Lecture Torsion Properties for Line Segments and Computational Scheme for Piecewise Straight Section Calculations

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1 Lecture Torson Propertes for Lne Segments and Computatonal Scheme for Pecewse Straght Secton Calculatons ths conssts of four parts (and how we wll treat each) A - dervaton of geometrc algorthms for secton propertes (cover quckly for sense of approach) B - dervaton of frst moment approach (for nfo - not covered) C - computatonal routne resultng from A (demo a few examples - routne avalable n lab) D - computatonal routne resultng from B (routne avalable n lab) sourced from secton.1 to.3 of Kollbrunner, Curt Fredrch, Torson n structures; an engneerng approach TA17.7.T.K811 19, and geometry. startng pont: a lne defned by two ponts, x1,y1 and x0,y0 ths assumes X.cg and Ycg known y y y 0 1 y 0 = lne passng through two ponts x x 0 x 1 x 0 y 1 y 0 (y 1 y 0 ) x 1 x 0 x 1 x 0 y = x1 x 0 x + y 0 x 0 (x 1 x 0 ) or... x = y1 y 0 y + x 0 y 0 y1 y 0 consder calculaton of ncrement of moment of nerta (relatve to centrod) b dx x b t x 1 y t ds ds = s = length = y y dx ( ) ( ) t ds = 0 cos α cos α 0 cos ( α ) x 0 x 1 x 1 x 0 x0 y 1 y 0 y 1 y 0 x + y0 x 0 x 1 x 0 dx smplfy 1 factor 3 y 0 + y1 y 0 + y 1 ( x 1 x 0 ) b t x 1 0 cos α y t dx = (x 1 x 0 ) y t ds = y1 ( ) ( ) 3 ( ) cos α ( ) + y 0 y 1 + y 0 x 0 I x = t ( s 1 s 0 ) y1 ( ) ( ) + y 0 y 1 + y 0 3 smlarly (by the symmetry of the expresson for the lne above): I y = t ( s 1 s 0 ) ( x1 ) + x ( ) 0 x 1 + x 0 3 cross moment of nerta b t x 1 t y 1 I xy = xyt ds = xydx = 0 cos ( α ) sn ( α ) xydy x 0 y 0 1 notes_15_torson_prop_calc.mcd

2 x 1 x 1 xydx = x + y x 1 x 0 x 0 x0 0 x0 y y 1 0 y y 1 0 x dx x 1 x 0 x 1 y y 1 0 y y 1 0 smplfy 1 x + y x 1 x 0 x 0 0 x 1 x 0 x dx factor ( x 1 x 0 ) ( x 1 y 1 + y 0 x 1 + x 0 y 1 + y 0 x 0) x0 t x 1 I xy = cos α t x 1 x 0 ( ) xydx = ( ) ( x x0 1 y 1 + x 0 y 0 ) + x 0 y 1 + x 1 y 0 cos α = we could calculate Iyx usng the same relatonshp but we know t s = Ixy ( ( x 1 y 1 + x 0 y 0 )... + x 0 y 1 + x 1 y 0 I yx = I xy to evaluate the warpng relatonshps: start wth lne passng through two ponts and obtan normal form of lne y y y y y y ( y 1 y 0) = or... y = x + y x x x x x x1 x ( x 1 x 0) multply by (x1 - x0) y ( x x 1 0 ) = ( y 1 y 0) x + y 0 ( x 1 x 0) x 0( y 1 y 0) rearrange => ( y 1 y 0 ) x + ( x 1 x 0) y + x 0 ( y 1 y 0) y 0 ( x 1 x 0) = 0 A = ( y 1 y 0 ) B = x 1 x 0 C = x 0 ( y 1 y 0 ) y 0 ( x 1 x 0) general form: Ax + By + C = 0 to reduce to normal form x*cos(β)+y*sn(β)=p dvde by y y x + x opposte of C. C 0 and β s angle between the x axs and the NORMAL to lne. denom ( y y x = denom = 1 0 ) + ( x 1 0 ) for the geometry shown: x0 > x1, y1 > y0 x1,y1 β x sn ( β ) ( 1 x 0 ) = denom ( ) ( ) where sgn s cos ( β) = denomsgn ( C) y 1 y 0 denom β x0,y0 Ax + By + C = 0 ( becomes ) Ax By = C y 1 y 0 x 1 x 0 denom x C + y = denom denom C ρ c = xcos ( β ) + ysn ( β ) = ρ denom c notes_15_torson_prop_calc.mcd

