BME-017_B-1(Unit_6)

77

Stress Distribution

in BeamsUNIT 6 STRESS DISTRIBUTION IN BEAMS

Structure 6.1 Introduction

Objectives

6.2 Section Modulus 6.2.1 Rectangular Section 6.2.2 Circular Section 6.2.3 I and T Sections 6.2.4 Triangular Section

6.3 Analysis of Partial Beam Section 6.3.1 Normal Force 6.3.2 Moment of Normal Force

6.4 Flitched Beams 6.4.1 Moment of Resistance 6.4.2 Equivalent Section

6.5 Beams of Uniform Strength 6.6 Shear Stress Distribution in Beams

6.6.1 Shear Stress Distribution in Rectangular Section 6.6.2 Shear Stress Distribution in Circular Section 6.6.3 Shear Stress Distribution in I and T Sections 6.6.4 Shear Stress Distribution in Triangular Section

6.7 Summary 6.8 Answers to SAQs

6.1 INTRODUCTION In Unit 5, we studied about the bending stress and the practical applications of the bending stress, design for bending and the determination of load bearing capacity of a beam of given cross-section.

In this unit, we are going to study about section modulus, flitched beams and shear stress distribution in beams of various cross-section. We shall also study about the normal force on a partial beam section and the moment of the normal force about the neutral axis of the beam.

Objectives

After studying this unit, you should be able to

• determine the section modulus for beams of various cross-section,

• analyse the partial beam section,

• calculate the normal force on a partial beam section and the moment of the normal force about the neutral axis of the beam,

• explain flitched beams and its equivalent section, and

• evaluate the shear stress distribution in beams of various cross-section.

6.2 SECTION MODULUS

Let M be the moment of resistance of a beam section and I be the moment of inertia of the beam section about the neutral axis.

78

Forces and Stresses in Beams

The bending stress, σ at any layer in the beam section, distance y from the neutral axis is

given by the expression σ =I

M× y. The maximum bending stress occurs at the outermost

layer.

Let ymax be the distance of the outermost layer from the neutral axis and the maximum bending stress at that layer be σmax.

Thus, σmax = I

M× ymax

∴ M = maxyI

× σmax

= Z × σmax

where Z is known as the section modulus. So the section modulus is the ratio of the moment of inertia of the beam section about the neutral axis to the distance of the outermost layer of the beam from the neutral axis.

If σmax is equal to the permissible bending stress, then M represents the greatest moment of resistance of the beam section.

6.2.1 Rectangular Section Solid Rectangular Section (Figure 6.1)

Let b be the breadth and d be the depth of a rectangular beam section.

The neutral axis coincides with the centroidal axis of the beam.

b

d / 2

d / 2

A N

d / 2

d / 2

σ

σ

Beam Section Bending Stress Distribution

Figure 6.1

Moment of inertia, I = 121

bd 3

Distance of outermost layer from the neutral axis, ymax = 2d

∴ Section modulus, Z = maxyI

= 121

bd 3 × d2

= 61

bd 2

79

Stress Distribution

in BeamsLet σ be the maximum bending stress at the outermost layer.

∴ Moment of resistance, M = σ × Z = 61

σ bd 2

We can see the bending stress distribution diagram in Figure 6.1. Hollow Rectangular Section

A hollow rectangular section with overall breadth B and depth D is shown in Figure 6.2.


A

b

d / 2

d / 2

B

D / 2

D / 2

N

D / 2

σ

σ

Figure 6.2

Let the breath and depth of the central rectangular hole be b and d, respectively.

∴ Moment of inertia, I = 1212

33 bdBD−

Distance of outermost layer from the neutral axis, ymax = 2D

∴ Section modulus, Z = maxyI

= 121

[BD 3 – bd 3] × D2

= ⎥⎥⎦

⎤

⎢⎢⎣

⎡ −D

bdBD6

33

Let σ be the maximum bending stress at the outermost layer.

∴ Moment of resistance, M = σ × Z = D6σ [BD 3 – bd 3]

The bending stress distribution is shown in Figure 6.2. Example 6.1

A rectangular beam of size 40 mm × 60 mm is subjected to bending. The maximum bending stress is limited to 75 N/mm2. Find the moment of resistance of the beam section.

Solution

Section modulus, Z = 61

bd 2

= 61

× 40 × (60)2 = 24 × 103 mm3

80


Moment of resistance, M = σ × Z

= 75 × 24 × 103

= 1800 × 103 N mm = 1.8 kN m

Thus, moment of resistance of the beam section = 1.8 kN m.

6.2.2 Circular Section Solid Circular Section (Figure 6.3)

Let the diameter of the solid circular section be d.

Moment of inertia about the neutral axis, 4

64dI π

=

Distance of outermost layer from the neutral axis, max 2dy =


Figure 6.3

Section modulus, Z =maxyI

= d2

64d 4

×π

= 32d3π


∴ Moment of resistance, M = σ × Z

= σ32d3π

The bending stress distribution is shown in Figure 6.3.

