University of Liège Aerospace & Mechanical Engineering
Aircraft Structures Beams - Bending & Shearing
Ludovic Noels
Computational & Multiscale Mechanics of Materials – CM3 http://www.ltas-cm3.ulg.ac.be/ Chemin des Chevreuils 1, B4000 Liège [email protected] Elasticity • Balance of body B – Momenta balance • Linear T • Angular – Boundary conditions • Neumann b • Dirichlet n • Small deformations with linear elastic, homogeneous & isotropic material
– (Small) Strain tensor , or
– Hooke’s law , or
with
– Inverse law l = K - 2m/3 2m
with
2013-2014 Aircraft Structures - Beams - Bending & Shearing 2 Pure bending of symmetrical beams • Assumptions – Symmetrical beams z z
– Cross-section remains plane and Mxx L perpendicular to fibers h x y • Bernoulli or Kirchhoff-Love theory • Only for thin structures (h/L << 1) Mxx • Limited bending: kL << 1 – Linear elasticity • Small deformations • Homogeneous material • Hooke ‘s law – For pure bending • Vertical axis of symmetry • Constant curvature of the
neutral plane
2013-2014 Aircraft Structures - Beams - Bending & Shearing 3 Pure bending of symmetrical beams • Kinematics
• z Mxx L h x • Linear elasticity z – Hooke’s law & stress-free edges Mxx
•
2013-2014 Aircraft Structures - Beams - Bending & Shearing 4 Pure bending of symmetrical beams • Resultant forces z – Tension z Mxx L h x y • Should be equal to zero
(pure bending) Mxx • So the neutral axis is defined such that the first moment of area
– The neutral axis passes through the centroid of area of the cross- z section z – As, here, Oz is a symmetry axis the neutral axis passes through y y the centroid of the cross-section
2013-2014 Aircraft Structures - Beams - Bending & Shearing 5 Pure bending of symmetrical beams • Resultant forces (2) z – Bending moment z Mxx L h x y
• With the Mxx second moment of area • As
2013-2014 Aircraft Structures - Beams - Bending & Shearing 6 Pure bending of symmetrical beams z t = 20 mm • Example f – I-beam subjected to a bending moment 30° • 100 kN.m M • Applied in a 30°- inclined plane y – Distribution of stress? h = 300 mm
– Neutral axis? tw = 25 mm
tf = 20 mm
b = 200 mm
2013-2014 Aircraft Structures - Beams - Bending & Shearing 7 Pure bending of symmetrical beams z t = 20 mm • Second moments of area f – Doubly symmetrical cross-section 30° origin of the axes is the centroid M y – h = 300 mm
tw = 25 mm
tf = 20 mm
b = 200 mm
Transport theorem
–
2013-2014 Aircraft Structures - Beams - Bending & Shearing 8 Pure bending of symmetrical beams z t = 20 mm • Bending moment components f – 30° – M y • Stress distribution h = 300 mm – Linear elasticity superposition tw = 25 mm
– tf = 20 mm
b = 200 mm
2013-2014 Aircraft Structures - Beams - Bending & Shearing 9 Pure bending of symmetrical beams z • Stress distribution (2) – 30° – For z = ± 0.15 m M y h = 300 mm
– For y = 0
b = 200 mm • Neutral axis z
76.4° – Such that sxx = 0 M y
2013-2014 Aircraft Structures - Beams - Bending & Shearing 10 Pure bending of unsymmetrical beams
• Assumptions z z – Same as for symmetrical bending q q • Bernoulli or Kirchhoff-Love theory y y • Only for thin structures (h/L << 1) • Limited bending: kL << 1 x Mxx Mxx • Linear elasticity • Bending moment in a plane being q-inclined Mxx
z – Let us assume q • The existence of a neutral axis a y – Being a-inclined
– Including the axes origin Mxx
2013-2014 Aircraft Structures - Beams - Bending & Shearing 11 Pure bending of unsymmetrical beams
