Lösungen Altklausuren Mechanik 2, Prüfungen von Mechanik

Nur auf Docsity: Lade Lösungen Altklausuren Mechanik 2 und mehr Prüfungen als PDF für Mechanik herunter! Klausur 9 Aufgabe 1 10RN/M ↓ Wcq = = · 10 . 83 . 10 = 2133,334 W W WCG 24 EI ↓20 W( = Eg . (3 - (f(2+ 138) . 3 + 13 : 8 +5.B3) = 1660 (WeF Iwx I wx + . (102 . 8 + 5 . 183) = 60 ↑C O = -Wcq + WCF-WCC 0 = - 2133, 33 + 16600 - 600C c = 24 , 12kN AK berechnen : EMB = 0 = 24 , 12 . 10 - 20 . 13 - A . 8 + 10 . 8 . 4 A = 37 ,7 RN EFy = 0 = 37, 7 + B + 24 , 12 - 80 - 20 B = 38, 18KN 37 ,7 po V t + I - 4,11 42, 3 60 18,9 - - I M + 70, 8 Aufgabe 2 A = 40 . 60 = 2400cm2 Iy=1 . 40 . 603 = 720000 cmY Iz= 12 . 60 · 403 = 320000 cm" 1200 + My . 100 . 15 = 0 => My = - 24000kNcM = - 240kNMA 2400 720000 1200 + Mz . 100 . 10 = 0 = Mz = - 16000kNCM = -160 KNMÖB = 2400 320000 Aufgabe 3 MA= 0,5: cos(30) Stosuchune = Drude F, = 0)5c0s(209-EA . cos(30°) Stredaung = Zug, 4 “0105 8 Fx=0 =- FAS? (4 cogta0) +4 cos*(60)+ cos(0)) +f P= A6000-0,05- /5 = AOCorN Aufaabe 4 a) AK berechnen : 2Ma= O= -5-2-40-44+D-8 D= 625 RN ZFy=0= -5-AO +625 +A A=8;t9RN 695 [7s y 6:25 Ly= 2B: 60% = 480001 S$ = AO'b-25=250b6 Vag? £:395-250b = 0,A2AS RNIcm AROcob Ronstan = Rein Integral _ BS- 2500 — Voc = TW sccn ©,052A RNicm T = 0, AZAS: LOO + 0,0524:200 = 34 ,42kN 24.42 c) Vwaged = = A,O8RN ¢ A.2RN d) Vag =0,A245 RNIom Vac = 60,0524 RN/am O\AZAS= A)2- aan =? deans 4%, Perm /wogel¢25en —-dDiffreve= 5\2em 2 Ly 25452 246.8 V detec 0,0524=4,2- =) dethiec £ 4eiPern/ Naga Anzahl der bagel reicht ous, aber andere Vertei Lang, Klausur 8 Aufgabe 1 AB = EA = A= A 12 = 0 , 61 -1 = En. = 0 , 4161 E = 0 ,461 Ea = 0, 11541E F2 = 061 EA = 0 , 12 a E ↑[Ma= 0 = - 50 . 4 . 2 + 0, 11541EA.3 . 2 + 0 , 121EA . 0 , 6 . 4 0 = -400 + 0 , 48 AEA 1 = 833, 3 Ame F1 = 96 , 2kN Fz= 100kN & Fyp = 0= Dy -96,2 . 0 , 8321-100 . 0 , 6 By = 140 KN & Fxp = 0 =-Dx + 96 ,2 . 0 ,5547 + 100 . 0 , 8 Dx= 133, 4kN EFx = 0 = Ax- 133 , 4 Ax= 133,4kN EFy = 0 = 140 + Ay- 50 . 4 Ay = 60 kN 100 96,2 Z Z GO 40 + I t -100 D D - - 2 80 -28 x = 1, 2 133, 4 40 60 Mmax= 60 . 1 , 2 . 0 , 5 = 36 KNM N V M b) Bereich A: Of£x%, EL Hix)= M- 2H. xa w'(a=- Ht 3H x, w'ly)= Hx, + Br” +C4 wlk,) = “Hatt Hira’ teyka tea wl0)=0 =>¢,=0 wilo)=O => c4=0 Bereich 2: O£%,2L Mug= M- 2B. x, w"(m)=- Ht The, w'ly)= Hx, * Bx, +e > Leiguung wh, )= “Hats Hig FCy% + Cy W(x,2L)= wlxg=O) => 2320 WOy4E0® wiligeo) =? cy => fu MI 2 = wyl%2=0) = a t SP welxg= l= yt 3er ee We ond we > Neigung = ° Aufgolbe 3 Senin = , wo ee 40) $:403 | A2BAG Wag> (24 ern. ex v 4o-4-410 = A600 er Wae =(g44)- 6 8-EL Ao: Ao: 40 _ AS) A.A _ SOOA waa ( aa 6 “eet fer — A294G _ AGO® _ G00A er et ex A= A8,86 RO 6433 ane Lave I + > = Vv 4 - 44,33, 35:58 4934 3444 Aufgaibe 4 Leim = = 0,45 RNIem* Kleber: Ve= 430 RNIM Nagel: vy=Az RN , d=20amu Hole: T= 0,35 RN/om* Ty= Fp AO-35° = 35429um" Stim= AO-AS- AO = ASOOcm? Stigny = AO-AO- A2,5 = A2SOcrn® Swager= 40-5 + AS =+50cm?3 — WsSOO -5 v= 0145 = 35429-10 =? V=AOFRN . ~ NAO 4A - 2-413 = as59q 2 =? Ve 74,3RN Soi. Ve =A8- S =A8 pwiem NB = Se = V=857RN v 0135 = 415° 350 =7 V=84,7RN Aufgobe 3 o=6RNicm™ bei 4O o=O0 bei 40 oD 30 2 positiv a ho x 202 yzmx+b m= 4O-AO = -0,5 -60-90 A0= -0,5-O+bD Ao=b => Y= -O,Sx%x+ 40 0) O=-O,5X410 =? Oy= ZOHPa 2 ) = FM18154 90-385 _ 29,a8om ©) 25> 2-33-4330 29,48 = -O;5xtd4O =? o,=3P,96HMP D c) -3,896= iw => v= 638, 9eN D aft. Tn 2 (30-3°4 233°) + 90 - (38,5- 23,49)? + 44- (48,5 - 29,48)" 4t54 om" Wy= 24754 = g30,8em° 23148 :\00 2a £282 +7030.8 => M=49)54 km Aufgabe 4 x bn Nip 3 yu