1. 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
2. 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
3. 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
4. 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
5. 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
6. 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
7. 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
8. 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 ߜݔ 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 ߜݔ
F = shear force at the section x
ܨ + ߜ ܨ = Shear force at the section
ݔ + ߜݔ
M = bending moment at the section x
ܯ + ߜܯ = Bending moment at the section ݔ + ߜݔ
Total load on the length ߜݔ = .ݓ ߜݔ 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 ߜ,ݔ equating vertical forces,
ܨ = ݔߜݓ + ሺܨ + ߜܨሻ
Or ݓ = −
ௗி
ௗ௫
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,
ܯ + .ܨ
ௗ௫
ଶ
+ ሺܨ + ߜܨሻ.
ௗ௫
ௗ௫ଶ
− ሺܯ + ߜܯሻ = 0
Neglecting the product and squares of small quantities, ܨ =
ௗெ
ௗ௫
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 ݓ = −
ௗி
ௗ௫
between two values of x,
ܨ − ܨ = ݔ݀ݓ
which is the area under the load distribution diagram.
Similarly, integrating equation ܨ =
ௗெ
ௗ௫
between two values of x,
ܯ − ܯ = න ݔ݀ܨ
This shows that the variation of bending moment between two sections is equal to the area under
the shear force diagram.
Also as ܨ =
ௗெ
ௗ௫
,
ݓ = −
݀ܨ
݀ݔ
= −
݀ଶ
݉
݀ݔଶ
9. 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 ݔݓ
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 ݔݓ to the right of
section can be assumed to be acting as a concentrated load at
a point at a distance ݔ
2ൗ from the free end.
Taking moments about the section, ܯ = .ݔݓ
௫
ଶ
=
௪௫మ
ଶ
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 =
௪మ
ଶ
at the fixed end.
10. 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
ܣ = ܴܫ − ܹ. ܽ = 0
Or ܴ =
ௐ
; Similarly ܴ =
ௐ
Shear force diagram
Portion BC: Consider a section at u distance ݔ 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 ݔ from the end B, the force to the right of section is
= ܴ − ܹ =
ௐ
− ܹ =
ௐିௐ
= −ܹ ቀ
ି
ቁ = −
ௐ
= ܴ (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, ܯ = ܴݔ =
ௐ௫
ܯ = 0; ܯ =
ௐ
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.
11. 8
Figure 3.3.2
ܯ = ܴݔ =
ௐ௫
; ܯ = 0; ܯ =
ௐ
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,
ܯ =
ௐቀ
మ
ቁቀ
మ
ቁ
=
ௐ
ସ
, 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 = ܹ݈ ; ܴ = ܴ =
ௐ
ଶ
Shear Force Diagram
At a section at a distance x from A
ܨ௫ = ܴ − ݔݓ =
݈ݓ
2
− ݔݓ = ݓ ൬
݈
2
− ݔ൰ ሺ ݈݅݊݁ܽݎሻ;
ܨሺ௫ୀሻ =
݈ݓ
2
; ܨሺ௫ୀሻ = −
݈ݓ
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.
ܯ௫ = ܴ. .ݔ
ݔ
2
=
݈ݓ
2
ݔ −
ݔݓଶ
2
=
ݔݓ
2
ሺ݈ − ݔሻ ሺ݈ܾܿ݅ܽݎܽሻ
ܯሺ௫ୀሻ = 0 ; ܯሺ௫ୀሻ = 0; ܨ =
ௗெ
ௗ௫
= 0 or
௪
ଶ
− ݔݓ = 0 ݎ ݔ =
ଶ
Thus maximum bending moment = ܯሺ௫ୀ
మ
ሻ
=
௪మ
଼
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 ܴ = ܴ =
ௐሺାଶሻ
ଶ
Shear force diagram
. Portion DA: ܨ௫ = −ݔݓ ሺ݈݅݊݁ܽݎሻ ܨௗ = 0;
ܨ
= −ܽݓ
12. 9
Figure 3.3.3
. Portion AB: ܨ௫ = −ݔݓ +
௪ሺାଶሻ
ଶ
ሺ݈݅݊݁ܽݎሻ;
ܨሺ௫ୀሻ =
௪
ଶ
; ܨሺ௫ୀାሻ = −
௪
ଶ
Portion BE: ܨ௫ = −ݔݓ +
௪ሺାଶሻ
ଶ
+
௪ሺାଶሻ
ଶ
= −ݔݓ +
ݓሺ݈ + 2ܽሻ ሺ݈݅݊݁ܽݎሻ
ܨሺ௫ୀାሻ = ;ܽݓ ܨሺ௫ୀାଶሻ = 0;
Shear force diagram is shown in Fig. 3.3.3 b.
