Bending Stresses in Beams

Department of Mechanical Engineering, K L University 1

CHAPTER – 5 Bending stresses in Beams.

5.1 Introduction

In Previous chapters we considered the stresses in prismatic

bars subjected to axial loads and Twisting moment. In this chapter we

consider third fundamental loading, bending. A beam is a structural

member that is subjected to loads acting transversely to the

longitudinal axis, as explained in the preceding chapter. Internal loads

develop in beams in the form of shear forces and bending moments to

resist the external loads. Shear stresses and Bending moments

develop with in the cross section due to internally develop shear forces

and bending moments respectively. In this chapter we will restrict

ourselves to study the bending stresses only due to bending moment.

The shear stresses due to shear forces will discuss in the next

chapter. Before stating the discussion on bending strains and stresses

we need to know the concept of Pure bending. Pure bending means a

beam or a portion of the beam under a constant bending moment,

which means that the shear force is zero.

Fig 5.1 Beam/Portion of beam in pure bending

To illustrate the concept pure bending, consider two simply supported

beams as shown in the Fig 5.1 (a) and (b). In Fig 5.1 (a), the region of

beam between the two point loads is constant and the value is ‘Pa’


and the shear force in this region is zero. Hence the central region is

in pure bending. In Fig 5.1 (b), the beam is loaded only by couples

that produce constant bending moment and zero shear force

throughout the beam. In this chapter we will calculate the normal

strains and stresses in pure bending.

5.2 Bending stress

The stresses caused by the bending moment are known as

bending stresses or flexure stresses. The relationship between bending

moment and bending stresses is called flexure stresses. The following

assumptions are made in deriving the flexure or bending stress

equation.

1. The beam is initially straight or has a very large radius of

curvature compared to its cross section dimensions.

2. The transverse cross section of the beam is symmetrical about

an axis passing through its centroid (in our case it is Y–axis)

and parallel to the plane of the bending.

3. The transverse section of beam, which is plane before bending,

will remain a plane after bending.

4. The cross section dimensions are small compared to its length.

5. The beam is in pure bending. i.e. the beam is subjected to only

bending moment.

6. The material of the beam is homogenous and isentropic.

7. The stress is in elastic and limit and obeys Hook’s Law.

Consider a portion of a beam ab which is in pure bending produced by

couples M as shown in the Fig. 5.2.

Fig. 5.2 A portion of beam in pure bending.


From the directions of the couples we can say that bending moments

are positive (ref. section 4.5 and Fig. 4.6) and the cross section is

symmetry about y-axis. Consider two plane transverse section mn, pq

separated by a distance of ‘dx’ apart. Under the action of the moments

the beam gets deflected in to a circular curve as shown in the Fig. 5.3.

Fig. 5.3 Deformation of a Beam in pure bending.

The length of the bottom fiber is elongated and top fibers are

contracted. Thus, bottom fibers are in tension and top fibers are in

compression. Somewhere between top and bottom of the beam there is

a surface which does not change in length. This surface indicated in

dashed line in Fig. 5.2 and 5.3, is called neutral surface. The

intersection of the neutral surface with any cross section plane is

called neutral axis of the cross section.

The planes mn, pq get deflected and occupies the positions

m1n1and p1q1 as shown in the Fig. 5.3, being inclined at angle dθ and

intersecting at ‘o’, the center of the curve. Let, R is the radius of

Curvature. The distance between n1, q1 is more than dx and between

m1, p1 less than dx, but along the neutral surface, the initial distance

dx is remains same. Hence,

R.dθ = dx …………………………………………5.1

To evaluate the strains, consider a fiber ‘ef’ at a distance of ‘y’

from the neutral axis as shown in the Fig. 5.2. Initially the length of


the fiber ef is Li=dx and after deforming the length of the fiber e1f1 is Lf

= (R + y) dθ. So, the strain in the fiber ef is

𝐿𝑓−𝐿𝑖

𝐿𝑓=

(R + y) dθ−𝑑𝑥

𝑑𝑥=

R dθ+y.dθ−𝑑𝑥

𝑑𝑥 (From eq.5.1)

=y.dθ

𝑑𝑥=

𝑦

𝑅 (From eq.5.1)

