand portion tends to slide downwards with respect to the right hand portion. Or in other words, all the downward forces to the section will cause negative shear. On the other hand, all the upward forces to the right of the section will cause negative shear and those acting downwards will cause positive shear. Sign Convention for Bending Moment (BM)

At sections, where the bending moment is such that it tends to bend the beam at that point to a curvature having concavity at the top as shown in Figure 4.15(b) is taken as positive, and where the bending moment is such that it tends to bend at that point to a curvature having convexity at the top as shown in Figure 4.15(b) is taken as negative. On the other hand, a loading which causes the beamto hog, will create negative bending moment, a loading which causes the beam to sag, will create positive bending moment.

Positive Bending Moment (Sagging)

Left Side

Section

Right Side

Left Side

Right Side

Negative Bending Moment (Hogging)

Section

Figure 4.15(b) : Sign Convention for Bending Moment (BM)

15

In other words, the BM is said to be positive, at a section when it is acting in an anticlockwise direction to the left and clockwise direction to the right of the section. The BM is said to be negative, at a section, when it is acting in clockwise direction to the left and an anticlockwise direction to the right of the section.


4.3.3 Relation between Loading, Shear Force and Bending Moment Consider a beam subjected to external loading of intensity w per unit length. Consider a small portion of the beam between sections 1-1 and 2-2, x apart, at a distance x from the left end support as shown in Figure 4.16. The load acting on the small portion is equal to w x. Let F and M be the SF and BM at left end to the element andF + F and M + M be the SF and BM at right end of the element. The forces and moments acting on the element of the beam are as follows : (i)

upward force F at section 1-1,

(ii)

downward force F + F at section 2-2,

(iii)

downward loadw x,

(iv)

anticlockwise moment M at section 1-1, and

(v)

clockwise moment M + M at section 2-2.

Since the element of the beamis in equilibrium, therefore the systemof forces and moments acting on the element, must obey the laws of equilibrium. Now equating the unbalanced vertical force at section 2-2, we get F + F =F w x

or

F x

. . . (i)

w

Thus, the rate of change of shear force (or in other words, the slope of shear force curve) is equal to the intensity of the loading. 2

1

x

w/unit length

2

1 x

2

1

M

w/unit length

M+ M F

1

x

2

F+ F

Figure 4.16

Taking moments about the section 2-2, M + M = M +F x

w( x )2

2

Ignoring higher powers of small quantities and simplifying the relation, we get M + M = M +F x

or 16

M x

F

. . . (ii)

Thus, the rate of change of bending moment (or in other words, the slope of the bending moment curve) is equal to the shear force at the section.


It is clear that SF curve can be obtained by integrating the loading curve and BM curve can be obtained by integrating the SF curve.

4.3.4 Maximum Value of Bending Moment If the BM diagramis a continuous curve, where

dM dx

=0, the SF will be zero, and BM

will be maximum positive or maximum negative. For maximum value of bending moment, M x

=0

In other words, for maximum bending moment, the shear force is zero. But in actual practice, the bending moment may be maximum where shear force changes sign. However, the above relation is very important to obtain the maximumvalue of bending moment over the beam. Equating the shear force at a distance of x from the support to zero, we get the point where the maximum bending moment will occur. Taking the moments at that section, maximum bending moment can be obtained.

4.4 SHEAR FORCE AND BENDING MOMENT DIAGRAMS A Shear Force diagramfor a structural member is a diagramwhich shows the values of shear forces at various sections of the member. A bending moment diagram for a structural member is a diagram which shows the values of bending moment at various sections of the member. The shear force and bending moment, at any section, can be obtained analytically. The values of shear force and bending moment are plotted as ordinates against the position of the cross section of the beam as abscissa. Such ordinates are drawn at important points of the beam and a straight line or a curve is drawn joining the tops of all such ordinates. These diagrams give the clear picture of the distribution of shear forces and bending moment along the length of the beam. Procedure for Drawing the SF Diagram (SFD) and BM Diagram (BMD) Step 1

Find the reaction of any one of the support, by taking moments about the hinged or pinned support and equating to zero ( M =0). Step 2

Find the reactions of other supports by means of static equilibrium equation ( V =0, H =0). Step 3

Draw a base line of length equal to the length of the beam to some scale. Step 4

Starting from one end (left end), calculate the shear force at all salient points. If there is a vertical load (including reaction) at a section, calculate the shear force just left as well as right of the vertical load (or reaction). The shear force diagram will increase or decrease suddenly at the point of application of load depending upon the direction of load, i.e. by a vertical straight line at the section. 17


Step 5

Calculate the bending moment at salient points. Step 6

Plot the positive values of shear force and bending moment above the base line and negative below it. The shear force diagram and bending moment diagram can be obtained by joining these ordinates. The following points should be kept in mind for drawing the shear force and bending moment diagrams : (a)

If there is no loading on the portion of the beam the shear force diagramwill be horizontal. The bending moment diagram will be inclined. The shear force diagram will remain unchanged between any two vertical loads provided there is no loading between the vertical loads.

(b)

If there is a uniformly distributed load between two sections, the shear force diagram will be inclined straight line and the bending moment diagram will be a curve.

(c)

The bending moment will be zero at the free end of the cantilever and at the two supports of simply supported beam.

(d)

The bending moment diagram will consist of either straight lines or smooth curves.

Cantilevers Cantilevers with a Point Load at the Free End

Let us consider a cantilevers beam AB of length l subjected to a point load W at the free end as shown in Figure 4.17. W

x

A

B x l x ( )

W

W

W

SFD Base Line

Wl

W.x

( ) BMD

Figure 4.17

Reaction at the support A, R A =W (since, V =0). Consider a section XX at a distance x from the free end. SF at this section is equal to the unbalanced vertical force either to the right or to the left of the section. Consider right side of the section, F x =+W (plus sign indicates right downward). The bending moment at this section, M x = Wx (minus sign indicates right anticlockwise). The SF just on the left side of the point load is +W . Since there is no loading between A and B, the shear force will remain sameover the length of the beam, i.e. the shear force diagram will be horizontal for the length AB. We know that M x = Wx, then we get 18

BM at B, when x is zero, M B =0 B


BM at A, when x is l, M A = Wl Since there is no other load between A and B, the bending moment diagram will be inclined, and the bending moment equation is straight line equation. Cantilever with a Uniformly Distributed Load

Consider a cantilever beam AB of length l carrying uniformly distributed load of w per unit run, over the entire length of the beamas shown in Figure 4.18. w/unit length x

A

B x L

Wl

x

( )

W.x SFD

wl2 2

w.x2 2

BMD

( )

Figure 4.18

Consider a section XX at a distance x from the right end B, SF at this section, F x =+w x (plus sign indicates right downward) SF at B, F B =0

(at x =0)

SF at A, F A =+w l

(at x =l )

Reaction at the support A, R A =wl. Since the loading is uniformly distributed load, the SF diagram will be inclined and also, the shear force equation is straight line equation. BM at section xx, M x = w

l

l

2

(minus sign indicates right

anticlockwise). SF at B, M B =0 SF at A, M A =

(at x =0)

wl 2

2

(at x =l )

The bending moment diagram will be in the form of parabolic curve. Cantilever with a Gradually Varying Load

Consider a cantilever beam AB of length l, subjected to a gradually varying load from zero at the free end tow per unit length at the fixed end, as shown in Figure 4.19. Reaction at the support A, R A

Total downward load 19

1 l 2


wl

w (area of loading diagram)

(plus sign indicates right downward)

2

w.x l

w/unit length

B A

x

L

wl 2

w.x2 2l

( ) SFD

wl 2 6

BMD

( )

w.x3 6

Figure 4.19

Consider a section XX at a distance  x from the free end. Shear force at section XX 1 2

F x

w l

x

x (area of loading between B and xx)

wx 2

2l SF at B,

F B =0, when x =0,

SF at A,

F A

B

wl

2

, when x =l.

