Homework7Solution

# Homework7Solution - HOHENDKK £7“"éébmmg 665 i...

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Unformatted text preview: HOHENDKK £7“ "éébmmg 665% i -- Smaafbkfm. Q“ /2i yam: ’5 i ‘P /g ‘55 LﬂéC/gf'f . Problem 9.2 The coefﬁcients of the stiffness matrix corresponding to these DOF are computed following Example 9.4. For instance, to obtain the ﬁrst column of the stiffness matrix, apply a unit displacement 141:1 while the other displacements u}. = 0, j: 2, 3, 6. Identify the resulting elastic forces and the stiffness influence coefficients. kl] [(21 k5! A k6] k3l k4) 324151/1.3 @% 541:1le MEI/L2 54E1/L2 324 EI/L3 By statics, 648E] — 324 E] k 2 : —~——— k = 0 11 L3 1‘21 L3 31 54 E1 — 54 E] k = k = q k = O 41 L2 51 L. 61 Similarly to obtain the third column, apply L43 = 1 with all other uj = O and identify the resulting elactic forces and the stiffness inﬂuence coefﬁcients. I I p]( ) P2( ) km [(23 [(53 [(63 ‘ a e . k k mL/3 mL/3 33 43 5451/1} L/3 I L/3 L/3 C GEI/L 6El/L i 1251/!“ 54131/1.2 v Part a The elastic properties of the beam (neglecting axial Other elements of the stiffness matrix are obtained deformation) are represented by six DOFs: two similarly. Apply a unit displacement ui = 1 while uj = O, KTaHSIaIional displacements and four TOlalional j i 1'. Identify the resulting elastic forces and by statics displacements obtain the stiffness coefﬁcients kij. u I u: The complete stiffness matrix is M5 143 M4 M6 648 ~324; 0 54L ~54L 0 ~ 324 648 — 54L 0 0 54L 5] " uy"6m:3417-3123”REP“ ” "6i?" " ""0 k 2 "L3" 541. 0; 6L3 24L2 0 6L3 —54L 0; 61} 0 12L2 0 0 54L? 40 61,2 0 121} (a) The stiffness matrix is partitioned: k k k z I: 1: r0] k0! k00 where the subscript 2‘ identifies the translational displacements, u, and 1,13, and the subscript 0 identifies the rotational displacements. 113., £14. 145 and L46. Part I) The DOF representing the inertial properties are the two translational displacements 141 and u2 associated with the concentrated masses. To obtain the coefficients of the mass matrix for these DOF, apply ﬁrst a unit acceleration Lil = 1, while ii: = 0. Next apply a unit acceleration ii: = 1, while til 2 0. Determine the associated inertia forces and identify the mass coefficients: 17,21, uzzo 271:0, £22.21 L1,: 1 L12: 1 t 1 MH mg] Inn 11122 A t 1 NIL/3 0 O mL/3 mL mL 11211 = -—:~ 17112 = 0 11232 = 3 i7123 = 0 Thus the mass matrix is mL 1 O m = c ale ll <> Part c The condensed stiffness matrix for the two vertical DOF is * ~l kit 2 kn '— k10 kOO k0: 3‘ A 162E] 8 —— 7 kl! : , ‘4 3L“ — 7 8 The force vector is given by a [71(1) 90) — {mm} (6) Substituting Eqs. (c), (d) and (e) in Eq. (9.2.12) with c = O Gives the equation governing the translational v a motion of the beam: mil 0' int V 162E] 8 ~7 u; :jpitt) 3 l0 1 £22} T 513 «7 8 u: lpztz) (f) Problem 9.6 m/2 132(2) 1 Q I: [91(2) ._., T J]: L = 2 12 777’” FM Part a The elastic properties of the shear frame (neglecting axial deformation) are represented by six DOFs: [WU horizontal displacements and four rotational displacements. The coefficients of the stiffness matrix corresponding to these DOF are computed following Example 9.7. The complete stiffness matrix is 48 ~24; 0 0 ~6h ~6h «24 24‘ 6]: 6h 6h 6h 5] ""6""6871137?"WHEN?) k2}? 