Thus, Normal strain in beam = 𝑦 𝑅

……………………………… 5.2

This equation shows that longitudinal strain in the beam is inversely

proportional to the radius of curvature and varies linearly with the

distance from the neutral axis. Thus, bending stress is maximum at

the outer surfaces which are at the greatest distance from the neutral

axis. If the distance is measured above the neutral axis, y is negative

and strain is also negative.

according to Hooke’s Law, Normal stress in beam

𝐸𝑦

𝑅 ……………………………………. 5.3

Thus, normal stresses acting on the beam vary linearly with the

distance from neutral axis. This type of stress distribution is shown in

the Fig. 5.4(a). Along the neutral axis bending stress is zero. Top fibers

are in compression (above the neutral axis) and bottom fibers are in

tension (below the neutral axis) for positive bending moment. For

negative bending moment top fibers are in tension and bottom fibers

are in compression.

Fig. 5.4 Distribution of Normal stresses on a cross section.


5.3 Flexure Formula

Let us consider the resultant force acting over the cross section.

To calculate the resultant force consider a small elemental area dA in

the cross section at a distance of ‘y’ from the neutral axis as shown in

the Fig. 5.4(b).The force acting on this small elemental area normal to

moment can be calculate from the equation 5.3. Since no external

axial force is acting on the beam, (assumption 5) the equilibrium of

the forces in the x direction leads to equation

∫ σ. dA = 0

∫Ey

R. dA = 0 (from equation 5.3)

Since modulus of elasticity E and radius of curvature R is constant we

can conclude that

∫ y. dA = 0 ………………………………………………..

5.4

This is the equation for first moment of area of cross section with

respect to the neutral axis. That is the neutral axis must pass through

the centroid of the cross section. This property can be used to locate

the position of the neutral axis for a beam of any cross sectional shape

which has symmetry about y- axis.

Let us consider next the moment of the resultant force acting

the over the cross section (Fig. 5.4(b)). The moment acting on this

y. The external moment acting on the

beam is M, the equilibrium of the moment leads to equation

∫ σ. dA.y = M

∫Ey2

R. dA = M (from equation 5.3) ………………………………..

5.5

But we know that ∫ y2. dA is the equation for second moment of

area of cross section with respect to the neutral axis, which is moment

of inertia of cross sectional area with respect to the neutral axis

I = ∫ y2. dA


Now the equation 5.5 is reduces to 𝐸𝐼

𝑅 = M

𝑀

𝐼=

𝐸

𝑅 ………………………………………………. 5.6

Combining the equation 5.3 and 5.6, we get the flexure formula

𝜎

𝑦=

𝑀

𝐼=

𝐸

𝑅 ……………………………………………. 5.7

From the above, we can write the equation for bending stress

𝑀𝑦

𝐼 ………………………………………………….. 5.8

Thus, bending stress is directly proportional to the bending moment,

inversely proportional to the area moment of inertia about neutral axis

and varies linearly with the distance from neutral axis as shown in the

Fig. 5.4 (a). The maximum stress due to bending moment occurs at

the fiber for which y is maximum, i.e., the extreme fiber from the

neutral axis. max = 𝑀.𝑦𝑚𝑎𝑥

𝐼. The

quantity 𝐼

𝑦𝑚𝑎𝑥 is known as section modulus, which is the property of

the cross section denote with a letter ‘z’.

The section modulus for simple cross sections are shown in the

Fig.5.5

For Rectangular cross section:

I = 𝑏ℎ3

12 and ymax =

ℎ

2

h z = 𝑏ℎ2

6

b

For Circular Cross section

I = 𝜋𝑑4

64 and ymax =

𝑑

2

z = 𝜋𝑑3

32

d


Fig. 5.5 Section modulus for simple cross sections

5.4 Composite Beams

Beams that are built of more than one material are called

composite beams. Examples are bimetallic beams, sandwich beams,

reinforced concrete beams as shown in the Fig. 5.6. Composite beams

can be analyzed by the same way as that of ordinary beam. The main

advantage with these beams is it can withstand with more bending

load within less space/cross sectional area compared to beam with

single material.

Fig. 5.6 Composite beam cross sections (a) bi metallic beam

(b) sand witch beam (c) reinforced concrete beam

Let us consider a beam made of two materials. Let 1 and 2 are

the suffixes used for material 1 and 2 respectively. At the common

surface, strain in both the material is same.

