Unsymmetrical bending (2nd year)

Unsymmetrical Bending

Dr Alessandro Palmeri <[email protected]>

Teaching schedule Week Lecture 1 Staff Lecture 2 Staff Tutorial Staff 1 Beam Shear Stresses 1 A P Beam Shear Stresses 2 A P --- --- 2 Shear centres A P Basic Concepts J E-R Shear Centre A P 3 Principle of Virtual

forces J E-R Indeterminate Structures J E-R Virtual Forces J E-R

4 The Compatibility Method

J E-R Examples J E-R Virtual Forces J E-R

5 Examples J E-R Moment Distribution -Basics

J E-R Comp. Method J E-R

6 The Hardy Cross Method

J E-R Fixed End Moments J E-R Comp. Method J E-R

7 Examples J E-R Non Sway Frames J E-R Mom. Dist J E-R 8 Column Stability 1 A P Sway Frames J E-R Mom. Dist J E-R 9 Column Stability 2 A P Unsymmetric Bending 1 A P Colum Stability A P 10 Unsymmetric Bending 2 A P Complex Stress/Strain A P Unsymmetric

Bending A P

11 Complex Stress/Strain A P Complex Stress/Strain A P Complex Stress/Strain

A P

Christmas Holiday

12 Revision 13 14 Exams 15 2

Mo@va@ons (1/2)

•  Many cross sec@ons used for structural elements (such us Z sec@ons or angle sec@ons) do not have any axis of symmetry

•  How does the theory developed for symmetrical bending can be extended to such sec@ons?

3

Mo@va@ons (2/2) •  The figure shows the

finite element model of a can@lever beam with Z cross sec@on subjected to its own weight, in which the gravita@onal (ver@cal) load induces lateral sway (horizontal), –  exaggerated for clarity

•  How can we predict this?

4

XY XZ

YZ

XY

X

Z

Y

Z

Learning Outcomes

•  When we have completed this unit (2 lectures + 1 tutorial), you should be able to:

– Determine the principal second moments of area AND the principal direc@ons of area for unsymmetrical beam’s cross sec@ons

– Evaluate the normal stress σx in beams subjected to unsymmetric bending

5

Further reading

•  R C Hibbeler, “Mechanics of Materials”, 8th Ed, Pren@ce Hall – Chapter 6 on “Bending”

•  T H G Megson, “Structural and Stress

Analysis”, 2nd Ed, Elsevier – Chapter 9 on “Bending of Beams” (eBook)

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Symmetrical Bending (1/4) •  Our analysis of beams in bending has been restricted so far

(part A) to the case of cross sec@ons having at least one axis of symmetry, assuming that the bending moment is ac@ng either about this axis of symmetry (a), or about the orthogonal axis (b)

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(a) xz

yaxis of symmetry

beam

’s axis

My

G

(right hand)

σ x > 0

σ x < 0

tensilestress

compressivestress

(b)

xz

yaxis of symmetry

beam

’s axis Mz

G

σ x > 0σ x < 0

tensilestress

compressivestress

Symmetrical Bending (2/4)

8

xz

yaxis of symmetry

beam

’s axis Mz

My

G

Right-‐Hand Rule

If the thumb point to the

posi0ve direc0on of the axis,

then the curling of the other

fingers give the posi0ve

direc0on of the bending

Noteworthy: Some0mes a double-‐headed arrow is used to represent a

moment (as opposite to a single-‐headed arrow used for a force)


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xz

yaxis of symmetry

beam

’s axis

My

G

(right hand)

σ x > 0

σ x < 0

tensilestress

compressivestress

xz

yaxis of symmetry

beam

’s axis Mz

G

σ x > 0σ x < 0

tensilestress

compressivestress

⬇ A posi@ve bending moment My>0, induces tensile stress σx>0 in the boiom fibres of the cross sec@on

A posi@ve bending moment Mz>0, induces tensile stress σx>0 in the right fibres of the cross sec@on (looking at it from the posi@ve direc@on of the x axis)

⬆

σ x =

My zIyy

Eq. (1)


•  The simplest case when the bending moment My acts about the axis y, orthogonal to the axis of symmetry z

•  Therefore, the beam bends in the ver@cal plan Gxz

•  The direct stress σx is given by:

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x z

y

axis of symmetry

beam

’s axis

My

σ x > 0

G

Unsymmetrical Bending (1/3)

11

•  The case of unsymmetric bending deals with: –  EITHER a bending moment ac@ng about an axis which is neither an axis of symmetry, nor orthogonal to it (le9)

–  OR a beam’s cross sec@on which does not have any axis of symmetry (right)

xz

y

beam

’s axis

My

G

xz

yaxis of symmetry

beam

’s axis

My

G


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•  The first case is trivial, and can be solved by using: –  Decomposi@on of the bending moment:

–  Superposi@on of effects:

Mp = My cos(α )

Mq = −My sin(α )

σ x (A) =MpIpp

⋅d2−MqIqq

⋅e

=Myd /2Ipp

cos(α )+ eIppsin(α )

⎛

⎝⎜⎞

⎠⎟


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•  Par@cular cases…

Bending about the strong axis

σ x (A) =MyIpp

⋅d2

Bending about the weak axis σ x (A) =

My

Iqq

⋅e

Product Moment of Area (1/3)

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•  Let’s introduce a new quan@ty, Iyz, called “Product Moment of Area”

–  Defined as:

•  If and only if Iyz =0, a bending moment ac@ng on one of these two axes will cause the beam to bend about the same axis only, not about the orthogonal axis (symmetric bending) –  I.e. a ver@cal transverse load will not induce any lateral sway and a lateral transverse will not cause any ver@cal movement

= ∫ dyzA

I y z A


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•  The product moment of area is defined mathema@cally as the integral of the product of the coordinates y and z over the cross sec@onal area

•  Similarly the second moments of area Iyy and Izz are the integrals of the second power of the other coordinate, z2 and y2

•  G is the centroid of the cross sec@on

Iyy = z2 dAA∫

Izz = y2 dA

A∫

Iyz = y zdA

A∫


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•  The “Parallel Axis Theorem” (also known as Huygens-‐Steiner Theorem) can be used to determine the product moment of area Iyz, as well as the second moments of area Iyy and Izz, provided that: –  The cross sec@on can be split into simple blocks, e.g. rectangular blocks –  The corresponding quan@@es for the central axes η (eta) and ζ (zeta), parallel

to y and z, are known

Iyy = Iηη

(i ) + zi2 A(i )

i∑

Izz = Iζζ(i ) + yi

2 A(i )i∑

Iyz = Iηζ(i ) + yi zi A

(i )

i∑z

yG

ηi

ζ i

Γ i

A(i)

yi (< 0)

zi (> 0)

Moments of Area: Worked Example (1/5)

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1.  Split the cross sec@on in rectangular blocks

2.  Calculate the area of each block

3.  If the posi@on of the centroid G is unknown –  Calculate the first moment of each block

about two arbitrary references axes

A(1) =30×30 = 900A(2) =30×50 =1,500

Qm(1) = A(1) ×15 =13,500 Qn

(1) = A(1) ×15 =13,500Qm(2) = A(2) ×45 =67,500 Qn

(2) = A(2) ×25 =37,500

?

?

m mn

n


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–  Calculate the posi@on of the centroid

4.  Calculate the two second moments of area (and the product moment of area, if needed) for each block

dm =Qm(i )

i∑A(i )

i∑ = 81,0002,400

=33.75

dn =Qn(i )

i∑A(i )

i∑ = 51,0002,400

=21.25

Iηη(1) = 30×303

12= 67,500 Iζζ

(1) = 30×303

12= 67,500 Iηζ

(1) = 0

Iηη(2) = 50×303

12=112,500 Iζζ

(2) = 30×503

12=312,500 Iηζ

(2) = 0

m mn

n

?

?


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5.  Calculate the coordinates of the centroid Γi of each block…

y1 =21.25−302

=6.25 >0

z1 = − 33.75− 302

⎛⎝⎜

⎞⎠⎟

= −18.75 <0

y2 = − 502

−21.25⎛⎝⎜

⎞⎠⎟

= −3.75 <0

z2 =30+302

−33.75

=11.25 >0


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6.  Apply the Parallel Axis Theorem for the two second moments of area…

Iyy = Iηη(i ) + A(i )zi

2( )i∑= 67,500+900× −18.75( )2

+112,500+1,500× 11.25( )2

= 686,250

Izz = Iζζ(i ) + A(i )yi

2( )i∑= 67,500+900× 6.25( )2

+312,500+1,500× −3.75( )2

= 436,250


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7.  … And the product moment of area

Iyz = Iηζ(i ) + A(i )yizi( )i∑

= 0+900×6.25× −18.75( )+0+1,500× −3.75( )×11.25= −168,750

Rota@ng the Central Axes

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QuesBon: What happens to second moment of area (Imm) and product moment of area (Imn) if we rotate the central axes of reference for a given cross sec@on?

m

n

m

nm

n

m

n

m

n

m

n

Imm

Imn

Y

Z

Product moment of area (+ve, -‐ve or null)

Second moment of area (always +ve)

Iyy

Iyz

-‐Iyz

Izz

Mohr’s Circle

Gy ≡

≡

z

Answer: The points of coordinates {Imm,Imn} will describe a circle

Mohr’s Circle (1/6)