The shear force diagram is in the formof parabolic curve as the shear force equation is parabolic equation. Bending moment at section XX 1 w 2 l

M x

x

x

1 x (Moment of the area loading between B and XX ) 3

wx 3

(i.e. area CG of loading diagram about XX )

6l

BM at B, M B =0

(when x =0), and

B

BM at A, M A

wl 2

6

(when x =l).

x w x l

20


Figure 4.20

The The bending ing moment diag iagram is in the for form of parabolic curve. Example 4.1

A canti cantillever beam of 8 mlength is subje bjected to point point loa loads ds of 10 kN, kN, 15 kN, kN, 25 kN and and 20 kN at dista distances nces of 2 m, 4 m, 6 m and and 8 m, res respe pecti ctivel vely y from from the fixed end. end. Draw Draw the shear shear forc force e di diagram agram and bendi bending ng moment ent diagram diagram. Solution

Reaction ction at the support support A = R A =10 +15 +25 +20 =+70 kN. Shear Force

SF at A =F A =+70 kN SF just left of C =+70 =+70 kN SF just right of C =+70 =+70 10 =+6 =+ 60 kN SF just left of D =+60 kN SF just right of D =+60 =+60 15 =+4 =+ 45 kN SF just left of E =+45 =+45 kN SF just right of E =+45 =+45 25 =+2 =+ 20 kN SF just left of B =+20 kN SF just left of B =+20 =+20 kN (consi (consideri dering ng righ rightt si sideof the section) ction) 15 kN

10 kN A

2m

2m

2m D

C

R A = 70 kN

10 70

25 kN

8m

20 kN 2m B

E

15

25

( )

20 SFD

( )

40 130

390 250

BMD

Figure 4.21

Bending Moment

BM at B, M B =0

(since (si nce the momen omentt at the free end end is equa equall to zero)

BM at E , M E = 20 2

(conside (consideri ring ng the the righ rightt side) side)

= 40 kN m BM at D, M D = (20 (20 4)

(25 (25 2)

= 130 kN m BM at C , M C 6) C = (20 6)

(25 4) 4)

(15 2) 2)

21

= 250 kN m


BM at A, M A = (20 8) 8)

(25 6) 6)

(15 4) 4)

(10 2) 2)

= 390 kN m. Example 4.2

Draw Draw the shea shear force f orce diag diagram (SF (SFD) and bending bending mom moment diag diagram (BM (B MD) for the beam shown in Figure 4.22. Solution

Reaction at the support A, R A =+4 =+4 +(1 +(1.5 6) =+13 =+13 kN. Shear Force

SF at A, F A =+13 =+ 13 kN (consi (considering ring righ rightt side) SF at C , F C C =+13

1.5

6

=+4 kN SF just left of D =+13 1.5 6 =+4 kN SF just right of D =+4 =+4 4 =0 1.5 kN/m A R A = 13 kN

C

6m

13 kN

4

( )

4 kN

2m

D

2m

B

4

SFD

( )

8

59

Figure 4.22

Bending Moment

BM at B, M B =0 (since (si nce,, the moment ent at the free end end is is equal equal to zero) BM at D, M D =0 (consideri (considering ng righ rightt side of D, there is no loading on BD) BM at C , M C 2 = 8 kN m ( ve sign due to righ right C = 4 anticlockwise) BM at A, M A = 4 8 1.5 6

6 = 59 kN m. 2

Example 4.3

A canti cantillever bea beam carries carries a uniformly unif ormly distri di stribute buted load of 2t /mover the enti entire re length of 6 m and point loads of 5 t , 3 t , 7 t and and 2 t at at a di distance of 2 m, m, 4 m, 5 m, and and 6 m, re respectively ctively from the fixed end. Draw Draw SFD and BMD BM D for f or the beam. Solution

Reaction at the support A, R A =+5 =+ 5 +3 +7 +2 +(2 6) =29t Shear Force : Starting from end A,

SF at A, R A =29 t 22


SF just left of C =+29 2 2 =+25 =+ 25 t SF just right of C =+25 =+25 5 =+20 =+ 20 t SF just left of D =+20 =+ 20 2 2 =+16 t SF just right of D =+16 3 =+13 =+ 13 t SF just left of E =+13 =+13 2 1 =+11 t SF just right of E =+11 =+11 7 =+4 =+ 4 t SF just left of B =+4 =+4 2 1 =+2 =+ 2 t SF just right of B =+2 =+ 2 t

(consi (considering ring righ rightt side) 2 t/m 5t

3t 2m

2m A R A = 29 t

C

1m

1m D

6m

2t

7t

B

E

5 29

3

( )

20

7

13

4

2

SFD

105

3

15

( ) 51

Figure 4.23

from B, Bending Moment : Starting from BM at B, M B =0 BM at E , M E = 2 1 2 1

1 = 3 t m 2

BM at D, M D = 2 2 7 1 2 2

2 = 15 t m 2 4 = 51 t m 2

BM at C , M C 4 7 C = 2

3 3 2 2 4

BM at A, M A = 2 6 7

5 3 4 5 2 2

6

6 = 105 t m. 2

SAQ 1 (a)

A canti cantillever beamof len length 8 m carrying carryi ng a u.d.l. u.d.l . of 3 kN/m kN/m over a length ength of 6 m from rom free end and 1.5 kN/m kN /m on a span span of 2 m at a distan distance ce of 2 m from the fixed end and a point load of 6 kN at a distance of 1 m from the fixed end. end. Draw Draw the SFD SFD and BMD BM D for f or the canti cantillever beam.

(b)

A cantil cantileve ever of 3 m length ngth is loade oaded with with a uni uniforml ormly varying varying load oad of intensity 2000 N/m at free end to 600 N/m at fixed end. Draw SFD and BMD for the cantilever.

23


(c)

A cantilever beam5 m long caries point loads of 2 kN, 3 kN, and 3 kN at 1 m, 3 m and 5 mrespectively from the fixed end. Construct SFD and BMD for the beam.

(d)

Draw SFD and BMD for the cantilever beam of 3 m long which carries a uniformly distributed load of 2 kN/m over a length of 2 m from the free end.

(e)

A cantilever beam 5 m long carries point loads of 3 kN, 3 kN and 3 kN at distances of 1.5 m, 3 m and 4.5 m respectively from the fixed end. In addition to this the beam carries an uniformly distributed load of 1 kN/m over the entire length of the beam. Draw SF and BM diagrams for the beam.

Simply Supported Beams

In case of simply supported beam, the point of contraflexure is very important. Where the BM changes sign at some point, the bending moment will be zero and that point is called the point of contraflexure. Simply Supported Beam with a Point Load at its Mid-span

Let us consider a simply supported beam AB, span l, carries a point load W at its mid-point C  as shown in Figure 4.24. w

x C

A RA = w/2

B RB = w/2

l/2

l/2 l

x 1000 w/2

x

( ) ( )

w/2

SFD

( )

Wl/4

BMD

Figure 4.24

Let us first find the reaction at the support B by taking moments about A, and equating to zero. R B

B

l

R B =+

l

W

2

=0

W

2

R A =W R B =W B

W

2

=+

W

2

(On the other hand, since the load is at the mid-point of the beam, the reactions at the supports R A = R B =W /2.) B

24

Consider a section XX between B and C at a distance  x from the end  B. The shear force equation is W

F x =


(  sign indicates right upward)

2

This equation is valid for all values of x from 0 to

l

2

. Since there is no loading

between B and C , the shear force diagram is horizontal, i.e. SF remains unchanged. Consider a section XX between A and C at a distance  x from the end  B. The shear force equation is W

F x =

2

+W =+

W

2

This equation is valid for all values of x from

l

2

to l. The shear force diagram is

horizontal. Alternatively, the shear force equation can also be written by considering a section between A and C at a distance  x from the end A. The bending moment at A and B is zero. I t increases by a straight line law, and is maximum at the centre of the beamwhere shear force changes sign. The bending moment equation is M x =+

BM at C ,

M C =

W

2

Wl

4

. x

(+ sign indicates right clockwise)

when x =

l

2

Simply Supported Beam with an Eccentric Point Load

Consider a simply supported beam AB of length l subjected to eccentric point load W at C at a distance a from the end A and B from the end B as shown in Figure 4.25. w

C

A RA =

x B b

a

Wb

RB =

l

l

Wa l

x

Wb

x

( )

l

( )

l

SFD

( )

Wa

Wab l

BMD

Figure 4.25

To find the reaction at the support B, taking moment about  A R B

B

l

R B =+

W a =0 Wa l

25


Wa

R A =W

=+

l

W (l

a)

l

=+

Wb l

(where l  a =b)

Consider a section XX between B and C at a distance x from the end B, i.e. x
Wa

( sign indicates right upward)

l

This equation is valid for the portion BC . The shear force diagram is horizontal. Consider a section XX between A and C at distance x from the end B, i.e. x >a. The shear force equation is as follows : F x =

Wa l

+W =

W (l

a)

l

=+

Wb l

(where l  a =b)

This is valid for the portion AC . Alternatively, the shear force equation for the portion AC may also be obtained by considering the section between A and C at a distance  x from the end A. The bending moment equation for the portion BC is as follows : M x =+

Wa

x (+sign indicates right clockwise)

l

The bending moment at A and B is zero. I t increases by a straight line law, and is maximumat C where the SF changes sign. BM at C ,

M C =+

Wab l

(at x =b)

Simply Supported Beam with a Uniformly Distributed Load

Consider a simply supported beam AB of span l carrying a uniformly distributed load of w per unit length over the entire length as shown in Figure 4.26. x

W/Unit Length C RA

B

A Wl 2

Wl 2

l/2

l/2

RB

l

Wl 2

x x ( ) ( )

SFD

Fx

Wl 2

Mx

Wl2 8 BMD

Figure 4.26

Let us find the reaction at the support B, taking moments about A and equating to zero, R B

l

R B =+

w

l

l

2

wl

2

R A =wl R B =wl B

26

=0

wl

2

=+

wl

2

Consider a section XX at a distance x from the end B.