Tmox= 7 A ny 2 Vinnen as Tmax A~ “Bz )de “b nw = 2b-tmax Ce- HPT, Ztrax: (h-a- 3) = 2Tmex' 44h -b us = -a4 tmox - Fn aA =? Vinnen= 76 ¥ Aufaaloe 5 My -409 o= Ww AK berechnen : SUp= OF - Pgu+ BL B= P A=%3P o Jule as|7> z gPe Ane gon ole sl Omax = Omox = 3 tmax M=3-45% w me M =3-ASon on ™ bh 6M = 45V h 6 gm up. us n s cy & n Tmax = 13° 2 Aufgoive 6 = 408° 22133 ma 24eL er er ~ 52. bY 2-62 42-64 _ 320 a2 eL carn | C44 +4) =4EE oly = MB 4 Mido = 6M 3€0 «Bel eT GM = 533,33 => M= 8, 89RHM er eL Av€gabe + Annahme : 2 vapfanle PR verurgacink Vefsonieoung | My=0 My Verorgacit Nerdr enone , P=0 Ad Pahle AL=0,0Um EQ=MQOOMNIM = N=P bei 12 Stiiben cl P= £000 -0,04- AL = 480NN D 0,005 - 5= 40 0,005: cos (30°) 5 z S,= AOOD- 0,005: 5 $g= 4000 -0,005* co830")-S Hebelorm = 5- cosl3o*) Wy = A000: 0,005 - 5-(2cos(o*)+ 4- cag 30°)# 4° Cos 60°) 5 = 750Mum Aufgobe 3 o = 2,S RVI cm W= I-20? = 785,4 cm? 32 MW: AOo 25= +85,4 => M=49,CURNM Hy = “ oh = 43,56 km Ma= 8, F8RNM Avfoabe 4 Yeo = Ox+D 2O+30 20-0 m= =25 Yea = 209x-30 Yeq = 215°R- 30 =-10 27 p= AORN 2 Yao =mxrb m=-40-40 __ 5 ety 3 2z-Zxtb Yao 3% 4oz- §'(-4) 4+ ese or"; £5442 yao 3%ts =-5.(6442 z et = - 23,33 = Popa 23,33RN 2 Arfoabe 5 AK berechnen: SMg=O= 20-40 4° 4-40 - AOA A>33,33R0 V(X) = 33,33 - Fx Ux = 33,33 —x? A Ty = jg 06> A9= 0,05m" S= 0,6-0,5- 0,25= 0,045m* 2,075 3 T= "G05 J (33,38- -x*)dx = 97k Aufgabe 6 21 = 10 . 10341667 zI 24 EE dz = 10 . 32 . 10 = 75 GUZ -12 EI ET w(x)= (2x2- x + L) 23 =W(0) = ML WZ 3EI Einspannung y= 0 L 0 = 416 , 67 - 75 + 3 , 333Ma Ma = -102 , 5 kNM Annahme falsch -> M = GUZ 55 , 75 102, 5 30 45 - + - - W M + - 55 , 752 4 4.2S Mmax = - 102 , 5 + 2 . 10 = 52 , 9RUm Aufgabe 7 3q - - 3q + - 3q 5q Ms = 3q Ns = 1000 + 49 - = 1000 +49 + 39 . 100 = 20 = q = 120kN/m D 198 2880 MB = 5q N = 0 , 75q 5q . 100 op = 0 1Es i + 1930 = 20 = q = 75 , 8kN/m Klausur 11 Aufgabe 1 a) AK berechnen : EMA= 0 = - F . 2 + B . 3L B = F A = F Bereich 1 : 01X124 ; El = 1 Bereich 1 : OX2L , El = 1 2 M(x1) = EFx , M(Xz) = 5FXZ Z W"(x) = - 5FXn W"(xz) = - 5Fx 2 w(x) = - 5Fx, +C w(x) = - -Fx13 + c1X1 + 22 WE-FX22 -S X + Cy2 w(x,=0) = 0 = c=G W(xz= 0)= 0 =>C=0 w(xy = 2) = W(xz = 2) w(x1= 2) = - w(xy=2) - BF . (2L(3 +42= - FF . ( + (z - -F . (2) + c = FL2 - cz - BF(3 + 2(4 = - 5F(3 + (32 (3 = FL2 - C1 2Lc 1 = EFL + (3L 4 = FL2 + EC3 => G = 5F(2+ E . (FL - c) 2c = FL2 4 = 4FL2 =(z= F22 W(Xz= 1)= fa = W(x1 = 0 = -T im Uz-Sin (B = W(Xz =0) = 5* im GUz-Sinn b)w(x)= (3 - x +3 W(x) = 0 o=3 - x +3 - o = E - x +3 1 . 27 0 = X2 - 2x +2 3 M1 = 21-- = L = 5L2 = LILV => . (1 - E) = 0, 4226L Wmax = w(x = 0, 42264) = 0, 06415 Me Klausur 12 Aufgabe 1 50RNM AAX = AAY EA A 1 = Sin (50 %) . AAY + Cos(500). 12= 13 = -Sin130 %) SAY + COS 130%E EA · 11 = 53, 58kN Fr = 4 1 67 Fz = EA . 12 = 16 , 67RN 3 Fz = EA . 13 = -22 , 72kN 3 , 46 GGB [Fy = 0 = . (Sin (50%. 1 AY + cos150%. ) sin /50%- -sin10%SAYToS 130 %.E) sin (30 % -404 , 67 Ay = 207 kNm EA EFy= 0 => F= 26 , 03 KN Aufgabe 2 A= 20 . 40 + 60 . 15 = 1700cm2 Eg = 800 . 20 + 900 . 47,2 = 34 , 559am 1700 Iy = , (20 . 403 + 60 . 153) + 800 . (20 - 34,559)2 + 900 . (47, 5 - 34 , 589)2 = 443840cm4 Sy = 60 . 15 . (7, 5 + 40 - 34 , 559) = 11647cm3 AK berechnen EM , = 0 = D . 10-10 . 10 . 5-20 . 3 D = 56 KN c = 64 RN 64 54 34 + 4 I V A B - 56 MA = 64 . 1-10 . 1 . 0 , 5 = 59RNM MB = 56 . 6 - 10 . 6 . 3 = 156 RNm V(x = 64 - 10X1 01 011647 . (64 - 10x)dx = 231kNTi = 0, 0044384 1 V(xz) = - 56 + 10x2 Tz = 01 011647. 