Hollow Circular Section (Figure 6.4)

Let the external and internal diameter of the hollow circular section be D and d respectively.

Moment of inertia about the neutral axis, I = 64π [D4 – d 4]

Distance of outermost layer from the neutral axis, ymax = 2D

d / 2

σ

σ

d

d / 2

A N

81

Stress Distribution

in Beams∴ Section modulus, Z =maxyI

4 4 2[ ]64

D dD

π= − ×

4 4[ ]32

D dD

π= −


∴ Moment of resistance, M = σ × Z

= D32

σπ [D4 – d 4]


D / 2

σ

σ

D / 2

D d

A N

Figure 6.4

Example 6.2

A circular beam of diameter 100 mm is subjected to bending. The maximum bending stress is limited to 50 N/mm2. Find the moment of resistance of the beam section.

Solution

Section modulus, Z =32

3dπ

= 32π

× (100)3 = 98.17 × 103 mm3

Moment of resistance, M = σ × Z

= 50 × 98.17 × 103

= 4908.5 × 103 N mm = 4.91 kN m

Thus, Moment of resistance of the beam section = 4.91 kN m.

6.2.3 I and T Sections

In these sections, first we have to find the centroid of the section. Then the moment of inertia of the section about the centroidal axis is determined. As the neutral axis coincides with the centroidal axis, the moment of inertia about the neutral axis is the same as that about the centroidal axis.

In unsymmetrical sections, the distance of the extreme layer, ymax, is not the same for the topmost and bottommost layers. Therefore, the topmost layer distance, yt, and the bottommost layer distance, yb, are found. The section modulus is found using the higher

82

value of yt or yb viz. the maximum value of y. Then, the moment of resistance of the section is equal to the product of the maximum bending stress and the section modulus.


Example 6.3

An I section in Figure 6.5 is used as a beam. The beam is subjected to a bending moment of 2.5 kN m at its neutral axis. Find the maximum stress developed in the beam.

σmin

σmax

yt = 79.2 mm

yb = 60.8 mm

100 mm

60 mm

100 mm

20 mm

20 mm

N A 20 mm


Figure 6.5

Solution

Let y be the distance of centroid from the bottom face.

Then, y = aay

∑∑

y = (100 20) 10 (20 100) 70 (60 20) 130(100 20) (20 100) (60 20)

× × + × × + × ×× + × + ×

= 60.8 mm

Moment of inertia of the section about the horizontal axis passing through the centriod,

I = 3 21 100 (20) (100 20)(60.8 10)12

⎡ ⎤× × + × −⎢ ⎥⎣ ⎦

+ 3 21 20 (100) (20 100)(70 60.8)12

⎡ ⎤× + × −⎢ ⎥⎣ ⎦

+ 3 21 60 (20) (60 20)(130 60.8)12

⎡ ⎤× + × −⎢ ⎥⎣ ⎦

= 1285.04 × 104 mm4

Topmost layer distance, yt = 140 – 60.8 = 79.2 mm

Bottommost layer distance, yb = 60.8 mm

∴ ymax = yt = 79.2 mm

From the relation, yI

M σ=

We get, σmax = I

M× ymax

Bending moment, M = 2.5 kN m

= 2.5 × 106 mm

83

Stress Distribution

in Beams∴ σmax = 4

6

1004.1285105.2×

×× 79.2 = 15.41 N/mm2

∴ The maximum bending stress in the beam = 15.41 N/mm2.

6.2.4 Triangular Section A triangular section with base b and height h is shown in Figure 6.6.

The position of neutral axis is at a distance h/3 from the base.

Moment of inertia about the neutral axis, I = 36

3bh

The topmost layer is at a distance of 3

2hfrom the neutral axis and the bottommost layer

is at a distance of 3h

from the neutral axis.

ytop = 32

h and ybottom = 31

h

b

h/3 G

N A

h

σmin

σmax

2/3 h

h/3


Figure 6.6

∴ Section modulus, Z =maxyI

=242

336

23 bhh

bh=×

Let σmax be the maximum bending stress at the topmost layer.

∴ Moment of resistance, M = σmax Z = σmax 24

2bh

SAQ 1

(a) A rectangular beam of size 60 mm × 100 mm has a central rectangular hole of size 15 mm × 20 mm. The beam is subjected to bending and the maximum bending stress is limited to 100 N/mm2. Find the moment of resistance of the hollow beam section.

(b) A water main of 1200 mm internal diameter and 15 mm thick is subjected to bending. The maximum bending stress is limited to 60 N/mm2. Find the moment of resistance of the water main.

