bending stresses - wordpress.com · 2017. 9. 7. · chapter 6: bending mechanics of material 7. th....

In-Class Activities:

•Follow up

•Concepts

• Theory, formula, steps

•Applications

• Problem Solving

• Quiz types

• Video Project discussion

Today’s Objective : 7TH September

To:

a) Understand the context, concept and derivation for bending stresses.

b) Be able to calculate max and min stresses due to symmetric and unsymmetric bending on a given body.

BENDING STRESSES

© 2008 Pearson Education South Asia Pte Ltd

Chapter 6: BendingMechanics of Material 7th Edition

Beams subjected toBending and Shear Earlier in the semester you were able to find

bending moment and shear force diagrams for these types of beams.



Example 1:

SFD

BMD



Example 2:

SFD

BMD



Example 3:

The shear diagram represents a plot of Eqs. 1 and 3

The moment diagram represents a plot of Eqs. 2 and 4



Bending Deformation of a Straight Member Cross section of a straight beam remains plane

when the beam deforms due to bending. There will be tensile stress on one side and

compressive stress on the other side.



Bending Deformation of a Straight Member Longitudinal strain varies linearly from zero at the

neutral axis. Hooke’s law applies when material is homogeneous.

Neutral axis passes through the centroid of the cross-sectional area for linear-elastic material.

εσ E=



The Flexure Formula Resultant moment on the cross section is equal to

the moment produced by the linear normal stress distribution about the neutral axis.

By the right-hand rule, negative sign is compressive since it acts in the negative x direction.

IMy

−=σσ = normal stress in the memberM = resultant internal momentI = moment of inertiay = perpendicular distance from the neutral axis

A condition for flexure formula is a symmetric cross-sectional area about an axis perpendicular to neutral axis (symmetric bending)

The flexure formula can also be applied either to a beam having x-sectional area of any shape

OR to a beam having a resultant moment that acts in any direction (Unsymmetric Bending – see later)

9



Example 6.15The simply supported beam has the cross-sectional area as shown. Determine the absolute maximum bending stress in the beam and draw the stress distribution over the cross section at this location.

Solution:The maximum internal moment in the beam is kNm 5.22=M



Solution:By symmetry, the centroid C and thus the neutral axis pass through the mid-height of the beam, and the moment of inertia is

( )( )( ) ( )( )( ) ( )( )

( ) 46

323

2

m 103.301

3.002.012116.0002.025.002.025.0

1212

−=

+

+=

+=∑ AdII

Applying the flexure formula where c = 170 mm,

( )( ) (Ans) MPa 7.12103.301

17.05.22 ; 6maxmax === −σσI

Mc



Example 6.17The beam has a cross-sectional area in the shape of a channel. Determine the maximum bending stress that occurs in the beam at section a–a.

Solution:The resultant internal moment must be computed about the beam’s neutral axis at section a–a. Since the neutral axis passes through the centroid,

( )( )( ) ( )( )( )( )( ) ( )( )

mm 09.59m 05909.0 25.002.0015.02.02

25.002.001.0015.02.01.02

==

++

==∑∑

AAy

y



Solution:Applying the moment equation of equilibrium about the neutral axis, we have

( ) ( ) kNm 859.4005909.00.124.2 ;0 =⇒=−+=+∑ MMM NA

The moment of inertia about the neutral axis is

( )( ) ( )( )( )

( )( ) ( )( )( )

( ) 46

23

23

m 1026.42

05909.01.02.0015.02.0015.01212

01.005909.002.025.002.025.0121

−=

−++

−+=I

The maximum bending stress occurs at points farthest away from the neutral axis.