1 2

But, 1 = 𝜎1

𝐸1 2 =

𝜎2

𝐸2

𝜎1

𝐸1=

𝜎2

𝐸2

𝜎1

𝜎2=

𝐸1

𝐸2

Thus, the ratio of stresses between material 1 and 2 is directly in the

ratio of their modulus of elasticity.


Let y1 and y2 are the distances of the farthermost fibers of

material 1 and material 2 respectively form the neutral axis.

[𝜀1

𝜀2]

𝑚𝑎𝑥=

y1

y2 (From equation 5.2)

σ1max

E1X

E2

σ2max =

y1

y2

σ1max

σ2max =

y1

y2X

E1

E2 ……………………………….. 5.9

But we know that 𝜎

𝑦=

𝑀

𝐼 (From equation 5.7)

M1 = σ1max. I1

y1 and M2 =

σ2max. I2

y2

The total moment of resistance

M = M1 + M2

= σ1max. I1

y1+

σ2max. I2

y2

= σ2max. I1

y2.

E1

E2+

σ2max. I2

y2 (From eq. 5.9)

M = σ2max

y2[

E1

E2I1 + I2]

Similarly M = σ1max

y1[

E2

E1I2 + I1]

So, for a composite beam the total resisting moment is

M = σ1max

y1[

E2

E1I2 + I1] =

σ2max

y2[

E1

E2I1 + I2] …………… 5.10

[E2

E1I2 + I1] is equivalent moment inertia for material 1 & [

E1E2

I1 + I2]

is equivalent moment inertia for material 2.


EXAMPLES EXAMPLES EXAMPLES EXAMPLES

5.1. A thin steel rule (E = 206.7GPa) having a thickness 0.734mm

and length 254mm is bent by couples as shown in Fig.P.5.

what is the maximum stress in steel rule.

Fig. P. 5.1

Sol: Given data: t = 0.734mm, E = 206.7x103MPa, l = 254mm,

θ=600=1.047

We know that flexure formula 𝜎

𝑦=

𝑀

𝐼=

𝐸

𝑅.

From the given data we can write the equation from flexure formula

𝜎 =𝐸𝑦

𝑅

Where y = maximum distance of a surface from neutral axis

= 0.734/2 = 0.367

From Fig. P. 5.12 we can write that l = Rθ

Radius of curvature R = 254/1.047=242.6mm

Bending stress 𝜎 =206.7𝑥103𝑥0.367

242.6

= 312.63MPa

5.2. A steel wire of diameter 4mm bent over a drum of radius

0.5m as shown in the Fig. P.5.2. Calculate the maximum

bending stress if E = 200GPa.

Fig. P. 5.2


Sol: Given data: d = 4mm, E = 200x103MPa, r = 0.5x103m=500mm

We know that flexure formula𝜎

𝑦=

𝑀

𝐼=

𝐸

𝑅.

From the given data we can write the equation from flexure formula

𝜎 =𝐸𝑦

𝑅

Where y = maximum distance of a surface from neutral axis

= 4/2 = 2mm

Radius of curvature R = r+(d/2) = 500+2 = 502mm

Bending stress σ =200x103x2

502

= 796.8 MPa.

5.3. A simple beam AB of length 6.7m supports a uniform load of

intensity 22KN/m and a concentrated load of 53.4KN as

shown in the Fig. P.5.3. The beam cross section is rectangle

with width is 222mm and depth is 686mm. Determine the

maximum compressive and tensile stresses in the beam due

to bending.

Sol: Given data: b = 222mm and d = 686mm.

The maximum bending stress occurs at the cross section due to

maximum bending moment. To find the maximum bending moment

let us construct the shear force and bending moment diagram as

shown in the Fig.P.5.3. From the Shear force and bending moment

diagram shown in the Fig.P.5.3, the maximum bending moment is

204.9KNm.

From the flexure formula, the bending stress is

𝜎 =𝑀𝑦

𝐼

M = Maximum bending moment = 204.9x106 N mm

y = Maximum distance from neutral axis. we know that neutral

axis passes through the centroid. Since the given cross section is

rectangle cross section, the centroid is coincides with the

geometric center.


y = 686/2 = 343mm.

Fig.P.5.3

I = Moment of Inertia of rectangle cross section

= 𝑏ℎ3

12=

(222)(686)3

12

= 6x109 mm4

The maximum bending stress

𝜎 =204.6𝑥106𝑥343

6𝑥109

= 11.7 MPa.

