E-CONTENT

ON

MECHANICS OF SOLIDS-I

BTME 1205

(L 3 - T 1 - P 0)

FOR

2ND SEMESTER

2012-13

DEPARTMENT OF MECHANICAL ENGINEERING

Centurion University of Technology & Management

PARALAKHEMUNDI GAJAPATI

ODISHA – 761211

16

BTME 1205 Mechanics of Solids-I (3-1-0)

Module-I (16)

General case of forces and moments and their resultant in a plane, Reduction to a single force at a point or to a single force and moment at a point. Condition of equilibrium, centroids of composite plane figures, Pappus theorem, Moment of inertia of plane figures: product moment of inertia; composite figures.

Friction and its application to screw jack, belts, winches, wedges, and simple machines.

Module-II (16)

Plane trusses and frames: Method of joints and sections. Principle of virtual work. Axially loaded members: stress – strain diagram; Hooke’s law, working stress, factor of safety. Composite bars in tension or compression, temperature stress, shear stress and shear strain, modulus of rigidity, complementary shear stress, bulk modulus, relation between elastic constants.

Module-III (16)

Type of supports, type of beams, type of loads, shear force and bending moment and their relationship, Shear force and bending moment for simple beams- with different support conditions ,loads and moments. Flexural and shear stress in beams and their distribution over rectangular, circular and I-sections

Text Books.

1. S.P.Timoshenko and D.H.Young: Engineering mechanics

2. Tayal,A.K: Engineering mechanics

3. S.P.Timoshenko and D.H.Young: Elements of strength of materials

4. G.H.Ryder: Strength of materials

5. S.S.Bhavikatti: Engineering mechanics,

6. S.S.Bhavikatti: Strength of materials

Table of contents

Lecture – 3.1 SUPPORT REACTIONS 1

Lecture – 3.2 SHEAR FORCE 4

Lecture – 3.3 SHEAR FORCE AND BENDING MOMENT 7

Lecture – 3.4 SHEAR STRESS IN BEAMS 10

Lecture – 3.5 CIRCULAR SECTION 13

Lecture – 3.6 HEXAGONAL SECTION 15

Lecture – 3.7 BENDING STRESS IN BEAMS 17

1

Figure 3.1.2

MODULE – 3

Lecture 3.1

Support Reactions

When a number of forces are acting on a body, and the body is 8upported on another body, then

the second body exerts a force known as reactions on the first body at the points of contact so

that the first body ¡s in equilibrium. The second body is known as support and the force, exerted

by the second body on the first body, is known as support reactions.

Types of Supports

Though there are many types of supports, yet the following are important from the subject point

of view:

(a) Simple supports or knife edge supports (b) Roller support

(e) Pin-joint (or hinged) support (d) Smooth surface support

(e) Fixed or built-in support.

Simple Support or knife edge support. A beam

supported on the knife edges A and B is shown

in Fig. 3.1.1 (a). The reactions at A and B in

case of knife edge support will be normal to the

surface of the beam. The reactions RA and RD

with free body diagram of the beam is shown

in Fig. 3.1.1 (b).

Roller Support. A beam supported on the

rollers at points A and B is shown in Fig 3.1.1

(a). The reactions in case of roller supports will

be normal to the surface as shown in Fig.3.1.1.

(b). Pin joint (or hinged) support. A beam, which is hinged (or pin-joint) at point A, is shown

in Fig. 3.1.3. The reaction at the hinged end may be either vertical or inclined depending upon

the type of loading. 1f the load ¡s vertical, then the reaction will also be vertical. But if the load is

inclined, then the reaction at the hinged end will also be inclined.

Figure 3.1.1

2

Figure 3.1 4

Figure 3.1 3

Figure 3.1.5

Figure 3.1 .6

Figure 3.1.7

Smooth Surface Support. Fig. 3.1.2 shows a body in

contact with a smooth surface. The reaction will always

act normal to the support as shown in Fig. 3.1.2 (a) and

3.1.2 (b). Fig. 3.1.3 shows a rod AB resting inside a

sphere, whose surface are smooth. Here the rod become

a body and sphere becomes surface. The reactions on the

ends of the rod (i.e. at point A and B) will be normal to

the sphere surface at A and B. The normal at any point

on the surface of the sphere will always pass through the

center of the sphere. Hence reactions RA and RB will

have directions AO and BO respectively as shown in Fig

3.1.3.