84


(c) A T-section has a flange of size 150 mm × 50 mm and web of size 50 mm × 150 mm. It is used as a beam. The beam is subjected to bending moment of 3 kN m at its neutral axis. Find the stresses at the topmost and bottommost layers of the beam.

(d) A triangular beam of base 30 mm and height 60 mm is subjected to a bending moment of 45 N m. Find the stresses at the top and bottom of the triangular beam.

6.3 ANALYSIS OF PARTIAL BEAM SECTION

Till now, we have seen forces and moments acting over the entire cross-section of the beam. Now, let us consider a part of the beam section and try to establish a relation for the normal force acting on the partial beam section and the moment of the normal force about the neutral axis.

6.3.1 Normal Force Consider a beam section as shown in Figure 6.7.

Let σmax be maximum bending stress at the topmost layer and ymax be the distance of the topmost layer from the neutral axis.

Let us find the normal force on a partial area A, shown shaded in Figure 6.7. In that shaded area, let da be an elemental area and y be its distance from the neutral axis.


Figure 6.7

The stress, σ on the elemental area is proportional to the distance y.

maxmax

yy

σ = σ ×

Normal force on the elemental area da= σ ×

maxmax

y day

= σ

Total normal force on the shaded area maxmax

y day

= ∑ σ

max

y y day

= ∑

σmin

σmax Area

da

yymax

N A

85

Stress Distribution

in BeamsLet y be the distance of the centroid of the shaded area from the neutral axis.

y da A y∑ =

∴ Normal force on a partial beam section max

maxA y

yσ

=

6.3.2 Moment of Normal Force Now, let us find the moment of this normal force about the neutral axis. Again have a look at Figure 6.7.

Normal force on the elemental area maxmax

y day

= σ

max

maxy da

yσ

=

Moment of this normal force about the neutral axis 2max

maxy da

yσ

=

∴ Moment of normal force on the shaded area about the neutral axis

2max

maxy da

yσ

= ∑

2max

maxy da

yσ

= ∑

But Moment of inertia of the shaded area about the neutral axis, I2y da∑ = s.

∴ Moment of normal force on a partial beam section about the neutral axis,

max

maxsI

yσ

= ×

Example 6.4

A rectangular beam has a width of 100 mm and a depth of 200 mm. This is used as a simply supported beam and the maximum bending stress is limited to 10 N/mm2. Then determine the following :

(a) total normal force on the left bottom corner area of size 40 mm × 60 mm, and

(b) moment of this normal force about the neutral axis.

Solution

Maximum bending stress, σmax = 10 N/mm2

Normal force on the shaded area, left bottom corner max

maxA y

yσ

=

10 (40 60) (40 30)100

= × × × +

Here, A = Shaded area = 40 × 60 mm2

y = Centroid of shaded area from the neutral axis

60(100 60) (40 30) 70 mm2

= − + = + =

86

Forces and Stressin Beams

es

100 mm

100 mm

A N

Figure 6.8

∴ Normal force on the partial beam section

10 2400 70 16800 N 16.8 N100

= × × = =

The partial area is subjected to a tensile force

∴ Normal force on the partial beam section = 16.8 kN (tension) Moment of the force about the neutral axis

max

maxsI

yσ

= ×

Is = Moment of inertial of the shaded area about the neutral axis

3 21 40 (60) (40 60) (70) 1248 10 mm12

= × × + × × = × 4 4

∴ Moment of normal force about the neutral axis

4 310 1248 10 1248 10 Nm 1248 m100

= × × = × =

SAQ 2 A rectangular beam of size 160 mm × 240 mm is used as a simply supported beam. The maximum bending stress is 9 N/mm2. Find the normal force on the shaded area shown in Figure 6.9. Also find the moment of this normal force about the neutral axis.

Figure 6.9

100 m m

40 mm

60 mm

⎯yymax

80 mm

160 mm

N

60 mm

240 mm

87

Stress Distribution

in Beams6.4 FLITCHED BEAMS

A wooden beam may be strengthened by the addition of mild steel plates. This composite beam is known as flitched beam. The mild steel plates may be added to the sides of the wooden beam, as shown in Figure 6.10.

b t t

d

A N

Mild Steel Plate Wooden Beam

Figure 6.10

The wooden beam and the mild steel plates are connected together so that there is no slipping between them. They act together as a single beam.

Let b be the width and d be the depth of wooden beam and let t be the thickness and d be the depth of each steel plate.

Let σw be the bending stress in wood and σs be the bending stress in steel plate, at a distance y from the neutral axis.

The strain in wood and steel at a distance y from the neutral axis will be equal, as they act as a single beam.

Let Ew be the modulus of elasticity of wood and Es be the modulus of elasticity of steel plate.

Then, Strain in wood = Strain in steel

w s

w sE Eσ σ

=

ss w

w

EE

σ = σ

∴ s wmσ = σ

where m is the modular ratio between steel and wood and equals s

w

EE

.