1 - 56 + 10x(dx = 23 , 6 kN 0, 00443846 T= 254 , 6kN a) b) c) Aufgabe 3 AK berechnen ZMg= O= 64-c-8 C=8kN ZUq=O= G4- 8-13 -3-S+HA Ha=t5kUM ZFy= OF -F- B+A A= AS RN as aa) Bereich 4: uv Y OL%,£5 ;ET=A Hix) =-45 + A5x, w"lal= $5- AS x4 wilig) = F5%q - 4b Aa? cy wWhel= Fx EX? teynet ez Wht, =0)=O w'y=0)F=0 3,20 Wlhxa= O)2O =?cy=0 W (x42 5)= W (x2=8) =?C2=0 =eg= 2488 S ¢3 whaz5)= 43:5"-%-S> = fog = W'le,= 5) = 483,5 fpo= w'lyg= 8) =- 4/2 Bpg= 197,5-3,2= 480. 2 w'lx2=0) = 248.8 =? ¥e= Bet aRNer ABI, ZBNme Bereich 2: O£*%,£8 , ET=A HOq\= 64- Sx. we" LG )= - 64 + xo wil) = - ba te +C3 We)=- 32x: +55 Kp 463% + Cy _ B25? L et im GU2-SiNn Ausgabe 4 Year+ Ox tc Pe)sOFOK+C tlo)=O 7% c=0 7(6)= 0,54 > 0S4=36at6b c'Ix)= 2oxtb t(o)=O 2» b=0 =? 0,s4=36a 1:36 a =Q015 =? F(x) = 0,015, =? Alx)= 1: (0,015x)” € =7A-E = ##50kN/m 4 a= 4. (J Ao +f 500 2 TOS Kr)? TT + (Q01Sx*)? Aufgabe 5 Wo = A@Dem? Wy = +O0em® — M:AOO _ %= “300 = 2° B= M100 = 20 10200 Ms doo _ C2 G00 *“° Oy = Mz40° = ro -~, opm Ms tk q => M=A40RNm =? M=200RNm => MeFORN — grépter positives Voment (unten 20g oben Druk) = => Mesooriim steht Negotives oment. (oben zug unten Druck) aufgabe 2 ASO “so 33:5 2 N AsO IN SEL aso OM 9 250 7 A700, A700 Hs= ASO kU Ngs= A400 kW = 450-400 _ “ . Op= re + 2380 43,9 Rb/em? Stitze , innen , oben _ AOO | 4s0-AOS =- 3,38 z Op=- "aR + Z3R0 138 RN/emn Hp= 250 kENm Ne= 325SRN 315 , 260-400 _ Op = 30 7 18:49 RN/em™ 325 2s0-AOO 2 * 02> - heat agag > 42,72 BNlem Balen, onien , mite Aufgatbe 3 A= 20-20- 19 -(8 = #6em™ Ty> Ta = &: (20-20°-18- 185) = 4 585,3 an" N Wyte + H2-400 «10 =-5 Ca= ag + 4se53 4s95.3 N, Myl00 i954 He -A0O 4 =- 4,8 Og= 367 45953 4s9s13 N _ Myl00 «9 = 6S Se= 367 45953 =? N=AG2 BW Drude. Aufgatce 4 AK berechnen : 2Mg,=O= 56-C°4 c= SRN 2Fy= O= -3-AOtA A= AR RN 2Mg,=O= -A8-6 +My Ha= AO8 RN a) ™ @ Bereich O£X,26 Ml) = - 408+ ABx, EL-w “lx,) = 408 - 18x, ET- w' (xa) = 08x, - Sx ea EL: wly)= suxt- 3x2 tcqxtce RB: Wly=0)=O =? C220 wly=0)=O =? c4=O0 =? wlx4)= = (s4x47-3%>) A296 NF b) wliy=6)= Ex 2 ¢) fo = w'liy= 6)= BERN im v2-Sinn = = A498 BNen® = W'(x,=0)= ~ A0b Ne” Ter wil ex 2 Afg = Sat RNe -(- AAD R BNer EL ex Bereich OLx, £4 MIx:)=- 408+ AB-(64+¥2)- AOK, = 8x2 ELT -w'lx,)= - Bx. ET: w'lr,) = - 4 xo" eq EL: wlx)= -3x,* +CyXK + C2 RQ: wlxy=6)= wix2=0) £296 = Co whys t=O => G=-3,8 2 wlia)= Zo (- 3%" AA, B49 +226 ) im 02 - Sinn \_ 443 8 Ne? d) fox wilky> 2)= ~ SABRNO S45: 8kNO eg Go2- Sinn eL aufgalbe 5 du= 6am do = z 2 MS Nig Vez u < Ve Ty= 4.40-25° = 13024 cm“ Y~ Ar - So = 40-5 (12,5 - 2:5.) = 500cms Su = 10-40 -(A2,5- 5) = 50cm _ 5SOoV _ SON Yo= “43024 Vo = A302 s00V. aA _ 25OV .& 302A. 2 A024 3 => aA=6cm Aufgalbe 6 Te 2 “Svcd dx T= % -0,6- A> = Q,05m" 3 § = 0,6: 025° 0,345 = Qosce5m AK berechnen + BMg=O= - A104 50-942010°5 A=AYO RN EFy=O= AO - 56-200 +6 B= AAORYD VOg) = AYO - 20x, N(x) #40 + 20x, 0.08625 . 2 8 v= G05 (s (240-20, )dx, + J (- Mot 2082 Axe ) = A,A25 - (AAO +40) = AGS, 2 RN aAufgatbe 2 Druck posihv Op =2S5N/mm = 2,5 kNian™ . 2 sinla) = 7o a = 53,A3° Tei TC: 20% = #854 ern! = Yy-s00 Hz -AOO -1)5= “a qasy 40° C0S(93,43") 2,5 = 0, A0A86- My - 0,04639-M2 0 = Hy-40019.co5(307)+ MEAZE JQ. sin (20°) 6= 0,44 027%. My + 0,06366- He GTR liefert + My=-4O,68kNm 7 Ue= 48,49 Rm aufgabe 3 © = AS Nicm® A=A00cm” ASRNIM Lid K Lt x hovast Forse Mmox= OF6L:$ -A9-§> eC ic ale Ty= lead’ - di) = % A5i" I .asais _ 45 4,3 W=¢ Asa =e ai hi 2 N00 = Te: (2 44)? - TW LE)* Ai= 6,5A44%cm Aufgabe 4 A= 8. 12 = 96cm2 w = 5 . 8 . 122 = 192cm3 AK berechnen : [MA = 0 = B . 4 - q . 5, 5 . 2, 75 B = 3,781q A = 1 ,7199 1,7199 1,5q 1 , 124q - + V I M - 2)281q 1/ 477q 1 / 4779 . 