Fixed or built.in Support. Fig. 3.1.4 shows the end A of a beam,

which is fixed. Hence the support at A is known as a fixed support

In case of fixed 8upport, the reaction will be inclined. Also the

fixed support will provide a couple.

Types of Loading

The following are the important types of loading:

(a) Concentrated or point load, (b) Uniformly distributed load, and (c) Uniformly varying load.

Concentrated or point load. Fig. 3.1.5 shows a beam

AB, which is simply supported at the ends A and B. A

load W is acting at the point C. This load is known as

point load (or concentrated load). Hence any load acting

at a point on a beam, is known as point load.

In actual practice, it is not possible to apply a load at a

point (i.e. at a mathematical point) as it must have some

contact area. But this area in comparison to the length of

the beam is very - very small (or area is negligible).

Uniformly distributed load. 1f a beam is loaded in such a

way, that each unit length of the beam carries same intensity of

the load, then that type of load is known as uniformly

distributed load (which is written as U.D.L.). Fig. 3.1.6 shows a

beam AB, which carries a uniformly distributed load. For

finding the reactions the total uniformly distributed load is

assumed to act at the C.C. of the load.

Uniformly varying load. Fig. 3.1.7 shows a beam AB, which

carries load in such a way that the rate of loading on each unit

length of the beam varies uniformly. This type of load is

3

known as uniformly varying load. The total load on the beam is equal to the area of the load

diagram. The total load acts at the C.G. of the load diagram.

Beam

A beam is a structural element which has one dimension considerably larger than other two

dimensions (generally width and depth) and it is supported at a few points. The distance between

the supports is called span. A beam is usually loaded normal to its cross-sectional areas. Every

cross section of a beam faces bending and shear when it is loaded. The load finally gets

transferred to supports. The system of forces (applied forces) and reactions keep the beam in

equilibrium.

When a horizontal beam is loaded with vertical loads, it bends due to action of the loads. The

internal shear stress and bending moment arc developed to resist bending. The amount of

bending in the beam depends upon the amount and type of the loads, length of the beam,

elasticity of the beam and dimensions of the beam. The best way of studying the deflection or

any other effect is to draw and analyze the shear force diagram (SFD) and the bending moment

diagram (BMD) of the beam.

Types of Beams

Beams can be any of the following types (Figure 3.1.8):

(1) Simply supported (2) Cantilever type (3) Overhanging (4) Hinged and roller supported

(5) Fixed (6) Continuous (having more than two supports)

Figure 3.1 8

4

Figure 3.2.2

Lecture 3.2

Shear Force

Shear force is the unbalanced vertical force on one side (to the left or right) of a section of a

beam and is the sum of all the normal forces on one side of the section. It also represents the

tendency of either portion of the beam to slide or shear) actually relative to the other. Remember

that a force at a section means a force of a certain magnitude acting at that point whereas the

shear force at a section means the sum of all the forces on one side of the section.

Consider the beam as shown in Fig. 3.2.1 a. It is simply supported at two points and carries four

loads. The reactions at the supports are R1 and R2. Now if the beam is imagined to cut at section

x-x into two portions (Fig. 3.2.1 b), the resonant of all the forces (loads as well as reaction of

support) to the left of the section is F (assuming upwards). Also, as the beam is in equilibrium,

the resultant of the forces to the right of x-x must also be F downwards. The force F is known as

shear or shearing force (S.F.)

Shear fore is considered positive when the resultant of the forces to the left of a section is

upwards or to the right downwards.

A shear force diagram (SFD) shows the variation of shear force along the length of a beam.

Bending Moment

Bending moment at a section of a beam is defined as the algebraic sum of the moments about the

section of all the forces on one side of the section.

1f the moment M about the section x-x of all

the forces to the left is clockwise (Fig.

3.2.2), then for the equilibrium, the moment

of the forces to the right of x-x must be M

counter-clockwise.

Bending moment is considered positive if

the moment on the left portion is clockwise

or on the right portion counter-clockwise.

This is usually referred as sagging bending moment as it tends to cause concavity upwards. A

bending moment causing convexity upward is taken as negative bending moment and is called

hogging bending moment.

Figure 3.2.1

5

Figure 3.2.3

A bending moment diagram (BMD) shows the variation of bending moment along the length of

a beam.