6.4.1 Moment of Resistance Let Mf be the moment of resistance of the flitched beam.

Let Mw and Ms be the moments of resistance of wood and steel, respectively.

f w sM M M= +

2 21 126 6w sb d t d⎛ ⎞ ⎛= × σ + × ×⎜ ⎟ ⎜

⎝ ⎠ ⎝⎞σ ⎟⎠

But σs = m σw, where m is the modular ratio.

88

Forces and Stresses in Beams ∴ 2 21 12

6 6f w wM bd m td= σ + × σ

21 [ (2 )]6 w b m t d= σ +

6.4.2 Equivalent Section In Section 6.4.1, we have seen that

21 [ (2 )]6f wM b m t d= σ +

The moment of resistance of the flitched beam section is equal to that of a wooden beam of width [b + m (2t)] and a depth d. This rectangular cross-section of width [b + m (2t)] and depth d is known as the equivalent wooden beam section. Figure 6.11 shows the flitched beam and the equivalent beam.

b t t

d

Mild Steel Plate

Wooden Beam

d

m t b m t

b + m (2 t)

Equivalent WFlitched Beam Section ooden Beam Section

Figure 6.11

The moment of resistance of the flitched beam nay also be found from the equivalent wooden section.

The following example illustrates the determination of moment of resistance of a flitched beam.

Example 6.5

A flitched beam consists of a wooden beam of width 120 mm and depth 250 mm and mild steel plates of thickness 10 mm and width 120 mm one on each side of the wooden beam. The maximum bending stress in wood is 7 N/mm2 and the modular ratio between steel and wood is 20. Find the moment of resistance of the flitched beam section.

Solution

Moment of resistance of the wooden beam 216 w bd= σ

21 7 120 (250)6

= × × ×

89

Stress Distribution

in Beams 38759 10 N-mm 8750 Nm= × =

Moment of resistance of the steel plates,

2

32 112 2

2

wm d tt b btd tt

⎛ ⎞σ ⎛ ⎞⎜ ⎟= × × + +⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠+⎜ ⎟⎝ ⎠

2

32 20 7 1 120 1010 120 120 10120 12 2 210

2

⎛ ⎞× × ⎛ ⎞⎜ ⎟= × × × + × +⎜ ⎟⎜ ⎟⎛ ⎞ ⎝ ⎠⎝ ⎠+⎜ ⎟⎝ ⎠

= 29167 Nm

Moment of resistance of flitched beam = Moment of resistance of wooden beam + Moment of resistance of steel plates.

= 8750 + 29167

= 37917 Nm

SAQ 3 A flitched beam consists of a wooden beam of width 120 mm and depth 240 mm and mild steel plates of thickness 10 mm and width 180 mm, symmetrically placed on either side of the wooden beam. The maximum bending stress in wood is 7 N/mm2 and the modular ratio between steel and wood is 20. Find the maximum bending stress in steel and the moment of resistance of the flitched beam.

120 mm

240 mm

A N

Mild Steel Plate

Wooden Beam

10 mm 10 mm

Flitched Beam Section

180 mm

Figure 6.12

90

Forces and Stresses in Beams 6.5 BEAMS OF UNIFORM STRENGTH

Usually, a beam has the same cross-sectional area throughout the span. But the bending moment due to loads on the beam is not the same at all sections of the beam. We design the beam for the maximum bending moment. Hence, maximum permissible stress occurs in the extreme layers of the beam only at the particular section where the bending moment is maximum. At all other sections, the stresses in the extreme layers of the beam are less than the permissible stress.

As the bending moment decreases towards the supports, the cross-section of the beam may also be reduced towards the support. This ensures that the stress in the extreme layers reaches the permissible stress at each and every section of the beam. Such a beam is called a beam of uniform strength.

The cross-section of a beam of uniform strength may be reduced by one of the following ways :

(a) width of beam is varied keeping the depth constant.

(b) depth of beam is varied keeping the width constant.

(c) width and depth are both varied.

The following example deals with a beam where the depth is varied keeping the width constant.

Example 6.6

A simply supported beam of span l carries a concentrated load P at midspan. If the beam is to be a beam of uniform strength, find the shape of the beam by keeping the width constant and varying the depth towards the supports.

Solution

Let the beam has uniform width b throughout the span and let d be the maximum depth at midspan.

Let dx be the depth of beam at any section XX, distance x from the support A.

Bending moment at section XX 2P x= .

Moment of resistance of the beam at section XX 21 ( )6 xb d= × σ × × .

Moment of resistance = Bending moment at that section

∴ 21 ( )6 2x

Pb d× σ × × = x

∴ 2 3x

Pd xb

=σ ×

∴ 3x

Pd xb

=σ ×

Hence, the depth at any section at a distance x from the left support A may be found. The variation of the depth is shown in Figure 6.13.

