Question

Determine the displacement at point D. Use the principle of virtual work. El is constant. 60 kN/m А B 3 m 5m Sm

Answer 1

Moc = Rcx – 60 x Moc= 300% – 30x2 (0<x<5m) MAB= RC 2- 60X10 CX-5) ► 8 Mp 300% - 600(2-5) 103X313m) MBD = RCX-1CX-5) ► 0.5% (X-5) Đ MBO = 0.5% - (2-5) EMB=0 GO KNM С (RCX60) = 60X1045 ARC 3m 5m 50 х X X Rc=300KN M= Bending moment en due to actual loading. moment in the portion pc due to adual loading. Moc = Bending MBD = Rcd-602² (5 5X5 lom) + MBO- 3006- 30 x2 Y 1 с EMB=0 1 O ARC Smx sm RCXL0 = 1X5 smt RC = 0.5 ma Bending moment due to the unit luad at o. * moc = 0.5% moc = Rocx MABS RicX - QL-5) MAB= 0:56-66-5) Deflection ato 13 5 LO MABMAB dx. MBD MBD moc moc dr dxt EL t AD ED E2 LO

Mpc moc dro (300% -30x2)0.5% dx EP ET 150L - 150 zdz EI pisan to pasou (15042-1523) dr de 1992 (150x3 1524 니 3 15625 4EP 10. 10 (300W-3022) (0,5%-(2-5)) di MBO MB DdX ER СТ 10 (150X-1542) - (300XC2-5)) + 30% CX-5) dx ER 16 | [150X– 15x = 3000 + 1500% + 302?- 150x) dx 16 [-285x2-15000) dx 10 28523 1500X e to F28929 2 = -139375 СТ

13 13 MAB MAB dx (300% - 60062-5)) x (0.58 -62–5)] dre EL EL 10 13 (1500 – 300 (1252) – 3000x252) + 600 (7-5) dx တင် Tox+25-10% 10 13 (15022-300% +15000) db СТ 10 13 to [15073_300x2+15000% 10 1 94500 EL 15625 1393795 94500 + AD= HET CP CD ED - KN Xm4 m2 Ao = 163875 KM m KM m2 4 EP EI - KN-m² we indicates downward deflection.