Relation between w, F AND M

Consider a small length �� cut out

from a loaded beam at a distance x

from a fixed origin O (Fig. 3.2.3). Let

w = mean rate of loading on the

Length ��

F = shear force at the section x

� + � � = Shear force at the section

� + ��

M = bending moment at the section x

� + �� = Bending moment at the section � + ��

Total load on the length �� = �. �� acting approximately through the center C (if the load is

uniformly distributed, it will be exactly acting through C).

For equilibrium of the element of length ��, equating vertical forces,

� = ��� + � + ���

Or � = − � �

That is, rate of change of shear force (or slope of the shear force curve) is equal to intensity of

loading.

Taking moment about C,

� + �. �� + � + ���. �

�� − � + ��� = 0

Neglecting the product and squares of small quantities, � = � �

I.e. rate of change of bending moment is equal to the shear force.

The point of zero bending moment, i.e. where the type of bending changes from sagging to

hogging is called a point of inflection or contra-flexure.

Integrating equation � = − � � between two values of x,

�� − �� = � ����� which is the area under the load distribution diagram.

Similarly, integrating equation � = � � between two values of x,

�� − �� = � �����

This shows that the variation of bending moment between two sections is equal to the area under

the shear force diagram.

Also as � = � � ,

� = − ���� = − ���

���

6

Figure 3.2.4

Figure 3.2.5

Shear Force and Bending Moment Diagrams for Cantilevers

A cantilever may carry concentrated or uniformly distributed loads.

Concentrated Loads

Assume a cantilever of length ¡ carrying a concentrated load W at its free end as shown in

Fig.3.2.4 a.

Shear force diagram Consider a section at a distance x from the free end. The force to (he right

of the section is W downwards and is constant along the whole length of the beam or for all

values of x. Therefore, the shear force will be considered

positive and the shear force diagram is a horizontal straight

line as shown in Fig. 3.2.4 b.

Bending moment diagram Taking moments about the

section, M = W. x

As the moment on the right portion of the section is

clockwise, the bending moment diagram is negative. The

bending moment can also be observed as hogging, and thus

negative. The bending moment diagram is thus an inclined

line increasing with the value of x (Fig. 3.2.4 c).

Maximum bending moment = �. � at the fixed end.

Reaction and the fixing moment. From equilibrium

conditions, the reaction at the fixed end is W and the fixing

moment applied at the fixed end = �. � Uniformly Distributed Load

Assume a cantilever of length � carrying a uniformly

distributed load w per unit length across the whole Span as

shown in Fig. 3.2.5 a.

Shear force diagram Consider a section at a distance x from

the free end. The force to the right of the section is ��

downwards and varies linearly along the whole length of the

beam. Therefore, the shear force is positive and the shear

force diagram is a straight line as shown in Fig. 3.2.5 b.

Bending moment diagram. The force �� to the right of

section can be assumed to be acting as a concentrated load at

a point at a distance � 2� from the free end.

Taking moments about the section, � = ��. �� = ���

�

As the moment on the right portion is clockwise, the bending moment diagram is negative

(hogging). The bending moment diagram is parabolic and increases with the value of x (Fig.

3.2.5 c).

Maximum bending moment =� �

� at the fixed end.

7

Figure 3.3.1

Lecture 3.3

Shear Force and Bending Moment Diagrams for Simply Supported Beams

A simply supported beam may carry concentrated or uniformly distributed loads.

Concentrated Loads

Assume a simply supported beam of length � carrying a

concentrated load W at a distance a, from end A as shown

in Fig. 3.3.1 a.

Let the distance CB be b.

First it is required to find the reactions at the supports.

Taking moments about end

! = "�# − �. $ = 0

Or "� = %� ; Similarly "� = %�

Shear force diagram

Portion BC: Consider a section at u distance � from the

end B. The force to the right of the section is "� upwards

and is constant along the length up to point Con the beam.

Therefore, the shear force will be negative and the shear

force diagram is a horizontal straight line as shown in

Fig.3.3.1b.

Portion AC: At a section at a distance � from the end B, the force to the right of section is

= "� − � = %� − � = %�&%

= −� ' &� ( = − %�

= "� (Downloads)

Thus, the shear force will be positive and the shear force diagram is a horizontal straight line as

shown in Fig. 3.3.1 b.

Note that the portion AB can also be taken first and the forces to the left of any section may be

considered. In that case, the force to the left is Ra and upwards and the shear force positive, i.e.

the same as before.

Bending moment diagram

Portion BC: Consider a section at a distance x from the end B.