• Bending moment components z – q – y
Mxx
• Pure bending z Mz – Signed distance from neutral axis
a y – Plane section remains plane d My
– Pure bending
• As the neutral axis still passes through the cross-section centroid
2013-2014 Aircraft Structures - Beams - Bending & Shearing 12 Pure bending of unsymmetrical beams
• Neutral axis z z Mz q – As y a y d My Mxx – It yields •
•
• In tensorial form
– So position of the neutral axis depends on • The geometry • The loading
2013-2014 Aircraft Structures - Beams - Bending & Shearing 13 Pure bending of unsymmetrical beams
• Principal axes z q – As y
Mxx
– The referential can be changed such that z • We are in the principal axes of the section Mz – This referential is load-independent • The neutral axis is then obtained by a y d My
• And the stresses are rewritten as
– So in the principal axes, everything happens as for symmetrical loading
2013-2014 Aircraft Structures - Beams - Bending & Shearing 14 Pure bending of unsymmetrical beams
• Example z’ – Beam subjected to a bending moment • 1.5 kN.m
• In the vertical plane b1 = 40 mm b2= 80 mm y’ – Neutral axis? – Maximum stress due to bending ? t = 8 mm
M
h = 80 mm
t = 8 mm
2013-2014 Aircraft Structures - Beams - Bending & Shearing 15 Pure bending of unsymmetrical beams
• Position of centroid z’ – New frame Cyz linked to centroid C z • b1 = 40 mm b2= 80 mm y’
t = 8 mmy C M
h = 80 mm
• t = 8 mm
2013-2014 Aircraft Structures - Beams - Bending & Shearing 16 Pure bending of unsymmetrical beams • Second moments of area – z
b1 = 40 mm b2= 80 mm 17.6 mm t = 8 mm 12 mm y C M
h = 80 mm –
t = 8 mm
2013-2014 Aircraft Structures - Beams - Bending & Shearing 17 Pure bending of unsymmetrical beams • Second moments of area (2)
– z
b1 = 40 mm b2= 80 mm 17.6 mm t = 8 mm 12 mm y C M
h = 80 mm
t = 8 mm
– As each part is symmetrical, Iyz with respect to C is equal to the sum of • Area part times • y-distance from C of section part to global section C times • z-distance from C of section part to global section C
2013-2014 Aircraft Structures - Beams - Bending & Shearing 18 Pure bending of unsymmetrical beams • Neutral axis
– As z
17.6 mm with q = p/2, so 12 mm 14.7° y C M
– Neutral axis in inclined by 14.7° as h = 80 mm
t = 8 mm • Stresses –
– By inspection the maximum stress is reached
2013-2014 Aircraft Structures - Beams - Bending & Shearing 19 Pure bending of unsymmetrical beams
• Bending deflection z z q – Let us consider q a y y • Unsymmetrical beam a
• Frame origin at the x Mxx Mxx z cross-section centroid
• A bending moment in M xx y a planed being q-inclined
• A resulting neutral-axis
being a-inclined
2013-2014 Aircraft Structures - Beams - Bending & Shearing 20 Pure bending of unsymmetrical beams
• Bending deflection (2) z z q – Due to the bending moments a y y • There is a deflection normal a
to the neutral-axis x Mxx • The centroid is deflected by x • The neutral surface has a curvature k with – Deflection components z x • q a y
M • xx
2013-2014 Aircraft Structures - Beams - Bending & Shearing 21 Pure bending of unsymmetrical beams
• Bending deflection (3) z q – As a y •
Mxx •
– The neutral-axis is obtained by
•
&
• An unsymmetrical beam – Deflects both vertically & horizontally even for a loading in the vertical plane
– Excepted in the principal axes (Iyz=0)
2013-2014 Aircraft Structures - Beams - Bending & Shearing 22 Bending of unsymmetrical beams • Shearing-bending relationship fz(x) z z – Linear balance equation Mz Tz dx Tz+ ∂xTz dx x a y d My My My+ ∂xMy dx – Momentum balance equation