Since the maximum bending moment is positive, the top fibers are in

compression and bottom fibers are in tension. The bending stress

distribution is as show in the Fig.P.5.3a


Fig.P.5.3a

5.4. Determine the maximum allowable length of simple beam of

rectangular cross section (Fig.P.5.4) subjected to uniformly

distributed load of 6.5KN/m, if the allowable stress is

8.2MPa.

Sol: Given data:

Fig.P.5.4

Since it is a simply supported beam with UDL the maximum bending

moment is WL2

8 = 0.8125L2. (See Fig. 4.14, Chapter 4)

From the flexure formula, the bending stress is

𝜎 =𝑀𝑦

𝐼

M = Maximum bending moment = 0.8125L2 N mm

y = Maximum distance from neutral axis. We know that neutral

axis passes through the centroid. Since the given cross section is

rectangle cross section, the centroid is coincides with the

geometric center.

y = 240/2 = 120mm.

I = Moment of Inertia of rectangle cross section

= 𝑏ℎ3

12=

(140)(240)3

12

= 1.61x108 mm4

8.2 = 0.8125𝐿2𝑥120

1.61𝑥108

Length of the beam L = 3680mm = 3.68m


5.5. The beam shown in the figure Fig.P.5.5 is subjected to

positive bending by couples M. Determine the ratio of the

maximum tensile and compressive stresses if the cross

section is (a) an equilateral triangle (b) a semi circle

Fig.P.5.5

Sol: (a) Equilateral triangle:

Fig.P.5.5a

For triangle the centroid is at a distance of 2h/3 from the big end and

h/3 from the small end (Fig.P.5.5a). The bending moment is positive,

so top fibers are in compression and bottom fibers are in tension.

Maximum compression stress at the top fibers:

M= M, y = 2h/3 and I = I

c = 𝑀.2ℎ

3𝐼=

2𝑀ℎ

3𝐼

Maximum tension stress at the bottom fibers:

M= M, y = h/3 and I = I

c = 𝑀.ℎ

3𝐼=

𝑀ℎ

3𝐼

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑠𝑡𝑟𝑒𝑠𝑠=

𝑀ℎ

3𝐼.

3𝐼

2𝑀ℎ = 0.5

(b) Semi Circle:

For Semi circle the centroid is at a distance of 4r/3π (0.212d) from the

big end and 0.282d from the small end. The bending moment is


positive, so top fibers are in compression and bottom fibers are in

tension.

Maximum compression stress at the top fibers:

M= M, y = 0.282d and I = I

c = 𝑀.(0.282𝑑)

𝐼

Maximum tension stress at the bottom fibers:

M= M, y = 0.212d and I = I

c = 𝑀.(0.212𝑑)

𝐼

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑠𝑡𝑟𝑒𝑠𝑠=

𝑀(0.212𝑑)

𝐼.

𝐼

𝑀(0.282𝑑) = 0.752

5.6. A beam ABC is loaded as shown in the Fig.P.5.6. The cross

section is a channel section as shown in the Fig. Calculate

the maximum tensile and compressive stresses in the beam.

Sol: To find the maximum bending moment, let us construct shear

force and bending moment diagram as shown in the Fig.P.5.6

Fig.P.5.6

From Fig.P.5.6, we can understand that the maximum bending

moment is 3.375KN.m and it is negative bending moment which

results top fibers are in tension and bottom fibers are in compression.


To locate the neutral axis, let us find the centroid, since neutral axis

is passes through centroid. For this divide the given cross section into

three areas as shown in the Fig.P.5.6.

A1 = 300x12 = 3600mm2, y1 =74mm,

A2 = 68x12 = 816mm2, y2=34mm,

A3 = 68x12 = 816mm2, y3=34mm.

Centroid �̅� = 𝐴1𝑦1+𝐴2𝑦2+𝐴3𝑦3

𝐴1+𝐴2+𝐴3 = 61.52mm.

To find the Moment of Inertia let us use parallel axis theorem

I = [(300)(12)3

12+ (300)(12)(74 − 61.52)2] +

[(12)(68)3

12+ (68 )(12)(34 − 61.52)2] + [

(12)(68)3

12+ (68 )(12)(34 − 61.52)2]

= 43200+560701.44+314432+617997.9+314432+617997.9

= 2468761mm4.