Taking moments about the section, � = "�� = %�� �� = 0; �* = %��

As the moment on the right portion of the section is counter-clockwise, the bending moment is

positive. The bending moment can also be observed as sagging, and thus positive. Therefore the

bending moment is linear and increases with the value of x (Fig. 3.1.1 c).

Portion AC: Consider a section at a distance x from the end A.

8

Figure 3.3.2

� = "�� = %�� ; �� = 0; �* = %��

The moment on the left portion of the section is clockwise, the bending moment is positive. The

bending moment can also be observed as sagging, and thus positive.

If the load is at the mid span, $ = + = �/2

The bending moment at the midpoint,

� = %'-�('-

�( = %

. , which is maximum for any

position of the load on beam.

Uniformly Distributed Load

Assume a simply supported beam of length carrying

a uniformly distributed load w per unit length as

shown in Fig. 3.3.2 a.

Total load = �� ; "� = "� = % �

Shear Force Diagram

At a section at a distance x from A

�� = "� − �� = ��2 − �� = � /�

2 − �0 �123$4�; ���56� = ��

2 ; ���5 � = − ��2

Shear force diagram is shown in Fig.3.3.2 b.

Bending moment diagram

The bending moment at a section is found by treating the distributed toad as acting at its center

of gravity.

�� = "� . �. �2 = ��

2 � − ���2 = ��

2 � − �� 7$4$+8�19�

���56� = 0 ; ���5 � = 0; � = � � = 0 or

� � − �� = 0 84 � =

�

Thus maximum bending moment = ��5-�� = � �

:

Bending moment diagram is shown in Fig. 3.3.2 e.

Uniformly Distributed Load with Equal Overhangs

Let w be the uniformly distributed load on the beam as shown in Fig. 3.3.3 a.

As the overhangs are equal "� = "� = % ;����

Shear force diagram

. Portion DA: �� = −�� �123$4� � = 0; �� = −�$

9

Figure 3.3.3

. Portion AB: �� = −�� + � ;���� �123$4�;

���56� = � � ; ���5 ;�� = − �

� Portion BE: �� = −�� + � ;���

� + � ;���� = −�� +

�� + 2$� �123$4�

���5 ;�� = �$; �<�5 ;��� = 0; Shear force diagram is shown in Fig. 3.3.3 b.

Bending moment diagram

. Portion DA �� = − ���� 7$4$+8�19�;

� = 0; �� = − �$�2

. Portion AB

�� = − ���2 + �� + 2$�

2 � − $� 7$4$+8�19�

�� = − �$�2 ; ���5 ;�� = − �$�

2

Portion BE: Bending moment will be reducing to zero

in a parabolic manner at E. It is convenient to consider

it from end E. Then � = −���/2. At midpoint C,

�*'�5�; �( = − �'$ + � 2� (�

2 + �� = 2$�2 . �

2

− �2 =$� + ��

4 + $� − ��2 − $�? = �

8 �� − 4$��

Loading and Bending Moment Diagrams from Shear Force Diagram

The loading and bending moment diagrams can easily be drawn from the shear force diagram if

the following points are kept in mind:

‘The shear force diagram consist of

• rectangles in case of point loads.

• inclined lines for the portion of uniformly distributed load and

• parabolic curve for the portion of triangular or trapezium load

The bending moment diagram consists of

• inclined lines in case of point loads.

• parabolic curve for the portion of uniformly distributed load and

• cubic curve for the portion of triangular or trapezium load

10

Figure 3.4.1

Figure 3.4.2

Lecture 3.4

Shear Stress in Beams

While discussing the theory of simple bending in the previous chapter, it was assumed that no

shear force acts on the section. However, when a beam is loaded, the shear force at a section is

always present along with the bending moment. It is, therefore, important to study the variation

of shear stress in a beam and to know its maximum value within safe limits. It is observed that in

most cases, the effect of shear stress is quite small as compared to the effect of bending stress

and it may be ignored. In some cases, however, it may be desirable to consider its effect also.

Usually, beams are designed for bending stresses and checked for shear stresses. This chapter

discusses the shear stress and its variation across the section.

A shear force in a beam at any cross-section sets up shear stress on transverse sections the

magnitude of which varies across the section. In the analysis, ¡t is assumed that the shear stress is

uniform across the width and does not affect the distribution of bending stress. The latter

assumption is not strictly true as the shear stress causes a distortion of transverse planes and they

do not remain plane.