100 1 , 5 = 192 => q= 1 , 95RN/M 0 . 8 = 1 , 5 . 2 ,2819 = q = 22, 45RN Aufgabe s a) N = F M 36 + W = 5 . 40 . 402 = 10667ca 3 F 36 . 100 0z = - 1600 + 10667 => F = 540kN 200 M . 100 b) -z = - 1600 t 10667 => M = 13 , 33kNm M(x) = x2 13 , 33 = x2 x = 3, 651 = X = 6 - 3, 651 = 2 , 349M 200 + 36. 100 c 0,1 - dZ 5 . b3 => d 49cm aufgaibe 6 _ Al fA nu AB 403. 238%, 2) _ BUSS. Woy lio)= EE (z-40 ~ e400 + 4 4o*) = eL Wer = Wee * for 4 tadew'le)= Ee (4x74 lx) =\)! = 4 z= ue ple) =w'(L) EC guel) oe ° &. AWSO , Wore Er (ee4a 6) + er 4 = 2-4 = 5760-2-4 = AAS2O eL Woe Be dh 1) = 23213336 eL =) B= BABSO*AMSZO — 1408, 8RN 33333 Aufgae 2 I= z “06 0,8? = 0,0256 m4 S = 0,45: 0,6: (0,4 - 0,025) =0,02325m* Mp= 400-3-20-3-1S= 240 kNm Vp= 400 - 20:°3= YORN OF ee “(0,4- 0,48) = 2 050 RN/m = 2,05 Nimmn* — 4O: 002925. Ue = Cy0256- 016 = 746.2 kN/m* = Hor? AOCOY = 06,0362 Niro Aufe gabe 3 F, = 1200 kw ZMea=OF Fo =7F2=0 My => F225 + & Mes ~1200 , fz, + My AS =0 45° 45 ~ = UPA On = = 1200 4 AS oO ot Be az.gA (-A200- 225+ Fy A,25) Fy= #32 RN Spammongsn linia, y durch My > bei Ha nicht toga > Ha=0 “rE Aufoobe 4 0) AK berechnen a ZMp=O= -A-C+ 20-6°3° 9-6 A= 20kN ZF) =0= 20-20°6-4 +B B=4O RN 210 4o b) OAS = 5° ey V = 20kN 4o c) 0,45=2,5° 15-(20#x) X= 6,64 = Fem a) v(x)= 20- x: x-3 =20- Sy? -30 = 20- 5x? Ks SUF Rereich: 5,4FF 2x56 e) T= 6-45-2399 = 24604cm" $= AS-#-AO = A050em? Ve = AEBS = 4907 RU /em 2 M703 = Ww 5 Vy 4:27 RN  a) 70 20 NX A M + 45 b) EI=A M(x)= - 30 + 30%- 5x? w'lx)= 5x? - 30x +20 wile) = $y - 15x74 30% Cy why) = Bx - SHS eK tle wlo)=O =7c,=0 wl6)=O =7c,=0 =? w(x) =< (éx' - 5+ sx’) c) EIwl3)= 33,45 dl) tasEtw!(0)= 0 Yor ELw'lé)= 0 e) 40 0 NE Klausur 17 Aufgabe 1 AK berechnen : [Ma = 0= B. 10 - 10 . 14,2 . 7 , 1 B= 100,82kN A= 41 , 18KN al 88, 2 - M + 84 , 8 M. 100 b) = 5 . 30 . 602 = 0 ,4 M= 72 kNM Bereich 1 : 0x110 M(x) = 41 , 18x - Sx2 =72 X1= 2 , 519 X + = 5 ,717 M(X) = 41, 18x - Sx2 = - 72 Xz = 9 , 718 Bereich 2 : 02X2[4 , 2 M(x)= - 5x2 = 724 M(x)+ 5x2 = -72 xy = 3,79 (M 72kNm = 2,519x 5, 717 -> 9 ,718 <X 10 , 41 Aufgabe Aufgolbe2 a) A= 20-307 908cm* OM 900 = 2-30'S5 + 20-x x =3emn b) Ig= i: 30-30° = 63500cm" Ty = 230 -20°- 2-43,5-2.0) = 49500em!4 Fy. 4080 - 6,333 Iq 678% > Verringerung um 26,77. 245% oye tT AaSta c) T2457 7VE TSS Vay = 3° T 30-30 = 6adem". & +S - t:r'b tHE 7 EE Tz 49500 cm" $= 5°30-22,5 + A0-3-5 = 2025em? 7 Flachen aufteilen Vz 249500 -3 = #3,33em?- t 205 Vr _ 43:33 . 0, 4222 Vo. 600 2 Verringesung om 8448 7/- Aufgate 3 a) ate Seb AA, a4= A 2, 43 = 0-0,6 Fy, Fy = SE“ = 0,3333EA4 2 Fasfy = MEE = O24EAA 2Fy=0= -50-50 + 2- O;2YEAa- 06 + 2- 0,3333 EAa A= 0435 EA =? Fy =Fy= 34,92 RN o> Foe F225, 44 RN b) 3422 + AGA + Aufgolbe 8 {Ld a”.”~(“—~CM ME 3 A/2 8 - Squ" a) Yrmox™ seyet welt 2a25=2,04 => Errahung von w om AO0F7- 4 = & £_ Bel? b) oF Minox 9-3 + PG = Ze ov? 2 A227 = 044 > Erhéhung von g om 47 4 c) w oa d= 4,2" aA = 12 => Erhahung von d om 207 2 d) ons d?= 4,2" d= A129 => Errah ng von d om 12,597 e) ong ) d= 4,20 o~ aan > 0,93333 => Vervingprong von o UM A6\/- f) wes d) d= ARIA Aa? _ iy = m vy = 1,2 > Erne von w om t ’ ung 27.57 Klausur 18 Aufgabe 1 a) AL & > B ↑98 EFy= 0= 90 - A. 0 ,6 A= 150 KN Druck EFx= 0 = -150 . 0, 8 + B B = 120kN Zug 0 ErA [Fy = 0 = -40 + 150 . 0.6 - C - 0 , 6 2 = 83 , 33 kN Zug EFx = 0 = 150 . 0. 8 + 83 , 33 . 0, 8 - D D = 186 , 67 RN Druck => max Fzug= 120KN => max FDruck= 186, 7 RN 186 , 7 b) 10= π . r2 =>V = 2 , 4 CM 120 20 = Tr2 => V= 1 , 38 CM ca = A Al = 120. 200. 4 = 0 , 25cm 21000 .T . 