The shear force equation, F x =

SF at B,

F B =

SF at C ,

F C =

SF at A,

F A =

B

wl

2

+wx

wl

2 wl

2 wl

2

(at x =0)

l

+w

2

+wl =+

wl

The shear force at B is equal to

=0

wl

2

(at x =

l

2

)

(at x =l)

and increases uniformly by a straight line law

2

wl to zero at C and continuous to increase uniformly to + at A.

2

The bending moment at section xx, M x =

=

wl

2

w x

x

wl

2

x

2

wx 2

x

2

BM at A and B is zero M A = M B =0 B

M C =

=

wl

l

wl

l

2

2

2

2

2

wl 2

wl 2

wl 2

4

8

8

Since the shear force is zero at mid point of the beam, the maximum bending moment occurs at C  M max = M C =+

wl

2

8

The bending moment diagram in the formof parabolic curve as the bending moment equation is parabolic equation. Simply Supported Beam with a Triangular Load Varying Gradually from Zero at both Ends to w per metre at the Centre

Consider a simply supported beam AB of length l, subjected to a triangular load, varying gradually from zero at both ends to w per unit length at centre as shown in Figure 4.27.

W/m x

2Wx l

27 B

A l/2

l /2

R

Wl

Forces and Stresses in Beams RA

Wl 4

Figure 4.27

Since the load is symmetrical, therefore, the reactions R A and R B will be equal. B

R A

1 2

RB

1 2

w

wl

l

4

Taking moments about A, R B

1 2

l

l

l

0

2

wl

R B R A

w

4 1 2

w

l

RB

wl

wl

wl

2

4

4

Consider a section XX at a distance  x from the end B. The shear force at XX , F x

SF at B, F B SF at C , F C

4

1 2

wl

wx 2

4

l

wl

wl

4

2wx

x

l

(at x =0)

wl

w

4

l

wl

wl

4

4

.

l

2

l

at x

2

2

0

The shear force diagram is in the formof parabolic curve as the shear force equation is parabolic equation. The bending moment at XX , M x

wl

4 wl

4

28

x

x

1 2 wx 3

3l

x

2wx

x

l

3

The BM at A and B is zero. The bending moment will be maximum, where the SF 1 changes sign. In this case, the maximum bending moment occurs at x , i.e. 2 at C . M max

M C

3

wl

l

w

l

4

2

3l

2

wl 2

wl 2

8

24

2wl 2 24


wl 2

12

Since, the bending moment equation is cubic equation, the bending moment diagram is in the form of cubic curve. Simply Supported Beam with a Gradually Varying Load, from Zero at One End to w per metre at the other End

Consider a simply supported beam AB of span l carries a gradually varying load from zero at one end to w per unit length at the other end as shown in Figure 4.28. Taking moments about A and equating to zero, R B

l

w

l

l

0

3

wl

R B R A

1 2 6

1 2

w

l

1 wl 2

RB

wl

wl

6

3 x

Wx 1

W/m

B

A Wl RA 3

l

RB

Wl 6

x x

Wl 3

0.5771 ( )

C

Fx ( )

Wl 6

SFD

Wl 2 9 3

Mx

BMD

Figure 4.28

The shear force at any section XX at a distance x from the end B. F x

SF at B, F B SF at A, F A

1 2

wl

6 wl

6

w l

x

x

wl

wx

6

2l

2

. . . (i)

(at x =0)

wl

wl

wl

6

2

3

(at x =l)

The bending moment at any section XX at a distance x from the end B. 29


wl

M x

6

M x

1 2

x

w

x

l

wlx

wx3

6

6l

x

x

3 . . . (ii)

The BM at A and B is zero (when x =l and x =0). To find the maximum bending moment, equate the shear force Eq. (i) to zero. wl

wx 2

6

2l

wx 2

wl

l

3

x 2 x

0

l2

3 l

M max

3

0.577 l

M C

3

wl

l

w

l

6

3

6l

3

wl 2

wl 2

2wl 2

wl 2

6 3

18 3

18 3

9 3

0.06415 wl 2

The shear force diagram is in the form of parabolic curve as given by in Eq. (i). Since, the power of  x in the moment equation is of order 3, the bending moment

diagram is in the form of cubic curve. Example 4.4

A simply supported beam of 7 m length carries point loads 2 kN, 4 kN and 6 kN at distances of 1 m, 2 m and 4 m from the fixed end respectively. Draw SFD and BMD and also calculatethe maximum bending moment that will occur. Solution

Taking moments about A to find R B, B

R B

Thus,

7 (6 4) (4 2) (2 1) =0

R B =4.857 kN, and R A =12 4.857 =7.143 kN

Shear Force (Starting from left end A)

SF at A, F A =+7.143 kN SF just left of C =+7.143 kN SF just right of C =+7.143 2 =+5.143 kN SF just left of D =+5.143 kN SF just right of D =+5.143 4 =+1.143 kN SF just left of E =+1.143 kN SF just right of E =+1.143 6 = 4.857 kN SF just left of E = 4.857 kN =Reaction at B Bending Moment (Starting from right end B)

BM at A, M A =0 BM at E , M E =+4.857 3 =+14.571 kN m BM at D, M D =+(4.857 5) (6 2) =+12.285 kN m 30


BM at C , M C =+(4.857 6) (6 3) (4 1) =+7.142 kN m Or

M C =7.143

1 =+7.143 kN m (considering left side)

BM at A, M A =0 2kN A

1m

4kN

6kN

1m

2m

C

RA = 7.143

3m

D

B

E

RB = 4.857 KN

7m

2 7.143

5.143

( )

4 1.143

1.143 ( )

4.587

4.587 SFD 14.571 12.285 7.143

( )

BMD

Figure 4.29

Maximum Bending Moment

Maximum bending moment will occur at a point where the shear force changes sign. Here, SF changes from positive to negative at E . The Bending Moment at E will be maximum bending moment. Thus,

M max =+14.571 kN m

Example 4.5

A 12 mspan simply supported beamis carrying a uniformly distributed load of 2 kN/m over a length of 6 m from the left end and point loads 6 kN, 3 kN and 4 kN at distances of 7 m, 8 m and 9 m, respectively. Draw SF diagramand BM diagram for the beamand find the maximum bending moment. 2 kN/m

6 kN

1m

1m

A

C

6m

3 kN

D

4 kN 3m

1m E

B

F

12m

13.5 ( )

1.5

4.5

( ) 7.5

SFD

11.5 46.5

45

42

34.5

( )

BMD

Figure 4.30

31


Solution

Taking moments about A, and equating to zero. R B

Thus,

12 (4 9) (3 8) (6 7) 2 6

6 =0 2

R B =11.5 kN, and R A =(2

6) +6 +3 +4 RB =25 11.5 =13.5 kN B

Shear Force (Starting from the left end)

SF at A, F A =+13.5 kN SF at C , F C =+13.5 2 6 =+1.5 kN SF just left of D =+1.5 kN SF just right of D =+1.5 6 = 4.5 kN SF just left of E = 4.5 kN SF just right of E = 4.5 3 = 7.5 kN SF just left of F = 7.5 kN SF just right of F = 7.5 4 = 11.5 kN SF just left of B = 11.5 kN =Reaction at B. Bending Moment

BM at A, M A =0 BM at C , M C

(13.5 6)

2

6

6 2

45 kNm (considering left side)

BM at D, M D =(11.5 5) (4 2) (3 1) =46.5 kN m (considering right side) BM at E , M E =(11.5

4) (4 1) =42 kN m

M F =(11.5

3) =34.5 kN m

BM at F ,

The maximum bending moment occurs at D where the shear force changes sign. M max =46.5 kN m.

Example 4.6

A simply supported beam of length 8 m is subjected to a uniformly varying load from zero at centre to 3 t/m at both ends. Draw the SFD and BMD. 32

Solution

Taking moment about A, R B


1 2 4 3 4 4 2 3

8

1 1 4 3 4 2 3

0

R B =6 t B

R A

1 4 3 2

2

12

RB

6 60

Alternatively, since the load is symmetrical, the reaction at the support will be equal to half of the total downward load. R A

6 t

RB

¾(4 x) x

3t/m A

B

C

4m

3t/m

4m

8m x

4 x x

Parabolic Curve 6

( )

Fx ( )

6

SFD

Cubic Curve

Mx 8t.m

Figure 4.31

Shear Force

SF at A,

F A =+6 t

SF at C ,

F C

SF at B,

F B

6

1 4 3 0 2 1 4 3 2

0

6 t =Reaction at B.