3 we could also have determned ths drect from the geometry x, y s a pont on a lne a dstance hc from the centrod: hc x,y x*cosβ y*snβ ( ) + ysn β xcos β ( ) = h c β β dω c = dω c = ρ c ds defnton of ω c = h c ds for a straght lne segment ρ c = constant ω c = ρ c L and s lnear along lne ρ c = p from normal form of lne ρ c s postve f centrod s to the left when vewng the element from to j (0 to 1) along the tangent lne alternatve form of lne (cos(α), sn(α) and p defned n terms of x1,y1 x0,y0 above n ths form p s the dstance from orgn to lne and β s angle NORMAL to lne makes wth x axs the ncrease n ωc due to ths lne segment s then s 1 s 1 s 1 y 1 x 1 ω c = ρ c L = h c ds = xcos ( β) ds + ysn ( β) ds = x dy y dx s0 s0 s 0 y0 x0 xcos ( β ) + ysn ( β ) = ρ c = h c ds*sn(β) =-dx = -dx*cos(α) y h D ds β ds*cos(β) =dy =ds*sn(α) β x D,y D β a h C tangent Parallel to tangent α b C β x 3 notes_15_torson_prop_calc.mcd

4 "t can be shown" s 1 y 1 x 1 x + x h c ds = x dy y dx = 1 0 s0 y0 x0 ( y y 1 0 ) (y 1 + y 0 ) ( x x 1 0 ) ω c = x 1 + x 0 y 1 + y 0 ( y 1 y 0 ) ( x 1 x 0 ) xm = md-pont ω c = xm( y) ym x x = x1 - x0 y = y1 - y0 b b I xωc = ω c y ds I yωc = ω c x ds 0 0 ω c s lnear wth s for a lne h c s constant => ω c = h c ds = h c 1 ds = h c s ntal value s ω0 and end value ω1 beng lnear wth s also mples lnear wth x and y.e. wth x () = ω c0 + (ω c1 ω c0 ) ( x x 0 ) ω c1 ω c0 (x x 0 ) ω c s = x + ωc0 ω x c0 1 x 0 x 1 x 0 x 1 x 0 whch s exactly lke wth y substtuted for ω y y 1 0 ( y 1 y 0) y = x + y x x1 x ( x 1 x 0) b so... just as I xy = xy da I xy = 0 t ( s 1 s 0 ) ( x 1 y 1 + x 0 y 0 ) + x 0 y 1 + x 1 y 0 b b ω c1 ω c0 (x x 0 ) I yωc = ω c x ds = 0 x 1 x 0 0 ( I yωc = x + ω c0 ω c0 x1 x 0 x ds x 1 ωc 1 + x 0 ωc 0 ) + x 0 ωc 1 + x 1 ωc 0 ( and... I xωc = y 1 ω c1 + y 0 ωc 0 ) + y 0 ωc 1 + y 1 ωc 0 ( ( notes_15_torson_prop_calc.mcd