Thus, the depth at midspan, 1at2

x⎛ =⎜⎝ ⎠

⎞⎟ , we get

1 12 23 3

2 2P l Pldb

⎛ ⎞ ⎛= × = ×⎜ ⎟ ⎜σ σ⎝ ⎠ ⎝

⎞⎟⎠

.

91

Stress Distribution

in Beams

x l / 2 l / 2

x

x

P

BA

RA = P/ 2 RB = P/ 2

Simply Supported Beam

x′

x

Elevation

b

Plan

Figure 6.13

SAQ 4 For the problem given in Example 6.6, find the shape of the beam by keeping the depth constant and varying the width towards the supports.

6.6 SHEAR STRESS DISTRIBUTION IN BEAMS

When a beam is loaded, bending moment and shear force are developed at all section of the beam. In Unit 4, you have already learnt the methods of determining shear force and bending moment in a beam section under the given load conditions. Previously, we have seen the bending stress distribution at any cross section of the beam. Now, we will study the shear stress distribution at any section of the beam.

The vertical shear force at any section of a beam produces shear stress at that section. This vertical shear stress is accompanied by a horizontal shear stress of equal magnitude, known as complementary shear stress. So at any point in a cross section of the beam, there is a vertical shear stress and a horizontal shear stress of equal magnitude. These two shear stresses cause the diagonal tension and compression inclined at 45° to the horizontal.

Let us consider two cross sections AB and CD at a distance dx apart, in a beam under transverse loading.

Let M and M + dM be the bending moments at the sections AB and CD, respectively, as shown in Figure 6.14.

Let da be an elementary area at a distance y from the neutral axis.

Let σ1 and σ2 be the bending stresses at the sections AB and CD respectively, on the elementary area.

∴ σ1 = IM

× y and σ2 = IdMM )( +

× y

92


Figure 6.14

The force on the elementary area at the section AB = σ1 × da =I

My× da

The force on the elementary area at the section CD = σ2 × da =I

ydMM )( +× da

The unbalanced force on the elementary area =I

dM× y × da

Considering any level EF, total unbalanced force above the level EF between the two sections AB and CD, is as follows :

IdM

∑ × y × da

=2

1

yy

yy

IdM =

=∑ × y × da

=I

dM A y

Here, A y is the moment of area above level EF about the neutral axis. To ensure that the part of beam above the level EF and between the sections AB and CD is in equilibrium, as equilibriant to the unbalanced force, the beam section at the level EF must offer a shear resistance. Let the width of beam at the level EF be b. The intensity of the horizontal shear stress at the level EF is τ.

Then, Unbalanced forceShear area

τ =

=I

dM A y × )(

1dxb ×

=dx

dM×

IbyA

As we have seen earlier,dx

dM = F, the shear force, and shear stress, τ =IbF

× A y

where A is the area of cross section of the beam above the level EF, and y is the distance of the centroid of the area above the level EF, from the neutral axis.

For a particular cross section of a beam,IF remains constant for any point along the

depth of the beam. Therefore, we can conclude that the shear stress at any point on a

cross section is proportional to byA .

N A

dx

B D

C A

M M + d M da

E

y1

y2 F y

d

Loaded Beam

b

Beam Cross Section

93

Stress Distribution

in BeamsFor sections of uniform width, the shear stress will have maximum value at the neutral axis, since A y is maximum at the neutral axis. The shear stress will have zero value at top and bottom layers of the beam cross-section, since A y is zero at these layers.

Now let us see the shear stress distribution in some cross sections.

6.6.1 Shear Stress Distribution in Rectangular Section Let us consider a rectangular section of width b and depth d subjected to shear force F.

Let τ be the shear stress at any level EF, as indicated in Figure 6.15.

Area ABFE, A = b ⎟⎠⎞

⎜⎝⎛ − y

2d

Distance, y = y + ⎟⎠⎞

⎜⎝⎛ − y

2d

21

= ⎟⎠⎞

⎜⎝⎛ + y

2d

21

E

y

A B

F

N

GX

⎯y

b

C D

A

d / 2

d / 2

Beam Cross-section

τmax

Shear Stress Distribution

Figure 6.15

Moment of area ABFE about the neutral axis,

A y = b ⎟⎠⎞

⎜⎝⎛ − y

2d

× ⎟⎠⎞

⎜⎝⎛ + y

2d

21

= ⎟⎟⎠

⎞⎜⎜⎝

⎛− 2

2

42ydb

∴ Shear stress, τ = IbF

× A y

=IbF

× ⎟⎟⎠

⎞⎜⎜⎝

⎛− 2

2

42ydb

= ⎟⎟⎠

⎞⎜⎜⎝

⎛− 2

2

42yd

IF

∴ Shear stress has a parabolic variation.

The maximum shear stress occurs when y = 0, at the neutral axis.

