straight bending beam

8/10/2019 Straight Bending Beam

1/17

Bending of straight beams

In mechanics of materials we cover symmetricalcross sections and bending in one plane. Now

we will consider the more general case

Moment perpendicular to a plane at an angle phi

from x-z plane (plane of loads). Centroidal axes.

Cantilever beam with an arbitrary cross section subjected to pure bending


2/17

Shear loading

We assume for now pure bending with notwist. This implies shear forces passing

through shear center

Cantilever beam with an arbitrary cross section subjected to shear loading


3/17

Symmetrical bending

Moments of inertia

Moments of inertia also a tensor so has

principal axes

AdxyI

AdrJ

AdxI

AdyI

xy

y

x

==

=

=

2

2

2


4/17

Symmetrical and anti-symmetrical

cross sections

Are these also principal axes?

Equilateral triangle Open channel section

Z- sectionAngle section


5/17

Symmetrical bending

Euler-Bernoulli beam theory, Leonhard Euler(1707-1783) and Daniel Bernoulli (1700-1782)

What are the assumptions?

For symmetrical cross section

Neutral axis

X

X

Y

Yzz

I

yM

I

xM+=

0=zz


6/17

Rectangular cross section

Maximum bending stress

2max

||6

bh

MX

=

Cantilever beam with rectangular cross section


7/17

Unsymmetrical bending

Equations of equilibrium

Plane sections remain plane

Combining it all

=

=

=

dAxM

dAyM

dA

zzy

zzx

zz

0

cybxaE

ycxba

zz

zzzz

zz

++=

=

++=

yIII

IMIMxIII

IMIMxyyx

xyyyx

xyyx

xyxxyzz

++

+= 22


8/17

Moments

Moment is perpendicular to the plane of the

loads. If the plane of the loads makes an angle with the x-axis,

cotcot

cossin

xy

x

y

yx

MMM

M

MMandMM

==

==


9/17

Neutral axis

For bending stress to be zero

cot

cottan

tan

tan

xyy

xxy

xyyyx

xyxyx

xyyyx

xyxyx

zz

II

II

IMIM

IMIM

xxIMIMIMIM

=

+

+=

=

++=

Pure bending of a nonsymmetrically loaded

cantilever beam


10/17

Example 7.3 A cantilever beam of length 3m as shown in the figure

has a channel section. A concentrated load P=12.0 kN

lies in the plane with an angle =/3 with the x-axis.

The plane of the loads passes through the shear center

C. Locate points of maximum tensile and compressive

stresses and find the magnitude of stresses.


11/17

Location of max stresses

460

462

10x73.300.82

010x69.3610000

:

mmImmy

ImmImmA

propertiesSection

y

xyx

==

===

rad

I

I

II

II

AxisNeutralLocate

y

x

xyy

xxy

6407.07457.0tan

5774.0

3

sin

3cos

3cot

3

cotcot

cottan

:

==

=

=

=

=

=

mkNMM

mkNPM

Moments

x .18.31sin

.0.3600.3

:

==

==

Since moment is negative, the part

above N.A is in tension and the bottom

part is in compression. Therefore

maximum tensile stress occurs at

point A and maximum compressive

stress at point B.


12/17

Maximum stresses

( )

( )( )

( )( )

x

x

x

Stress at A(-70,-118)

M tan

tan

M 118 70 tan133.7MPa

tan

Stress at B(70,82)M 82 70 tan

105.4MPatan

zz

x xy

Ax xy

B

x xy

y x

I I

I I

I I

=

= =

= =

.


13/17

Deflections

Determine separatelyxand y

components of displacement.

Here we show ycomponent

Curvature

)()tan(

11

22

2

2

2

xyyx

xyyyx

xyx

x

yy

zz

y

IIIE

IMIM

IIE

M

dz

vd

dz

vd

Rwhere

RR

+=

=

=


14/17

Total displacement

cos

tan

22 vvu

vu

=+=

=

Components of deflection of a

nonsymmetrically loaded beam.


15/17

Example 7.6

A simply supported beam of length 3m has a channel section. A

concentrated load P= 35.0 kN applied at the center of the beam, lies in

a plane with an angle = 5/9 with the x-axis. Locate points ofmaximum tensile and compressive stresses and magnitude of

stresses. Find the maximum deflection. E=72 GPa.

46

0

462

10x73.300.82

010x69.3610000

:

mmImmy

ImmImmA

propertiesSection

y

xyx

==

===


16/17

Solution

MPaII

M

MPaII

M

II

xyM

xyx

xB

xyx

xA

xyx

xzz

2.87tan

)tan)70(82(

8.63tan

)tan)70(118(

tan

)tan(:Stress

=

=

=

=

=

Locate Neutral Axis:

tan cot 5 9

tan 0.2277 0.2239

= =

= =

x

y

I

I

rad

mmvvu

mmvu

mmEI

PLv

EI

x

95.6cos

54.1tan

78.648

sin48

PL

:beamsupportedSimplyDeflection

22

3

3

==+=

==

==

=

mkNMM

mkNPL

M

x .85.25sin

.25.264

:Moment

==

==


17/17

Reading assignment

Sections 7.3-5: Question: Consider a horizontal cantilever

beam under a tip vertical load. What condition is required so

that tip will move both down and sideways?

Source: www.library.veryhelpful.co.uk/ Page11.htm

straight bending beam

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