Bending moment diagram
. Portion DA ܯ௫ = −
௪௫మ
ଶ
ሺ݈ܾܿ݅ܽݎܽሻ;
ܯௗ = 0; ܯ = −
ܽݓଶ
2
. Portion AB
ܯ௫ = −
ݔݓଶ
2
+
ݓሺ݈ + 2ܽሻ
2
ሺݔ − ܽሻ ሺ ݈ܾܿ݅ܽݎܽሻ
ܯ = −
ܽݓଶ
2
; ܯሺ௫ୀାሻ = −
ܽݓଶ
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 ܯ = −ݔݓଶ
/2.
At midpoint C,
ܯ
ቀ௫ୀା
ଶ
ቁ
= −
ݓቀܽ + ݈
2ൗ ቁ
ଶ
2
+
ݓሺ݈ = 2ܽሻ
2
.
݈
2
−
ݓ
2
ቆܽଶ
+
݈ଶ
4
+ ݈ܽ −
݈ଶ
2
− ݈ܽቇ =
ݓ
8
ሺ݈ଶ
− 4ܽଶሻ
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
13. 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 ߜݔ apart. Considering the
complimentary shear stress ߬ at a distance y0 from the neutral axis, let ,ܨ ܨ + ߜܨ ܽ݊݀ ,ܯ ܯ +
ߜܯ be the shear forces and the bending moments at the two sections. ݖ is the width of the cross-
section at this position.
The force due to
complimentary shear
stress on the area
ܴܳܵܶ(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 ߪand ߜߪ are the normal stresses on an
elemental area ߜA at a distance ݕ from the neutral axis at
the two transverse sections, then for the equilibrium,
߬. .ݖ ߜݔ = න ݀ߪ. ݀ܣ
14. 11
Figure 3.4.3
Bending stresses at the two sections at distance y from the neutral axis,
ߪ =
ெ௬
ூ
and ߪ + ߜߪ =
ሺெାఋெሻ௬
ூ
Therefore, ߜߪ =
ሺఋெሻ௬
ூ
߬. .ݖ ߜݔ = ቀ
ఋெ.௬
ூ
ቁ . dA or ߬ =
ఋெ
ఋ௫.௭.ூ
.ݕ ݀ܣ = ቀ
ఋெ
ఋ௫
ቁ
ଵ
௭.ூ
. ݕܣത
But
ఋெ
ఋ௫
ܨ ; ∴ ߬ = .ܨ
௬ത
௭ூ
In the above relation,
Z is the actual width of the section at the position where ߬ is to be calculated
I is the total moment of inertia about the neutral axis
ݕܣത 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). ߬ = .ܨ
௬ത
ூ
= .ܨ
ቂቀ
మ
ି௬ቁቃቂ௬ା
భ
మ
ቀ
మ
ି௬ቁቃ
.൬
್య
భమ
൰
= .ܨ
ቀ
మ
ି௬ቁ.
భ
మ
ቀ
మ
ା௬ቁ
.൬
್య
భమ
൰
=
ி
ௗయ
. ቀ
ௗమ
ସ
− ݕଶ
ቁ
This indicates that there is parabolic variation of shear
stress with y.
At neutral axis ሺݕ = ܱሻ, shear stress =
ଷ
ଶ
.
ி
ௗ
;
it is the maximum shear stress.