2, 42 Aufgabe Aufgabez 51 = 840kN Z 1 = 4 a) AL = 84020 . 400 = 0,8 cm 0 , 8 b) sp = sin (30% AP = 1/ 6 CM 2 12 = sin (15%. 1 , 6 = 0, 414 CM 13 = 0 14 = 0, 414CM A 5 = 0 , 8cm S2 = 01414 . 21000 . 20 = 434 , 7kN Zug +Streckung400 Sy = 434, 7RN Druck Sz = 0 55 = 0, 8 · 21000. 20 = 840 KN Druck-Stauchung 400 d) EFy = 0 = - F + 840 . sin (30) . 2 + 434 , 7. sin (159) . 2 F= 1065kN Aufgabe 8 Y = V5 I , b Ronstant M= 93.h = 9 M(x) = * . x . E . -x q= 4 = Ex= h I qx3 64 IV(x) = Ex . x E I &x 2 V(52)= (L) = 8 V(54) = 4 . (5) = 4 M(5L)=, M(EL) = 4922 81 a) VB=Va Sa = b - Eh . Th = S1 = bn => 8 Sa =be SB = b. E h . h = SB= bh2 => Sp= 2 ↳ 2 D 85A = SB SB = ESA EB= b) VB= -97 V= Eq VB = EVc PVB =V Sc = SB - = 4 Aufgabe 3 er=4 Mix)= 4 wi(x)=-4 wilx) = -Mx + C4 w (x)= - SMX CA¥+ Co W(0)=O =?¢2=O w(8)= O =7¢,= 4M => wl_)=- 5 My? 4xM wy) =-SM-4244-4M =3N wl-a)= - 84 -@H=-BN-07-4o a= A,653m Klausur 21 Aufgabe 1 A = = v(X) - O A(X) r(x) = mx+b m= 0 ,24-0, 50 = - 0 , 04 8 - 0 r(x) = - 0 ,04x + 0, 56 A(x) = π . (E)- 0 , 04x + 0,56))2 800 1200 = 0, 36cm a = 20 . Sπ . (1- 0,04x + 56)/2 O Aufgabez Feder A = F2 F = 11 . R 11= Fz = 2A . k 6 F = 5 A1 . k Fy = 511 . k Fs = E11 . R F = 11 . R [Mo = 0= 4600 - RA . (6 + E . 3 + 5 - 1) . 2 RA1 = 300 Fz= E . 300 = 150kN 300 - 1 = 10000 = 0,03m y = tan - 10, 3 = 0 , 286 Aufgabe 11 = Ay . 0 , 8 + AX . 0 , 6 = 7 , 5 12= 1y . 08 - AX . 0 , 6 = -3 => AX= 8 ,75mm Ay = 2 , 81mm Aufgabes Aufgabe i = E : b E =E V(X) = 94 . x . E = 9x V(L)= # man z L V(L)= bl 2 SB= b . E . = - 2 Sa = b . 4 . 3 = 30h32 TA E - 32 = 27 [B I 1 · 4 18 Aufgabe 9 a) WIE) = 0 b) AK berechnen : EMB= 0 = A . L - M-ZM -M A = 4M L M M - - M t + M e C EI = 1 M(x)= M - 4x w"(X) = - M+ 1MX L M(x) = - x + M EI = 1 w"(x) = - M +x W(x) = - Mx + 2 Ex3 +4 w(x) =- X+Ex + (1X + 22 w(0) = 0 => c= 0 w(E) = 0 =>1 =MML w(x) = E . (3X- Ex + EX7) w()= I w(π) = 0 Avfgace AO AAC = AA r 4 a) Wmax ~ ct A Age = 0683 =? Verringerung, 3A 7 A b) Cmox= Wo? Omax™~ a ( ded 22 bore sich rows) A wae = 734 =? Verringerung 24,97 u Co —_— ) Wma, 7a 3 be LY ay Lye WIL? i ly =4f%L =) Erhshung 43,67- L a) omax= = L : wt = key Lys APL al Lya=4334L => Echsheng 13, A+. 2) Omox= ‘ad = 0,444 =? Verringerung 24,4 7 f) ware 334% = 4,644 => €Erhdhung 64,17 Avfgabe | A nyla)= Be (Sx?- GLE 424 do A. 93 _ 243,333 - AO Yea- er 24 = er wile) = A+ (Sx? x45) mM Your? ax (3g 07-0456) = 2M €L wy id= fe. (-x4+QL) forma = eq (0428) = 4M € Aufgolbe 5 a) AK berecinnen BFy=O= - 240+ 95,5-2428 A=24 ARN b) Mg= 2411 -F- 40-7-3,5=-763 RIM > Hox Mgg = 244° 12+ 95,9-5-10-12- 6 =48,F bm > Mar, c) 60*=307+h" h = 54,96cm = 233806 _ 3 Wom a sage 35" y = 233806 = 6350cm> % -5A%6 max o> B&4CS =0.565 RMVer* Lin & coer) max og = WHA? = 07574 Ru lem? (in Feldmitte 2 unter ) +50 Aufgolbe 6 A= 40: 60= 2400 cm* T=" 10-60°= 720000cm* Th Onz = ig 40°60 = 320000em" _ 2400 4+ 120-400 2 s0 2400 720000 GTR Uefert: 2 AOem % _ 2400 , 880-460 . 2400 * B20000 YY =O GTR Uefert: y= 3, Hom P, (3.6410) B(0140) as 10 = 245 364 p=40 B= 175ytA0 Aufgatce ¢ AK berechnen ZH =O= B:8-6:8-9°5°8 B= Bev Vila -8 4 $e-x = -3t3x? T= 4: 46?-24 = 8422 em’ = 9 -214-4= 468 cm? VXI=O =? 4,613 8 =~ 01000768 Fy 9 3.2 006008192 ds’ 84 gx')dx =234 RN 2, Az T= 3 [a2-423- 3-B-4)- 2: 9 -(45-6)*-2- 8 - (40,5-6)” = 9F2em! T= sce 8 = 4,39 kN /emn™ Aufgabe 9 al e M + M b) w(x = E) = 0 c) M(x) = M EI= 1 w"(X)= - M w((x) = - Mx +21 w(x) = - Ex2 +crX + Cz w(0) = 0 = (y = 0 w(E) = 0 => 1 = ELM => w(x)=7 X+ 4xM) I w(x=E)= + Aufgabe 10 F= 100 X =18 1 = 18 w(10) = 9213t a = Bb = F= 150 X = 8 W(8)= 149337 L = 18 a = 12 b= 6 F= B x = 8 w(8)= 118,52B4 1 = 18 a = 8 b= 10 0 = - 118, 52B + 9213 + 14933 B = 9213 + 14933 = 203 , 7 118,52 c = 37,2kN Avfgabe 3 Tox = 215 RW/cm? Ta -V:S 5=.