Chapter 11 Bending

Chapter Objectives

To generalize the procedure by formulating equations that

can be plotted so that they describe the internal shear and

moment throughout a member.

To use the relations between distributed load, shear, and

moment to draw shear and moment diagrams.

To determine the stress in elastic symmetric members

subject to bending.

To develop methods to determine the stress in unsymmetric

beams subject to bending.

Copyright 2011 Pearson Education South Asia Pte Ltd

1. Reading Quiz

2. Applications

3. Shear and moment diagrams

4. Graphical method for construction of shear and

bending moment diagrams

5. Flexural formula

6. Unsymmetric bending

7. Stress Concentrations

8. Concept Quiz

In -class Activities


READING QUIZ

1) Provided that the bending formation of a straight

member is small and within elastic range, which of

the following statements is incorrect?

a) Plane section remains plane

b) Cross section remains perpendicular

c) The length of the longitudinal axis remains unchanged

d) In-plane distortion of cross section to the longitudinal axis is

not negligible


READING QUIZ (cont)

2) Which of the following statements is

incorrect for bending of a straight member?

a) Bending stress is proportional to

b) Bending stress is inversely proportional

c) Bending stress is inversely proportional to the moment of

inertia of the section

d) Bending stress is not a function to the second moment of

area of the section of the location


READING QUIZ (cont)

3) Which of the following statements is not true?

The differences between the flexure formula for a curved beam from that for a straight beam are:

a) The normal strain in a curved beam does not vary linearly with

depth

b) The neutral axis in a curved beam does not pass through the

centroid of the cross section

c) Plane section remains plane

d) In-plane distortion within cross section cannot be ignored


APPLICATIONS


APPLICATIONS (cont)


SHEAR AND MOMENT DIAGRAMS


Shear is obtained by summing forces

up to the end of the segment.

Moment is obtained by

summing moments about

the end of the segment.

Note the sign conventions are

opposite when the summing

processes are carried out with

opposite direction. (from left to right vs from right to left)

Please refer to the website for the animation: Shear and Moment Diagrams

EXAMPLE 1


Draw the shear and moment diagrams for the beam shown in

Fig. 6 4a.

EXAMPLE 1 (cont.)


Solution

The support reactions are shown in Fig. 6 4c.

Applying the two equations of equilibrium yields

2 2

022

;0

1 2

02

;0

2xLxw

M

Mx

wxxwL

M

xL

wV

VwxwL

Fy

EXAMPLE 1 (cont.)


Solution

The point of zero shear can be found from Eq. 1:

2

02

Lx

xL

wV

8222

22

max

wLLLL

wM

From the moment diagram, this value of x represents the point on the

beam where the maximum moment occurs.

GRAPHICAL METHOD FOR CONSTRUCTING

SHEAR AND MOMENT DIAGRAMS


Regions of distributed load:

dxxwV

dxxVM

Change in moment = area under shear

diagram

Change in shear = area under distributed

loading


Regions of concentrated force and moment:

FV

VVFV

0

0

0

0

MM

MxVMMM

GRAPHICAL METHOD FOR CONSTRUCTING

SHEAR AND MOMENT DIAGRAMS (cont)

EXAMPLE 2


Draw the shear and moment diagrams for the beam shown in

Fig. 6 12a.

EXAMPLE 2 (cont.)


The reactions are shown on the

free-body diagram in Fig. 6 12b.

The shear at each end is plotted first,

Fig. 6 12c. Since there is no

distributed load on the beam,

the shear diagram has zero slope

and is therefore a horizontal line.

Solution

EXAMPLE 2 (cont.)


The moment is zero at each end,

Fig. 6 12d. The moment diagram

has a constant negative slope of

-M0/2L since this is the shear in the

beam at each point. Note that the

couple moment causes a jump in the

center, but it does not affect the

shear diagram at this point.

Solution

EXAMPLE 3


Draw the shear and moment diagrams for each of the beams

shown in Figs. 6 13a and 6 14a.

EXAMPLE 3 (cont)


Solution

BENDING DEFORMATION OF A STRAIGHT

MEMBER


Assumptions: 1. Plane section remains plane

2. Length of longitudinal axis remains unchanged

3. Plane section remains perpendicular to the longitudinal axis

4. In-plane distortion of section is negligible

FLEXURAL FORMULA


Assumptions: Material behaves in a linear-

i.e. =E.

I

My

I

My

I

Mc

dAyc

M

dAc

yydAyydFM

MM

A

AA

ZZR

max

2max

max

;

EXAMPLE 4


The simply supported beam in Fig. 6 26a has the cross-

sectional area shown in Fig. 6 26b. Determine the absolute

maximum bending stress in the beam and draw the stress

distribution over the cross section at this location.

EXAMPLE 4 (cont)


The maximum internal moment in the beam, 22.5 kN m,

occurs at the center.

By reasons of symmetry, the neutral axis passes through the

centroid C at the mid-height of the beam, Fig. 6 26b.

Solution

(Ans) MPa 7.12103.301

17.0105.22 ;

m 103.301

3.002.016.002.025.002.025.02

6

3

46

3

12123

121

2

bB

BI

My

AdII

EXAMPLE 4 (cont)

A three-dimensional view of the stress distribution is shown in Fig. 6 26d.

At point B,

Solution

MPa 2.11103.301

15.0105.22 ;

6

3

BB

BI

My


UNSYMMETRIC BENDING


Moment applied along principal axis

dAyMM

dAzMM

dAFF

AZZR

AyyR

AxR

0 ;

0 ;

0 ;

(The integral is called the product of inertia)

UNSYMMETRIC BENDING (cont)


Moment arbitrarily applied

Alternatively, identify the orientation of the principal axes (of which one is the neutral axis)

Orientation of neutral axis:

y

y

z

z

I

zM

I

yM= +

= +

tantany

z

I

I

EXAMPLE 5


The rectangular cross section shown in Fig. 6 33a is subjected to a

bending moment of 12 kN m. Determine the normal stress developed at

each corner of the section, and specify the orientation of the neutral axis.

EXAMPLE 5 (cont)


The moment is resolved into its y and z components, where

The moments of inertia about the y and z axes are

Solution

mkN 20.7125

3

mkN 60.9125

4

z

y

M

M

433

433

m 10067.14.02.012

1

m 102667.02.04.012

1

z

y

I

I

EXAMPLE 5 (cont)


For bending stress,

The resultant normal-stress distribution has been sketched

using these values, Fig. 6 33b.

Solution

(Ans) MPa 95.4102667.0

1.0106.9

10067.1

2.0102.7

(Ans) MPa 25.2102667.0

1.0106.9

10067.1

2.0102.7

(Ans) MPa 95.4102667.0

1.0106.9

10067.1

2.0102.7

(Ans) MPa 25.2102667.0

1.0106.9

10067.1

2.0102.7

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

3

E

D

C

B

y

z

z

z

I

zM

I

yM

EXAMPLE 5 (cont)


Solution

m 0625.02.0

95.425.2z

zz

The location z of the neutral axis (NA), Fig. 6 33b,

can be established by proportion.

We can also establish the orientation

of the NA using Eq. 6 19, which is used

to specify the angle that the axis makes

with the z or maximum principal axis.

(Ans) 4.791.53tan102667.0

10067.1tan

tantan

3

3

y

z

I

I

STRESS CONCENTRATIONS


Stress concentrations occur at locations where cross

section suddenly changes; e.g.

Chapter 11 Bending

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Transcript of Chapter 11 Bending