To find the form of SFD, consider a section XX at a distance  x from the end B. F x

F x

6

1 2

6

1 x 2

6

x

8

x

3 12

3 (4 x) 4 3(4 x) 4

(12 12 3x )

48

24x 8

3x 2

Since, the SF equation is parabolic equation, the SFD is in the form of parabolic curve. Bending Moment

33

BM at A and B, M A = M B =0


B

1 2 4 3 4 2 3

(6 4)

BM at C , M C

24 16 8 t m

Maximum bending moment occurs at centre, since, the SF is zero at centre of the beam. Bending moment at XX , 3 (4 x) (2 3) 4 (24 3x) 3 8 (4 x) 3 4

x

6x

M x

6 x

6 x

6 x

6 x

M x

x

8

(24 3x)

x 2

24 x 2

24 x

3(4 x) 24 3(4 x) 12

x

3

x

3

(24 3x)

3(4 x ) 24 3(4 x) 12

(24 3x)

36 3x 24 3 x

2

24

(36 3x)

6 x

3 x 2 24

3x3 24

6x

3 x 2 2

x3

8

This equation indicates that the bending moment diagram is in the formof cubic curve. Example 4.7

Construct the SFD and BMD for 10 m span simply supported beam subjected to a system of loads as shown in Figure 4.32. 1.2 kN/m

2 kN/m 1 kN

A 11.6 kN

C

2m

3m

5 kN

4 kN

D

E

10m

3 kN

F

2m

2m

B 15.4 kN

11.6 9.2 6.2 ( )

4.6 0.4

( )

0.4 4.4

8.4 11.4

SFD 40

15.4 39.6 26.8

20.8

( )

Figure 4.32

34

Solution

Taking moment about A and equating to zero. R B

10 2 4

6

R B =

154 =15.4 kN 10

4 2


3 8 4 6 5 5 1.2 5

5 1 2 0 2

R A =(1.2 5) +1 +5 +4 +3 +(2 4) R B R A =27 15.4 =11.6 kN.

Shear Force (starting from left end)

SF at A, F A =+11.6 kN SF just left of C , F C =+11.6 (1.2 2) =+9.2 kN SF just right of C , F C =+9.2 1 =+8.2 kN SF just left of D, F D =+8.2 (1.2 3) =+4.6 kN SF just right of D, F D =+4.6 5 = 0.4 kN SF just left of E , F E = 0.4 kN SF just right of E , F E = 0.4 4 = 4.4 kN SF just left of F , F F = 4.4 (2 2) = 8.4 kN SF just right of F , F F = 8.4 3 = 11.4 kN SF just left of B, F B = 11.4 (2 2) = 15.4 kN B

Bending Moment

BM at A and B, M A = M B =0 B

2 2

BM at F , M F

(15.4 2)

2 2

BM at E , M E

(15.4 4)

(3 2)

2 4

BM at D, M D

(11.6 5)

(1 3)

1.2 5

BM at C , M C

(11.6 2)

1.2 2

26.8 kN m=+39.6 kN m 2 2

39.6 kN m

5 40 kN m 2 (considering left side)

2 2

20.8 kN m (considering left side)

Since the SF changes sign at D, the maximum bending moment occurs at D. M max = M D =+40 kN m

SAQ 2 (a)

A beam6 m long is simply supported at the ends carries a uniformly distributed load of 2 kN/m over the middle 2 m length and point loads of 1 kN and 4 kN at distances of 1 m and 5 m from the left end respectively. Draw SFD and BM D and determine the magnitude and position of the maximum BM.

(b)

A beam simply supported at its ends has a span of 6 m. It is loaded with a gradually varying load of 750 N/m from the left hand support to 1500 N/m 35

to the right hand support. Construct the SF and BM diagrams and find the magnitude and position of the maximum BM over the beam.


(c)

A simply supported beam of 6 m span is loaded with a uniformly distributed load of 1.5 kN/m over the entire span and concentrated load of 4 kN and 5 kN at distances of 2 mand 4 mfrom the left hand support respectively. Find the magnitude and position of the maximum BM.

(d)

Draw the shear force and bending moment diagram for the simply supported beamshown in figure below. I ndicate the numerical values at all salient points. 1000 N

A

1.5 m C

2000 N

1500 N/m

3000 N

1.5 m D 1.5 m

B

E 1.5 m

6m

Figure For SAQ 2(d)

Overhanging Beams

An overhanging beam is a beamwhich overhangs from the support either one side or both sides. The overhanging portion of the beam will be treated as cantilever. Point of Contraflexure

For the purpose of shear force and bending moment, the overhanging beamwill be analyzed as a combination of simply supported beamand acantilever beam. We have discussed that the BM is negative for the cantilever and positive for the simply supported beam. It implies that in an overhanging beam, there is a point at which the BM changes from positive to negative and vice-versa. Such a point where the BM changes sign is known as contraflexure. There may be one or more points of contraflexure. Example 4.8

Draw the shear force and bending moment diagrams for the overhanging beam shown in Figure 4.33. x

0.8 kN/m

5 kN

A

6 kN

D 1m

2m

E

6m

6

1.93

( ) 3.07

( ) 3.07

5.47

SFD

3.86

0.81

( ) ( ) BMD 12

Figure 4.33

Solution

Taking moments about A and equating to zero, 36

2m

C

( )

6

x

x

1.93

B

3m

R B

6 (6 8)

0.8 3

3

3 2

R B =11.47 kN and R A =5 +(0.8

(5 2)


0

3) +6 11.47 =25 11.5 =1.93 kN

Shear Force (Starting from the left end A)

SF at A, F A =+1.93 kN SF just left of D, F D =+1.93 kN SF just right of D, F D =+1.93 5 = 3.07 kN SF at E , F E = 3.07 kN SF just left of B, F B = 3.07 (0.8 3) = 5.47 kN B

SF just right of B, F B = 5.47 +11.47 =+6 kN B

SF just left of C , F C =+6 kN =L oad at E Bending Moment

BM at C , M C =0 BM at B, M B = 6 2 = 12 kN m (11.46 3)

BM at E , M E

(6 5)

0.8 3

3 2

0.81kN m

BM at D, M D =+1.93 2 =+3.86 kN m BM at A, M A =0 Maximum Bending Moment

Maximum positive bending moment occurs at D and maximumnegative bending moment occurs at B. M max (positive) =+3.86 kN m M max (negative) = 12 kN m

Point of Contraflexure

Since the bending moment changes sign between E and B, consider a section XX between E and B at a distance x from C . BM at section XX , M x = 6 x +11.47 ( x

2) 0.8 ( x

= 6 x +11.47 x

22.94 0.4 ( x

2)

( x

2) 2

2)2

= 0.4 x2 +7.07 x 24.54 By equating this equation to zero, we get the point of contraflexure, 0.4 x2 +7.07 x x

2

24.54 =0

17.67 x +61.35 =0

Solving by trial and error, x =4

Value of ( x2 17.675 x +61.35) =6.65

x =4.9

Value of ( x2 17.675 x +61.35) = 1.2475

x =4.8 Value of (x2 17.675 x +61.35) = 0.45 x =4.7

Value of ( x2 17.675 x +61.35) =0.3675

x =4.75

Value of ( x2 17.675 x +61.35) = 0.04375

37

Valueof ( x2 17.675 x +61.35) = 0.00285

x =4.745


Point of contraflexure is at a distance of 4.745 m from the end C . Example 4.9

Draw the shear force and bending moment diagramfor 10 m span overhanging beam having overhanging portion of 4 m subjected to a system of loads as shown in Figure 4.34. Calculate the maximum bending moment and also locate the point of contraflexure. Solution

Taking moments about A and equating it to zero, R B

6 (8 10)

3 4

4 2

6

(20 4)

(10 2)

0

R B =46 kN R A =10 +20 +(3

B

x

10 kN

x

3 kN/m

20 kN

2m

2m A

4) +8 R B =50 46 =4 kN

2m

D

C

B

E

8 kN

4m

40 x 20 4

( )

( )

4 6

( ) 26

26 SFD ( )

( )

4

56

Figure 4.34

Shear Force (Starting from the end A)

SF at A, F A =+4 kN SF just left of D, F D =+4 kN SF just right of D, F D =+4 10 = 6 kN SF just left of E , F E = 6 kN SF just right of E , F E = 6 20 = 26 kN SF just left of B, F B = 26 kN B

SF just right of B, F B = 26 +46 =+20 kN B

SF just left of C , F C =+20 kN =Load at C Bending Moment

BM at C , M C =0 38

6

(0.3 10)

