Lecture Kollbruner Section 5.2 Characteristics of Thin Walled Sections and.. Kollbruner Section 5.3 Bending without Twist

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1 Lecture Kollbruner Section 5. Characteristics of Thin Walled Sections and.. Kollbruner Section 5.3 Bending without Twist thin walled => (cross section shape arbitrary and thickness can vary) axial stresses and shear stress along center of wall govern normal (to curved cross section) stress neglected position determined by curvelinear coordinate s along center line of cross section St. Venant torsion not a player as K ~ t^3 use shear flow: y, η z => ζ x => ξ z, ζ a) equilibrium of wall element: x σ ds dx τ = τ xs = τ sx s q t x q + dq/ds*ds σ + dσ dx*dx στ σ + d σ dx t ds σ t ds + q + d ds qds dx dx qdx = 0 using q = τ t as it includes τ(s) and t(s) d q + d σ t = 0 ds dx 1 notes_1_bending_wo_twist.mcd

2 s ds b) compatibility (shear strain) d d u + v = γ ds dx v is displacement in s direction u is displacement in x direction du/ds*ds=du du/ds dv/dx dv/dx*dx=dv dx x view looking to surface c) tangential displacement (δv) in terms of η, ζ and φ α dv dη α dv dζ y component z component hp*dφ dφ hp. p dv dφ rotation component dv = v is displacement in s direction η is displacement in y direction ζ is displacement in z direction and φ is rotation all at point s on cross section. δ v is differential over distance dx δη is component of δv in y δζ is component of δv in z direction direction and hp*δφ is component due to differential rotation between x and x+dx superposition => dη cos ( α) ( ) + dζ sin α + h p dφ η, ζ and φ depend on s and x while α and hp are independent of x (prismatic section) rewrite as: notes_1_bending_wo_twist.mcd

3 δv δη δζ = cos ( α ) + sin ( α ) + hp δφ δx δx δx δx further assumptions: 1) preservation of cross section shape => ζ = ζ(x); η = η(x) φ = φ(x) ) shear though finite is small ~ 0 => 3) Hooke's law holds => σ = E δu δx d u = d v ds dx axial stress equilibrium for the cross section: σ d = N x σ y d = M z σ z d = M y τ h p d = qh p ds = T p Nx = axial force Mz and My are bending moments wrt y and z respectively note integral expressions Tp is torsional moment wrt cross sectional point P Qx and Qy shear forces d = q cos α ds = V y τ cos ( α ) ( ) d = q sin α ds = V z τ sin ( α ) ( ) 5.3 Bending without twist σ d = 0 qh p ds = 0 possible only if lateral loads pass through P 3 notes_1_bending_wo_twist.mcd

4 δv δη δζ = cos ( α ) + sin ( α ) + hp δφ from above, δx δx δx δx δφ = 0 ds dx δx using d u = d v with no twist => becomes: δu dη = cos α δs dx dx dζ ( ) sin ( α ) which can be integrated to become: dζ ( ) ds ζ' sin α ds + u 0 ( x ) where = ζ' (prime is control F7) dx u = η' cos α ( ) δη and is η', and comes outside the integral due to our prismatic assumption δx dy (=dy) =ds*cos(α), dz (=dz) =ds*sin(α), where Y and Z refer to a coordinate system with an arbitrary origin, whereas x and y are defined centroidal (refer to center of area) => ux ( ) = η' 1 dy ζ' 1 dz + u 0 ( x) ds dz α dy y,η u = η' Y ζ' Z + u 0 ( x ) z,ζ which says longitudinal displacement u is distributed linearly across cross section (plane sections remain plane) u 0 is the constant of integration which is f(x) axial strain = du/dx => u' = η'' Y ζ'' Z + u' 0 ( z) and σ = E u' = E η'' Y ζ'' Z + Eu' 0 ( z) now with σ d = 0 σ d = Eu' da = E η'' Y d E ζ'' Z d + Eu' ( )d = 0 => 0 z E η'' Y d E ζ'' Z d + Eu' 0 ( z) 1 d = 0 => 4 notes_1_bending_wo_twist.mcd

5 Y d determines Eu' ( ) in stress σ => Eu' 0 ( z ) = E η'' + E ζ'' Z d 0 z Y d σ = E η'' Y E ζ'' Z + E η'' Z d + E ζ'' rearranged becomes σ = E η'' Y Y d Z d E ζ'' Z but... Y d Z d is the definition of the y position of the centroid and the z position => Y d Z d y = Y and z = Z and σ = E η'' y E ζ'' z where y and z are the position of the point s in the centroidal coordinate system. η'' and ζ'' are determined by the equilibrium conditions σ y d = M z σ z d = M y σ = E η'' y E ζ'' z σ y d = ( E η'' y E ζ'' z ) y d = η'' yyd E ζ'' E zyd = Mz 5 notes_1_bending_wo_twist.mcd

6 E E σ z d = ( η'' y E ζ'' z ) z d = η'' yzd E ζ'' zzd = My noting that zzd = Iy, zyd = Iyz and yyd = Iz and solving the two equations for two unknowns E η'' and E ζ'' leads to => Given Eη'' I z Eζ'' I yz = M z Eη'' I yz Eζ'' I y = M y solving two equations two unknowns I yz M y + M z I y I Find( Eη'', Eζ'' ) yz + Iy I z I yz M z M y I z I yz + Iy I z Eη'' := (I yz M y + M z I y ) Eζ'' := ( I yz M z M y I z ) I z reversed signs now substituting back into the relation for axial stress ( σ := E η'' y E ζ'' z ) => σ := Eη'' y Eζ'' z σ ( I yz M y + M z I y ) I yz M z M y I z y z I yz + I y I z I yz + I y I z or... rearranging in terms of My and Mz => σ := M y ( I y y + I yz z ) M z + ( I z z I yz y ) M y 6 notes_1_bending_wo_twist.mcd