Fig.P.5.6 (a)

Maximum compressive stress (point B in Fig. P.5.6)

y = 61.52mm, M = 3.375x106 N mm and I = 2468761mm4

𝑀𝑦

𝐼=

3.375𝑥106𝑥61.52

2468761 = 84.10 MPa (Compressive)

Maximum tensile stress (point A in Fig. P.5.6)

y = 18.48mm, M = 3.375x106 N mm and I = 2468761mm4

𝑀𝑦

𝐼=

3.375𝑥106𝑥18.48

2468761 = 25.26 MPa (Tensile)

The stress distribution and neutral axis is shown in the Fig.P.5.6 (a)

5.7. Determine the maximum bending stress caused by a

concentrated load 5.4KN on a simple beam of cross section as

shown in the Fig.P.5.7


Sol: The given beam is a simply supported beam with a point load.

The maximum bending moment is 𝑊𝑎𝑏

𝐿=

5.4𝑥103𝑥1.2𝑥1.8

3.0 = 3888Nm.

Position of neutral axis

A1 = 25x75=1875mm2, y1 = 12.5mm, A2 = 1875mm2, y2 = 62.5mm

Centroid �̅� = 𝐴1𝑦1+𝐴2𝑦2

𝐴1+𝐴2 = 37.5mm.

Fig.P.5.7

Moment of Inertia I = [(75)(25)3

12+ (75)(25)(37.5 − 12.5)2] +

[(25)(75)3

12+ (75 )(25)(37.5 − 62.5)2]

= 97656.25+1171875+878906.25+1171875

= 3320312.5 mm4 =33203125x10 -12 m4

Since the maximum bending moment is positive top fibers are in

compression and bottom fibers are in tension.

Maximum tensile stress (point B in Fig. P.5.7)

y = 37.5x10-3 m, M = 3888 Nm and I =332.312.5x10 -12 m4

𝑀𝑦

𝐼=

3888𝑥37.5𝑥10−3

3320312.5x10−12 = 43.91 MPa (Tensile)

Maximum compressive stress (point A in Fig. P.5.7)


y = 62.5x10-3 m, M = 3888 Nm and I =332.312.5x10 -12 m4

𝑀𝑦

𝐼=

3888𝑥62.5𝑥10−3

3320312.5x10−12 = 73.2 MPa (Compressive)

5.8. An overhang beam ABC of T cross section supports a

concentrated load as shown in the Fig.P.5.8. Calculate the

maximum permissible value of load P based upon allowable

stress in material 40MPa in tension and 70MPa in

compression.

Sol: To find the maximum bending moment let us construct the shear

force and bending moment diagram as shown in the Fig.P.5.8. The

maximum bending moment is – P.


A1 = 20x80=1600mm2, y1 = 40mm,

A2 = 100x20=2000mm2, y2 = 90mm

Centroid �̅� = 𝐴1𝑦1+𝐴2𝑦2

𝐴1+𝐴2 = 67.8mm.

Fig.P.5.8


12+ (20)(80)(67.8 − 40)2] +

[(100)(20)3

12+ (100 )(20)(67.8 − 90)2]

= 853333.33+1236544+66666.67+985680

= 3142224 mm4 = 3142224x10-12m4


Since the maximum bending moment is negative, top fibers are in

tension and bottom fibers are in compression.

Maximum tensile stress (Point B in Fig.5.8)

y = 33.3x10-3 m, M = P Nm and I =3142224x10-12m4

𝑀𝑦

𝐼=

𝑃𝑥33.3𝑥10−3

3142224x10−12 = 0.01054P MPa (Tensile)

Given that maximum tensile stress = 40MPa

0.01054 = 40

P = 3.774KN.

Maximum compressive stress (Point A in Fig.5.8)

y = 67.7x10-3 m, M = P Nm and I =3142224x10-12m4

𝑀𝑦

𝐼=

𝑃𝑥67.7𝑥10−3

3142224x10−12 = 0.0215P MPa (Compressive)

Given that maximum compressive stress = 70MPa

0.0215P = 70

P = 3.26 KN.

So, The maximum possible load is 3.26KN

5.9. A cantilever beam AB, loaded as shown in the Fig. P.5.9 is

constructed of a section as shown. Find the maximum

compressive and tensile stresses in the cross section.

Sol: To find the maximum bending moment let us construct the shear

force and bending moment diagram.