BM at B, M A

BM at E , M E =+(4 4)

10 2

1 10

47 kN m


(10 2) = 4 kN m

BM at D, M D =+(4 2) =+8 kN m Maximum Bending Moment

Since SF changes sign at D and B, the maximum positive bending moment will occur at D and maximum negative bending moment will occur at B. M max (positive) =+8 kN m M max (negative) = 56 kN m


BM changes sign between D and E . Therefore, consider a section XX at a distance x from the end A. BM at section XX , M x =+4 x

10 ( x

2)

Equating this to zero, we get 4 x

10 ( x

2) =0

4 x

10 x +20 =0 6 x +20 =0 x =3.333 m

The point of contraflexure is at a distance of 3.333 m from the left end A. Example 4.10

An overhanging beam of 15 mspan is carrying an uniformly distributed load of 1 kN/m over the length of 10 mat a distance 5 mfrom the left free end and point loads 7 kN and 4 kN, at free end and at a distance 2 mfrom the free end, respectively. Sketch the SFD and BMD for the beam. Locate the point of contraflexure. Solution

Taking moments about B and equating to zero, R A

10 (7 15)

(4 13)

1 10

10 2

0

R A =20.7 kN

and

R B =7 +4 +(1

10)

B

20.7 =0.3 kN

Shear Force (Starting from the left end C )

SF just right of C , F C = 7 kN SF just left of D, F D = 7 kN SF just right of D, F D = 7 4 = 11 kN SF just left of A, F A = 11 kN SF just right of A, F A = 11 +20.7 =9.7 kN SF just left of B, F B =+9.7 (1 10) = 0.3 kN =Reaction at the support B. B

Bending Moment

39


BM at B, M B =0 BM at A, M A

(0.3 10)

1 10

10 2

47 kN m

BM at D, M D = (7 2) = 14 kN m BM at C , M C =0 Maximum Bending Moment

Consider a section XX at a distance x from the end B as shown in Figure 4.35. Shear force at section XX , F x = 0.3 +1( x)

For maximum bending moment, the shear force is zero. 0.3 + x =0 x =0.3 m

We have, M x =+0.3( x)

1 ( x)

x

2

=0.3 x

x 2

2

(0.3) 2 =+0.045 kN m 2

M max =0.3 (0.3)

1 kN/m 7 kN C

x

4 kN 2m

3m

B

A

D

10 m x x 9.7 ( ) 0.3

7 ( ) 11

11

0.045 ( ) 14

( )

47

Figure 4.35

It is seen that the maximumnegative bending moment occurs at A and maximum positive bending moment occurs at a distance 0.3 m from the right end B. M max (positive) =+0.045 kN m M max (negative) = 47 kN m


It is observed that the BM changes sign between A and B, 40

BM at any section XX ,


x 2

M x =0.3 x

2

Equating this to zero,

x 2

0.3 x

2

=0

0.6 x x2 =0 x =0.6 m

So, the point of contraflexure is at a distance of 0.6 m from the right end B. Example 4.11

Draw the shear force and bending moment diagrams for 15 m span overhanging beam, which overhangs on both sides. It is subjected to a u.d.l. of 5 kN/m on left side overhanging portion of length 5 m and a u.d.l. of 4 kN/m on right side overhanging portion of length 4 m. Indicate the numerical values at all salient points. Solution To Find Reaction at B

Considering right side, the moments about A, M A

RB

M A =6 R B

6

4 4

6

4 2

128

Considering left side, take moments about A, 5 5

M A

5 2

62.5

Equating these two values, we get 6 R B 128 = 62.5 6 R B =65.5 R B =10.92 kN R A = (5

5) +(4 4) RB

R A =41

10.92 =30.08 kN

B


SF at C , F C =0 SF just left of A, F A =0 5 5 = 25 kN SF just right of A, F A = 25 +30.08 =+5.08 kN SF just left of B, F B =+5.08 kN B

SF just right of B, F B =+5.08 +10.92 =+16 kN B

SF at D, F D =+16

(4 4) =0

5 kN/m

C

4 kN/m

A 5m

D

B 6m

4m 16

41


Figure 4.36

Bending Moment

BM at C and D, M C = M D =0 BM at A, M A = 5 5

5 = 62.5 kN m 2

BM at B, M B = 4 4

4 = 32 kN m 2

Example 4.12

Sketch the shear force and bending moment diagrams for the beam overhanging on both sides as shown in Figure 4.37. Find the magnitude of maximum positive and maximum negative bending moment and also locate the point of contraflexure, if any. Solution

Considering right side, take moments about A, M A = R B

12 2 12 10

M A =12 R B

240

Considering left side, take moments about A, M A = 6

4 = 24 kN m

Equating these two values, we get the reaction at the support B, 12 R B 240 = 24 R B =18 kN R A =6 +(2

12) R B = 30  18 = 12 kN


SF at C , F C =  6 kN SF just left of A, F A = 66 kN SF just right of A, F A = 6 +12 =+6 kN SF at D, F D =+6 kN SF just left of B, F B =+6 (2 8) = 10 kN B

SF just right of B, F B = 10 +18 =+8 kN B

SF at E , F E =+8 (2 4) =0. x

42

2 kN/m

6 kN C

A

D

B

E


Figure 4.37

Bending Moment

BM at C and E, M C = M E =0 BM at B, M B

2 4

BM at D, M D

(18 8)

4 2

16 kN m

2 12

12 2

0

BM at A, M A = (6 4) = 24 kN m Maximum Bending Moment

Consider a section XX at a distance  x from the end E . SF at section XX , F x = 18 +2 x

For maximum bending moment, F x should be zero. 18 +2 x =0 x =9 m

BM at any section XX between D and B, M x =18 ( x

=18 ( x M max =18 (9

4)

2 x

x

2

4) x2 4)

92 =9 kN m

Maximum negative bending moment = 24 kN m Maximum positive bending moment =+9 kN m Points of Contraflexure

Let M and D be the points of contraflexure where the BM changes sign. To find the position of M , equate the moment equation between D and B to zero.

43

4) x2 =0

18 ( x


2 x +18 x 2

or,

x

we get,

72 =0

18 x +72 =0

x =6 m and x =9 m.

Thereare two points of contraflexure, one at a distance of 6 m and the other at a distance of 9 m from the end E .

SAQ 3 (a)

Draw shear force and bending moment diagrams for the beam shown in figure below. Indicate the numerical values at all salient points. 1000 N/m 8000 N

A

4000 N

5m

E

B

D

C 5m

1600 N/m

5m

5m

Figure for SAQ 3(a)

(b)

A simply supported beamwith overhanging ends carries transverse loads as shown in figure below. w/m

W

C

A l

E 10 m

W

B

D l

Figure for SAQ 3(b)

If W =10 w, what is the overhanging length on each side, such that the bending moment at the middle of the beamis zero? Sketch the shear force and bending moment diagrams.

Characteristics of the Diagrams

Shear force diagrams (SFDs) and bending moment diagrams (BMDs) are very useful which give the clear picture of the distribution of shear force and bending moment along the length of the beam. Some important points regarding these diagrams are as follows :

44

(i)

Where the rate of loading is zero, the SF curve will have constant ordinates and BM curve will vary linearly. In other words, if there is no increase or decreases in SF curve between any two points, i.e. SF line is horizontal and consists of rectangle, it indicates that there is no loading between the two points.

(ii)

If there is a sudden increase or decrease, i.e. a vertical line of SF diagram, it indicates that there is either a point load or reaction of the support at that point or in other words, if there is any point load or reaction, it will cause sudden increase or decrease in the SF diagramwith respect to upward or downward direction of point load or reaction, and it will not cause any distinct variation in the BM diagram.

(iii)

Where the intensity of load is constant, i.e. uniformly distributed load, the SF curve will vary linearly and the BM curve will be a parabola. In other words, if the SF line is an inclined straight line between any two points, and the BM curve is a parabolic curve between any two points, it indicates that there is uniformly distributed load between the two points.

(iv)

Where the loading curve is varying linearly, i.e. uniformly varying load, the SF curve will be a parabola and theBM curve will be a cubic parabola. In other words, if the SF line is a parabolic curve between any two points and BM curve is a cubic parabola between any two points, it indicates that there is a uniformly varying load between the points.

(v)

SF and BM diagrams can be drawn by successive integration for complex case of loading.