0 6h: 1h2 10h2 0 2h2 (3) ~6h 6h; 2h2 0 cm2 1122 ~6h 6h' 0 2h2 11:2 6h2 This matrix can be written in partitioned form as follows: k : [kn km] k0! kOO Part b The DOFs representing the inertial properties are the two translational displacements uI and 1.42. (b) m/2 Q “"2 “u! (C) Part c The condensed stiffness matrix is a k f! E] m l‘ 0511" “i '72 23— }+ i 37.15 ~15.12 _1‘_5_I_ ha 1 -l ‘ kn '- krO k00 k0: — 15.12 10.19 Substituting Eqs. (c) and (d) in Eq. (9.2.12) gives 37.15 — 15.12 — 15.12 1019 ll “1 “2 H (d) P10) P20) } Problem 9.13 Pan a The elastic properties of the umbrella (neglecting axial deformation of the elements) are represented by six DOFs: three translational displacements and three rotations. The coefﬁcients of the stiffness matrix corresponding to these DOFs are computed following Example 9.4. For instance, to obtain the first column of the stiffness matrix, apply a unit displacement ux =1 while the other displacements are zero, i.e., uj 2 O,j= 2, 3, 6; identify the resul ting elastic forces, and by statics obtain the stiffness coefﬁcients: (d) of the 10) 20‘) 3(1) (6) g axial DOFS: 3nding ,4. For natrix, other ientify "7‘ (SE/IL2 “1251/8 12E] kn — L3 1‘21 = 0 k3l ‘3 0 6E] kai’jé" [(51:30 [(6120 Other columns of k are determined similarly. The complete stiffness matrix is 2 0 0; 6L 0 0 0 12 0E ~6L —6L 0 E: -9__--9__glutamqua kg"; 5 2 2 2 (a) L 6L ~6L 6L512L 2L 2L 0 —6L 0; 2L2 4L2 0 o 0 6L; 2L2 0 4L2 The stiffness matrix can be written in partitioned form as follows: k:[ku kw] (b) k0! k00 where the subscript t identifies the translational displacements, u], uz and u3, and the subscript 0 identiﬁes the rotational displacements 144, u5 and ué. Part b The DOFs representing the inertial properties are the three translational displacements u,, u2 and M3. M3 “2 137 “l u = [1,: 142 “3 To obtain the coefﬁcients of the mass matrix for these DOF, first apply a unit acceleration ii1 = 1, while £22 2 O and £33 2: O. m11 = 5m Next apply a unit acceleration iiz 7- 1, while ii] = O and L13 = O. "112:0 Finally, apply a unit acceleration 1'23 = 1, while ill = 0 and ii? = 0. "113:0 "123:0 "'33:"1 Thus, the mass matrix is 5 m = m l (c) aMWWAMWMWW,WWMWWWMW Part c The condensed stiffness matrix for the three translational DOFs is A -i kn : kit " kIO kOO k0! “I 28 6 ~6 = 3E3 6 7 3 (d) W «6 3 7 The equations governing the translational DOFs are .4 5 “1 28 6 ~6 141 0 ..t 3E1 m 1 uz + 3 6 7 3 u2 = 0 1 11% W —6 3 7 “3 0 (e) (i) If the excitation is horizontal ground motion ago), the total and relative displacements are related as follows: 1 “1 ll] l L15, = L12 + 0 ugx (T) I u 0 L13 3 Substituting Eq. (0 in Eq. (e) gives 5 ii} 28 6 -6 M; 5m ,. 3E1 .. m 1 112 +-———— 6 7 3 L12 =—— 0 ugxm .. 