Usually, ܾ݀/ܨ is known as the mean stress and thus
߬௫ = 1.5߬
15. 12
Figure 3.4.5
I-Section
In the flange, at a distance y from neutral axis (Fig. 3.4.4 a),
߬ = .ܨ
ݕܣത
ܫݖ
= .ܨ
ܤ ቀ
ܦ
2
− ݕቁ ቂݕ +
1
2
ቀ
ܦ
2
− ݕቁቃ
.ܤ ܫ
ܨ
ܫܤ
. ܤ ൬
ܦ
2
− ݕ൰
1
2
൬
ܦ
2
+ ݕ൰൨ =
ܨ
2ܫ
. ቆ
ܦଶ
4
− ݕଶ
ቇ
At ݕ = ܦ
2ൗ , ߪ = 0
At ݕ = ݀
2ൗ ,
߬ =
ܨ
8ܫ
ሺܦଶ
− ݀ଶሻ
In the web, at a distance y from neutral axis (Fig.3.4.4 b),
߬௪ = .ܨ
ݕܣത
ܫݖ
= .ܨ
ܤ ቀ
ܦ
2
−
݀
2
ቁ ቂ
݀
2
+
1
2
ቀ
ܦ
2
−
݀
2
ቁቃ + ܾ ቀ
݀
2
− ݕቁ ቂݕ +
1
2
ቀ
݀
2
− ݕቁቃ
ܾ. ܫ
=
ܨ
ܾܫ
. ܤ ൬
ܦ − ݀
2
൰ ൬
ܦ + ݀
4
൰ + ܾ ൬
݀
2
− ݕ൰
1
2
൬
݀
2
+ ݕ൰൨
=
ܨ
8ܫ
ܤ
ܾ
ሺܦଶ
− ݀ଶሻ + ሺ݀ଶ
− 4ݕଶሻ൨
Maximum shear stress =
ி
଼ூ
ቂ
ሺܦଶ
− ݀ଶሻ + ݀ଶ
ቃ at the neutral axis (y=0)
At the top of web, ݕ = ݀ 2⁄
ி
଼ூ
ቂ
ሺܦଶ
− ݀ଶሻቃ =
. ݂߬
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
16. 13
Figure 3.5.1
Figure 3.5.2
Lecture 3.5
Circular Section
Refer fig 3.5.1
ݖ = 2ඥݎଶ − ݕଶ ݎ ݖଶ
= 4ሺݎଶ
− ݕଶሻ ݎ 2. .ݖ ݀ݖ
= −8ݕ݀ݕ ݎ ݕ݀ݕ = −.ݖ ݀ݖ 4⁄
Now, ݕܣത = Moment of shaded area about neutral axis
Consider a strip of thickness ߜݕ at a height y from neutral
axis and parallel to it,
Area of the strip=.ݖ ߜ,ݕ
Moment of elementary area about neutral axis= .ݖ ߜ.ݕ ݕ
Moment of whole of the shaded area about neutral axis,
ݕܣത = න .ݖ .ݕ ݀ݕ
௬
= −
1
4
න .ݖ .ݖ ݀ݖ
௭
=
1
4
න ݖଶ
݀ݖ
௭
=
ݖଷ
12
߬ = .ܨ
1
ݖܫ
ݖଷ
12
= .ܨ
1
ܫ
ݖଶ
12
=
ܨ
3ܫ
ሺݎଶ
− ݕଶሻ
Thus shear stress variation is parabolic in nature.
߬୫ୟ୶ሺ௬ୀሻ = .ܨ
݀ଶ
4ൗ
3ሺߨ݀ଶ 64⁄ ሻ
=
4
3
ܨ
ሺߨ݀ଶ 4⁄ ሻ
=
4
3
߬௩
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
ߠ from vertical (Fig. 3.5.2). Let the shear
stress at A and B be ߬. Now, the
complimentary shear stress is on
longitudinal planes and is balanced by
difference of normal stresses on the area
subtended by angle 2ߠ. 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.
17. 14
Figure 3.5.3
Thus,
For a length ߜݔ of the beam.
