unsymmetrical bending (2nd year)
DESCRIPTION
Lecture slides on the calculation of the bending stress in case of unsymmetrical bending. The Mohr's circle is used to determine the principal second moments of area.TRANSCRIPT
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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
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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?
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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
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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
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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)
7
(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
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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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Symmetrical Bending (3/4)
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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)
⬆
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σ x =
My zIyy
Eq. (1)
Symmetrical Bending (4/4)
• 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:
10
x z
y
axis of symmetry
beam
’s axis
My
σ x > 0
G
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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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Unsymmetrical Bending (2/3)
12
• 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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Unsymmetrical Bending (3/3)
13
• Par@cular cases…
Bending about the strong axis
σ x (A) =MyIpp
⋅d2
Bending about the weak axis σ x (A) =
My
Iqq
⋅e
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Product Moment of Area (1/3)
14
• 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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Product Moment of Area (2/3)
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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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Product Moment of Area (3/3)
16
• 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)
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Moments of Area: Worked Example (1/5)
17
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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Moments of Area: Worked Example (2/5)
18
– 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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Moments of Area: Worked Example (3/5)
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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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Moments of Area: Worked Example (4/5)
20
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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Moments of Area: Worked Example (5/5)
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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
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Rota@ng the Central Axes
22
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
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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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Mohr’s Circle (2/6)
24
• 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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Mohr’s Circle (3/6)
25
• 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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Mohr’s Circle (4/6)
26
• 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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Mohr’s Circle (5/6)
27
• 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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Mohr’s Circle (6/6)
28
• 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
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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
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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
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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):
31
My
z
yG
q
pααyp
Mp = My cos(α )Mq = −My sin(α )
⎧⎨⎪
⎩⎪
My(> 0)
M p(>0)
M q(<0)
ααyp
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Normal Stress due to Unsymmetrical Bending: General Procedure (2/4)
• 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
• Similarly for the case of the bending moment Mz (due, for instance, to some lateral forces):
32
My
z
yG
q
pααyp
Mp = Mz sin(α )Mq = Mz cos(α )
⎧⎨⎪
⎩⎪
M p(>0)
α
Mz (> 0)
M q(>0)αzq
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Normal Stress due to Unsymmetrical Bending: General Procedure (3/4)
• Once Mp and Mq are known, the normal stress σx (+ve in tension) can be computed with the expression:
33
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)
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Normal Stress due to Unsymmetrical Bending: General Procedure (4/4)
• 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
⎧
⎨⎪⎪
⎩⎪⎪
34
Eq. (4)
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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
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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
36
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Normal Stress Calcula@ons: Worked Example (1/3)
37 37
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
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Normal Stress Calcula@ons: Worked Example (2/3)
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y
z
Gαyp
My
Iyy =686,250 Ipp =771,254Izz = 436,250 Iqq =351,246Iyz = −168,750 αyp =26.7°
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
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Normal Stress Calcula@ons: Worked Example (2/3)
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y
z
GMy
Iyy =686,250 Ipp =771,254Izz = 436,250 Iqq =351,246Iyz = −168,750 αyp =26.7°
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
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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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