Fig.P.5.9


From the Fig.P.5.9, maximum bending moment is 1.8x106Nmm and it

is negative. That is top fibers are in tension and bottom fibers are in

compression.

Position of neutral axis:

A1 = 400x120=48000mm2, y1 = 200mm,

A2 = (π/4)x602=2827.4mm2, y1 = 300mm

Centroid �̅� = 𝐴1𝑦1−𝐴2𝑦2

𝐴1−𝐴2 = 193.74mm.


12+ (400)(120)(200 − 193.74)2] −

[(𝜋)(60)4

64+ (𝜋 4⁄ )(602)(300 − 193.74)2]

I = 6.4x108 + 1.88x106 – 6.3x105 – 3.2x107

= 6.1x108 mm4

Maximum compressive stress (Point B in Fig.P.5.9)

y = 193.7 mm, M = 1.8x106Nmm and I =6.1x108 mm4

𝑀𝑦

𝐼=

1.8𝑥106𝑥193.7

6.1x108 = 0.57 MPa (Compressive)

Maximum tension stress (Point A in Fig.P.5.9)

y = 206.3 mm, M = 1.8x106Nmm and I =6.1x108 mm4

𝑀𝑦

𝐼=

1.8𝑥106𝑥206.3

6.1x108 = 0.61 MPa (Tensile)

5.10. A beam of I section shown in Fig.P.5.10 is subjected to a

bending moment of 10KNm at its neutral axis. Find the

maximum stress induced in the beam.

Sol: Bending stress = 𝑀𝑦

𝐼

Where M = bending moment = 10x106 N.mm.


Fig.P.5.10


A1 = 100x20=2000mm2, y1 = 10mm,

A2 = 20x100=2000mm2, y1 = 70mm

A3 = 60x20=1200mm2, y3 = 130mm,


𝐴1+𝐴2+𝐴3 = 60.77mm


12+ (100)(20)(10 − 60.77)2] +

[(20)(100)3

12+ (100)(20)(70 − 60.77)2] +

[(60)(20)3

12+ (60)(20)(130 − 60.77)2]

= 66666.67+5155185.5+1666666.7+170385.8+40000+5751351.5

= 12850256 mm4

Since the bending moment is positive, top fibers are in compression

and bottom fibers are in compression.

Maximum tensile stress (at Point B in Fig.P.5.10)

y = 60.77mm, M = 10x106 N.mm, I = 12.58x106 mm4

𝑀𝑦

𝐼=

10𝑥106𝑥60.77

12.58x106 = 48.31 MPa (Tensile)

Maximum compressive stress (at Point T in Fig.P.5.10)

y = 79.23mm, M = 10x106 N.mm, I = 12.58x106 mm4

𝑀𝑦

𝐼=

10𝑥106𝑥79.23



5.11. An I section beam has following dimensions. Top flange 6cm

wide, 1cm thick. Bottom flange 12cm wide, 1cm thick. Web

1cm thick. Total depth of the section is 12cm. The beam is

5m long simply supported over a span 3m, overhanging both

supports by the same amount and it carries a point load of

2KN each end. Find the maximum stress in the material due

to bending.

Sol: The beam with loading and the cross section dimensions are

shown in the Fig.P.5.11. The maximum bending moment is 2KNm and

is positive. The top fibers are in compression and bottom fibers are in

tension.


A1 = 120x10=1200mm2, y1 = 5mm,

A2 = 10x100=1000mm2, y1 = 60mm

A3 = 60x10=600mm2, y3 = 110mm,


𝐴1+𝐴2+𝐴3 = 47.14mm


12+ (120)(10)(47.14 − 5)2] +

[(10)(100)3

12+ (100)(10)(47.14 − 60)2] +

[(60)(10)3

12+ (60)(10)(47.14 − 110)2]

= 10000+2130935.5+833333.33+165379.6+5000+2370827.8

= 5.5x106 mm4


Fig.P.5.11

Maximum tensile stress (at Point B in Fig.P.5.11)

y = 47.14 mm, M = 2x106 N.mm, I = 5.5x106 mm4

𝑀𝑦

𝐼=

2𝑥106𝑥47.14

5.5x106 = 17.14 MPa (Tensile)


y = 72.86mm, M = 2x106 N.mm, I = 5.5x106 mm4

𝑀𝑦

𝐼=

2𝑥106𝑥72.86


5.12. The bending moment acting on the triangular cross section

of a beam is 3.6KNm. Determine the maximum tensile and

compressive stresses on the cross section.