5 now we can locate the shear center: (assume for the tme beng that these values are the results for a more complete secton - we'll te ths together later) from prevous lecture (I yωc I z I yz I zωc ) ( I zωc I y + I yz I yωc ) y D := z D := I I y I z I yz I y I z I yz the coordnate system s changed from y,z to x,y changng y to x (frst) and then z to y (I xωc I y I xy I yωc ) ( I yωc I x + I xy I xωc ) x D := y D := I I x I y I xy I x I y I xy now we can calculate ωd by frst calculatng Ω (ths s warpng referenced to shear center wth an arbtrary coordnate system. β s the same as our angle α h D ds = h C ds ( ) ( ) x D sn α ds + y D cos α ds dω D = dω D = h D ds => Ω D s s s () = h D ds = h C x D sn ( α ) + yd cos ( α ) 0 0 ds s s s y x ds y D cos α ds = ωc() s x D 1 dy + y D 1 dx y0 x0 Ω D () s = h C ds x D sn ( α ) ( ) where α constant as... dscos ( α ) ( ) = dx dssn α = dy Ω D () s = ωc x D ( y 1 y 0 ) + y D ( x 1 x 0 ) f we set Ω D0 = 0 at the start of a lne segment, then Ω D1 = Ω D0 + Ω D we fnd the centrod as we would X and Y (t's lnear therefore cg of each segment s (Ω1+Ω0)/). Then the normalzed ΩD.e.ωD s. ΩD-ΩDo and the moments are calculated as above calculate "centrod" of warpng wrt shear center: Q ΩD = Ω D + Ω D 1 Ω Dcg = + Q ΩD A ω Dj = Ω Dj Ω Dcg 5 notes_15_torson_prop_calc.mcd

6 I ω = (ωd 1 ) + ωd 0 ωd 1 + (ωd 0 ) ( I yωd = I xωd = 3 x 1 ωd 1 + x 0 ωd 0 ) + x 0 ωd 1 + x 1 ωd 0 ( ( ( y 1 ω D1 + y 0 ωd 0 ) + y 0 ωd 1 + y 1 ωd 0 ( these should be thought of n terms of contrbutons from the segment as we'll see n the overall scheme the total should = 0 why? b b b b b b σx da = E φ'' ωx da = ωx da = 0 σy da = Eφ'' ωy da = ωy da = 0 as bendng moments = 0 frst moment approach assume Xcg and Ycg known s 1 Q x = y da s0 s 1 dx x b t x 1 ytds ds = s = length = ytds = s0 cos ( α ) cos ( α) 0 cos α ( ) x0 y dx x 1 y 1 y 0 y 1 y 0 smplfy 1 x + y x 1 x 0 x 0 dx 0 x 1 x 0 x0 factor ( x 1 x 0 ) ( y 1 + y 0 ) => s 1 ytds = s0 t y + y 1 0 y + y 1 0 y 1 + y 0 (x ( ) 1 x 0 ) = t ( s 1 s 0) = a cos α ths should have been obvous as cg s md pont and moment of ares y cg * area ym = y + y x + x xm = now for the moments of nerta: we saw that: now ths presents a small problem: b y 1 Q s lnear only where x or y s constant otherwse t's parabolc I x = y t ds = Q x t dy ths can be handled easly f we calculate the values at the mdponts and use 0 y0 Smpson's rule for ntegraton: t s exact for a parabolc varaton (lnear s a subset order = 1) we wll get *n_elements + 1 values ths tme we'll keep a runnng total notes_15_torson_prop_calc.mcd

7 ncrease s calculated usng the approach above over each half (hence 1/ of area and 1/ of endponts) of the segment: a a a Q x := ( y + ym ) Q x+1 := ( ym + y 1 ) Q y := ( x + xm ) Q y+1 := ( xm + x ) k := 1.. n_elements Q x0 := 0 Q xk = Q xk + Q 1 xk Q 1 y0 := 0 Q yk = Q yk + Q yk 1 ntegraton s over entre element usng mdpont and endpont values => n_element values y 1 y 1 ( I x = y t dy = y 1 Q x t dy = t y0 y0 y 0 ) Q + Q +1 x +1) Smpson's rule x + Q x ( snce ths result wll be useful later on we'll put t asde: smlarly for Iyy Q x_bar = Q x + Q x+1 + Q x Q ( +1) y_bar = Q y + Q y+1 + Q y ( +1) I xx = Q_bar x y I yy = Q_bar y x = 0 = 0 the cross moments of nerta are: b b I yz = Q y dz = Q z dy 0 0 I xy = Q_bar x x I yx = Q_bar y y = 0 = 0 derved above: and n lecture b b x + x y + y I 1 0 ω c = ( 1 0 y y 1 0 ) ( yω = Q y dω = Q ω dx x x 1 0 ) 0 0 b b I xω = Q x dω = Q ω dy ω c = xm( y) ym x notes_15_torson_prop_calc.mcd