As every shear Stress is accompanied by an equal complimentary shear stress, shear stress on

transverse planes has complimentary shear Stress on longitudinal or horizontal planes parallel to

the neutral axis.

Variation of Shear stress

Figure 6.1 shows two transverse sections of a beam at a distance �� apart. Considering the

complimentary shear stress A at a distance y0 from the neutral axis, let �, � + �� $2� �, � + �� be the shear forces and the bending moments at the two sections. B is the width of the cross-

section at this position.

The force due to

complimentary shear

stress on the area

C"DE(Fig. 6.2) tends to

slide the block above

the area which is

resisted by the

difference of the

longitudinal forces over the area A of the two transverse

sections. If Fand �F are the normal stresses on an

elemental area �A at a distance H from the neutral axis at

the two transverse sections, then for the equilibrium,

A. B. �� = � �F. �!

11

Figure 3.4.3

Bending stresses at the two sections at distance y from the neutral axis,

F = �IJ and F + �F = �;K��I

J

Therefore, �F = K��IJ

A. B. �� = � 'K�.IJ ( . dA or A = K�

K�.M.J � H. �! = 'K�K� ( N

M.J . !HO

But K�K� � ; ∴ A = �. QIO

MJ

In the above relation,

Z is the actual width of the section at the position where A is to be calculated

I is the total moment of inertia about the neutral axis

!HO is the moment of the shaded area about the neutral axis.

Shear Stress Variation in Different Sections

Variation of shear stress in different type of sections is discussed below:

Rectangular Section

At a distance y from neutral axis (Fig. 6.3). A = �. QIORJ

= �. S�'T�&I(USI;V

�'T�&I(U

�./WTXV� 0

= �. �'T�&I(.V�'T

�;I(�./WTX

V� 0 = Y�� X . ' �

. − H�(

This indicates that there is parabolic variation of shear

stress with y.

At neutral axis H = Z�, shear stress = [� . �

� ; it is the maximum shear stress.

Usually, �/+� is known as the mean stress and thus

A\�� = 1.5A\<�_

12

Figure 3.4.5

I-Section

In the flange, at a distance y from neutral axis (Fig. 3.4.4 a),

A` = �. !HOB# = �. a 'b2 − H( SH + 12 'b2 − H(U

a. #

�a# . ca /b

2 − H0 12 /b

2 + H0d = �2# . =b�

4 − H�?

At H = b 2� , F = 0

At H = � 2� , A = �

8# b� − ���

In the web, at a distance y from neutral axis (Fig.3.4.4 b),

A� = �. !HOB#

= �. a 'b2 − �2( S�2 + 12 'b2 − �2(U + + '�2 − H( SH + 12 '�2 − H(U+. #

= �+# . ca /b − �

2 0 /b + �4 0 + + /�

2 − H0 12 /�

2 + H0d

= �8# ca

+ b� − ��� + �� − 4H��d

Maximum shear stress = �:J Se

� b� − ��� + ��U at the neutral axis (y=0)

At the top of web, H = � 2⁄ �:J Se

� b� − ���U = e� . Ag

As shear stress has to follow the direction of the boundary (Section

1.3), the shear stress in the flanges follows the horizontal pattern as

shown in Fig. 6.5. As a result, the complimentary shear stress in the

flanges is on longitudinal planes perpendicular to the neutral axis.

Width z in the expression for shear stress is then replaced by the

flange thickness.

In practice, it is found that most of the shear force is carried by the

web and the stress in the flanges is very small. Also the variation of

stress over the web is comparatively small (about 25% to 30%).

Many times for design purposes an assumption is taken that all the

shear force is carried by the web uniformly over it.

Figure 3.4.4

13

Figure 3.5.1

Figure 3.5.2

Lecture 3.5

Circular Section

Refer fig 3.5.1

B = 2h4� − H� 84 B� = 44� − H�� 84 2. B. �B= −8H�H 84 H�H = −B. �B 4⁄

Now, !HO = Moment of shaded area about neutral axis

Consider a strip of thickness �H at a height y from neutral

axis and parallel to it,

Area of the strip=B. �H,

Moment of elementary area about neutral axis= B. �H. H

Moment of whole of the shaded area about neutral axis,

!HO = � B. H. �HiI

= − 14 � B. B. �BM

6= 1

4 � B��BM6

= B[12

A = �. 1#B

B[12 = �. 1

#B�12 = �

3# 4� − H��

Thus shear stress variation is parabolic in nature.