as second order terms vanish
fy(x) y – Eventually y T + ∂ T dx My Ty dx y x y • a x z Mz • d Mz Mz+ ∂xMz dx
2013-2014 Aircraft Structures - Beams - Bending & Shearing 23 Bending of unsymmetrical beams • Bending approximation: deflection equations
– As &
– Deflection equations read
• Euler-Bernoulli equations for x in [0; L] z f(x) Tz Mxx uz =0 x duz /dx =0 M>0 L • Low order boundary conditions
– Either on displacements &
– Or on shearing
• High order boundary conditions – Either on rotations &
– Or on couple
2013-2014 Aircraft Structures - Beams - Bending & Shearing 24 Bending of unsymmetrical beams • Bending approximation: deflection equations with energy method – Same results can be obtained with an energy method
– Let us assume we are in the principal axes and Mz = 0
z f(x) Tz Mxx • As uz =0 x duz /dx =0 M>0 L • Internal energy variation
• Work variation of external forces
2013-2014 Aircraft Structures - Beams - Bending & Shearing 25 Bending of unsymmetrical beams • Bending approximation: deflection equations with energy method (2) – Energy conservation z f(x) Tz Mxx uz =0 x duz /dx =0 M>0 L
• Integration by parts of the internal energy variation
• Work variation of external forces
2013-2014 Aircraft Structures - Beams - Bending & Shearing 26 Bending of unsymmetrical beams • Bending approximation: deflection equations with energy method (3) – Energy conservation (2) z f(x) T • As duz is arbitrary: z Mxx Euler-Bernoulli equations uz =0 x duz /dx =0 M>0 • on [0, L] & L
•
• Remark: if displacement or rotation constrained at x=0 or x=L
– duz =0 is no longer arbitrary at this point – The boundary condition
» Becomes uz = value or/and uz,x = value » Instead of being on shear or/and couple
2013-2014 Aircraft Structures - Beams - Bending & Shearing 27 Bending of unsymmetrical beams • Bending approximation: shearing – As
•
• – There are no shearing loads only if the bending moment is constant • Cross section remains plane only if constant bending moments
• Beam with cross-section dimensions << L L – If bending variation is reduced (Saint-Venant) h – Shear stress O(T/bh) ~ bending stress x h/L – Shear stress can be neglected
• For thin-walled cross sections (as plates/shells) L – Shear stress O(T/th) cannot be neglected – If bending variation is reduced (Saint-Venant) t Deflection is primarily due to bending strain h
2013-2014 Aircraft Structures - Beams - Bending & Shearing 28 Bending for thin-walled sections z • Second moments of area for thin-walled sections t – Thin walled section • Thickness t << cross-section dimensions y • Example Iyy of a U-thin walled cross-section h
t t
b
z t • This result is obtained directly if the section is represented as a single line y h
t t
b 2013-2014 Aircraft Structures - Beams - Bending & Shearing 29 Bending for thin-walled sections • Second moments of area for thin-walled sections (2) z’ z t – Centroid & z’C= 0 – Second moments of area
• & y’ y h C t t
b
• null as Cy is a symmetry axis
2013-2014 Aircraft Structures - Beams - Bending & Shearing 30 Shearing of beams z • Shear stress L T – A naïve solution is a uniform stress Tz z h x •
z • By reciprocity this would lead to s = t zx t • Impossible as the surface is stress-free t y – So the shear stress has to be tangent to the contour x For thick cross section For thin-walled section
z z L z T H Tz Tz h x t y
H – How to compute these profiles?