Sol: The bending moment is a positive bending moment. So, the top

fibers are in compression and bottom fibers are in tension.


Fig.P.5.12

The Moment of Inertia = (60)(120)3

36 = 2.88x106 mm4

Maximum tensile stress (at point B) (Fig.P.5.12)

M = 3.6x106 N.mm, y = 40mm, I = 2.88x106 mm4

𝑀𝑦

𝐼=

3.6𝑥106𝑥40

2.88x106 = 50 MPa (Tensile)


y = 80mm, M = 3.6x106 N.mm, I = 2.88x106 mm4

𝑀𝑦

𝐼=

3.6𝑥106𝑥80

2.88x106 = 100 MPa (Compressive)

5.13. A timber beam of rectangular cross section of length 8m is

simply supported. The beam carries a UDL of 12KN/m over

the entire length and a point load of 10KN at 3m from the

left support. If the depth is two times the width and the

stress in the timber is not to exceed 8MPa, find the suitable

dimensions of the section.

Sol: Let the width of the cross section is b and thus, depth of the

cross section is 2b.


Fig.P.5.13

From the above, Fig. P.5.13 shear force and bending diagram, M =

111.586 KNm.

Moment of Inertia I = b(2b)3

12 = 0.67b4

Maximum bending stress:

M = 111.586x106 N.mm, y = b mm, I = 0.67b4mm4

Bending stress = 8 = 𝑀𝑦

𝐼=

111.586𝑥106𝑥𝑏

0.67b4

Width b = 275mm

Depth 2b = 550mm

5.14. A steel beam having an I- section as shown in the Fig.P.5.14

is 4m long and is simply supported at the ends. If the safe

stress in tension for the beam is 30MPa, determine the

permissible uniformly distributed load acting on the whole

span of the beam.

Sol: Given data: , l = 4m. Let the value of the UDL is

w/unit length. So, the maximum bending moment is M = wl2

8 = 2w.


Since the given cross section is symmetry about both the axis, the

centroid is at the center of the web.


Fig.P.5.14

Moment of Inertia

I = 2. [(200)(20)3

12+ (200)(20)(290 − 150)2] + [

(20)(260)3

12]

= 1.57x108 + 29.29x106

= 1.86x108 mm4

Maximum bending stress:

M =2w, y = 150 mm, I = 1.86x108 mm4

30 = 𝑀𝑦

𝐼=

2𝑤𝑥150

1.86x108

UDL value w = 18.6x106 N/mm

W = 18.6 KN/m

5.15. A beam of 2m length is simply supported at the ends and

carries a UDL of 30KN/m over its entire length. If the cross

section of the beam as shown in the Fig.P.5.15, determine

the maximum tensile and compressive stresses in the beam.

Sol: Given data: l = 2m, w = 30KN/m.

The maximum bending moment M = wl2

8 = 15x106 N.mm


A1 = 160x120 = 19200mm2, y1 = 80mm

A2 = 80x60 = 4800mm2, y2 = 110mm


Centroid �̅� = 𝐴1𝑦1−𝐴2𝑦2

𝐴1−𝐴2 = 70mm


12+ (120)(160)(80 − 70)2] −

[(80)(60)3

12+ (80)(60)(110 − 70)2]

= 40.96x106+1.92x106 – 1.44x106 – 7.68x106 = 33.76x106 mm4

Fig.P.5.15

The maximum bending moment is a positive. So, the top fibers are in

compression and bottom fibers are in tension.

Maximum tensile stress (Point B in Fig.P.5.15)

M =15x106, y = 70 mm, I = 33.76x106 mm4

𝑀𝑦

𝐼=

15𝑥106𝑥70

33.76x106 = 31.10MPa (Tensile)

Maximum compressive stress (Point T in Fig.P.5.15)

M =15x106, y = 90 mm, I = 33.76x106 mm4

𝑀𝑦

𝐼=

15𝑥106𝑥90

33.76x106 = 39.98MPa (Compressive)

5.16. Determine the ratio of weights of three beams having same

lengths, made of same material, subjected to same maximum

bending moment and having same maximum normal stress, if

their cross sections are (i) a rectangular with height equal to

twice the width (ii) a square (iii) a circle.