( )( ) (Ans) MPa 2.161026.42

05909.02.0859.46max =

−== −I

Mcσ



Unsymmetric BendingMoment Arbitrarily Applied We can express the resultant normal stress at any

point on the cross section in general terms as

= +

y

y

z

z

IzM

IyM+−=σ

σ = normal stress at the pointy, z = coordinates of the point measured from x, y, z axes

My, Mz = resultant internal moment components directed along y and z axes

Iy, Iz = principal moments of inertia computed about the yand z axes



Unsymmetric Bending

Orientation of the Neutral Axis The angle α of the neutral axis can be determined by

applying σ = 0,

θα tantany

z

II

=

= +



Example 6.19 A T-beam is subjected to the bending moment of 15 kNm. Determine the maximum normal stress in the beam and the orientation of the neutral axis.

Solution:Both moment components are positive,

( )( ) kNm 50.730sin15

kNm 99.1230cos15=°=

=°=

z

y

MM

For section properties, we have

( )( )( ) ( )( )( )( )( ) ( )( ) m 0890.0

2.003.004.01.02.003.0115.004.01.005.0=

++

==∑∑

AAz

z



Using the parallel-axis theorem, the principal moments of inertia are thus

Solution:

( )( ) ( )( ) ( )

( )( ) ( )( )( )

( )( ) ( )( )( ) ( ) 4623

23

4633

m 1092.13089.0115.003.02.003.02.0121

05.0089.004.01.01.004.0121

m 1053.202.003.012104.01.0

121

−

−

=

−++

−+=

=+=

y

z

I

I

2AdII +=

The largest tensile stress at B and greatest compressive stress at C.

( )( )

( )( )

( )( )

( )( ) (Ans) MPa 3.901092.13

089.099.121053.2002.05.7

MPa 8.741092.13

041.099.121053.20

1.05.7

66

66

−=−

+−=

=+−

−=

+−=

−−

−−

C

B

y

y

z

z

IzM

IyM

σ

σ

σ



y must represent the axis for minimum principal moment of inertia, and z must represent the axis for maximum principal moment of inertia.

Solution:

( )( )

°=

°

= −

−

6.68

60tan1092.131053.20tan 6

6

α

α



Composite Beams In reality, Beams constructed of two or more

different materials are referred to as composite beams.

The transformation factor is a ratio of the moduli of the different materials that make up the beam.



Example 6.21 (Not examined)A composite beam is made of wood and reinforced with a steel strap located on its bottom side. It has the cross-sectional area as shown. If the beam is subjected to a bending moment of 2 kNm, determine the normal stress at points B and C. Take Ew= 12 GPa and Est = 200 GPa.

Solution:We will transform the section into one made entirely of steel.

( ) mm 915020012

=== wst nbb

The location of the centroid (neutral axis),

( )( )( ) ( )( )( )( )( ) ( )( ) m 03638.0

15.0009.015.002.015.0009.0095.0150.002.001.0

=++

==∑∑

AAy

y

The transformed section is as shown.



Solution:The moment of inertia about the neutral axis is therefore

( )( ) ( )( )( )

( )( ) ( )( )( )

( ) 46

23

23

m 10358.9

03638.00095.015.0009.015.0009.0121

01.003638.002.015.002.015.0121

−=

−++

−+=NAI

Applying the flexure formula, the normal stress at B’ and C is

( )( )

( )( ) (Ans) MPa 87.2710358.9

03638.002

MPa 6.2810358.9

03638.017.02

6

6'

==

=−

=

−

−

C

B

σ

σ

The normal stress in the wood at B is ( ) (Ans) MPa 71.156.2820012

' === BB nσσ



Quiz 1: Where will the max bending stresses occur for the following beam

a) Right hand side top faceb) Mid point bottom facec) Left end bottom faced) Left end top face

SFD & BMD



Quiz 2: For a linear elastic material, neutral axis will pass through the

a) Centre of gravity of the cross sectionb) Centre of gravity along longitudinal axisc) Through the shear centred) Centre of mass



Quiz 3: The angle of inclination in unsymmetric bending is calculated bya) x

b)

c)

d)

θα tantany

x

II

=

θα tantanz

y

II

=

θα tantany

z

II

=

θα tantanIxI y=

bending stresses - wordpress.com · 2017. 9. 7. · chapter 6: bending mechanics of material 7. th....

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