23

•  Named aser the German civil engineer Chris@an Oio Mohr (1835-‐1918), the Mohr’s circle allows determining the extreme values of many quan@@es useful in the stress analysis of structural members, including minimum and maximum values of stress, strain and second moment of area

m

n

m

nm

n

m

n

m

n

m

n

Imm

Imn

Y

Z

Product moment of area (+ve, -‐ve or null)

Second moment of area (always +ve)

Iyy

Iyz

-‐Iyz

Izz

Mohr’s Circle

Gy ≡

≡

z


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•  We can draw the Mohr’s circle, once its centre CI and its radius Ri are known: –  The centre is always on the

horizontal axis, whose posi@on is the average of the second moments of area about two orthogonal axes, e.g. Iyy and Izz

–  From simple geometrical

considera@ons (Pythagoras’ theorem), the radius requires the product moment of area as well

Imm

Imn

Y

Z

Iyy

Iyz

-‐Iyz

Izz

y

z

G

m

n CI ≡ Iave,0{ }

Iave =Iyy + Izz2

RI =Iyy − Izz

2⎛⎝⎜

⎞⎠⎟

2

+ Iyz2

CI Iave

RI

=561,250

=210,004


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•  Points Y and Z in the Mohr’s circle, representa@ve of the central axes y and z in the cross sec@on, are the extreme points of a diameter

•  A rota@on of an angle α of the central axes in the cross sec@on corresponds to an angle 2α in the Mohr’s circle (in the same direc@on), i.e. twice the angle in the Mohr’s plane

Imm

Imn

Iyy

Iyz

-‐Iyz

Izz

y

z

G

m

n

Y

Z

CI Iave

RI

α

2α

M


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•  We can determine the maximum and minimum values of the second moment of area for a given cross sec@on:

•  The axes p and q associated with the extreme value of I are called “principal axes of iner@a” –  They are orthogonal each other –  In this example:

Ipp= Imax è p-‐p is the strong(est) axis in bending Iqq= Imin è q-‐q is the weak(est) axis in bending, e.g. to be used when calcula@ng the Euler’s buckling load

Imm

Imn

Y

Z

Iyy

Iyz

-‐Iyz

Izz

y

z

G

CI Iave

RI

Imin = Iave −RI

Imax = Iave +RI

Q

Imin P Imax

=771,254

=351,246


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•  We can also evaluate the inclina@on of the principal axes p and q with respect to reference axes y and z

•  In this example:

–  In general, you don’t know whether p is the strong axis or the weak axis, but it’s for sure one of the two extreme values

Imm

Imn

Y

Z

Iyy

Iyz

-‐Iyz

Izz

y

z

G

CI Iave

RI

αyp =α zq =12sin−1 Iyz

RI

⎛

⎝⎜⎜

⎞

⎠⎟⎟

Q

Imin P Imax

αyp αzq=αyp

2αyp

2αzq

=26.7°


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•  For any beam’s cross sec@on, the principal axes p and q always sa@sfy the mathema@cal condi@on

–  That is, their representa@ve points P and Q in the Mohr’s circle belong to the horizontal axis

•  An axis of symmetry is always a principal axis of the area Imm

Imn

Y

Z

Iyy

Iyz

-‐Iyz

Izz

y

z

G

CI Iave

RI

Q

Imin P Imax

αyp αzq=αyp

2αyp

2αzq

Ipq =0

Mohr’s Circle: Par@cular Cases •  If for a given cross sec@on

Imin=Imax, then all the central axes m will have the same second moment of area, i.e. Imm=Imin=Imax, and all the central axes m will be principal axes of area, i.e. Imn=0

–  This is the case, for instance, of both circular and square shapes

–  The neutral axis (where σx=0) will always coincide with the axis about which the bending moment is applied

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z

y G

x

Mm

m

m

z

y G

x

Mm

m

m

Bending about Principal Axes •  In general, a bending

moment Mp ac@ng about the principal axis p will cause the beam to bend in the orthogonal Gxq plane

•  The simple formula of direct stress σx due to pure bending can be resorted to:

–  Similar to Eq. (1) 30

x

beam

’s axis

G

q

p

Mp

principal axis

σ x > 0

σ x < 0

tensilestress

compressivestress

σ x =MpqIpp

Eq. (2)

Distance (with sign) to the neutral axis

Normal Stress due to Unsymmetrical Bending: General Procedure (1/4)

•  If the bending does not act along one of the principal axis (p and q), then the bending moment can be decomposed along the principal axes

•  In the figure, My is the bending moment about the horizontal axis (due, for instance, to the dead load):

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My

z

yG

q

pααyp

Mp = My cos(α )Mq = −My sin(α )