2013-2014 Aircraft Structures - Beams - Bending & Shearing 31 Shearing of symmetrical beams
• Assumptions f (x) z z – Thick section z Tz dx Tz+ ∂xTz dx – Symmetrical beam x y • My = ||Mxx|| My • Mz = 0 My+ ∂xMy dx
* A* N x * * • Shear stress N x+ ∂xN x dx – From equilibrium (previous slide): – Let us consider the lower part of the beam (cross section A*) • Normal force acting on this lower part extremities
• With the reduced moment of area
2013-2014 Aircraft Structures - Beams - Bending & Shearing 32 Shearing of symmetrical beams
• Shear stress (2) fz(x) z – We found Tz dx Tz+ ∂xTz dx x My – But lower part of the beam is at equilibrium Rdx My+ ∂xMy dx
– So the upper part should apply * N x N* + ∂ N* dx • On the lower part x x x • A force Rdx • With z
b(z) y – Shear stress t • On the cut surface, assuming uniform shear stress on b(z) x t A*
2013-2014 Aircraft Structures - Beams - Bending & Shearing 33 Shearing of symmetrical beams
z • Deformation z – Section shear strain in linear elasticity b(z) h y t • h t • But shear stress is not uniform t A* z – g = 0 on top and bottom surfaces Tz+ ∂xTz dx – at neutral axis x g • The average shear strain g is defined as T gmax – The angle of the deformed neutral axis z with its original direction dx
2013-2014 Aircraft Structures - Beams - Bending & Shearing 34 Shearing of symmetrical beams • Deformation (2) z – Average g obtained by energy method Tz+ ∂xTz dx • Work of external forces g dx x – g
gmax – If the shear stress was uniform Tz one would have – To account for the non-uniformity, a dx corrected area A’ is used
• Internal energy variation
–
– As
2013-2014 Aircraft Structures - Beams - Bending & Shearing 35 Shearing of symmetrical beams • Deformation (3) z – Average g obtained by energy method (2) Tz+ ∂xTz dx • Work of external forces = Internal energy variation g dx x g
gmax Tz
dx – Eventually • Average shear strain
• With
• Under the assumption of a constant shear stress on b(z)
2013-2014 Aircraft Structures - Beams - Bending & Shearing 36 Shearing of symmetrical beams • Deformation (4)
– Shearing equations z qy • Shearing is the difference between – Neutral fiber deflection x g qy – Orientation of the cross section g » Angle qy
dx • Curvature is then computed from the variation in cross-section direction
z f(x) T z M • Timoshenko equations on [0, L] xx uz =0 x du /dx =0 M>0 – As z L
– As
2013-2014 Aircraft Structures - Beams - Bending & Shearing 37 Shear effect • Bernoulli assumption z qy – Timoshenko beam equations
x qy g
– Euler-Bernoulli equations z f(x) Tz Mxx uz =0 x duz /dx =0 M>0 – It has been assumed that L the cross-section remains planar and perpendicular to the neutral axis
– uz,x = -qy – Validity: (EI)/(L2A’m) << 1
2013-2014 Aircraft Structures - Beams - Bending & Shearing 38 Shearing of symmetrical beams • Example: shear stress for a rectangular section z
– Using b(z) h y t • Reduced moment of area x t A*
• Shear stress z
t h
• Maximum shear stress reached at z=0: – Exact solution (elasticity theory): • Maximum shear stress reached at y=z=0
h/b 2 1 1/2 a 1.033 1.126 1.396
• Shear stress is not uniform on b, but this is a correct approximation for h>b
2013-2014 Aircraft Structures - Beams - Bending & Shearing 39 Shearing of symmetrical beams • Example: shear strain for a rectangular section z
– Reduced moment of area b(z) h y t x – Corrected area t A*
– So if shear stress is uniform on b (ok for h>b)
2013-2014 Aircraft Structures - Beams - Bending & Shearing 40 Shearing of unsymmetrical beams • Extending these equations to unsymmetrical beams
– The equations remain valid z f(x) Tz Mxx uz =0 • If we are in principal axes as Iyz = 0 x duz /dx =0 M>0 as everything happens as if uncoupled • In linear elasticity, as superposition L principle holds, the same equations
can be derived for uy – But if the loads do not pass through a point called the shear center S there will be a twist of the beam • This point is difficult to obtained for unsymmetrical non-thin-walled sections – This problem is not relevant for light aeronautic structures based on thin- walled sections • Only thin-walled sections will be considered
