Question

Q.1) For the beam shown below (E = 200 GPa, I = 216 x 100 mm*), find the deflection of point B (a) use discontinuity functions, (b) use superposition method. 80 kN 140 kN/m 50 kN/m 1.5 m 1.5 m 1.5 m 1.5 m

Please explain steps if possible. I want to understand the material. Thank you.

Answer 1

Solution E = 200 Goa = 2x105 l/mm2: I = 216 x 106mm 4. a) use. discontinuéty. Fimchion s na if usa tras - (n-a> = të iftca. ruokkim lokn 50 kNfm varianta 15m 15 m 1.5m *ve E fy=0 vatre = 80+50% (15715) = 230 kW. -0 EMQ -0 = 40 +80 x 105+50 X3X4'5 =V6 4620 ve = 655 KN. E Me=0= 40 - 80 x 4.5 80x3x1.5 tvaxt=0 va = 925 KN 6. Double integration Method Et on FM et -1407725x4 - 80 x (<2–1.59'-spxlen-35 Eray = 4500+ 125m2 x = 332 -50 (4-335379 €TM) = -Mox?: 1254°- 80 (E4 –10571 & 36

(<2-3394+ 0,*+62 at *=0 Y=o [6=0] at x=6 7 y=0 0 = - 140x8" x 425x63 - 90.445? - 50x34+.61x6 2 36 24 Deflection at B (x=1.5) - E725x153 x1.5 t -140x1.52 2 CI (9B) = 36 -6435 YB = -6435 32 EI € = 2x105 N/mm² = ax 10 x103 (163,2 kN/m2 = 2X108 kN/m² I = 216* tell momento = 216 x 16 x(103)4 mo4 = 2.16x joy m4 YB = -6435 32X2X108 x 2.168104 2 -193 20150 30720 YB = -40655 mm

0 он Supe рынкет, пъһы - Jeo N - -а 3m Ч 140 kN/m во д - а) с » " » С уха- зла») $ - ток 15 (2х- зx6xts is-) nox - 3LY + GLET 6х6 хє? . Т.: 2005 • ° CET gokn vs'я sh (= Em) - - . - Fь, - - - or Cosx48) X 4а1 БLET fo - cox +5 x 15 (6-5--) ТэЯ X9 soh - = ZEI

50 lenim (111) song E cfom tar vam VatVc = 50x3 = 150 KN 70 Emato 50 x 3x 4.5 = V(XG =Vc = 225 kN va = 150 - 225 KN Es done PM - (-23) €144 ? s(-3»):+G

EI (O). 57589 - 50 (64-30) '+62+(2. at 150 = 4:0 =2207 at xeo6 => y=0 0 = x63 - 50x34+986 la = -1575 Egyp = 1575x105 YA = 2205 SEI - 405 QEI -8775 - 32 EI 6435 32 EI 48 5-4.655mm.