8 ω c = xm y ym x I xωc = Q x_bar ω c I yωc = Q y_bar ω c = 0 = 0 as above: (I xωc I y I xy I yωc ) ( I yωc I x + I xy I xωc ) x D := y D := I I x I y I xy I x I y I xy now we can calculate the warpng parameters: as above: calculate ΩD and centrod Ω D () s = ωc x D ( y 1 y 0 ) + y D ( x 1 x 0 ) f we set Ω D0 = 0 at the start of a lne segment, then Ω D1 = Ω D0 + Ω D we fnd the centrod as we would X and Y (t's lnear therefore cg of each segment s (Ω1+Ω0)/). Then the normalzed ΩD.e.ωD s. ΩD-ΩDo and the moments are calculated as above calculate "centrod" of warpng wrt shear center: Q ΩD = Ω D + Ω D 1 Ω Dcg = + Q ΩD A ω Dj = Ω Dj Ω Dcg nstead of drect ntegraton based on the lnear relatonshp as above we calculate the value at the md-ponts and the Q ω D + ω D + 1 ωm D = now for the moments of nerta: we saw that above (ths was coped an x and y changed to ω: b ω 1 I ω = ω t ds = Q ω t dω 0 ω0 ncrease s calculated usng the approach above over each half (hence 1/ of area and 1/ of endponts) of the segment: a Q ω := (ω + ωm ) Q ω+1 := (ωm + ω + 1 ) 8 notes_15_torson_prop_calc.mcd

9 k := 1.. n_elements Q ω0 := 0 Q ωk = Q ωk + Q ωk 1 ntegraton s over entre element usng mdpont and endpont values => n_element values ω 1 ω 1 ( I ω = ω t dω = ω 1 Q ω t dω = t ω0 ω0 ω 0 ) Q ω + Q ω + +1 Q ω +1) ( Smpson's rule snce ths result wll be useful later on we'll put t asde: Q ω_bar = Q ω + Q ω+1 + Q ω ( +1) I ωω = Q ω_bar ω = 0 the cross moments are: I xω = Q x_bar ω I yω = Q y_bar ω = 0 = 0 9 notes_15_torson_prop_calc.mcd

10 Computatonal Scheme for Cross-Sectonal Quanttes n_elements := 3 X, Y, 0... n_elements as get extra when start wth 0 A 0... nelements -1 X := nput Y := nput n_elements := nput a := nput or... t := nput := 0.. n_elements 1 a n_elements := 0 a n_elements := 0 t n_elements := 0 j := 0.. n_elements t n_elements := X := 5 Y := 0 t := elements Y we wll use these later X X := X 1 X x := X Y := Y 1 Y y := Y + + calculate aref necessary a := f + a = 0, t, a A := a A = 15 ( X ) ) ( Y calculate centrod n X and Y coordnate system and coordnates n centrodal system: Q Y := ( X + X + 1 ) X cg := Q Y A X cg =.17 x j := X j X cg Q X := ( Y + Y + 1 ) Y cg := Q X A Y cg = 10 y j := Y j Y cg 10 notes_15_torson_prop_calc.mcd