7 as a check on this development let: I yz := 0 and σ := ( I yz M y + M z I y ) ( I yz M z M y I z ) y z z σ M z M y y + z I z I y which matches our previous understanding on bending c) Shear Stress qs, x) = q 1 x s integration of d q + d σ t = 0 (the equilibrium relationship above) along s leads to : ds dx where q 1 ( x) is f(x) and represents the shear flow ( ( ) d σ t ds at the start of the region. it is 0 at a stress free q 1 ( x) = 0 dx boundary which is convenient for an open 0 section: d σ = d ( I yz M y + M z I y ) y ( I yz M z M y I z ) z dx dx I yz + Iy I z I yz + Iy I z Vz Vz+dVz/dx*dx y where only My and Mz are x x dependent and My My+dMy/dx*dx Vy Vy+dVy/dx*dx d Mz = V y and d M y = V z => dx dx x z Mz Mz+dMz/dx*dx M z ( x ) := V y x M y ( x ) := VV z x I yz := I yz d ( I yz M y ( x ) + M z ( x ) I y ) ( I yz M z ( x ) M y ( x ) I z ) I yz V z V y I y I yz V y + V z I z y z y z dx I yz + Iy I z I yz + Iy I z I yz + Iy I z I yz + Iy I z s s (I yz V z V y I y ) ( I yz V y + V z I z ) z t ds qs, ( x) = d σ t ds = dx 0 0 y I yz + Iy I z copy and substitute s s (I yz V z V y I y ) ytds ( I yz V y + V z I z ) ztds qs, ( x) = I yz + I y I z notes_1_bending_wo_twist.mcd

8 if we designate the integrals which are the static moments of the cross section area: Qy and Qz: s s Q z = ytds Q y = ztds 0 0 qs, ( x) := Q (I yz V z V y I y ) Q z ( I yz V y + V z I z ) Q y I yz + I y I z or rearranging as we did for axial stress ( 1 qs, x) collect, V y, V z ( I y Q z + I yz Q y ) V y ( I yz Q z I z Q y ) V z I yz + I y I z I yz + I y I z qs, ( x) := (I yz Q z I z Q y ) V z + ( I y Q z + I yz Q y ) VV y or if the axes are principal (Ixy = 0) I yz := 0 qs, ( x) := (I yz Q z I z Q y ) V z + ( I y Q z + I yz Q y ) VV y V z V y qs, ( x) simplify collect, Q y Qy + Q z ( qs, x) := Q y V z Q z V y + I y I z I y I z I yz := 0 d) Shear Center the above relationships apply for bending without twist i.e. when σ d = 0 qh p ds = 0 possible only if lateral loads pass through P q I zy + I y I z (I zy Q z I z Q y ) V z + ( I y Q z + I zy Q y ) V y from above (reset in hidden area 8 notes_1_bending_wo_twist.mcd

9 this point P was designated (by Maillart in 191, see page 106 Kollbrunner) as the shear center. In the centroidal coordinate system, this is located at y D and z D. the second condition applies for a shear force V. The center of action must pass through P. Divide Q into components Vy and Vz thus from (Qy), the moment the moment equilibrium wrt C (in centroidal coordinates) is: qv y ) h c ds + V y z D = 0 ( q Vy hc y qv y ) h c ds = V y z D ( Vz yd,zd qv y ) is q with Vz set to 0 and hc is z ( perpendicular distance from centroid to line of action (i.e. y(s) and z(s)). set V z := 0 reset q := q I zy + Iy I z ( I y Q z + I zy Q y ) V y qv y ) h c ds = V y z D = (I yz Q z I z Q y ) V z + ( I y Q z + I yz Q y ) VV y substitute ( ( I y Q z + I zy Q y ) V y h c ds I zy + I y I z V y V y z D = ( I y Q z + I zy Q y ) h c ds I zy + Iy I z or... z D = I y Q z h c ds + I zy Q y h c ds 1 ( I y Q z + I zy Q y ) h c ds = I zy + I y I z I zy + I y I z moment from V.z qv z ( ) h c ds = V z y ( ) h c ds V z y D = 0 qv z D 9 notes_1_bending_wo_twist.mcd

10 ( ) qv z is q with Vy set to 0 and hc is perpendicular distance from centroid to line of action (i.e. y(s) and z(s)). reset reset V z := V z set V y := 0 q := Q y (I yz Q z I z Q y ) V z + ( I y Q z + I yz Q y ) V y q ( I zy Q z I z Q y ) V z substitute I zy + I y I z qv ( ) h c ds = V z y D = z I zy + I y I z ( I zy Q z I z Q y ) V z h c ds Izy Q z h c ds I z Q y h c ds V ( I z y D = V z zy Q z I z Q y ) V z h c ds = y D = I zy + Iy I z I zy + Iy I z I zy Q z h c ds I z Q y h c ds I zy + Iy I z for principal axes (Iyz = 0): I zy := 0 y D := I zy Q z h c ds I z Q y h c ds I y Q z h c ds + I zy Q y h c ds Izy + Iy I z z D := I zy + I y II z Q y h c ds Q z h c ds z = y D = D I I z y 10 notes_1_bending_wo_twist.mcd