94

Forces and Stresses in Beams ∴ τmax = ⎟

⎟⎠

⎞⎜⎜⎝

⎛− 0

42

2dI

F =I

Fd8

2

= 3

2 128 bd

Fd×

= bdF

×23

= 32 Cross - sectional area

F= ×

The shear stress is zero at the top and bottom layers, i.e. at y = 2d

± .

22

min 02 4 2F d dI

⎡ ⎤⎛ ⎞τ = −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

=

Average shear stress Shear forceArea of cross section of beam

=

τaverage = bdF

∴ τmax = bdF

×23

τmax = 1.5 × τaverage

Example 6.7

A simply supported beam has a width of 100 mm and a depth of 150 mm. It is loaded with uniformly distributed load over the entire span of 3 m. If the permissible shear stress is 3 N/mm2, find the value of the uniformly distributed load on the beam.

Solution

Let w be the uniformly distributed load on the beam.

Then, maximum shear force, F at the support = wwwl23

23

2=×=

Maximum shear stress at the neutral axis =bdF

×23

This value equals 3 N/mm2.

Thus, ∴ bdF

×23 = 3

1501001

23

23

××× w = 3

∴ w = 20000 N/m

∴ Value of uniformly distributed load on the beam = 20 kN/m.

6.6.2 Shear Stress Distribution in Circular Section Now we shall consider a circular section of radius R.

Let dy be the thickness of an elementary strip at a distance y from the neutral axis.

95

Stress Distribution

in BeamsMoment of this trip about the neutral axis = (b dy) y

Moment of area above the plane EF about the neutral axis, A y = ∫=

=

Ry

yy(b × y) dy

τma


E F

O N A

b′

y

dy

Beam Cross-section

R

Figure 6.16

But b = 2 × 22R y−

or b2 = 4 (R2 – y2)

On differentiating, we get,

2b db = 4 (– 2y) dy

= – 8y dy

∴ y dy =4db b

−

When y = y, b = b and y = R, b = 0.

Now changing the integration variable from y to b,

A y = ⎟⎠⎞

⎜⎝⎛−∫ 4

0 db bbb

= db bb∫ −0

)(41 2

= 123

)(41 33 0

b

bb=⎥

⎦

⎤⎢⎣

⎡−

∴The shear stress, τ =IbF

× A y

=I

FbbIbF

1212

23=×

= )(412

22 yRI

F−×

= )(3

22 yRI

F−

Thus, shear stress has a parabolic variation. The shear stress is maximum when y = 0, at the neutral axis.

96

Forces and Stresses in Beams ∴ τmax = 2

3R

I F

×

= 4

2 43 R

FRπ

× since, I = 4

4Rπ

= 234

RF

π×

43 Cross - sectional area

F= ×

The shear stress is zero at top and bottom layers, i.e. at y = R,

∴ τmin = )(3

22 RRI

F− = 0

Average shear stress 2Shear force

Area of cross section of beamFR

= =π

∴ τmax = 234

RF

π×

= ×34

τaverage

Example 6.8

A circular beam of 100 mm diameter is subjected to a shear force of 5 kN. Find the value of the maximum shear stress and also draw the shear stress distribution diagram.

Beam Cross-section

100 mm N

O A

τmax = 0.849 N/mm2


Figure 6.17

Solution Diameter of beam = 100 mm

Area of cross section of beam = 4π

× (100)2 = 7854 mm2

Shear force, F = 5 kN = 5000 N We have,

max43 Cross-sectional area

Fτ = ×

∴ 2max

4 5000 0.849 N/mm3 7854

τ = × =

97

Average shear stress = 0.6366 N/mm2 Stress Distribution

in Beams Minimum shear stress = 0.849 N/mm2

6.6.3 Shear Stress Distribution in I and T Sections I-Section

Let us take an I-section with flange width B and overall depth D. Let b and d be the thickness of web and depth, respectively.

N

Beam Cross-section

y

A

b

B

E F

d D τmax


Figure 6.18

Shear Stress Distribution in the Flange

Width of section at a distance y from the neutral axis = B

Area above the plane EF = B ⎟⎠⎞

⎜⎝⎛ − yD

2

Centroidal distance of this area from the neutral axis

= ⎟⎠⎞

⎜⎝⎛ − yD

221 + y = ⎟

⎠⎞

⎜⎝⎛ + yD

221 .

Moment of the area above the plane EF about the neutral axis,

A y = B ⎟⎠⎞

⎜⎝⎛ − yD

2 × ⎟

⎠⎞

⎜⎝⎛ + yD

221

= ⎟⎟⎠

⎞⎜⎜⎝

⎛− 2

2

42yDB

∴ Shear stress, τ = IBF

× A y

= IBF

× ⎟⎟⎠

⎞⎜⎜⎝

⎛− 2

2

42yDB

= ⎟⎟⎠

⎞⎜⎜⎝

⎛− 2

2

42yD

IF

Thus, the shear stress has a parabolic distribution in the flange.