(vi) The area of loading curve in elementary length will be equal to change in shear force and area of shear force curve in elementary length will be equal to change in bending moment. (vii) The BM will be equal to zero at the free end of the cantilever and at the simply supported ends. (viii) At the intermediate supports, the bending moments are always negative. (ix) The maximum bending moment occurs at the point where the shear force changes sign and the point of contraflexure is the point where the BM changes sign. Sometimes, instead of load diagram, a SF diagram will be given. In such cases, a loading diagramis drawn first. After drawing the load diagramfor the beam, the BM diagrammay be drawn as usual. Example 4.13

Shear force diagram for a loaded beamis shown in Figure 4.38. Determine the loading on the beamand hence, draw the bending moment diagram. L ocate the point of contraflexure, if any. Solution

Let us analyse theshear force diagramgiven in Figure 4.38. At A

The shear force diagram increases suddenly from 0 to 6.875 kN in upward direction, at A. This indicates that there is support at A, having magnitude of reaction 6.875 kN. Between A and C

The SF diagram is an inclined straight line between A and C . It indicates that there is a uniformly distributed load between A and C . The load increases from 6.875 kN to 3.875 kN (6.875 kN  3.875 kN = 3 kN). Thus, the beam 3 carries a uniformly distributed load of 2 kN/m between A and C . 1.5 At C

The shear force diagram suddenly decreases from 3.875 kN to 1.875 kN. It indicates that there is a point load of 2 kN (3.875 kN  1.875 kN) acting in downward direction at C . 6.875

( )

A

3.875 1.875

1.875

1.5

B

E

D

C

1.5

1.125

( ) 1.5

1.5 9.125

45


Figure 4.38

Between C and D

Since the shear force diagram is horizontal between C and D, there is no load betweenC and D. Between D and E

The SF diagram is an inclined straight line between D and E . It indicates that there is a uniformly distributed load. Load decreases from +1.875 kN to  1.125 kN. Therefore, the beam carries a uniformly distributed load of (+ 3 2 kN/mbetween D and E . 1.875 +1.125 =3 kN) 1.5 At E

The shear force diagram has sudden decrease from  1.125 kN to  6.125 kN. It indicates that there is a point load of 5 kN ( ) at E . Between E and B

The SFD decreases from  6.125 kN to  9.125 kN by an inclined straight

line, which shows that the beam carries a u.d.l. of

3 1.5

2 kN/m between E

and B. At B

Since there is a sudden increase from  9.125 kN to 0 at B, there is a support at B of reaction 9.125 kN. Bending Moment

BM at A, M A =0

46

BM at C , M C

(6.875 1.5)

BM at D, M D

(6.875 3)

2 1.5

1.5 2

2 1.5 2.25

8.06 kN-m ( 2 1.5)

10.875 kN-m

BM at E , M E

(9.125 1.5)

2 1.5

1.5 2


11.44 kN-m


Consider a section XX between D and E at a distance x from the end B. SF at section XX , 9.125

F x

x

5 2x

4.125

2x

(for maximum BM)

0

2.0625m (9.125 2.0625)

M max

5 (2.0625 1.5)

2 2.0625

2.0625 2

11.75kN-m Example 4.14

The shear force diagram for the overhanging beam is shown in Figure 4.39. Draw the loading diagramand bending moment diagram. Find the magnitude of maximum bending moment and locate the point of contraflexure. 440 N

A 400 N

1m

480 N

( )

( )

( ) B 560 N

( ) 6m

C 520 N

SFD

D 3m

160 N/m 400 N A

1m B

6m

C

x

2.5

3m

D

125 N.m M2 ( )

M1 ( )

( )

480 N.m 5m

720 N.m BMD

Figure 4.39

Solution

Let us analyse theshear force diagramgiven in Figure 4.39. At A

The shear force diagram suddenly decreases from 0 to  400 N. It indicates that there is a downward point load of 400 N at A. Between A and B

The shear force diagram is an inclined straight line and decreases from  400 N to  560 N. It indicates that there is a uniformly distributed load of 160 1 =160 kN/m between A and B. (560  400 = 160) At B

There is a sudden increase from  560 N to + 440 N at B. It indicates that there is a support reaction of 1000 N at B.

47


Between B and C

Since the shear force diagram varies linearly from + 440 N to  520 N between B and C . It indicates that a u.d.l. of (440 +520 =960)

960 =160 kN/m is acting between B and C . 6 At C

At C , the shear force diagram increases suddenly from  520 kN to 480 N. I t indicates that there is a support reaction of 1000 N (520 +480) atC . Between C and D

The shear force diagram is an inclined straight line which indicates that there is a uniformally distributed load of 480/3 =160 kN/m from C to D. Bending Moment

BM at A and D, M A = M D =0. BM at B, M B

(400 1)

BM at C, M C

160 3

160 1 3 2

1 2

480 N m

720 N m


SF at any section XX between B and C , F x = +1000  400  160 x

For maximum bending moment, F x should be equal to zero. 600  160 x =0 x =3.75 m

(1000 2.75)

M max

(400 3.75)

160 3.75

3.75 2

=+125 N m Maximum positive bending moment occurs at a distance of 3.75 m from the end A, where SF changes sign. Maximum negative bending moment occurs at a support C where SF changes sign. M max (negative) = 720 N m

Points of Contraflexure

Let M 1 and M 2 be the points of contraflexure, where BM is zero. But at any section XX between B and C at a distance x from the end A. M x

or

1000 (x

1)

400x

1000 x  1000  400 x  80 x2 =0 80 x2  600 x +1000 =0 8 x2  60 x +100 =0 2 x2  15 x +25 =0 (2 x  5) ( x  5) = 0

48

Thus,

160 x

x =2.5 m and 5 m.

x

2


4.5 BEAMS SUBJECTED TO COUPLES A beammay be subjected to a clockwise (called positive couple) or anticlockwise couple (called negative couple) at a section. To calculate the reactions at the support, the magnitude and the direction of the couple is considered. The bending moment at the section of the couple changes suddenly in its magnitude. While drawing the bending moment diagram, the bending moment, just left of the section of the couple and just right of the section of the couple, should be calculated, with the help of the reactions at both the ends. Since, the couple does not involve any load, the shear force does not change at the section of the couple. Some examples of equivalent couples have been given in Figure 4.40. d2 2

P1

d1 2

= d1 2

P1

d2 2

M1 =P 1.d1

M2 =P 2.d2

P2

P2 P1 d2 2

d 1 2

= d1 2

d2 2

P1

M1 =P 1.d1

M2 =P 2.d2

P2

P1 d1 2

d1 2

M1 =P 1.d1

P1

Figure 4.40

Cantilever Beam Subjected to Couple

Let us consider a cantilever beam AB of length subjected to an anticlockwise couple at a distance a from the fixed end as shown in Figure 4.41.

C

A a

B

l a l

49 ( )


Figure 4.41

Since the moment is applied at C , the portion BC is not subjected to any moment. Bending Moment (Starting from the End B)

BM at B, M B =0 B

BM just right of C , M C =0

(considering right side)

BM just left of C , M C =+


BM at A, M A =+


Since there is no load between A and C , the bending moment diagram will be horizontal line between A and C . However, the beam is not subjected to any load, the SF on the beam is zero. Simply Supported Beam Subjected to Couple at Mid-span

Let us consider a simply supported beam AB of length l, subjected to a clockwise couple ( ) at mid-span. The tendency of the couple will be to uplift the beam from its support A and to depress the beam at its support B. Therefore, the reaction at A, will be downward and at B will be upward. B

C

A l 2

l 2 l

( )

l

l

SFD

2

( )

( )

2

BMD

Figure 4.42

Taking moment about A and equating to zero, R B

R B

0

l

l

R A =Total downward load

R B

0

RB

0

l

l

(Negative sign indicates that the reaction at A is downward) Shear Force (Starting from the Left End A)

50

SF at

A

SF just left of B,

F B

F A


l

(Reaction at the support B)

l

Bending Moment (Starting from Left End A)

BM at A and B,

M A = M B =0 B

l

BM just left of B, M B

2

l

BM just right of B, M B

2

2

2 l

BM just right of B, M B

2

l

2


Supply Supported Beam Subjected to an Eccentric Couple

Consider a simply supported beam AB of span l, subjected to an anticlockwise couple at a distance a from the left end A as shown in Figure 4.43. C

A

B

b

a l

( )

l

l

SFD

a ( )

l

b ( ) l

BMD

Figure 4.43

Taking moments about A and equating to zero, R B

R B

0

l

l

(Negative sign indicates the reaction at B is downward)

R A =Total downward load

R B

0

l

l

Shear Force (Staring from the Left End A )

SF at A,

F A

SF just left of B,

F B

l l

Reaction at the support B.

Since, there is no load between A and B, the shear force diagram is horizontal.

