Bending of straight beams• In mechanics of materials we cover symmetrical

cross 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

Shear loading• We assume for now pure bending with no

twist. This implies shear forces passing through shear center

Cantilever beam with an arbitrary cross section subjected to shear loading

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

Symmetrical and anti-symmetrical cross sections

• Are these also principal axes?

Equilateral triangle Open channel section

Z- sectionAngle section

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

yMI

xM+−=σ

0=zzσ

Rectangular cross section• Maximum bending stress

2max||6

bhM X=σ

Cantilever beam with rectangular cross section

Unsymmetrical bending• Equations of equilibrium

• Plane sections remain plane

• Combining it all

∫∫∫

−=

=

=

dAxM

dAyM

dA

zzy

zzx

zz

σ

σ

σ0

cybxaE

ycxba

zz

zzzz

zz

++=∈=

′+′+′=∈

σσ

yIII

IMIMx

IIIIMIM

xyyx

xyyyx

xyyx

xyxxyzz ⎟

⎟⎠

⎞⎜⎜⎝

⎛

−

++⎟

⎟⎠

⎞⎜⎜⎝

⎛

−

+−= 22σ

Moments

• Moment is perpendicular to the plane of the loads. If the plane of the loads makes an angle φwith the x-axis,

φφ

φφ

cotcot

cossin

xyx

y

yx

MMMM

MMandMM

−=⇒−=

−==

Neutral axis

• For bending stress to be zero

φφ

α

α

ασ

cotcot

tan

tan

tan

xyy

xxy

xyyyx

xyxyx

xyyyx

xyxyxzz

IIII

IMIMIMIM

xxIMIMIMIM

−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

+

+=

=⎟⎟⎠

⎞⎜⎜⎝

⎛

+

+=

Pure bending of a nonsymmetrically loaded cantilever beam

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

has a channel section. A concentrated load P=12.0 kNlies in the plane with an angle φ = π/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.

Location of max stresses

460

462

10x73.300.82

010x69.3610000:

mmImmy

ImmImmApropertiesSection

y

xyx

==

===

rad

II

IIII

AxisNeutralLocate

y

x

xyy

xxy

6407.07457.0tan

5774.0

3sin

3cos

3cot

3

cotcotcot

tan

:

−=⇒−=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

=⎟⎟⎠

⎞⎜⎜⎝

⎛⇒=

−=−−

=

αα

π

πππφ

φφφ

α

mkNMMmkNPM

Moments

x .18.31sin.0.3600.3

:

−==−=−=

φ

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.

Maximum stresses

( )

( )( )

( )( )

x

x

x

Stress at A(-70,-118)M tan

tan

M 118 70 tan133.7 MPa

tan

Stress at B(70,82)

M 82 70 tan105.4 MPa

tan

zzx xy

Ax xy

Bx xy

y xI I

I I

I I

ασ

α

ασ

α

ασ

α

−=

−

− − −= =

−

−= = −

−

.

Deflections

• Determine separately x and ycomponents of displacement. Here we show y component

• Curvature

)()tan(

11

22

2

2

2

xyyx

xyyyx

xyx

x

yy

zz

y

IIIEIMIM

IIEM

dzvd

dzvd

Rwhere

RR

−

+−=

−−=

≈∈

=−

α

Total displacement

αδ

α

cos

tan

22 vvu

vu

=+=

−=

Components of deflection of a nonsymmetrically loaded beam.

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 φ = 5π/9 with the x-axis. Locate points of maximum tensile and compressive stresses and magnitude of stresses. Find the maximum deflection. E=72 GPa.

460

462

10x73.300.82

010x69.3610000:

mmImmy

ImmImmApropertiesSection

y

xyx

==

===

Solution

MPaII

M

MPaII

M

IIxyM

xyx

xB

xyx

xA

xyx

xzz

2.87tan

)tan)70(82(

8.63tan

)tan)70(118(

tan)tan(

:Stress

−=−−

=

=−

−−−=

−−

=

αασ

αασ

αασLocate Neutral Axis:

tan cot 5 9

tan 0.2277 0.2239

= − ⇒ =

= ⇒ =

x

y

II

rad

α φ φ π

α α

mmvvu

mmvu

mmEI

PLv

EI

x

95.6cos

54.1tan

78.648

sin48PL

:beam supportedSimplyDeflection

22

3

3

==+=

−=−=

==

=∆

αδ

α

φ

mkNMM

mkNPLM

x .85.25sin

.25.264

:Moment

==

==

φ

Reading assignmentSections 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?

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