At the top of I-section, i.e. at y = 2D , the shear stress, τ = 0.

At the junction of the flange and web, i.e. at y = 2d

,

98

Forces and Stresses in Beams Shear stress, τ = )(

844222

22dD

IFdD

IF

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−

Shear stress distribution in the Web

Width of section at a distance y from the neutral axis = b.

y

N

Figure 6.19

Area above the plane EF = area of flange + area of web upto EF

= ⎟⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ − ydbdDB

22

Centroidal distance of the flange area from the neutral axis,

= ⎟⎠⎞

⎜⎝⎛ +

2221 dD

Centroidal distance of the web area from the neutral axis,

= ⎟⎠⎞

⎜⎝⎛ + yd

221

Moment of both the areas about the neutral axis,

A y = ⎟⎠⎞

⎜⎝⎛ +×⎟

⎠⎞

⎜⎝⎛ −+⎟

⎠⎞

⎜⎝⎛ +

×⎟⎠⎞

⎜⎝⎛ − yDyDbdDdDB

221

2221

2

= ⎟⎟⎠

⎞⎜⎜⎝

⎛−+− 2

222

42)(

8ydbdDF

∴ Shear stress, τ = IbF

× A y

= IbF

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−+− 2

222

42)(

8ydbdDB

∴ The shear stress has also a parabolic distribution in the web.

A

b

B

E F

Beam Cross-section

d

D

99

Stress Distribution

in BeamsAt the junction of the flange and web, i.e. at y = 2d

,

∴ Shear stress, τ = IbF × )(

822 dDB

−

Now, it can be noticed that at the junction of the flange and web, the shear

stress suddenly changes from )(8

22 dDI

F− to

IbF

× )(8

22 dDB− .

At the neutral axis, i.e. at y = 0, the shear stress is maximum and it is as follows :

∴ τmax = IbF

⎥⎥⎦

⎤

⎢⎢⎣

⎡×+−

42)(

8

222 dbdDB

The shear stress distribution can be seen in Figure 6.18.

From the shear stress distribution diagram, we conclude that most of the shear stress is taken up by the web. This is very important in the design of beams using I and T sections.

T-Section

The method of finding the shear stress distribution in T-sections is similar to that of I-section. As the T-section is not symmetrical about the neutral axis, the shear stress distribution also will not be symmetrical.

Example 6.9

A T-section has a flange of size 150 mm × 50 mm and web of size 50 mm × 150 mm. This section is subjected to a vertical shear force of 100 kN. Find the maximum shear stress and draw the shear stress distribution diagram with values at important points. Centroid of section is 125 mm from the bottom face and moment of inertia about the centroidal axis is 5312.5 × 104 mm4.

N A

150 mm

50 mm

150 mm

125 mm

50 mm

Beam Cross-section Shear Stress Distribution

τMax = 14.71N/mm2

4.71

14.12

Figure 6.20

Solution

Maximum shear stress occurs at the neutral axis.

Thus, τmax = IbF

× A y

100

Forces and Stresses in Beams 4

100 1000 [(150 50) 50 (50 25) 12.5]5312.5 10 50

×= × × + ×

× ××

= 14.71 N/mm2

Shear stress in the flange at the junction of flange and web,

4100 1000 (150 50 50)

5312.5 10 150×

= × ×× ×

×

= 4.71 N/mm2

Shear stress in the web at the junction of flange and web,

4100 1000 (150 50 50)

5312.5 10 50×

= × ×× ×

×

= 14.12 N/mm2

The shear stress distribution is shown in Figure 6.20.

6.6.4 Shear Stress Distribution in Triangular Section

Let us consider a triangle of base b and h. Its centroid is at a distance 3h

from the base.

Beam Cross-section

2/3 h

N

h/3

h

E F

b

h′

y

A

h/2

b′ τmax = 3 Fbh

τNA = Fbh

83


Figure 6.21

Let the shear stress at the plane EF, at a distance y from the neutral axis be q.

Let b′ and h′ be the width and height of the triangle above the plane EF.

