04. mech3001y - unsymm.bending (3)

MECH 3001Y

Week4

Mechanics of Materials

& Machines III

Unsymmetrical

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Unsymmetrical

Bending

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� INTRODUCTION TO SOLID MECHANICS

IRVING H. SHAMES, JAMES M. PITARRESI

� MECHANICS OF MATERIALS (5th Edition in SI Units)

F. P. BEER, E. R. JOHNSTON, J. T. DeWOLF, D. F. MAZUREK

Textbooks & website

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Wk3 - Problem1

A beam of rectangular cross section supports aninclined load P having its line of action along a diagonalof the cross section (see figure). Show that the neutralaxis lies along the other diagonal.

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Wk3 - Problem1

1) The beam has a rectangularcross-section.

2) Axes y and z lie along the planesof symmetry of the section. yand z are principal axes.

3) Suppose it is a cantileveredbeam of span T (cf. figure)

T

P

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beam of span T (cf. figure)4) Resolve the load (or the bending

moment) along the principalaxes.

5) Apply the overall bendingequation.

Wk2 - Problem1 / SolutionLocation of neutral axis

.cos .cos

.sin .sin

. . For neutral axis, 0

z y

y z

y zb b

y z

L P M P T

L P M P T

M z M y

I I

θ θ

θ θ

σ σ

= ⇒ = − ×

= − ⇒ = − ×

= − =

Overall bending equation

γγγγ

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. .z (equation of NA)

. (t

y z

yz

y z

yz

y z

I I

MIy

I M

Mdy I

dz I M

=

=

3 3 2

2

angent of NA)

12 12

zz y

y

bh hb I hI I

I b= = =

NA

θθθθ

Wk2 - Problem1 / SolutionLocation of neutral axis

2

2

.cos 1 1

.sin tan

2 2

.

y

z

M P T b

h bM P T h

dy h b h

dz h bb

θ

θ θ

− ×= = = =

− ×

= =

γγγγ

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2 tan

2

The neutral axis lies along the other diagonal.

h h

b bγ = =

∴

Observe that,NA

θθθθ

Ch.1 - Unsymmetrical Bending

Assumptions

a) Beam cross-section has anarbitrary shape.

b) Assume loading is such that

y

θ

z

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b) Assume loading is such thatthere is no twisting of thecross-section, i.e loads actthrough the centroid. Line of action of load passes

through centroid (no torsion)

z


NB

Observe that we use a right-handed coordinate system.

y

θy

x

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θ

z

z

How to identify a right handed coordinate system

How are the axes z & y chosen ?


NB

Observe that we use a right-handed coordinate system.

y

θy

x

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θ

z

z

How to identify a right handed coordinate system

z & y are chosen such that they are principal axes for

the section


Principal Axes

a) For a beam section possessing at least one plane ofsymmetry, one principal axis will coincide with theplane of symmetry. There will be a 2nd principal axis,perpendicular to the 1st one.

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b) A beam section with 2 planes of symmetry will have 2principal axes coinciding with these planes.

However, any given section possesses principal axes, even if the section is not

symmetrical.


Principal Axes - Moments of Inertia

Second moments of area (area moment of inertia) of across-section about its principal axes are found to bemaximum and minimum values.

The product second moment of area about the principalaxes is found to be zero.

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axes is found to be zero.

All plane sections, whether they have an axis ofsymmetry or not, have two perpendicular axes aboutwhich the product second moment of area is zero.

Principal axes are thus defined as the axes about which

the product second moment of area is zero.


Principal Axes

2

2

.

.

x x

A

y y

A

I y dA

I x dA

−

−

=

=

∫

∫

Moment of inertia about x-axis

Moment of inertia about y -axis

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.

A

xy

A

I xy dA= ∫ndProduct 2 moment of area

. 0xy

A

I xy dA= =∫Condition for x & y

to be principal axes


Road map to solving unsymmetrical bending problems

Beam bending problem

Unsymm.Bending

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Symmetrical section

Load applied along axis of symmetry

Load applied through centroid and at an angle to axes of

symmetry

Unsymmetric section


Unsymmetrical bending of beams having symmetricsections

1. Identify principal axes

2. Resolve load along principal axis directions

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axis directions

3. Find bending moments / stresses for each principal direction

4. Superimpose solutions for each direction to obtain the overall solution of the problem


Bending of beams having unsymmetric sections

Case 1: (Principal axes are given and load applied at anangle to principal axes)


y

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2. Calculate moments of inertia of section with respect to principal axes