= 38° Tp 54> 56-3 3-3: 45-4: A- 015-2 =48,Sern5 6.7 2-3:45= Tem’ by = 3cm ba= den Ty BGG - 3-822-2- 4.-23) - 3-(6- 418)". 2 = 340,67 em" 3ig= V4RS => v= 73,75RN 340, 643 35-22 => v= 88/32 RN 340, 64°72, Vyue™ F378 27 A 340.6% Gq = AS RNicem je Schweifnont Aufgave 4 a) AK berechnery ZFy= O= GO-AG-AO+ Ay Ay= 400 RN €Uq=0= Mp + 60 -AO~- 46-40-S Ha= 200 Rum Umax = 442,5R0m b) Mx)= - 200+ A00x - Bx? er=4 W"'(x) = 200- AOO« + 8x” wilx) = 200x- Sor +!x? +, w (x) = A0b1*~ $2x° + Ex" + CAKxtCe wlo)=O0 %,=0 w'lo)=0) »c,=0 a A w(40) = (400: 40% - $2.40? + 5 AO") - ee =O ° * SAm x Wmox= Bb6eNm? er Aufgobe 5 a) dA= dx: (0,6- 06x?) “ Tr Jx?- 0:6-(A- x?) dx =0,46m" 4 ») Aso =H ole Wéhe~ 4 o a Ww 424 AB hat ha = A2a> =? Vergréperung, 22,447- Klausur 19 Aufgabe 1 - Op = 10 + M . 10 = 15 = M= 4, 64kNm 44 Op = 120 t M+ 100 = 15 => M = 16, 15kNM 27 153 + 0z = - 120 + 4.. 100 = 25 => M . = 45 , 05kNm 27 153 -z = - 120 + M . 100 = 25 = M + = 12, 96kNmt 27 44 - 4 , 64 kNm > M 12, 96 KNM Aufgabez F = 4 . (AcEc + AgEs) -g = E . Es 42 = E . 21000 E = 0 , 002 Fc = EF => F = E . AcEc = 0 , 002 . 3000 . (5000 - As) Fs = 4 F => Fs = E . As Es = 0 , 002 . 21000.As 0, 002 . 21000.Asi = 5 = 0 , 002 . 3000 . (5000 - As) => As= 227cm2 => Fc= 28638kN => F = 38180kN Fs =A .As Aufgabe Aufgolse a) AK berechnen BFy =O= -270 + £06,75-2 424 A=28,25 RN b) Mg= 28:25-8- 40-8-4=-34 RDM > Umax. oben 2g, oNIeN ‘Droce Mg = 23:25: 43,5 +406,45- 5,5 — A0- 13,5 6,45 =543 RUM > Umax+ coer Drvde, unin 2vg c) 50*=257+h™ h=43,3cm Wo= Aes = F8A2cm> 4.43; 3 Wy = 442754 2 3906em? 2% 45.3 QU-AOO _ 244 RNiceme Lin G oA) BI06 MOx O2 = sage = AAF RNIcm? (in Feldmitte 2 unten) MX Sp= Klausur 1 Aufgabe 1 5.333 a) 11 = 6, 6669 = 0 ,81 1 2 = 0, 6 A N = AL . EA L Fr = 0,85EA = 0,121 E F = 01 61 . EA = 0 ,121 · EA 5 &EMA = 0 = - 200 . 2 + 0, 6 . 0 ,121 · EA . 4 + 0,8 . 0, 121 · EA . 4 0 = - 400 + 0, 6721EA 1 = 595 EA -> F = Fz = 71 , 4kN b) t 7 D 100 · - 100 10O N W M Aufgabe 2 a) NEMB =0 = C . 6 - 1 . 8 . 4 C = 5, 333kN [Fy = 0 = 5, 333 + Ay - 1 . 9 ,5 Ay = 4 , 167kN ↑(Ma=0 = Ma + 5,333 . 7,5 - 1 . 9 , 5 . 4 ,75 Ma = 5 , 128 kNM 4,167 5, 128 2 4 , 167 = 7, 5 I X 7, 5 - - - V - M x = 4 , 167 - - 3 ,333 Mmax = - 5,128 + 8, 68 = 3, 552 kNM b) A= 20 . 40 + 20 . 60 = 2000cm2 Es = 800 . 20 + 1200. 50 = 38cm 2000 Iy = (20 . 403 + 60 . 20% + 800 . (38 - 20)2 + 1200 . (38 - 50)2 = 578700cm" ↓ = 578700 = 26305 cm3 O 22 Wu = 57870 = 15228cm3 # O = W 5 , 128 . 100 of = = 0 , 01949kN/cm2 Punkt A oben 26305 3 , 552 . 100 Of = 15228 => 0 , 02333kN/m2 Feld unter Op = 5 , 128 . 100 = 0, 03367 kN/cm2 Punkt A unter 15228 2RN/m2 Ch 9zu) I 0 , 03367kN/cm2 = 59 , 4 KN/m 0,5RN/cm2 9Zu = = 21 , 4 RN/m 0 , 02333RN/cm2 9min (59 , 4 ; 21 , 4) -> 9zu) = 21 , 4 RN/m Aufgobe 5 ch Aa et _ 4c)? a) Ty= 4 7282 - £on-c2. (0 3x ) = Rect _ grt 8 on or F-&) = ouoeert X/ b+ L_,* _———— Sqt" Wena 33MEL oo) alter Wert = Never Wert StL Sql tren) 0.5 yereEL 3eueL t ' X= ‘ 48 =u" AT O84OOL =L, =7 Ly muss um AS,947% verringert werden bb) Sg". SglY = Qo. asy-E-b-hi- 2 32 Eph Agi" = tov 64 Eb? 4 Sql 64Ebh> = -32E bh a <= oy ahs 3y-2hn? wy => 32 40h? M + ha = 4,026 => h moss UM 42,67. erndht werden C) Pas resultierende Moment hat eine Neigong von 4:3 Chorizontal av verbal). Die maximolen Spannungen treten senkrecht daw aufen am Querschnitt auf, Hz {40?430? =50 kW/m - oO My = M- costet) tan’ *(33) 40 = S0-cost«) l= 26,97" l= 3687° > 90° 36,87°=53,43° -4 4O cos” * 35 = 36,84° Klausur 3 Aufgabe 1 a) 3 = 1 -> a=E + E= E = 0 , 01 . 