Aklm I56� = �. �� 4�3n�� 64⁄ � = 4

3�

n�� 4⁄ � = 43 A�p

Thin Circular Tube

If the thickness of a circular tube is small,

then the fact that the shear stress follows

the direction of boundary can be used to

find the same.

Let the bending be about XX and A and B

two symmetrically placed positions at angle

q from vertical (Fig. 3.5.2). Let the shear

stress at A and B be A. Now, the

complimentary shear stress is on

longitudinal planes and is balanced by

difference of normal stresses on the area

subtended by angle 2q. The force due to complimentary shear stress on the area at A and B tends

to slide the block above which is resisted by the difference of the longitudinal forces over the

area above AB.

14

Figure 3.5.3

Thus,

For a length �� of the beam.

2A. r. �� = � �F�!s&s

= � �F. "�t�. rs&s

Or A = N� � Ku

K�s

&s ". �t�

But �F = K�.IJ

∴ A = � K�K�

s&s ". �t� .I

J = N� � �". �t� .I

Js

&s � = K�K�

#� = 12 . v8�$4 �8�32r 8g 1234r1$

= 12 × !43$ × �3$2 4$�1xy�� = 1

2 . 2n"r. "� = nr"[

Hence

A = �"2# � H. �ts

&s= �"

2nr"[ � "98y t . �ts&s

= �2n"r � cos t �ts

&s= �

2n"r }sin t�&ss = �y12 qn"r

Or A = 2 × �3$2 yℎ3$4 yr43yy

Square with a Diagonal Horizontal

Refer Fig. 6.8,

#�� = 2 �a. a 2⁄ �[12 � = a.

48

A = �. QIOMJ = �

e� .:⁄ N�I . N

� 2H, H� 'e� − �

[ H(.

= �a. 48⁄

H2 . /a

2 − 23 H0

= 4�Ha. 3a − 4H�,

i.e. it is parabolic.

At H = 0, A = 0

At neutral axis H = e� , A = ��

e�

If b is the side of the square, + = a√2

At neutral axis A = ���√���� = �

�� = ��i<� = A\<�_

For maximum value, � I =

I 3aH − 4H�� = 0 or 3a − 8H = 0 84 H = [: a

15

Figure 3.6.1

Figure 3.5.4

A\�� = 4�Ha. 3a − 4H� = 4� '38( a

a. /3a − 4 × 38 a0

= 9�4+� = 9

8 A\<�_

Triangular Section

Refer Fig. 3.5.4.

A = �. !HOB# = � /12 . H. +Hℎ 0 '23 ℎ 23 H(

+ℎ[36 . +Hℎ

= 12�H+ℎ[ ℎ − H�

For maximum value, �A�H = �

�H ℎH − H�� = 0 84 ℎ − 2H = 0 84 H = ℎ 2⁄

A\�� = 12� ℎ 2⁄+ℎ[ /ℎ − ℎ

20 = 3�+ℎ = 3

2 . �+ℎ 2⁄ = 1.5A\��

Lecture 3.6

Hexagonal Section

Refer Fig. 3.6.1

#�� = 2 S��

XN� + � X

N� U = �Y +�[ = �

Y + '√[� +([ = �√[

NY +.

B = 2+ − 2 H√3 = 2

√3 �√3+ − H�

Now, AHO = Moment of shaded area about neutral axis

Consider a strip of thickness �y at a height y from neutral

axis and parallel to it,

Area of the strip = B. �H

Moment of elementary area about neutral axis = B. �H. H

Moment of whole of the shaded area about neutral axis,

!HO = � �√3+ − H�H. �H/√[ �� 0� I

�√[ S√3 + I�

� − IX[ UI

/√[ �� 0� = �

√[ S√3 . [: +[ − [√[

�. +[ − √3 + I�� + IX

[ U

16

2√3 �√3

4 +[ − √3 + H�2 + H[

3 �

A = �. QIOMJ = � �

√Xc√X� �X&√X

� �I�;�XX d

�√XV� ��× �

√X�√[ �&I� = � �√Xc√X

� �X&√X� �I�;�X

X d�����√[�&I�

At neutral axis, H = 0, A = .��√[ �� = 0.4618 �

��

Average shear stress = �.'V

�W� (;�� = �

[�

= �3+�√3 +/2� = 2�

3√3 +� = 0.385 �+�

F_.�F�p� = 0.46180.385 = 1.2

Shear stress at H = �� , H = � 2

√3c√34 +3−√3

2 +'+2(2+1

3'+3(3d

58+4'√3+ −+

2( = �0.5 − 0.25 + 0.048�

0.77+� = 0.387 �

+� $2� FI56.��F�p� = 0.387

0.385 ≈ 1

17

Figure 3.7.1

Lecture 3.7

Bending Stress in Beams

Introduction

If a constant bending moment (no shear force) acts on sonic length of a beam, the stresses set up