⎧⎨⎪

⎩⎪

My(> 0)

M p(>0)

M q(<0)

ααyp


•  If the bending does not act along one of the principal axis (p and q), then the bending moment can be decomposed along the principal axes

•  Similarly for the case of the bending moment Mz (due, for instance, to some lateral forces):

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My

z

yG

q

pααyp

Mp = Mz sin(α )Mq = Mz cos(α )

⎧⎨⎪

⎩⎪

M p(>0)

α

Mz (> 0)

M q(>0)αzq


•  Once Mp and Mq are known, the normal stress σx (+ve in tension) can be computed with the expression:

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My

z

yG

q

pααyp

p and q here are the distances from the principal axes of the point where the stress σx is sought q

p

G

σ x

q

p

x

σ x =

Mp qIpp

−Mq pIqq

Eq. (3)


•  As an alterna@ve, the following binomial formula can be used

–  where the coefficients beta (β) and gamma (γ) are given by:

Mzz

G

My

y

x

σ x = β y + γ z

β = −Mz Iyy +My IyzIyy Izz − Iyz

2

γ =My Izz +Mz IyzIyy Izz − Iyz

2

⎧

⎨⎪⎪

⎩⎪⎪

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Eq. (4)

Neutral Axis (1/2) •  Along the neutral axis the normal stress

σx is zero, that is: –  The centroid G≡{0,0} belongs to the neutral axis, and indeed yG=0 and zG=0 sa@sfies the above equa@ons

–  We need a second point N≡{yN,zN} to draw the straight line GN represen@ng the neutral axis: we can choose a convenient value for the coordinate zN, e.g. the boiom edge of the cross sec@on, and the associated value of yN is given by:

z

Gy

x

Mz

N

zN

yN

elastic neutral axis

σ x = β y + γ z =0

β yN + γ zN = 0 ⇒ yN = − γ zN

β35

Neutral Axis (2/2)

z

Gy

x

Mz

N

zN

yN

elastic neutral axis

•  Although the bending

acBon is about the verBcal

axis z, the neutral axis is

not verBcal

•  The two flanges are parBally

in tension, parBally in

compression tension

compression

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Normal Stress Calcula@ons: Worked Example (1/3)

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y

z

Gαyp

A≡{-‐8.75,-‐33.75}

B≡{21.25, 26.25}

My

My =106

Mz =0⎧⎨⎪

⎩⎪

Iyy =686,250 Ipp =771,254Izz = 436,250 Iqq =351,246Iyz = −168,750 αyp =26.7°

β = −Mz Iyy +My Iyz

Iyy Izz − Iyz2 = 0.623

γ =My Izz +Mz Iyz

Iyy Izz − Iyz2 =1.610

⎧

⎨⎪⎪

⎩⎪⎪

σ x (A) = β yA + γ zA= −0.623×8.75−1.610×33.75= −59.80

σ x (B) = β yB + γ zB = +55.51


38 38

y

z

Gαyp

My


Mp = My cos(αyp ) = 893,092Mz = −My sin(αyp ) = −449,874⎧⎨⎪

⎩⎪

σ x (A) =Mp qA

Ipp

−Mq pA

Iqq

= 894,092× (−23.00)771,254

− (−449,874)× (−26.21)351,246

= −59.80

σ x (B) =

Mp qB

Ipp

−Mq pB

Iqq

= +55.51


39 39

y

z

GMy


yN = dn =21.25

σ x (N) = β yN + γ zN

= 0.623×21.25+1.610× zN = 0

⇒ zN = −13.241.610

= −8.22

N

tension

compression

point of max tensile stress

point of max compressive stress

Assume: Calculate:

(which gives the neutral axis GN)

elas0c neutral axis

Key Learning Points (1/2) 1.  The simple formula of bending stress, σx=Myz/Iyy, is valid if

and only if y is a principal axis for the cross sec@on –  That is, if and only if the product moment of iner@a is Iyz=0 –  This is the case, for instance, when y and/or z are axis of symmetry

2.  To calculate Iyz one can split the cross sec@on in elementary blocks, sum the contribu@on from each block and use the parallel axis theorem –  Important: Iyz can be nega@ve, posi@ve or null

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Key Learning Points (2/2) 3.  Knowing Iyy, Izz and Iyz , one can draw the Mohr’s circle for the

second moment of area, which allows determining the extreme values (Imin and Imax) and their direc@ons

4.  In the general case of unsymmetrical bending, the normal stress is given by the formula –  σx= β y + γ z

•  where β and γ depend on the components of the bending moments (My and Mz) as well as on Iyy, Izz and Iyz

5.  The above formula allows determining the inclina@on of the neutral axis

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Unsymmetrical bending (2nd year)

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