0, 8 + 0, 012 - 0 1 6 = 0, 00405 32 + 2.252 E = 0 , 01 = 0 , 00333 E =0, 01 . 0, 6-0. 012 . 0 1 8 = -0 , 00720 5 Stabkräfte : Fr = 0 , 00405 . 30000 = 121 &N Z F = 0, 00333 . 30000 = 100 kN Z Fz= -0, 00720 . 30000 = 21 , 6 KN D Gleichgewicht : [Fx= 0 = H - 121 . 0 , 6 -21 , 6 . 0 . 8 H = 90, 24 KN [Fy = 0 = - V + 121 . 0 , 8 + 100 - 21 , 6 . 0 , 6 V = 184 , 32 kN b) y =ax+ bx +C y= 2ax +b Ebloloag -00os => b(x) = - 0 , 009x2 + 0, 18 x b(2 , 5) = 0 , 39375 + 3-2 . 0 , 39375 = 2 , 2125 b17 , 5) = 0 , 84375 -> 3 - 2 . 0 ,84375 = 1 , 3125 AL = 5 ↑ ( 2G + 80 ( = 0 , 008566 m = 8 , 566mm 30000 1 , 3125 2, 2125 Aufgaioe 2 AK berechnen: 17Z. Maz O= B-S- A405. 2,5 -600-5- 60-44 50-4 a) b) Br 950rW =A &o 600 r ry] sO D D 350 uw C= atW dlz A= 20-60= A200cm? W= 4: 20-60*= 42c00cm? Drudespannung = ee Oe = 2AF RNiem™ __52 4 23715-4060 F2= "M00 12.000 G, = 4.94 kNlem? Spmax= 2AF RNIcm? Bugsponnong So + BAS -AOO 50 Hmox™ 350:2,5-5 -200 = 237,5 kWm 0 be \re0 m0 [42> ar 23nS 350 (Coberhatls Punt A; auBen) (in Feldmitie AG; oben) Cams” Garo * azeoo ~ 2102 RN/cm? schubspannung + T= 45: 22S = 0.4395 RWlem* Aufgabe 5 5. 1 3 H2 F 64 g F T= E. = +3 I 16 BHifB I= 4 . B . H3 b = B v = 3 .# . (+ ) = B. 32 32 5 . 2 MY . z I F218 . # =3 Fz↑B . H3 oF F E . B - H3 z = H SMy = 0 = -E Ein0 = - MA = FL - 5. 3 V . == T · Konstante . S - S quadratisch - [ quadratisch-D 5 . 4 V . S T= I . b = Konstante . V -V Konstant -> [ konstant 5 . 5 O = My · z = Konstante . z E Linear -> o linear Iy 5. 6 Nur der Verlauf der Spannung 2 verändert sich , weil die Querkraft nicht mehr Konstant sondern linear verläuft . Klausur 2 Aufgabe 1 SO a) 30 40 D D b) Al= . (30+ 50) . 4. + 40 . 3-ea x = E (160 + 240) EAAL = 400 kNM Aufgabez F = 11 . R 11 = 12 Fz = 2A . k 6 - F = 5 A1 . k Fy = 511 . k Fs = E11 . R F = 11 . R [Mo = 0 = 9200 - A . (6 + E . 3 + 5 - 1) . 2R RA1 = 600 => F1=600RN Aufgabe al tan(0, 231 %= = DB = 0 , 00806M tan(0 , 2319= S = 0, 02016 -1 = 0 , 00806 . = 0, 01677 m 12 = 0, 02016. = 0, 01037 m Fr = 0. 01677Et - 100000 = 186 , 1 kN T3 Or ·Z0137EA . 100000 = 355, 7kN Fz = 134 b) [Fx= 0 = Ax - 186 , 1. - 355, 7 . F Ax = 408kN Aufgabe b(x) = ax+bx+ C b(0) = 0 = c= 0 b(0) = 0 =>b = 0 b(20) = 10 = a = To = D(x)= x A = 40 . 10-2. Fox' dx = 266, 67 cm2 0 Iy = 11429 - 266 , 67 . (0-6) : = 1829cm" Aufgabe Aufgolbe 5 uv ore re " xIl> els x n a Mix)= 3.x 3- HL4L) = 4: (Gu) 5 = a (80) i) 162 - abe at HiL) = 6b 6 a a. Soa = a 4. 2 = 43,5 oe ap = 462 Aufgatse 6 n HU) =215x-* 3° 9% = 3x3 “ee 1 2)5x=h ce 2oo H:AOS _ G2=~ 3400 tT ~200 6 H= Sktum Mix)=B =? x= 2,648 =? 4- 2,648 = 4;322m von unten Aufgolbe t A= 2: 40-30: 4 = 600cm* Ty =tH, -4O- 30° + 600 2(32)* =180000 em4 Ty =24,- 20°-60 + 2.600 +(22)* =80000 em" o= 30. 400 AS - AGO +490 . an 430000 80000 => 4,25 RNlom? = A215 RNlem? 2ug Klausur 4 Aufgabe 1 a) [Fy = 0 = 0 , 2 F + 0 . 852 - 0 ,651 - 0 , F = 0 , 852 - 0 ,65 - [Fy = 0 = - F + 0 , 652 + 0, 851 F = 0 , 6 52 + 0 ,851 si = -F - 452 -> - 0 , 2F = 0, 852 - 0 , 6 . (F - 452) 52 = 0 , 44 F S1 = 0 , 92F = 0, 92F = 20 -> F= 434 , 8kN 1 20 0 = 01 44F = 20 -> F = 454 , 5RN FzuL = min(434 , 8 ; 454 , 5) = 434 , 8 kN b) · Definiere AH und AV als Unbekannte · Drücke Längenänderung , damit die Dehnung und Stabkräfte als Funktion von Ap und Ar aus · Gleichgewicht in Knoten -> AH und Ar als Funktion von F · Einsetzen in Stabkräfte Weiter wie bei al Avfgobe 2 a) AK