2߬. .ݐ ߜݔ = න ݀ߪ݀ܣ
ఏ
ିఏ
= න ݀ߪ. ሺܴ݀߮ሻ. ݐ
ఏ
ିఏ
Or ߬ =
ଵ
ଶ
ఋఙ
ఋ௫
ఏ
ିఏ
ሺܴ. ݀߮ሻ
But ߜߪ =
ఋெ.௬
ூ
∴ ߬ =
ఋெ
ఋ௫
ఏ
ିఏ
ሺܴ. ݀߮ሻ
.௬
ூ
=
ଵ
ଶ
ܨሺܴ. ݀߮ሻ
.௬
ூ
ఏ
ିఏ
ܨ =
ఋெ
ఋ௫
ܫ௫ =
1
2
. ܲݎ݈ܽ ݉ݐ݊݁݉ ݂ ݅݊݁ܽ݅ݐݎ
=
1
2
× ܽ݁ݎܣ × ሺ݉݁ܽ݊ ݏݑ݅݀ܽݎሻଶ
=
1
2
. 2ߨܴ.ݐ ܴଶ
= ߨܴݐଷ
Hence
߬ =
ܴܨ
2ܫ
න .ݕ ݀߮
ఏ
ିఏ
=
ܴܨ
2ߨܴݐଷ
න ܴܿݏ ߮ . ݀߮
ఏ
ିఏ
=
ܨ
2ߨܴݐ
න cos ߮ ݀߮
ఏ
ିఏ
=
ܨ
2ߨܴݐ
ሾsin ߮ሿିఏ
ఏ
=
݊݅ݏܨ ߠ
ߨܴݐ
Or ߬ = 2 × ݉݁ܽ݊ ݏℎ݁ܽݎ ݏݏ݁ݎݐݏ
Square with a Diagonal Horizontal
Refer Fig. 6.8,
ܫ௫௫ = 2 ቈ
.ܤ ሺܤ 2⁄ ሻଷ
12
=
ܤସ
48
߬ = .ܨ
௬ത
௭ூ
=
ி
ర ସ଼⁄
ଵ
ଶ௬
.
ଵ
ଶ
ሺ2,ݕ ݕሻ ቀ
ଶ
−
ଶ
ଷ
ݕቁ.
=
ܨ
ܤସ 48⁄
ݕ
2
. ൬
ܤ
2
−
2
3
ݕ൰
=
4ݕܨ
ܤସ
ሺ3ܤ − 4ݕሻ,
i.e. it is parabolic.
At ݕ = 0, ߬ = 0
At neutral axis ݕ =
ଶ
, ߬ =
ଶி
మ
If b is the side of the square, ܾ = 2√ܤ
At neutral axis ߬ =
ଶி
൫√ଶ൯
మ =
ி
మ
=
ி
= ߬
For maximum value,
ௗఛ
ௗ௬
=
ௗ
ௗ௬
ሺ3ݕܤ − 4ݕଶሻ = 0 or 3ܤ − 8ݕ = 0 ݎ ݕ =
ଷ
଼
ܤ
20. 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).
21. 18
Figure 3.7.2
Let R = radius of curvature of the neutral surface
ߠ = angle subtended by the beam length at centre O
ߪ = 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 ܤܣ =
ିேே
ேே
(original length of filament AB is NN)
Or
ఙ
ா
=
ሺோା௬ሻఏିோఏ
ோఏ
=
௬
ோ
Or
ఙ
௬
=
ா
ோ
Also ߪ = ݕ
ா
ோ
∝ ݕ
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 ߜܣ 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 ߪ. ݀ܣ = 0
Or
ா
ோ
.ݕ ݀ܣ = 0 ݎ
ா
ோ
.ݕ ݀ܣ = 0
Or .ݕ ݀ܣ = 0
22. 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 ܯ = ሺߪ. ݀ܣሻݕ =
ா
ோ
.ݕ ݀.ܣ ݕ =
ா
ோ
ݕଶ
݀ܣ =
ாூ
ோ
Where ܫ = ݕଶ
݀ܣ and is known as the moment of inertia or second moment of area of the
section.
∴
ఙ
௬
=
ெ
ூ
=
ா
ோ
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 ratioܫ
ݕൗ 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.
ܼ =
ݐ݊݁݉ܯ ݂ ݅݊݁ܽ݅ݐݎ ܾܽݐݑ ݈݊݁ܽݎݐݑ ܽݏ݅ݔ
݁ܿ݊ܽݐݏ݅ܦ ݂ ݂ܽݐݎℎ݁ݐݏ ݐ݊݅ ݂݉ݎ ݈݊݁ܽݎݐݑ ܽݏ݅ݔ
Thusܯ = ߪ
ூ
௬
= ߪݖ
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 resistanceሺܯሻ