From similar triangles,

b hb h

=′ ′

bhbh

′′ =

23

h h⎛ ⎞′ = −⎜ ⎟⎝ ⎠

y

23

bb hh

⎛ ⎞′ = −⎜ ⎟⎝ ⎠

y

Centroid of the triangle above the plane EF, from the neutral axis = ⎟⎠⎞

⎜⎛ ⎝

′+

3hy

101

Stress Distribution

in BeamsOn substituting the value of h′ = ⎟⎠⎞

⎜⎝⎛ +=⎟

⎠⎞

⎜⎝⎛ −+

332

32

31 hyyhy

Moment of this triangular area about the neutral axis,

⎟⎠⎞

⎜⎝⎛ +×

′′=

332

2A hyhby

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ +×⎥

⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −×⎟

⎠⎞

⎜⎝⎛ −×=

332

32

32

21 hyyhyh

hb

⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛ −=

332

3

2 hyyhhb

∴ Shear Stress, τ bIF

′= × A y

⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛ −×

⎟⎠⎞

⎜⎝⎛ −

×=33

23

32

1 2 hyyhhb

yhhbI

F

⎟⎠⎞

⎜⎝⎛ +⎟

⎠⎞

⎜⎝⎛ −×=

332

3hyyh

IF

⎟⎟⎠

⎞⎜⎜⎝

⎛+−×=

92

33

22 hyhy

IF

For shear stress τ to be maximum, 0=dydτ

,

or ⎟⎠⎞

⎜⎝⎛ − yh 2

3 = 0

∴ y = 6h

The height from the base of the triangle is 263hhh

=+

∴ Maximum shear stress is at a distance 2h

form the base of the triangle, which is also

at a distance of 6h

from the centroidal axis.

Substituting y = 6h

in the equation for the shear stress, we get,

τmax = IFhhhhh

IF

1292

6633

222

=⎥⎥⎦

⎤

⎢⎢⎣

⎡+⎟

⎠⎞

⎜⎝⎛−×

=bhF

bhFh 3

3612

3

2=

×

The shear stress at the neutral axis, i.e. y = 0, is as follows :

τNA =I

FhhI

F27

29

23

22=⎟

⎟⎠

⎞⎜⎜⎝

⎛

102

Forces and Stresses in Beams =

bhF

bhFh

×=×

38

3627

23

2

The shear stress distribution diagram can be seen in Figure 6.21.

Example 6.10

A beam of triangular cross section of base width 120 mm and height 80 mm is subjected to a shear stress force of 3 kN. Find the maximum shear stress and the shear stress at the neutral axis.

Solution

Maximum shear stress is at a distance 2h

from the base of the triangle.

Maximum shear stress, τmax = bhF3

2max

3 3000 0.94 N/mm120 80

×τ = =

×

Shear stress at the neutral axis,

NA83

Fbh

τ = ×

28 3000 0.83 N/mm3 120 80

= × =×

SAQ 5

(a) A simply supported beam of rectangular cross section 120 mm × 200 mm has a span of 4 m. Find the uniformly distributed load it can carry if the maximum bending stress and the maximum shear stress are not to exceed 10 N/mm2 and 0.6 N/mm2 respectively.

(b) A simply supported beam of circular cross section 120 mm in diameter has a span of 4 m and carries a central concentrated load of 16 kN. Find the maximum shear stress and its location.

(c) An I-section has an overall depth of 240 mm with horizontal flanges each measuring 120 mm × 20 mm and a vertical web 200 mm × 20 mm. It is subjected to a vertical shear force of 200 kN. Find the maximum shear stress and its position. Draw the shear stress distribution diagram.

6.7 SUMMARY

We conclude this unit by summarizing what we have covered in it. We have

(a) studied the section modulus of beams of various cross section.

(b) found the normal force for partial beam section and its moment about the neutral axis.

103

Stress Distribution

in Beams(c) obtained the moment of resistance of a flitched beam and its equivalent

section.

(d) describe about the beams of uniform strength.

(e) determined the shear stress distribution in beams of various cross-section.

6.8 ANSWERS TO SAQs

SAQ 1

(a) M = 9.98 kN m

(b) M = 1030.92 kN m

(c) σtop = 4.23 N/mm2 and σbottom = 7.06 N/mm2

(d) σtop = 10 N/mm2 and σbottom = 5 N/mm2

SAQ 2

Normal force = 32.4 kN

Moment of normal force = 3024 Nm

SAQ 3

σs (maximum) = 105 N/mm2 (Refer the Figure for Answer to SAQ 3 given below)

Moment of resistance of flitched beam = 19404 Nm.

120 mm 90 mm

7 N/mm2

5.25 N/mm2

Bending Stress Distribution in Wood

Figure for Answer to SAQ 3

SAQ 4

x I / 2 I / 2

X

X P

d

Elevation

x

b bx

x

Plan

Depth

Width

Figure for Answer to SAQ 4

104

SAQ 5 Forces and Stresses in Beams

(a) From bending consideration, w = 4 kN/m

From shear consideration, w = 4.8 kN/m

Maximum uniformly distributed load, w = 4 kN/m

(b) Maximum shear stress = 0.943 N/mm2

Position: along the neutral axis passing through the center.

(c) Maximum shear stress = 50.86 N/mm2

Position : along the neutral axis passing through the centroid.

(Refer the figure given below)

6.15

τmax = 50.86 N/mm2

36.88



Figure for Answer to SAQ 5(c)

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BME-017_B-1(Unit_6)