Question

A two-story building is modeled as shown in the figure. The columns have an identical rigidity k1 = k2 = 3 and the masses of the two stages are also identical m1 = m2 = 1


m2 = 1 k2 = 3 U2 (t) m = 1 ki = 3 Uit) Figure 1


Assuming that a wind load F(t) = sin(t) is applied to each floor (Do not take into account the effect of the weights masses in the static state)

a) Write the two differential equations of each of the masses m1 and m2.

b) Assuming zero initial conditions for the two masses, and using the Laplace transform method, find the solutions U1(t) and U2(t) as a function of the constants of the partial fractions, i.e. do not not calculate the values ​​of the constants of the partial fractions.

Answer 1

clc;clear all;close all;
num=[1 0 9];
u=[1 0 1];
v=[1 0 9 0 9];
den=conv(u,v);
[r,p,k]=residue(num,den)

a) 314,-2) E m,-)f>u, D'Alembert's principle: 30 ZFx = make Free body diagram 3e, -364,-42) = mü; ü, +64,-342= sint D'Alembert's prisciple: m2 =1) 1>u Sfx = incik 3 ( 401) F- 3142-4,) = üz Freebody diagram üz +342-34, = first 4,10), (o) = 6210) = C210)=0. O=) Lşü } +62{41} -3.2{42} =L& sinky $yis) - sulo) - ulo) +6 (0U,()] 34,6)= 8²+1 => 18++b) uçs) - 30,(3) - 3+। ③ =) L{4}+3L442}-3444,3 zuşinty > (+3) U,(5) - 30,15) = si s

- (5) 30 - (د) , (مجی) ده - 3 (9) + (43) (5) کی ۔ رو 3 (53 =) = (s الا 9 - (د) (+) 3 ریپ (so(43) + دیں (بای) 3- ( د ده s+۱ مد 21ی - 3 و+ی + 3+1 واں (۹- (44) (3) (5) ل (1-18 + 3+، +52) د (2ء 4 2+۹ 2+1 S +42 + ۹) - (5) وحی s, =) . (5) 2ی +۹ partial fraction from the matlas .(s) (+15+) دریا را + F اوار0.00- زواا00 ،ہ اک802 .2 - ي اک2,802 +S ايها7 3۔ اها اکو07، + ي نی و ما 31 ایہہ. ۱- کہ ات | #S

r =

0.0000 - 0.0044i
0.0000 + 0.0044i
0.0000 + 3.7483i
0.0000 - 3.7483i
-0.0000 - 4.0000i
-0.0000 + 4.0000i

p =

0.0000 + 2.8025i
0.0000 - 2.8025i
-0.0000 + 1.0705i
-0.0000 - 1.0705i
-0.0000 + 1.0000i
-0.0000 - 1.0000i

k =

[]

>>

0.012331+0.012331 4.0125 +4.0125 > U2CS) = + s² +(2.80251)? s? +(1-0705) s²+1 = U2 (5) - 0.024662 8.02S 5+(2-80251) + s? + (1.0705) sti =) & 12) = 0.24662 sin(2-802512) 2.8025) 025 sin (1.0705 t) + 8 sine 1,0105 = u(t) = 0.0879996 sin(2.802516) 7.496467 sin(1.0705 t) tersins t =) Üg (t) = -0.69115 sin(2-802516) + 8.590728 Sin (1.0705€) -8 sin Ł

@=) 34, --sint +0.0819996 sin(2.802516) -954964967 sin (1.0705t) +8 sint. ③-) 34,= - sint 0.69115 sin (2.80251t) +8.590728 sin (1.0705t) -8 Sint- 3(0.088 sin (2.80251t)- 7-4965 sin(1.07656) + & sint ) U, - -0.3333 sink – 0.2 3038 Sin (2.802516) + 2.86365incl.010st) - 2.6667 sinta -0.088 sin(2.802516) +7-4965 sin (1.0705t) - 8 sint M = -0.31838 sin(2.802516) +10.3601 sindi-otosta -11 sint