51


Bending Moment (Starting from the Left End A)

BM at A,

M A =0

BM just left of C , M C

a

a

l

l

(a

a

MB just right of C , M C

l

(l

l b

b)

a


b

l

a)

l

(Q l

l

BM just right of C , M C

l)

Example 4.15

Draw the shear force and bending moment diagrams for the cantilever beam shown in Figure 4.44. 2 kN/m

A 11 kN

4 kN.m

1 kN.m

3 kN

4 kN 2m

2m

B

C

2m

D

2m

E

2m

2m

F

9.125 7

7

11

( )

4

4

SFD

8

( )

27

17

59 31

41

BMD

Figure 4.44

Solution

Reaction at the support A, R A

(2 2)

3

4

11kN

Shear Force (Starting from the Left End A)

SF at A,

F A

11kN

SF at B,

F B

11 (2 2)

SF just left of D,

F D

7 kN

SF just right of D, F D

7 3

SF just left of F ,

4 kN = load at F

F F

7 kN

4 kN

Bending Moment (Starting from the Right End F)

52

BM just right of E , M E

4 2

8 kN-m

BM just left of E , M E

8 1

9 kN-m

BM at D,

4 4 1

M D

17 kN-m

9

BM just right of C , M C

(4 6)

BM just left of C , M C

31 4

1 (3 2)


31kN-m

27 kN-m

BM at B, M B

(4 8)

BM at A, M A

(4 10) 1 (3 6)

1 (3 4)

4

41kN-m

4

2 2

2 2

59 kN-m

Example 4.16

Draw the SFD and BMD for a simply supported bema of 15 mspan loaded as shown in Figure 4.45 9 kN.m 7.5 kN

qx = 0.5x kN/m

D

C

A

1.5 kN/m

E F

6m

1m 1m

B

2m 6m

1m

15 m Parabolic curve 2.5

( ) 3.5

3.5

( ) 4

4 13 17

60.5 57

55

51

51.5 ( )

Figure 4.45

Solution


15 1.5 6

R B

13 kN

R A

1 6 3 2

9

7.5

6 2

(7.5 8)

(1.5 6)

1 2

9

25.5 13

RB

2 3

6 3

6

0

12.5 kN


SF at A,

F A

12.5 kN

SF at C ,

F C

12.5

SF just left of E ,

F E

3.5 kN

1 6 3 2

SF just right of E , F E

3.5

SF at F ,

F F

4 kN

SF just left of B,

F B

4 (1.5 6)

7.5

3.5 kN

4 kN

Bending Moment (Starting from Right End B)

13 kN = Reaction at B.

53


BM at B,

M B =0

BM at F ,

M F

(13 6)

BM at E ,

M E

(13 7) 1.5 6

B

BM just right of D, M D

(13 8) 1.5 6

BM just left of D, M D

51.5 9

BM at C ,

(12.5 6)

M C

6 2

1.5 6

51kN-m 6 2

1 2

55 kN-m

6 2

7.5

51.5 kN-m

60.5 kN-m

1 3 6 2

1 6 3

57 kN-m

Example 4.17

Draw a SFD and BMD for the beam shown in Figure 4.46 x 15.2 kN

7.2 kN/m

qx = 0.6x kN/m 24 kN.m D

A

C

B 1m

2m

F

E

12 m

9 kN.m

2m

1m

x x 21.6

14.8

( )

( ) ( )

15.2

( )

15.2 SFD

28.4

47.7 ( )

8.8 M1

M2

+12

M3

M4

M5

M6

( ) 3.6 15.2

16.8 21.6

BMD

Figure 4.46

Solution Determination of Reaction at the Support B

Considering left side, take moments about A, R B M A

12 7.2 3 12RB

12

3 2

15.6

1 2 12 7.2 12 2 3

621.6

Considering right side, take moments about A, M A

54

(15.2 3)

24

21.6 kN-m

M A

Equating these two equations, 12 R B

621.6

12 R B

600

R B

50 kN

R A

15.2


21.6

1 12 7.2 2

(7.2 3)

80 50 30 kN

RB


SF at C ,

F C

15.2 kN

SF just left of A,

F A

15.2 kN

SF just right of A,

F A

15.2

SF just left of B,

F B

14.8

SF just right of B,

F B

28.4

50

SF at F ,

F F

21.6

(7.2 3)

30

14.8 kN

1 12 7.2 2

28.4 kN

21.6 kN 0

Bending Moment (Starting from the Right End F )

BM at F ,

M F =0

7.2 1

BM just right of E , M E BM just left of E , BM at B, M B

3.6 15.6

M E

3 2

7.2 3

BM at A, M A

(15.2 3) 24

BM at C ,

M C =0

BM just left of D,

M D

BM just right of D, M D

15.6

1 2

3.6 kN-m 12 kN-m

16.8 kN-m

21.6 kN-m (considering left side)

15.2 1 15.2

24

15.2 kN-m 8.8 kN-m


The SF changes sign at A, between A and B and at B. Consider a section XX at a distance x from the end C as shown in Figure 4.46 SF at section XX , F x

1 (x 2

30 15.2

3) 0.6 ( x

3)

Equating this equation to zero, 14.8

0.3( x

0.3 x 2

or

0.3 x 2

1.8x 1.8x

3)2

0

12.1 0 12.1 0

55


x 2

or

6x

40.333

0

On solving by trial and error, we get x =10 m. (30 7)

M max

1 1 7 7 7 0.6 2 3

24 (15.2 10)

47.7 kN-m Maximum positive bending moment =+47.7 kN-m Maximum negative bending moment = 21.6 kN-m Points of Contraflexure

Let M 1, M 2, . . . and M 6 be the points of contraflexure, where the bending moment changes sign. To find the position of M 2, consider a section XX at a distance x from the end C . 15.2x

M x

15.2 x

24

24

0

1.579 m

x2

To find the position of M 3 and M 4, consider a section XX at a distance x from the end E . M x M x

15.2x

24 30 (x

15.2x

3)

1 (x 2

90

0.1( x

24 30x

14.8 x

66

0.1( x

3)3

14.8 x

66

0.1[ x 3

(3

14.8 x

66

0.1( x3

9x 2

0.1 x3

0.9x 2

2

12.1x

3) 0.6 (x

( x

3) 3

3) 3

3)

27x

3)

(3x

9)

33]

27)

63.3

On changing the sign and equating it to zero, we get 0.1 x3

0.9x 2

12.1x

63.3 0

Solving by trial and error, we get

and

x1

4.4814 m

x2

14.357 m

x3

9.84 m

Since, the value of x3 is negative, it should be ignored. To find the position of M 3, consider a section XX at a distance x from the end E . 7.2 x

M x

x

2

On equating it to zero, we get 3.6 x 2 x

56

15.6

2.082m

0

15.6

The points of contraflexure are at distance of 1 m, 1.579 m, 4.4814 m and 14.357 m from the left end X and at distances of 1 m and 2.082 m from the right end F .


SAQ 4 (a)

Draw the shear force and bending moment diagramfor the overhanging beamshown in Figure for SAQ 4(a) given below. 3 kN/m

6 k N.m

A

B

6 kN/m

2m

D

C

E

1m

6m

Figure for SAQ 4(a)

(b)

Draw the SFD and BMD for the beamshown in Figure for SAQ 4(b). 6 kN

2 kN/m

A

3 kN/m

B

7.5 kN

1m

E

D

C

2m

4 kN/m

3 kN/m

2.5 m

1.5 m

Figure for SAQ 4(b)

(c)

A simply supported beam, 16 m length, is subjected to a systemof loads and couples as shown in Figure for SAQ 4(c) given below. Draw SFD and BMD and find the magnitude of maximum bending moment. Take the distances between the points of application of loads and couples as equal. 20 kN

A B

60 k N m

80 k N m

20 k N m

C

6 kN

24 kN

D

E

F

40 k N m

G

H

J

Figure for SAQ 4(c)

4.6 BEAMS SUBJECTED TO OBLIQUE LOADING So far, we have been discussing the shear force and bending moments for beams subjected to an external load system at right angles to the axis of the beam. Whenever a beam is subjected to inclined loads, these inclined loads are resolved in horizontal and vertical direction. The vertical components (at right angles to the axis of the beam) of the loads on the beamwill introduce shear force and bending moment in the member. The horizontal components (along the axis of the beam) of the loads on the beam will introduce axial force of thrust, i.e. pulls or pushes in the beam depending upon its end position. 57


Thrust Diagrams

If one end of the beamis hinged, whereas the other end is simply supported on rollers, the hinged end will be subjected to horizontal thrust equal to the unbalanced horizontal force due to inclined loads. An axial force diagram, which represents horizontal thrust in the beam, is drawn like shear force diagram. Sign convention used to draw the thrust diagram is tensile force (i.e. force acting towards left) as positive and compressive force (i.e. force acting towards right) as negative. Example 4.18

Draw the shear force, bending moment and thrust diagrams for the beamshown in Figure 4.47. 3 kN

4 kN

o

o

A

60

45

C

2m

4m

2.828 kN 2.828 kN A

4.516

5 kN

C

2m

D 1.5 m

E

2.598 kN

5 kN

D 1.5 m

E

B

2.5 m

1.5 m 4m

B

2.5 m

1.7236

( )

1.7236 0.8744 SFD

5.8744

( ) 15.9976

5.8744 14.686

9.1032

( )

BMD 1.328

( )

1.328 ( )

1.5

1.5

Trust Diagram

Figure 4.47

Solution

Vertical component of 4 kN at C , Horizontal component of 4 kN at C , Vertical component of 3 kN at D, Horizontal component of 3 kN at D,

4 sin 45o

2.828 kN (

4 cos 45o

2.828 kN (

58

10 5 7.5

R B

5.8744 kN

R A

2.828

3 cos 60o 1.5 kN (

R A

4.5516 kN

2.598 6

2.598

5

)

3 sin 60o 2.598 kN ( )

Taking moments about B, R B

)

RB

2.828 2

0

10.426 5.8744

)



SF at A,

F A

4.5516 kN

SF just left of C ,

F C

4.5516 kN

SF just right of C , F C

4.5516

SF just left of D,

1.7236 kN

F D

2.828


1.7236

SF just left of E ,

0.8744 kN

F E

SF just right of E , F E

0.8744

1.7236 kN

2.598

0.7844 kN

5.8447 kN = Reaction at B.

