3.1

Chapter 3

STRESSES IN LOADED BEAMS

3.1

PURE BENDING

Bending DeformationsBeam with a plane of symmetry in pure

bending: member remains symmetric

bends uniformly to form a circular arc

cross-sectional plane passes through arc centerand remains planar

length of top decreases and length of bottom increases

a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change

stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it

3.2

Strain Due to Bending

Consider a beam segment of length L.

After deformation, the length of the neutral surface remains L. At other sections,

(i) Strain Due to Bending

Consider a beam segment of length L.

Before and after deformation, the length of the neutral surface remains L. At other sections,

( )( )

mx

mm

x

cy

c

c

yyL

yyL

yL

=

==

===

===

=

or

linearly with y)ries(strain va

L

Strain in the x-direction

If we define the max numerical strain as: m

(i)

Strain in the x-direction

3.3

Stress Due to Bending

For a linearly elastic material,

linearly) varies(stressm

mxx

cy

EcyE

=

==

For static equilibrium,

=

===

dAyc

dAcydAF

m

mxx

0

0

First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass through the section centroid.

For static equilibrium,

IMy

cy

SM

IMc

cIdAy

cM

dAcyydAyM

x

mx

m

mm

mx

=

=

==

==

==

ngSubstituti

2

(ii)

3.4

Common I values: rectangular X-section, width b, height h:

Ibh

z =3

12

SIC

= Ch

=2

= bh

2

6 Circular X-section, radius r, diameter d:

Ir d

= = 4 4

4 64

SIC

= C = r

32

,4

33 dr =

3.5

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 1123

008 01212 1152 10

36 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

3.6

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

MxzkN m

= =

10 2 5 22210 .

Using xxMyI=

At top fibre xx MPa=

=

10 103 061152 10 6

521(0. ).

.

At bottom fibre xx = 52.1 MPa Plotting the stress distribution

3.7

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

27 5

2

3.8

We have

( )

AAM y

I

MPa

=

=

=

7 5 10 0 021152 10

13

3

6. .

.

AA

E=

=

=

13 10175 1074 3 10

6

9

6.

(c) Since

m

y

y

xx

xx

269103.7402.0

6

=

=

=

=

3.9

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 centroidfrom bottom = .y

( )y A A A y A y

y

mm

1 2 1 1 2 2

20 70 60 3020 60

50

+ = +

=++

=

(60) (20)(60) (20)

Example 2

3.10

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

Ibh

Ad

mm

= +

= +

+ +

=

32

3 2

3 2

4 4

121

1260 20 20 60 20

112

20 60 20 60 20

136 10

R1 = 5 kN. R2 = 10 kN

A1

A2

20

20

From equilibrium

3.11

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 W0

Represented by W0

W

3.12

kNm

(1)

Load Intensity

Shear force:

( ) 11001 433103445 ++= xxxxxxw

kNxxxxx

wdxF000 433103445 ++=

=

3.13

Plotting

X = 4 F = -5 + 12 - 10 = -3 kN, also F = -5 + 12 - 10 + 3 = 0 kN

Shear force:

X = 0, F = -5 kN, also F = 0 kN

X = 3, F = -5 + 4x3 = 7 kN, also F = -5 +12-3 = -3 kN

From Eq. (1)

1 m 3 m

A D

Fxy kN

7

B

-3

1.25 m

-5

x

3.14

Bending Moment:

++=

=000 433103445 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

kNxxxxxF 000 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 zeroshear force occurs at x < 3

Hence the Bending Moments are :

( )M

kN m

=

=

5 3 432

3

2

.

M xx

kN m

=

=

5 42

3125

2

. .

At x = 1.25 m

At x = 3 m

Note: x < 3 m

= MyI

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

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

3.15

max. ( ).

/

.

top fibre N mm

MPa

=

=

3125 10 30136 10

68 9

62

( )max. ( )

.