2013-2014 Aircraft Structures - Beams - Bending & Shearing 41 Shearing of thin-walled cross section beams
• General relationships z L – Consider thin-walled sections y • Symmetrical or not t x • Closed or open h – Assumption • Shear is uniform on the thickness t Definition of the shear flow – Balance equations • ss + ∂sss ds q + ∂sq ds sxx + ∂xsxx dx q + ∂xq dx x ds s dx • q sxx q ss
L
2013-2014 Aircraft Structures - Beams - Bending & Shearing 42 Shearing of thin-walled cross section beams
• Shear flow z z Tz – Assumption y • No twisting of the beam T z Ty x y Shear loads pass through S C T Ty y a particular point called
shear center S Tz • If this is not the case there is a torsion combined to the shearing – Equation • Direct bending stress obtained from • ! We use pure bending theory: section assumed to remain planar (not true)
2013-2014 Aircraft Structures - Beams - Bending & Shearing 43 Shearing of thin-walled cross section beams
• Shear flow (2) z T – Equations (2) z
• y C S Ty q s – Assuming the variation of moment is larger than the variation of section
• But , &
• Eventually
2013-2014 Aircraft Structures - Beams - Bending & Shearing 44 Shearing of thin-walled open section beams
• Shear flow for an open section beam z Tz – For an open section q(s) = 0 as y on the boundary sxz(0) = szx(0) = 0 C S Ty q s • In the principal axes
z • To be compared to the thick cross section approximation b(z) h – y t – Has been obtained for symmetrical beam x and assuming constant stress on b t A* – For unsymmetrical beams, the same approximation leads to
– Accurate only for h/b > 1
2013-2014 Aircraft Structures - Beams - Bending & Shearing 45 Shearing of thin-walled open section beams
• Shear center z Tz – Definition: • No twisting of the beam if shear loads pass through y a particular point called shear center C S S Ty – Practically q • Shear loads are represented by loads passing s through S and by a torque – Location • If axis of symmetry: S lies on it • If cruciform or angle sections, as q is along these section, S is at the intersection z t z z y h C=S h h C y C y t t t t t t t t t S S b b b • Generally speaking: obtained by considering the moment equilibrium
2013-2014 Aircraft Structures - Beams - Bending & Shearing 46 Shearing of thin-walled open section beams
• Example z’ z t – U-thin walled cross section Tz • From previous exercise y’ y S h C Ty t t
b
– Location of the shear center S?
2013-2014 Aircraft Structures - Beams - Bending & Shearing 47 Shearing of thin-walled open section beams
• Location of the shear center z’ z t – By symmetry: q Tz • On Cy – Horizontal location? q y’ y h • Consider a shearing Tz (Ty is useless S C as we know the vertical location) t t • Ensure moment equilibrium q – Shear flow b s •
• As
• With
2013-2014 Aircraft Structures - Beams - Bending & Shearing 48 Shearing of thin-walled open section beams • Shear flow – Shear flow on bottom flange
• As z’ z t q Tz
q y’ y S h O’ C t t • Maximum shear flow q b s • Moment around O’ induced by the shearing of bottom flange
+
2013-2014 Aircraft Structures - Beams - Bending & Shearing 49 Shearing of thin-walled open section beams • Shear flow (2) – Shear flow on the web
• As z’ z t q Tz
q y’ y S h O’ C t • “Boundary” shear flow t q b s
• Maximum shear flow
• Moment around O’ induced by the shearing of the web +
2013-2014 Aircraft Structures - Beams - Bending & Shearing 50 Shearing of thin-walled open section beams • Shear flow (3) – Shear flow on top flange
• As z’ z t q Tz
q y’ y S h O’ C t • Moment around O’ induced by the t shearing of the top flange q b s +
2013-2014 Aircraft Structures - Beams - Bending & Shearing 51 Shearing of thin-walled open section beams • Shear center – Moment due to the shear flow:
+ z’ z t q Tz – It has to be balanced by the shearing q y’ y S h O’ C t t q b s
2013-2014 Aircraft Structures - Beams - Bending & Shearing 52 Shearing of thin-walled closed section beams • Shear flow for a closed section beam z T – Equation z