berechnen: IZM =O = - 820-7 450-55 4250-44 5-4-2 tAx-4 Ay = ZA,3RN = Bx Py =320 RN = By a 2N = O =N+t 0,6-234,3+ 08-320 =- 394,8 BN = 394,8RN D 35 Bs 2013 4o 12, AN EE VANS N aN IN M 35 35 3348 3948 b) oaths a DrwdRspannung : aaa 35-AOO _ . . Tomax = ans? aaeo 7 THES RN lom® (stitze ,auBen, unmiteloar unter Port @) Zogspo.nnung : Oz,max = O* Ace = 646 RN lem™ (Kragarm , oben, unmitielbar links von B) Avfgabe 3 a) AK berechnen: 2H ge O= B-AO~ 10-13-69 B= 84,5RN ZFy=O = B45- 20-1341 Az45,5 RN 45 os Hmoxr 4-455 °455 2103,5 RUM b) ET=A Bereich 43 Of x40 Bereich 2: O£%263 Mu) = 4515 x4~ 5a HUd= -5x2 Ws 5x42 = 45,5x.4 witlel= 5x2 w'lx) = Exp - SEEK, 4 c4 w'l)= $x? +c, wlx) = fat SEE Pa cxter Wind = Ext texte, W(O)=O =? cQ=0 WUIO)EO =? c= 344,67 W(3)=O =? 3eg4Cy= - 33,95 FO =? cy= 34d, -W'[xy210) = w'lxg=3) =? ¢g= 22467 =? Cys- 698,25 A we2(;5°0%4 224,67-0- 688,35) = 69875 4 er al Klausur 5 Aufgabe 1 al 10RN/m in ~ IWBq W - iIWBB "B 10 . / . 102 - 13. .H+W BG (10) = EI I 24750 ET- WBB = B.103 3EI O = WBq-WBB 24758 = 1000B EI 3EF B = 7 4 , 25RN EFy = 0 = A + 74 , 25 - 10 . 13 A = 55 , 75 102, 5 45 - - M Mmax = - 102 , 5 + 55,752 = 52 , 9RM + 52 ,9 X 5, 58m x Aufgabe 2 a) x = Yv(x) · EA(X) N1 = 100 Nz= 400 A(x) = 0, 5. . EXx . 2 x =E 100 +400 · E . EXX = (1131 + 1875). = 3006 E b) Af = 0 , 5 . E . F . 2 = 0 , 25 Az = 0 , 5 . 4 . 5 . 2 = 0 , 433 3= 100 . 2 + 40033 : 2 E . 0,25 = 2648 E Aufgobe 3 a= OLAz 4:0,2 wy a ALO=A ALC+ 4-0/6 Fo= 0,8A-R > 240RN Fy= 4a-k > 200kN Fe= O6A-R 9 “po kN ZFY=O= 0,8 AR-O8 + Ae + O/6Ak 04-600 07 2Ak-600 = 202 ~"R EMp_=O= 100-25 - 300 -2,5-240-0,8-S+A8O-0,6°5+M H= 920 RNM Aufgobe 4 a) AK berechnen : BMy= OF -20-24- 10,5 4-18 + [00-8 C= 8,06 RN ZFy=O= A+ F2,064 [00 - 10-24 A= 34.94 RN S454 34194 30 wet 4s BN X yo v aA A M &eob 4E, 06 5402 Four 2 b) A=4932em zor 40-86-4O- 2-11-10. ca 4992 55440) = 36,24em vu o 4 I-4 .20-80° - 27s +5600 «(Yo - 36,84) = 200T -(65~ 36,84) = 2528636 cm Whe 2829630 — cesea cm? ° 43,46 2829630 _ 3 = S25 = 6 FE38em Wu 36:84 “ op= rr =O,424 kNicem* oben; 4, 24m Unes von C oy = FANFWO = 0,403 eNlem? 68638 op =e 100 2 oO, YokNicem™ open; & 58 s8F ows e) TFT V=54,94bN b= 30cm oe $= 70-45-25, 66- 1 Aot- (H+ 28,16 ) = 29263em® v= muss = 0,087 ebfem™  250 2F:A00 -+ on = => F525 kN Os sag” saa 7h 2290 4 $5F-A0S 2735 =? F=4S2RN 8446 AASO Aufgoe 3 Fe AL=—=— dug postu «= U2 + Ft OM + + Hy 0/0045 - QD = 36 rad l= 4 Fpl - 4M ~ py 0,0020 - A000= 2 Fai Om +4, -Yy Ey= 400 - (-44 3,64 2) = 460pN Fp= 400: (-4~ 3,6) = - #60 kN Fp= 400 -(-44+3,6-2)7=-240 kN ZHag>O= Bf, +160-16-240-(6 => Pp =ACORN Zap = 249-40-€- 100 + 160-10 => Pa=H4ookN) ZF = -IG0+ #604210 ~ YOO-160-F, => Py=Z2PRN Aufgobe 4 ve Vg Vn = Atoscherbroft vez WS T AK berechnen E Maz O=- 400.264 C-10 C* £00k A~%00RN 300 v [=| Aco Schhubfluss * v=300 T= 83604 2-75 - 1,6°- 20 + 32° 0,8 - 16,6)" 432° ( 32,4- 16,6)" = 24354 cm" § = 416-20 - AS, 8 = 505, bern? V,,= 300-5056 _ LH3SA 612283 RN/em 2 6 ,2289= \Ws- 25 Vs = 47,86 RN Vgc = 120-0518 29 095 L43sa 10463 RNlcm 210763 = \s - = Vs = 54,94 RN Aufgabe 5 ↓F WBF IWAB ~ B F- 13 . 5 . 8(1 +5- 8) = # . 87,037 WBF = GEI B . 8 . 10 . 8 . (1 + 1 - )= . 118 , 518 WBB = GEI WBF = WBB F El 87, 037= 118 , 518 B = 0 ,7344F [MA= 0 = 0 , 7344 F . 8 - F - B + C - 18 c = 0, 3958F EFy = 0 = A = - 0 , 1302F 0 , 6042 F + - 0, 1302F ossof 1,042 F - t 1 , 979 F Aufgabe 6 3 Iy = 2 . 1 , 54 . a + 0 ,7592(5a - a) + 0, 75a2. (a+ 5a - a)2 = 9436 Iz = 4 · 10,75ala + 0, 75 a . (3. 0 ,73a-0,75a) + 0 , 75a2. (0 ,75 a + 5 - 0,75 a - 0 , 75 a) 2 36 = a