5

Bending Moment (Starting from the Right End B)

BM at A and B,

M A = M B =0 B

BM at E , M E

(5.8744 2.5)

BM at D, M D

5.8744 4

BM at C , M C

(4.5516 2)

14.686 kN-m

5 1.5

15.9976 kN-m

9.1032kN-m (considering left side)


It will occur at D where SF changes sign. Thus,

15.9976 kN-m

M max

Thrust Diagram

Let us determine the horizontal reaction at A (being a hinged end). H

0 2.828 1.5 0

H A

1.328kN

H A

The section AC is subjected to 1.328 kN (tensile force). The section CD is subjected to 1.5 kN (compressive force). (2.828 1.328 1.5) The beam from D and B is not subjected to any axial force. The axial force diagram is shown in Figure 4.47. Example 4.19

Analyse the beamshown in Figure 4.48 and draw the SFD, BND and thrust diagram. Locate the point of contraflexure, if any. 4t

2t o

30

o

60

2t / m A

D

C

1m

2m

E

2t A

3.464

1t

F

B

1m

1.732t

3t

4m

x

2m

59 2t / m

F


Figure 4.48

Solution

4 cos 60o

Vertical component of 4 kN at C ,

2 kN ( )

4 sin 60o 3.464 kN (

Horizontal component of 4 kN at C ,

)

2 cos 30o 1.732 kN ( )

Vertical component of 2 kN at D, Horizontal component of 2 kN at D,

2 sin 30o 1kN (

)


8 (3 10)

R B

10.6495kN

R A

2 1.732

2 4

(2 4)

4

3

4 2

RB

(1.732 3)

(2 1)

14.732 10.6495

4.0825 kN


60

SF at A,

F A

4.0825 kN

SF just left of C ,

F C

4.0825 kN

SF just right of C , F C

4.0825

SF just left of D,

2.0825 kN

F D

2


2.0825 1.732

SF at E ,

F E

0.3505 kN

SF just left of B,

F B

0.3505

2.0825 kN

(2 4)

0.3505kN

7.6495 kN

SF just right of B, F B

7.6495 10.6495

SF just left of F ,

3 kN = load at the end F .

F F

Bending Moment (Starting from the F)

3kN

0

BM at F , M F =0


BM at B, M B

(3 2)

BM at E , M E

(10.4695 4)

BM at D, M D

(4.0825 3)

BM at C , M C

(4.0825 1)

6 kN-m

(3 6)

4 2

2 4

(2 2)

8.599 kN-m

8.2475 kN-m

4.0825 kN-m

BM at A, M A =0 Maximum Bending Moment

Maximum bending moment will occur at B and between B and E . Consider a section XX at a distance x from the end F . 10.6495

F x

3

2 (x

2)

For maximum bending moment, F x should be equal to zero. 10.6495 3 2 x

4

0

5.82475 m 5.825 m

x

BM at section XX , 10.6495 (x

M x

10.6495 ( x

2) 2)

10.6495(5.825

M max

3x 3x

2 (x

2) 2

2) 2

(x

2)

( x

2)

(5.825 2) 2

3(5.825)

8.629 kN-m

Maximum positive bending moment =+8.629 kN-m Maximumnegative bending moment = 6 kN-m. Point of Contraflexure

Equating the BM at section XX , to zero. 10.6495 ( x

or

10.6495 x

or

x

2

or

x 2

2)

3x

21.299

2) 2

(x

2

3x

x

11.6495x

25.299

0

11.6495x

25.299

0

0 4x

4

0

Solving by trial and error, we get x =2.9 m. Point of contraflexure is at a distance of 2.9 m from the end F . Thrust Diagram

Horizontal reaction at A, H A

3.464 1 0

H A

2.464

H A

0

2.464 kN (  indicates that the reaction is towards right)

The portion AC is subjected to a compressive force of 2.464 kN. The portion CD is subjected to a tensile force of 1 kN (i.e. 3.464  2.464 = 1).

61


SAQ 5 Analyse the beam as shown in Figure 4.49 is given below. 6 kN

8 kN

4 kN

A

o

60

o

o

30

45

C D

1m

1.5 m

2m E

2m

1.5 m

Figure 4.49

4.7 SUMMARY A beamis a structural member, subjected to a systemof external forces (including inclined load) to produce the bending of the member in an axial plane. The shear force at the cross-section of the beam is defined as the unbalanced vertical force either to the right or to the left of the section. The bending moment at the cross section of the beam is defined as the algebraic sumof moments of all the forces acting on the beam either to the right or to the left of the section. All the upward forces to the left of the section and all the downward forces to the right of the section cause positive shear force. All the downward forces to the left of the section and all the upward forces to the right of the section cause negative shear force. The BM is said to be positive, when it is acting in an anticlockwise direction to the left of the section and clockwise direction to the right of the section. The BM is said to be negative, when it is acting in clockwise direction to the left of the section and an anticlockwise direction to the right of the section. While drawing SF and BM diagrams, all the positive values are plotted above the base line and the negative values below it. The maximum BM occurs, where the SF is zero or changes sign. If the SF diagramline is horizontal between two points, the BM diagramis inclined. It indicates there is no load between two points. If the SF diagramis inclined between two points, the BM diagramis a parabola of second degree. It indicates that there is a uniformly distributed load between the two points. If the SF diagram is in the form of parabola of second degree between two points, the BM diagramis in the form of parabola of third degree (cubic parabola). I t indicates that there is a uniformly varying load between the two points. The point of contraflexure is a point where the BM is zero or changes sign. If a beamis subjected to a couple, the SF does not change. But the bending moment suddenly changes in magnitude equal to that of the couple. If there is a point load or reaction, the BM does not change at that point. But the SF suddenly changes (either decreases or increases) in magnitude equal to that of point load or support reaction. 62

If a beam is subjected to inclined loads, their components at right angles to the axis of components (vertical components) will cause shear force and bending moment and their components along the axis of the beam(horizontal components) will cause axial force or thrust in thebeam.


While drawing axial forces diagram, tensile forces are taken as positive and compressive forces are taken as negative.

4.8 ANSWERS TO SAQs SAQ 1

(a)

Shear Force Diagram Ordinates

SF at A, F A =+27 kN SF just left of C, F C =+27 kN SF just right of C, F C =+21 kN SF at D, F D =+21 kN SF at E, F E =+12 kN SF at B, F B =0 B

Bending Moment Diagram Ordinates

BM at B, M B =0 BM at E, M E = 24 kN m BM at D, M D = 57 kN m BM at C, M C = 78 kN m BM at A, M A = 105 kN m (b)

Maximum Shear Force =10500 N Maximum Bending Moment =13500 N m

(c)

MaximumShear Force =+8 N MaximumBending Moment = 26 kN m

(d)

Maximum Shear Force =+4 N MaximumBending Moment = 8 kN m

(e)

MaximumShear Force =+14 N MaximumBending Moment = 39.5 kN m

SAQ 2

(a)

Maximum Bending Moment =+7.6 kN m at 3.25 m from the left hand support

(b)

Maximum Bending Moment =+5077.5 kN m at 3.165 m from the left hand support.

(c)

Maximum Bending Moment =+15.70 kN m at 3.22 m from left the hand support

(d)

Maximum Bending Moment =+12750 N m at D

SAQ 3

(a)

Maximum Positive Bending Moment =+37500 N m at C MaximumNegative Bending Moment = 5000 N m at B

(b) SAQ 4

l =1.25 m

63

SFD BMD

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