• T Ty y C q s still holds • But q(s=0) is now ≠ 0 – Method • The cross section is virtually cut at s=0 • With the open section contribution
• q(s=0) is computed to balance the momentum – Shear loads pass through a given point T (not necessarily shear center S) – We will see that, for closed sections, shear and torsion stresses have the same form, so we do not really have to pass through the shear center S as it becomes useless to decompose shearing and torque
2013-2014 Aircraft Structures - Beams - Bending & Shearing 53 Shearing of thin-walled closed section beams • Shear flow for a closed section beam (2) z Tz – Evaluation of q(s=0) • Momentum balance (if q & s anticlockwise)
Ty T y
with p the distance from the wall tangent to C C p • If A is the area enclosed by the section mid-line h q s – z Tz
• At the end of the day Ty T y
C dAh ds p q s
2013-2014 Aircraft Structures - Beams - Bending & Shearing 54 Shearing of thin-walled closed section beams • Shear flow for a closed section beam (2) z T – Final expression z • T Ty y • q (s) is computed as for an open section o C • q s
– Remarks • The q(0) is related to the closed part of the section,
but there is a qo(s) in the open part which should be considered for the shear torque • This last expression assumes q, s anticlockwise • If momentum and distance p are related to the lines of action of the shear load T, this simplifies into • By cutting the section we are actually substituting the shearing by – A shear load passing through the shear center of the cut section » Which depends on the cut location – A torque leading to constant shear flow q(s=0) » Which depends on the cut location
2013-2014 Aircraft Structures - Beams - Bending & Shearing 55 Shearing of thin-walled closed section beams • Shear center z T – As we have already used the momentum balance, z
how to determine the shear center S ? T Ty y – As loads passing through this point do not lead to C section twisting, we have to evaluate this twist q s
2013-2014 Aircraft Structures - Beams - Bending & Shearing 56 Shearing of thin-walled closed section beams • Shear deformations
– As loads passing through shear center do not lead to z Tz section twisting, we have to evaluate the twist in the T Ty y general case C • Shear strain in the local axes q s
n • Remark x dux ds du s s dx
2013-2014 Aircraft Structures - Beams - Bending & Shearing 57 Shearing of thin-walled closed section beams
• Twisting and warping z – Closely Spaced Rigid Diaphragm (CSRD) y assumption x • The cross section can deform along Cx • The shape of the cross-section in its own plane remains constant • These 2 motions correspond to warping & twisting z – So in the Cyz plane only a rigid rotation is seen • Aeronautical structures
– You do not want to change the airfoil shape us – So structures are rigid enough for this R q p assumption to hold R y – Let us call R the center of twist C • When warping is constrained the center of twist is different from the shear center (see lectures on structural discontinuities)
2013-2014 Aircraft Structures - Beams - Bending & Shearing 58 Shearing of thin-walled closed section beams • Center of twist z – Equations
• pR is the distance from the wall tangent to R us R q p R y C
z • p is the distance from the wall tangent to C
us R q p Y R y C p
2013-2014 Aircraft Structures - Beams - Bending & Shearing 59 Shearing of thin-walled closed section beams
• Center of twist (2) z – Twist
• us R q p Y • Let & be respectively the components R y along Cy and Cz of the displacement of C due C p to the twist dq – – • As z us R q p Y R y – Remarks C C q -uz • Valid for closed and open sections C C C uy • uz , uy & q depend on x
• us depends on x & s
2013-2014 Aircraft Structures - Beams - Bending & Shearing 60 Shearing of thin-walled closed section beams
• Center of twist (3) z – Location
• As us R q p Y R y C p
• Eventually
• Remarks – Remains valid for a point other than the centroid C – Equations valid for open thin-walled sections but what about CSRD assumptions for such sections ?
