3.1

Chapter 3

STRESSES IN LOADED BEAMS

3.1 PURE BENDING

3.2

3.3

Points to note:

it is called the:

note:

3.4

For a fibre JK the new length L’:

(the change in length of JK)

and the max numerical strain is given by: m

(i) Strain Due to Bending

3.5

3.6

(ii)

3.7

Common I values:

rectangular X-section, width b, height h:

Ibh

z

3

12

SI

C C

h

2

bh2

6

Circular X-section, radius r, diameter d:

Ir d

4 4

4 64

SI

C C = r

32

,4

33 dr

3.8

Example 1

Consider a beam of rectangular X-section with

load of 5 kN/m (take E = 175 GPa) determine

(a) Max. tensile and compressive stress at mid-span

(b) normal stress and strain at A

(c) radius of curvature at B.

I bh

m

of section 1

123

008 01212

1152 103

6 4. .

.

A

D B

C

Note: A is 20 mm from NA, B is on NA, D is on top fibre, E is on bottom fibre

C is on NA

3.9

(a) At mid-span, B.M. = ?

MxzkN m

10 2 522

210 .

Using I

My

x

At top fibre MPax

1.5261052.11

)06.0(31010

At bottom fibre x = 52.1 MPa

Plotting the stress distribution

3.10

Note: A positive bending moment causes compressive (-ve)

axial stresses above the neutral axis and tensile (+ve)

stresses below the N.A.

- Stress distribution is linear.

- Max. stresses are induced at top and bottom fibre i.e. fibre

furthest from N.A.

(b) At a section thro’ A

B M

KN m

. . ( )( )

. .

10 15 1

2

7 5

2

We have

AAMy

I

MPa

7 5 10 0 02

1152 10

13

3

6

. .

.

A

A

E

13 10

175 10

74 3 10

6

9

6.

3.11

(c) Since

m

y

y

x

x

269

103.74

02.06

Note:

ρ is radius of curvature of the neutral axis

x

y

(a)

3.12

Example 2

An overhanging beam of T-shaped cross-section is loaded as shown

in the Fig. Determine the max. tensile and compressive bending

stresses.

X-sectional area is divided into A1 & A2. Let distance of centroid

from bottom = .y

y A A A y A y

y

mm

1 2 1 1 2 2

20 70 60 30

20 60

50

(60) (20)

(60) (20)

Example 2

3.13

Ibh

Ad

mm

3

2

3 2

3 2

4 4

12

1

1260 20 20 60 20

1

1220 60 20 60 20

136 10

R1

= 5 kN. R2

= 10 kN

A1

A2

20

20

From equilibrium

If you have a distributed load that does not act

along the whole length of the beam, you have to

introduce another distributed load of equal

magnitude but acting in the opposite direction

beyond the right end limit of the given distributed

load.

Equivalent System

0 a Lx

0 a Lx

Represented by

W<x0>0

Represented by

W<xa>0

W

3.14

kNm

(1)

Load Intensity

Shear force:

11001433103445

xxxxxxw

kNxxxxx

wdxF

000433103445

F = -5 + 12 -10 = -3 kN

3.15

Bending Moment:

000433103445 xxxxx

FdxM

(2)

To locate position of maximum bending moment

Consider 0 < x < 3

Max. B.M. occurs at positions of zero shear force

For zero shear force, 0 = -5 + 4x, x = 1.25 m

kNxxxxxF000

433103445

M Fdx

xx x

x x kNm

5 42

43

210 3 3 4

2 2

.

Note: From the shear force plot it is noted that zero

shear force occurs at x < 3

(B.M.)

3.16

Hence the Bending Moments are :

M

kN m

5 3 43

2

3

2

.

M xx

kN m

5 42

3125

2

. .

At x = 1.25 m

At x = 3 m

Note: x < 3 m

My

Ilargest bending stress occurs at max. y (top and bottom fibre)

i.e. at y = 30 mm, -50 mm.

3.17

max. ( ).

/

.

top fibre N mm

MPa

3125 10 30

136 10

68 9

6

2

max. ( )

.

.

bottom fibre

MPa

3125 10 50

136 10

114 9

6

4

max. ( ) /

.

top fibre N mm

MPa

3 10 30

136 10

66 2

6

4

2

c bottom fibre

MPa

( )

.

3 10 50

136 10

110 3

6

4

Hence At x = 1.25 m

At x = 3 m

Hence

max. tensile stress occurs at x = 1.25 m

max. comp. stress occurs at x = 3 m.

i.e. Stresses at locations of zero shear must be investigated

Check

1 m 3 m

A D

Fxy

kN

7

B

-3

1.25 m

-5

x

Check for values of maximum

bending stresses here

3.18

3.2 SHEAR STRESSES IN BEAMS

Mxz

X

dx

Mxz

Mxz

+ (dMxz

/dx)dx

Stresses in a beam

Mxz + δMxz

3.19

Stresses in a Beam

P

Fxy

Mxz

σxx comp top, tensile bottom

xygives

P

National University of SingaporeNational University of SingaporeNational University of Singapore

Bending moment induces normal bending stresses in the direction

of the beam axis. Shear force induces shear stresses.