Sol: We know that bending stress =𝑀𝑦

𝐼. Since, all the beams are

having same lengths, same material, same maximum bending

moment and same normal stresses we can say that all the beams

must have same 𝑦

𝐼 values.

Fig.P.5.16

[y

I]

rectangle=

a

(a)(2a)3

12

=1.5

a3

[y

I]

square=

b2

(b)(b)3

12

=6

b3

[y

I]

circle=

c2

(π)(c)4

64

=10.18

c3

1.5

a3 =6

b3 =10.18

c3

b = 1.587a and c = 1.89a

Weight of rectangular beam: Weight of square beam: Weight of

circular beam is

(ρ).(2a2.l) : (ρ).(b2.l) : (ρ).(𝜋

4c2 .l)

2a2 : 2.52a2 : 2.82a2

1:1.26:1.41


5.17. Find the dimensions of the strongest rectangular beam that

can be cut out of log of a wood 180mm diameter.

Sol: Let‘d’ is the diameter of wood and b, h are dimensions of

rectangular cross sections. Given that d = 180mm.

Fig.P.5.17

For the strongest beam the induced bending stress should be very

less. That means 𝑦

𝐼 value should be small. For the above rectangular

cross section y = ℎ

2 and I =

𝑏ℎ3

12

𝑦

𝐼 =

6

𝑏ℎ2 .For small values of 𝑦

𝐼 ,bh2 value should be high.

Let Z = bh2, but d = √𝑏2 + ℎ2

h2 = d2 – b2.

Z = (b)( d2 – b2)

Z = bd2 – b3

Let us take 𝑑𝑍

𝑑𝑏= 0

0 = d2 – 3b2

Width b = 𝑑

√3 = 103.92mm

and height h = √d2 – b2 = 146.96mm


Problems for Practice

5.1. A rectangular beam 20cm deep by 10cm wide is subjected to

maximum bending moment of 500KNm. Determine the maximum

stress in the beam. If the value of E for the material is

200GN/m2, find the radius of curvature for that portion of the

beam where the bending moment is maximum

[750MN/m2, 26.67m]

5.2. The moment of inertia of a symmetrical section of a beam about

its neutral axis is 2640cm4 and its depth is 20cm. Determine the

longest span over which, when simply supported, the beam would

carry a UDL of 6KN/m run without the stress due to bending

exceeding 120MPa. [5.93m]

5.3. A T-section beam having flange 2 cm X 10 cm and web 10 cm X 2

cm is simply supported over a span of 6m. It carries a UDL of 3

KN/m run including its own weight over its entire span; together

with a load of 2.5 KN at its mid span. Find the maximum tensile

and compressive stresses occurring in beam section.

[12.94MPa (comp.), 25.87MPa (tensile)]

5.4. Find the dimensions of the strongest rectangular beam that can

be cut out of log of a wood 450mm diameter.

[breadth = 259.8mm, depth = 367.43mm]

5.5. A CI beam of I- section with top flange 8 cm X 2 cm thick, bottom

flange 16 cm X 4 cm thick and the web is 20 cm deep and 2 cm

thick. The beam is freely supported on a span of 5m. If the tensile

stress is not to exceed 20MPa, find the safe uniformly distributed

load which the beam can carry. Find also the maximum

compressive stress. [6.82KN/m,37.34MPa]

5.6. Compare the bending strength of a solid circular section with that

of hollow with internal diameter equal to 2/3 the external

diameter, if both sections have the same cross sectional areas

[1:1.938]

5.7. A horizontal beam of section shown in Fig.P.5.7 is 3 m long and is

simply supported at the ends. Find the maximum UDL it can


carry if the compressive and tensile stresses must not exceed

56MPa and 30MPa respectively.

5.8. A beam having a cross section in the form of channel (Fig.P.5.8) is

subjected to bending moment. Calculate the thickness of the

channel in order that the bending stress at the top and bottom of

the beam will be 7:3. [50.8mm]

5.9. Fig.P.5.9 shows the section of a beam. Determine the ratio of

its resistance to bending in y-y plane to that in the x-x plane

if the maximum bending stresses is remains same in both the

cases. (Hint: Ratio between Moment of Inertias about y- and x-

axis) (0.33)

Fig.P.5.9

Bending Stresses in Beams

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