.

bottom fibre

MPa

=

=

3125 10 50136 10

114 9

6

4

( )max. ( ) /

.

top fibre N mm

MPa

=

=

3 10 30136 10

66 2

6

42

( )c bottom fibre

MPa

( )

.

=

=

3 10 50136 10

110 3

6

4

Hence At x = 1.25 m

At x = 3 m

Hence

max. tensile stress occurs at x = 1.25 mmax. 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

3.16

3.2

SHEAR STRESSES IN BEAMS

Mxz

X

dx

Mxz

Mxz + (dMxz/dx)dx

Stresses in a beam

Mxz + Mxz

3.17

Stresses in a Beam

P

Fxy

Mxzxx comp top, tensile bottom

xygives

P

National University of SingaporeNational University of SingaporeNational University of Singapore

3.18

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), 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.

3.19

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

MI

y=

3.20

( )

F

M M yI

dAMI

y dAyxxz xz

zA

xz

zA=

+ +

=Az

xzyx dAyI

MF

Dividing by x and letting x 0

=A

xz

z

yx dAydx

dMIdx

dF 1

Now y dA A y

A =

(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

dxFxz xy

dFdx

FI

A yyx xyz

=

3.21

yAI

Fbdx

dFb z

xyyxxy .

1.1 ==

- Shear force per unit length

- shear flow

The shear stress

Since

dxdFyx

dxdF

b

Areayx

yx

.1

forceshear

=

=

xyyx =

3.22

ExampleDetermine the shear stress distribution in a beam of rectangular cross-section (b x h) subjected to the loading shown below.

1

Consider the cross-section at y-y.

3.23

xy yxxy

z

F A yI b

= =

F Pxy = at section y-y

Ah

y b=

2

+= yhyy

221

I bhz =1

123

b b=

bbh

yhybyhPxy

3

121

221

2

+

=

xyP

bhh

y=

6

232

2

3.24

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).

3.25

3.3 RELATIVE MAGNITUDES OF BENDING AND SHEAR STRESSES

For a rectangular cross-section (bxh) simply-supported beam with a central point load,

M P L PL

F P

xz

xy

(max)

(max)

= =

=

12 2

14

12

Max. bending stress:

xxxz

z

MI y

PL

bhh

(max)(max)

max= =

141

1223

xx

PLbh(max) =

32 2

3.26

Max. shear stress: Using the previous example with y = 0 (for max. shear stress) and P replaced by 12 P; xy

Pbh(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.

3.27

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

3.28

Using the method shown in Chapter 2, the shear force and bending moment diagrams can be constructed.

1.5

-4

7

x

Fxy (kN)

-3

-8

6

4.5

1.5

Mxz (kNm)

x

3.29

To determine ty

++

++=

2201008020

26040)6020(2204080)604080120( ty

mmy

y

t

t

7.567200

10408

1017610168106472003

333

=

=

++=

At pt. E Fxy = - 3 kN

yt

yt

Beam cross-section

3.30

36

39

104.117

10)207.56(4080

)207.56(04.008.0

m

m

yA

=

=

=

xyxyF A yIb

N m

MPa

=

=

=

3 10 117 4 1010 5 10 2 0 020 839

3 6

62.

. ( . ).

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

( )

xxMyI

MPa

=

=

=

4 5 10 56 7 40 1010 5 10

7 16

3 3

6. .

..

For max. shear stress, Max. shear force = 7 kN , 2 < x < 4

3.31

( )

A y

mm

= +

= +

=

80 40 56 7 20 2 20 16 756 7 40

280 40 36 7 2 20 16 7 8 4 10

123 10

9 3

6 3

( . ) .( . )

. . .

( )MPa

IbyAFxy

xy

05.202.02105.10

101231076

63

=

=

=

3.32

For max. bending stress, Max. B.M. = - 8 kN.m at x = 4

( )

maxmax

..

.

=

=

=

M yI

MPa

8 10 63 3 1010 5 10

48 2

3 3

6

(compression) Note: xy (max) occurs at N.A. xx (max) occurs at bottom fibre (at support B).