3.20

Pure bending without shear induces only normal stresses

in the direction of the beam axis. When a shear force is

present (i.e. the bending moment varies along the length

of the beam), shear stresses are also induced. (In practice,

it is very uncommon to encounter pure bending in a

beam).

Consider an elemental length of a beam where the shear

force is constant but there is a variation in the bending

moment. e.g. a simply-supported beam with a central

point load.

3.21

Consider the portion of the beam element above a vertical

distance y from the centroid of the cross-section; i.e. we

have made an imaginary horizontal cut at y and chosen

the upper element which has a surface exposed by the cut.

For equilibrium in the axial direction;

xx yxA

xxA

dA F dA1 2 0

Using xxxz

z

M

Iy

3.22

F

M M y

IdA

M

Iy dAyx

xz xz

zA

xz

zA

Az

xzyx dAy

I

MF

Dividing by x and letting x 0

A

xz

z

yxdAy

dx

dM

Idx

dF 1

Now y dA A yA

(first moment of area A about the Z-axis)

A - area of the cross-section isolated by the horizontal

cut; i.e. above the location of the shear stress being

determined (i.e. above y)

y - vertical distance between the centroidal axis and the

CG of A

Since dM

dxF

xz

xy

dF

dx

F

IA y

yx xy

z

-this term is known as the

‘shear force per unit

length’ or ‘shear flow’

3.23

yAI

F

b

dx

dF

b

z

xy

yxxy

.1

.1

dx

dF

b

Area

yx

yx

.1

forceshear

xyyx Since

We have,

shear stress

3.24

Example 1

Determine the shear stress distribution in a beam of

rectangular cross-section (b x h) subjected to the loading

shown below.

Consider the cross-section at y-y.

3.25

xy yx

xy

z

F A y

I b

F Pxy at section y-y

Ahy b

2

y

hyy

22

1

I bhz 1

123

b b

bbh

yh

ybyh

P

xy3

12

1

22

1

2

xyP

bh

hy

6

23

2

2

At y = 0, bh

Pxy

2

3max

3.26

The shear stress distribution is parabolic; maximum at the

centroidal axis and zero at the top and bottom. (Contrast

this with the normal stress distribution caused by bending,

where the maximum stresses occur furthest from the

centroidal axis).

3.27

3.3 RELATIVE MAGNITUDES OF BENDING

AND SHEAR STRESSES

For a rectangular cross-section (bxh) simply-supported beam

with a central point load,

The maximum bending moment and shear force (numerical

values) are:

M P

LPL

F P

xz

xy

(max)

(max)

12 2

14

12

Max. bending stress:

xx

xz

z

M

Iy

PL

bh

h(max)

(max)

max

141

122

3

xxPLbh(max)

32 2

3.28

Max. shear stress:

Using the previous example with y = 0 (for max. shear

stress) and P replaced by 12P;

xyPbh(max)

34

xx

xy

Lh

(max)

(max) 2

Hence, if the beam length is much greater than its depth

(e.g. L 10h), the maximum shear stress will be at least

an order of magnitude smaller than the max. bending

(normal) stress.

3.29

Example 1

A box beam is loaded as shown in the Fig., I about N.A. =

10.5 x 10-6m4. Draw shear force and B.M. diagrams and

calculate

a) Shearing and bending stresses at point E,

b) Max. shear and bending stresses.

From equilibrium

RA = 3 kN, RB = 11 kN

Note: Point E is 40 mm from the top fibre

Beam cross-section

3.30

The shear force and bending moment diagrams can be

constructed as follows:

1.5

-4

7

x

Fxy

(kN)

-3

-8

6

4.5

1.5

Mxz

(kNm)

x

3.31

To determine ty

2

201008020

2

6040)6020(2204080)604080120( ty

mmy

y

t

t

7.567200

10408

101761016810647200

3

333

At pt. E Fxy = - 3 kN

yt

yt

Beam cross-section

3.32

36

39

104.117

10)207.56(4080

)207.56(04.008.0

m

m

yA

xyxyF A y

Ib

N m

MPa

3 10 117 4 10

10 5 10 2 0 02

0 839

3 6

6

2.

. ( . )

.

Bending stress, at E, Mxz = 4.5 kN.m

xxMy

I

MPa

4 5 10 56 7 40 10

10 5 10

7 16

3 3

6

. .

.

.

For max. shear stress,

Max. shear force = 7 kN , 2 x 4

3.33

A y

m

m

80 40 56 7 20 2 20 16 756 7 40

2

80 40 36 7 2 20 16 7 8 4 10

123 10

9 3

6 3

( . ) .( . )

. . .

MPa

Ib

yAFxy

xy

05.202.02105.10

101231076

63

3.34

For max. bending stress,

Max. B.M. = - 8 kN.m at x = 4

max

max

.

.

.

M y

I

MPa

8 10 63 3 10

10 5 10

48 2

3 3

6

(compression)

Note: xy (max) occurs at N.A.

xx (max) occurs at bottom fibre (at support B).