11 calculate moments of nerta these are contrbutons from segment = 0, t*(s1 - s0) = area of segment a I x = ( s 0 ) ( ) + y 0 y 1 + y I y = y1 ( 0 ) t s 1 3 t ( s 1 s 0 ) 3 ( ) x1 + x 0 x 1 + x ( 0 ) I x := a 3 I x = ( y 1) I + y y + 1 ( ) y := a 3 ( x 1) + x x + x I 1 y = 31.5 y + + ( ) + + I xy = ( x 1 y 1 + x 0 y 0 ) + x 0 y 1 + x 1 y 0 ( x 1 y 1 + x y ) + x y 1 + x 1 y Ixy = 0 I xy := ( I yx := I xy calculate ωc ths s a runnng total: ω c = x 1 + x + x y + y frst calculate each ncrement ωc := y ( x ) x y + y 0 ( 1 0 y y 1 0 ) ( x x 1 0 ) x + x y + y ωc 0 := 0 ωc + 1 := ωc + y ( x ) ωc + 1 := ωc + ωc calculate warpng moments wrt centrod: I yωc = I xωc = ( x 1 ωc 1 + x 0 ωc 0 ) + x 0 ωc 1 + x 1 ωc 0 ( ( y 1 ω c1 + y 0 ωc 0 ) + y 0 ωc 1 + y 1 ωc 0 ( contrbuton from each segment I yωc := x ωc x ωc ) + x ωc 1 + x ωc 1 13 ( Iyωc = y 1 ωc 1 + y ωc ) + y ωc 1 + y 1 ωc Ixωc I xωc := ( = notes_15_torson_prop_calc.mcd

12 from torson propertes: (I xωc I y I xy I yωc ) ( I yωc I x + I xy I xωc ) x D := y D := I x I y I xy I x I y I xy 1 x D =.333 y D = now we can calculate warpng Ω relatve to an arbtrary orgn Ω0 = 0 Ω D () s = ωc x D ( y 1 y 0 ) + y D ( x 1 x 0 ) f we set Ω D0 = 0 at the start of a lne segment, then Ω D1 = Ω D0 + Ω D Ω D := ωc x D y + y D x Ω D0 := 0 Ω D := Ω D + Ω D + 1 calculate "centrod" of warpng wrt shear center: Q ΩD := Ω D + Ω D 1 Ω Dcg := + Q ΩD A Ω Dcg = 35 ω Dj := Ω Dj Ω Dcg now we can calculate the normalzed warpng functons (relatve to the shear center) I ω = (ωd 1 ) + ωd 0 ωd 1 + (ωd 0 ) ( 3 I yωd = I xωd = x 1 ωd 1 + x 0 ωd 0 ) + x 0 ωd 1 + x 1 ωd 0 ( ( ( y 1 ω D1 + y 0 ωd 0 ) + y 0 ωd 1 + y 1 ωd 0 ( 1 notes_15_torson_prop_calc.mcd

13 I ω := a 3 ω D 1 + ω D ω D 1 + ω D I ω = I yωd := + ω D 1 + x ω D + x ω D 1 + x 1 ω D x I yωd = I xωd := + ω D 1 + y ω D + y ω D 1 + y 1 ω D I xωd = y Output: X cg =.17 Y cg = 10 0 elements, centrod, shear center 1 x D =.333 y D = Y 15 Ω Dcg = 35 Y cg 10 I x = I yx = 0 y D +Y cg I y = 31.5 I xy = 0 5 I xωc = I yωc = I ω = I xωd = I yωd = X, X cg, x D +X cg Note: the coordnate system n ths plot s X, Y therefore xd and yd needs to have Xcg and Ycg added back n 13 notes_15_torson_prop_calc.mcd