3. Find bending moments / stresses for each principal direction

4. Superimpose solutions for each direction to obtain the overall solution of the problem

θ

z

z & y principal axes



Case 2.1: (Principal axes not given)

1. Find moments of inertia (including product moment ofinertia) of section with respect to a set of known axes(if not given)**

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(if not given)**

2. Find the principal axes


4. Calculate moments of inertia of section with respectto principal axes**

5. Find bending moments/stresses for each principaldirection

6. Superimpose solutions for each direction to obtain theoverall solution of the problem

** document sent by mail



Case 2.2: (Principal axes not given)

1. Use the Generalized Flexure Formula

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Case 2.1: (Principal axes not given) – Example

A couple of magnitude M0=1.5kN.m acting in a verticalplane is applied to a beam having the Z-shaped cross-section shown. Determine (a) the stress at point A,

(b) the angle that the neutral axis forms with thehorizontal plane. The moments and product of inertia of

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horizontal plane. The moments and product of inertia ofthe section with respect to the y and z axes have beencomputed and are as follows:

Iy = 3.25x10-6m4

Iz = 4.18x10-6m4

Iyz = 2.87x10-6m4



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Principal axis ??

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1- Principal axis

Relate to

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Relate to stress /strain transformation

Used to find principal axes



Principal axis

y2.

tan2 yz

y z

I

I Iθ

−=

−

u

θθθθ1111=40.4=40.4=40.4=40.4°

θθθθ2222=130.4=130.4=130.4=130.4°

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z

( )

6

6

2 2.87 10

3.25 4.18 10

2 80.8 , 260.8

40.4 , 130.4

θθ

−

−

− × ×=

− ×

= ° °= ° °

v



Loading ??

The applied couple M0 is

resolved into components

along the principal axes.

yu

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along the principal axes.

Mu=M0.sinθ1=1500 x sin 40.4°

= 972 N.m

Mv=M0.cosθ1=1500 x cos 40.4°

= 1142 N.m

z

vθθθθ1111=40.4=40.4=40.4=40.4°

ΜΜΜΜ0000



(a) – Stress at A

y

z

u

A.

. . u vM v M u

σ = −

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v

. . u v

u v

M v M u

I Iσ = −

Iu , Iv , u , v ??



(a) – Stress at A

y

z

u

A.

( ) ( )( )

6 6

+ .cos2 sin22 2

3.25 4.18 10 3.25 4.18 10 + .cos 80.8 ...

y z y z

u yz

I I I II Iθ θ

− −

+ −= −

+ × − ×= ° −

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v

( ) ( )( )

( )

( ) ( )

6

6 4

6

3.25 4.18 10 3.25 4.18 10 + .cos 80.8 ...

2 2

2.87 10 .sin 80.8

0.81 10

.cos2 sin22 2

3.25 4.18 10 3.25 4.18

2

y z y z

v yz

m

I I I II Iθ θ

−

−

−

+ × − ×= ° −

× °

= ×

+ −= − +

+ × −= − ( )

( )

6

6

6 4

10.cos 80.8 ...

2

2.87 10 .sin 80.8

6.62 10 m

−

−

−

×° +

× °

= ×



(a) – Stress at A

y

u

A

( ) ( )A

y .cos .sin

50.cos 40.4 74.sin 40.4

A Au zθ θ= +

= ° + °

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z

v

uAvA

( ) ( )

( ) ( )A

50.cos 40.4 74.sin 40.4

86.0

y .sin .cos

50.sin 40.4 74.cos 40.4

23.9

A A

mm

v z

mm

θ θ

= ° + °

=

= − +

= − ° + °

=



(a) – Stress at A

. . u v

u v

M v M u

I Iσ = −

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3 3

6 6

972 23.9 10 1142 86.0 10

0.81 10 6.62 10

28.68 14.84

13.84

MPa MPa

MPa

− −

− −

× × × ×= −

× ×

= −

=



(b) – Neutral axis

. .0

.

u v

u v

u v

M v M u

I I

M Iu v

= −

=

yuNA

ββββ

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.

u v

v u

u vM I

=

z

v

Let ββββ be the angle between the NA and principal axis v

6

6

. 972 6.62 10tan 6.96

. 1142 0.81 10

81.8

40.4

4

u v

v u

M I

M I

Angle between neutral axis and the horizontal

β

β

β

−

−

× ×= = =

× ×

= °

∴ = − °

= 1.4°

04. mech3001y - unsymm.bending (3)

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