Question

Apply the principle of virtual work to compute the position(s) of static equilibrium of the following systems.

3. Apply the principle of virtual work to compute the position(s) of static equilibrium of the following systems. The natural lengths of the springs ky and ky are hy and hz, respectively. The two uniform links have mass equal to M. k, A Funds ------------- tex t

Answer 1

ram UK U ALO tung suum E & Rp t n The system has one degue of frudom, because specifying the value of single coordinate e completely determine configration of systemes The forces due to moses m & M are active forces because they are doing work. And reaction at A & B ie, RAA RB are not actures force. forces due to sforma are also acture forces.

Work done when coordinates are increcesed possities infantesmall amounts de Pont C is moved down of Sy and Pt A & Bare mored sn. Then Mass Mof rigid link is moned by sy The principh of virtual work says, total work must add to zero TATIANA £ Kz (Sy)² - 1 kp (Su)- Iki (Sn)? + mg (Sy) + Mg (Sy) = 0 for this case - coso - G-hi smo. Hha And also taking Sy=SH & Sa-So

Se the equation becomes po - Kz (SH)? - 5K, ($9)" IK, (SQ)? ting (SH) + Mg (SH) - 0 - 0 Hh = L sine & 8-h,= L coso 8 To relate SH to se e using ordinary formulal from calculs of differential &H - Laoso do – 0 SQ - -L sino do - 3 Putting Eq" ② & 3 in ear 0 - IK₂ Le cas o d'o – IK, L ano do - KiL'sino d'o + ma loode + Mg Lcoso do=0 do K₂ L cos'o + 1 k, I sen²O) - mg L600 do - Mg Lioso do =0

L² Now Putting values of coso a sino in equ do [ K₂ L (6-) + 1 K, L2 (4) -hz ) ] • - mg LQ-hi) - Mg LQ-h.) = 0 d@[1K2(-hi)'+kı ( 15 - h.)] -doing (8-hi) + Mg (8-hol=0 solving this egn we will get value of o. it which syster will be at static equieren abrium.