2013-2014 Aircraft Structures - Beams - Bending & Shearing 61 Shearing of thin-walled closed section beams
• Warping n – In linear elasticity x • Shear flow dux ds • Shear strain du s s dx
– Shear flow
• As z
us R q p Y R y C p
2013-2014 Aircraft Structures - Beams - Bending & Shearing 62 Shearing of thin-walled closed section beams • Warping (2) z – Shear flow integral
us R q p Y R y C p
z Tz
• As is the area
swept by p with the Ty T y area enclosed by the section mid-line
C dAh ds p q s
2013-2014 Aircraft Structures - Beams - Bending & Shearing 63 Shearing of thin-walled closed section beams • Warping (3) – As
• Performing the integral all around the section
z
• The wrapping displacement reads
us R q p Y R y C p • If axes origin corresponds to twist center R
– With
2013-2014 Aircraft Structures - Beams - Bending & Shearing 64 Shearing of thin-walled closed section beams • Warping (4) z – If axes origin corresponds to twist center R
• us – u (s=0) ? R q x Y pR • Symmetrical sections y C p – If origin of s lies on a symmetry axis ux(0)=0 • Unsymmetrical sections
– Linear response sxx(s) ÷ ux(s)-ux(0) – Axial load due to shear or torsion is equal to zero
2013-2014 Aircraft Structures - Beams - Bending & Shearing 65 Shearing of thin-walled closed section beams • Shear center z Tz – To compute yS
• Assume a shear load Tz passing through yS • So there is no twist
• As S y
C p ds q s
• Eventually
– With
• Shear center (Point T = S here) position using
2013-2014 Aircraft Structures - Beams - Bending & Shearing 66 Shearing of thin-walled closed section beams
• Example z’ – Simply symmetrical thin-walled closed section A • Constant t and m on all walls 8a – Shear center ? t y’ B D
9a 6a
C
2013-2014 Aircraft Structures - Beams - Bending & Shearing 67 Shearing of thin-walled closed section beams • Second moments of area z’ Tz – Simply symmetrical A • Centroid lies on Dy’ = 0 8a • Iyz t y’ • Shear center lies on Dy’ B D S – So we have only to consider 9a 6a • A vertical shear load
• Iyy = Iy’y’ C
2013-2014 Aircraft Structures - Beams - Bending & Shearing 68 Shearing of thin-walled closed section beams • Shear flow z’ Tz – Origin of s on D A • 8a t • With y’ B D s S 9a 6a
– Open shear flow on line DA C • z’ Tz A
q • o y’ D s B S
C
2013-2014 Aircraft Structures - Beams - Bending & Shearing 69 Shearing of thin-walled closed section beams
• Shear flow (2) z’ Tz – Open shear flow on line AB A
•
D y’ qo s B S
C z’ Tz A •
– Open shear on BC & CD by symmetry D y’ qo s • B S
• C
2013-2014 Aircraft Structures - Beams - Bending & Shearing 70 Shearing of thin-walled closed section beams
• Shear flow (3) z’ Tz – Open shear flow A
D y’ qo s B S – Constant shear flow for zero twist
• C • Using symmetry properties & as mt is constant
–
–
2013-2014 Aircraft Structures - Beams - Bending & Shearing 71 Shearing of thin-walled closed section beams • Shear flow (4) – Constant shear flow for zero twist (2)
z’ • As Tz A &
y’ q D s B S • Eventually
C
2013-2014 Aircraft Structures - Beams - Bending & Shearing 72 Shearing of thin-walled closed section beams
• Shear center z’ Tz – General expression A • Shear flow balances the applied torque (Point T = S here)
y’ q D s B S
• With C •
z’ • & A
17a 8a t y’ B p • But one has also 9a D 6a
C 2013-2014 Aircraft Structures - Beams - Bending & Shearing 73 Shearing of thin-walled closed section beams z’ • Shear center (2) A 17a – Torque due to open shear flow 8a t y’ B p
9a D 6a
C
– Shear center location
2013-2014 Aircraft Structures - Beams - Bending & Shearing 74 References • Lecture notes – Aircraft Structures for engineering students, T. H. G. Megson, Butterworth- Heinemann, An imprint of Elsevier Science, 2003, ISBN 0 340 70588 4 • Other references – Books • Mécanique des matériaux, C. Massonet & S. Cescotto, De boek Université, 1994, ISBN 2-8041-2021-X
2013-2014 Aircraft Structures - Beams - Bending & Shearing 75