on any cross-section on that part of the beam constitutes a pure couple, the magnitude of which is

equal to the bending moment. The internal stresses developed in the beam are known as flexural

or bending stresses. In Fig. 5.la, the portion (‘D of the simply supported beam is under pure

bending as there does not exist any shear force. If the end sections of a straight beam are

considered to remain plane, the beam under the action of bending moment bends in such a way

that the inner or the concave edge of cross-section comes in compression and the outer or the

convex edge in tension. There is an intermediate surface known as neutral surface, at which the

stress is zero. An axis obtained by inter-section of the neutral surface and a cross-section is

known as neutral axis

about which the

bending of the surface

takes place (Fig. 5.lb).

Theory of Simple

Bending

The following theory is

applicable to the beams

subjected to simple or

pure bending when the

cross-section is not subjected to a shear force since that will cause a distortion of the transverse

planes. The assumptions being made are as under:

(i) The material is homogeneous and isotropic, i.e. it has the same values of Young’s modulus in

tension and compression.

(ii) Transverse planes remain plane and perpendicular to the neutral surface after bending.

(iii) Initially the beam is straight and all longitudinal filaments are bent into circular arcs with a

common center of curvature which is large compared to the dimensions of the cross-section.

(iv) The beam is symmetrical about a vertical longitudinal plane passing through vertical axis of

symmetry for horizontal beams.

(y) The stress is purely longitudinal and the stress concentration effects near the concentrated

loads are neglected.

Consider a length of beam under the action of a bending moment M as shown in Fig. 5.2a. NN is

considered as the original length of the beam. The neutral surface is a plane through XX. In the

side view NA indicates the neutral axis, O is the center of curvature on bending (Fig. 5.2b).

18

Figure 3.7.2

Let R = radius of curvature of the neutral surface

q = angle subtended by the beam length at centre O

F = longitudinal stress

A filament of original length �� at a distance y from the neutral axis will be elongated to a

length AB

The strain in !a = Qe&���� (original length of filament AB is NN)

Or u� = �;I�s&�s

�s = I�

Or uI = �

�

Also F = H �� ∝ H

Thus stress is proportional to the distance from the neutral axis XX. This suggests that for the

sake of weight reduction and economy it is always advisable that in the cross- section of beams,

most of the material is concentrated at the greatest distance from the neutral axis. Thus there is

universal adoption of the I-section for steel beams.

Now let �! be an element of cross-sectional area of a transverse plane at a distance y from the

neutral axis XX (Fig. 3.7.2).

For pure bending,

Net normal force on the cross-section = ()

Or � F. �! = 0

Or � �� H. �! = 0 84 �

� � H. �! = 0

Or � H. �! = 0

19

This indicates the condition that the neutral axis passes through the centroid of the section.

Also, bending moment = moment of the normal forces about x-x

Or � = �F. �!�H = � �� H. �!. H = �

� � H��! = �J�

Where # = � H��! and is known as the moment of inertia or second moment of area of the

section.

∴ uI = �

J = ��

Conventionally, y is taken positive when measured outwards from the center of curvature.

The relation derived above is based on the theory of pure bending. In practice, however, a beam

is subjected to bending moment and shear force simultaneously. But it will also be observed

(section 6.2) that in most of the cases of continuous loading, the greatest bending moment occurs

where the shear force is zero which corresponds to the condition of simple or pure bending. Thus

the theory and the equations obtained above can safely be used with a reasonable degree of

approximation for the design of beams and structures.

Section Modulus

The ratio# H� where y is the farthest or the most distant point of the section from the neutral axis

is called section modulus. It is denoted by Z.

� = �8�32r 8g 1234r1$ $+8xr 23xr4$� $�1yb1yr$293 8g g$4rℎ3yr 7812r g48� 23xr4$� $�1y

Thus� = F JI = FB

Moment of Resistance

The maximum bending moment which can be carried by a given section for a given maximum

value of stress is known as the moment of resistance�i