10L3 1 u;; —6 3 7 143 o (9,) (ii) If the excitation is vertical ground motion ug),(z), 1 “1 “1 0 t 142 z 142 + 1 Mg), (h) l u l u3 3 Substituting Eq. (h) in Eq. (e) gives a matrix equation with its left hand side same as Eq. (g) and the right side is 0 Peff(1) = — m digym (i) m (iii) If the excitation is ground motion ugbd (I) in the direction bad, 138 u; u1 l/x/E ui = “2 + 1N5 ugbd ti) L1; L13 l/x/E Substituting Eq. 0) into Eq. (e) gives a matrix equation with its left hand side the same as Eq. (g) and the right hand side is 5771/5 m/ﬁ agbdm (k) m/ﬁ (iv) If the excitation is ground motion ugbcm in the pgff (I) : —' direction b—c, 141’ M1 *l/JE ué = “2 + 1N5 an 0) ug 143 l/ﬁ Substituting Eq. (1) into Eq. (e) gives a matrix equation with its left hand side the same as Eq. (g) and the right hand Side is — Sm/JE m/JE um (I) (m) m/JE (v) If the excitation is rocking ground motion deﬁned by counterclockwise rotation u g6 (in radians) in the plane of Penalz“ the structure, u1 ul - L “2 ~— uz + L age (n) u; 143 - L Substituting Eq. (it) into Eq. (e) gives a matrix equation with its left hand side the same as Eq. (g) and the right hand side is -— SmL mL iigg (t) (0) -— mL peff (I) z " 3. Determine the inﬂuence matrix. Problem 9.19 4. Write the equations ofmotion.. mii + ku = pew) where 1.Determine the stiﬁ‘ness matrix.. 2/3 1/3 h (I) .. u 1 “1 “2 "31 “32 peff(t) : 'mlugm z ’m[1/3 2/3:Hﬁg (0} 2 i i l l - g 2 -1§-—1 0 [k kg] k ,1 0 -1 Problem 9.20 r : “““ *7 *** " kg kgg —1 1 0 0 ~1‘ 0 1 ugx ‘4 “52 where A E] m ‘ . Maw... k ~ k 2 __1 o n. ‘ -1 2 }.__£_..+__.J:__..1 k z k "1 O 1.F0rmulate the stiﬂness matrix.. 8 0 —1 I O 7 fl! kgg : k 0 1 2. Determine the mass matrix. m = m 1 J2 - (mi 1:. ” 48E] 6E1 A 6E1 "11 = 7:3‘ 150 Coordinate transformation: ‘1 H r r—-«> s! u “82 tr =1 u; = (1 —1/2 4/2) as, L182 Stiffness matrix in DOFs u , ugl and L482: 1 k k] 651 T 5’ = ~1/2 ——<1 ~1/2 -1/2> [k8 kg: __]/2 L3 1 ~1/2 ~1/2 = :‘175‘77r7/2 —1/2 1/4 1/4 6E1 “ 2 [‘23—] 6151 kg = 13%— 1/2 *1/2) _ E 1/4 1/4 gg — L3 1/4 1/4 2. Write the mass matrix” m = m 3. Determine the inﬂuence matrix. =1 E uszl/Z By kinematics 1 = (1/2 1/2) Alternatively, L3 6E1 1 z —k"‘k = ——-—-—~ -12 —12 8 6E] L3 < / / > = (1/2 1/2) 4‘ Write the equation ofmotion. .. 6E1 ﬁgm) mu + ~—u = —~r711212 .. L3 / >{ugz(t) bOLUT/Ohig ! 2 .l g . ,(Etiéiasiﬁseilzu AVG ._ “.zyi... / 4 “mg w ¢ 1.; .a Aw #3 0 GM M W 61 We M i M 4 , .iazxé Kw} _ _ ‘,,l, a 0%.? ., .Gcé Mm» mWi a o 0“ mm WK? ..&L.M,CO._ 6 K o a; a w mm m . . N O 0 25333:?» O Q. : g 4; 7 3 k r 4 a §§§§§§§§§§ Eéé § W WW h ., a a. 553...} 3 g 7, 2;: ﬁx, MC AU r 3 ~ 4? w C 22% k ,b 3w; ,3 .M w & 0: 5.0 .a ﬁve , M m «(V Av. Q3, 7, 4 “v.43 % , w f ﬂ > O x r . N 4r. K O mu 0 a H .0 my a z m”, in . v w O {f V * (,5 x , r: w h 4‘ 1 . : a {o Mm AW 85 . O W 3 é my . a1“ Wow 0 a I . M 7 (21.25% 7y r M\ O . a 5 _ \ + . C .2 Q J; . a g _ é ﬂ : ﬂ Q as {0 iii}? a g 3 , a , g _ :35 § Q a «I I5; z» m I; g as A, . x» t , M/Io 3 ﬂ: Q E % ﬁg 0 m Q r: m O G a 4.- a o ‘ OH 9 O _ . I a {V J ,6 0 C a? f} 3%.: /3 f ,v. _ , p3 a s x ax! 5x112 ............. .32 L C M Q .A o m Hm, Q ﬂ 1 w 4. Go A»; Ave 0: Au Wag», ﬂ 3 . w a L éAyﬂg m ., ‘ _ Lava, W”. AU 0 rd AU m M a. x . w r? a; raéiiaiiisiaii. ’ m: L (133...“ riaIIEIL A, L a : .E m“ 2 5+ fwx A.» CCN 3.91%“ ,v mESE CE Q Em inﬂux . 331m 3. \$73 ._ . Mow ’5 i Mai 41 f i 5:» 5/“. a I \V O 4// C” G 7 f O [Mbﬁjﬁésézg :ML ,7 l A 54/1 :3 g/ﬁLz {gig : “E 2 Mg Mg; § § mg 30 ...
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