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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.
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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)
6
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
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)
Symmetrical Bending (3/4)
9
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)
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
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
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 )
⎛
⎝⎜⎞
⎠⎟
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
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
Product Moment of Area (2/3)
15
• 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∫
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)
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
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
?
?
Moments of Area: Worked Example (3/5)
19
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
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
Moments of Area: Worked Example (5/5)
21
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
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
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
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
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
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
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°
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
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
29
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):
31
My
z
yG
q
pααyp
Mp = My cos(α )Mq = −My sin(α )
⎧⎨⎪
⎩⎪
My(> 0)
M p(>0)
M q(<0)
ααyp
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
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)
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)
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
36
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
Normal Stress Calcula@ons: Worked Example (2/3)
38 38
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
Normal Stress Calcula@ons: Worked Example (2/3)
39 39
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
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
40
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
41