14 frst moment approach repeat centrod calculatons: to hold values from above I xx := I x I yy := I y we wll use these later X := X 1 X x := X Y := Y 1 Y y := Y + + calculate aref necessary a := f + a = 0, t, a A := a A = 15 ( X ) ) ( Y calculate centrod n X and Y coordnate system and coordnates n centrodal system: Q X := ( X + X + 1 ) X cg := Q X A X cg =.17 x j := X j X cg Q Y := ( Y + Y + 1 ) Y cg := Q Y A Y cg = 10 y j := Y j Y cg frst moment approach get mdponts and values for Q at end and mdponts: ym := y + y x + x xm := Q x := ( y + ym ) Q x+1 := ( ym + y 1 ) Q y := ( x + xm ) Q y+1 := ( xm + x 1 ) + + k := 1.. n_elements we have a total of *n_elements + 1, k = 1...*n_elements and 0 Q x0 := 0 Q xk := Q xk 1 + Q xk 1 Q y0 := 0 Q yk := Q yk 1 + Q yk 1 k1 := 0.. n_elements Q yk1 0 Q xk k1 k1 1 notes_15_torson_prop_calc.mcd

15 ntegrate usng Smpson's rule wth mdpont values Q x_bar := Q x + Q x+1 + Q x Q ( +1) y_bar := Q y + Q y+1 + Q y ( +1) I x := Q x_bar y I y := Q y_bar x = 0 = 0 I x = I xx = I y = 31.5 I yy = 31.5 cross moments of nerta I xy := Q x_bar x I yx := 1 Q y_bar y = 0 = 0 I xy = 0 I yx = warpng moments relatve to the centrod: ω c := xm y ym x I xωc := Q x_bar ω c I yωc := Q y_bar ω c = 0 = 0 I xωc = I yωc = as above calculate shear center x D := ( I xωc I y I xy I yωc ) x D =.333 y D := ( I yωc I x + I xy I xωc ) I x I y I xy I x I y I xy y D = now as above we can calculate the warpng parmeters now we can calculate warpng Ω relatve to an arbtrary orgn Ω0 = 0 Ω D () s = ω c x D ( y 1 y 0 ) + y D ( x 1 x 0 ) f we set Ω D0 = 0 at the start of a lne segment, then Ω D1 = Ω D0 + Ω D Ω D := ω c x D y + y D x Ω D0 := 0 Ω D := Ω D + Ω D notes_15_torson_prop_calc.mcd

16 calculate "centrod" of warpng wrt shear center: Q ΩD := Ω D + Ω D 1 Ω Dcg := + Q ΩD A Ω Dcg = 35 ω j := Ω Dj Ω Dcg ωm := ω + ω + 1 Q ω := (ω + ωm ) Q ω+1 := (ωm + ω 1 ) + k := 1.. n_elements Q ω0 := 0 Q ωk := Q ωk + Q ωk 1 1 Q ω_bar := Q ω + Q ω+1 + Q ω ( +1) ω := ω + 1 ω 0 I ωω := Q ω_bar ω I ωω = = 0 0 Q ωk1 the cross moments are: 0 I xω := 1 Q x_bar ω I xω = = 0 k1 I yω := 13 Q y_bar ω I yω = = 0 1 notes_15_torson_prop_calc.mcd

17 Output: X cg =.17 Y cg = 10 0 elements, centrod, shear center 15 x D =.333 y D = Y 15 Ω Dcg = 35 I x = I yx = Y cg y D + Y cg 10 I y = 31.5 I xy = 0 5 I xωc = I yωc = I ω = I xω = I yω = X, X cg, x D +X cg Note: the coordnate system n ths plot s X, Y therefore xd and yd needs to have Xcg and Ycg added back n f the example s as n the startng pont: the results can be compared wth Shames: example 11.0 X := 5 Y := 0 t := e = t 1 b h bt + t 1 3 for channel shape where e = dstance from web as shown t flange := t 0 tweb := t 1 b := 5 t 1 := t flange t := t web h := 0 e := t 1 b h bt + t 1 3 e = 1.5 from web as defned compares to dstance from cg e rel_cg := e + ( 5 X cg ) X cg =.17 x D =.333 e rel_cg = notes_15_torson_prop_calc.mcd

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