Question

Consider the two-beam system below. The beams are pin jointed at B and simply supported at their other ends at the base of the system). A spring of stiffness, k, connects the two beams to prevent the system collapsing. The unloaded length of the spring is h/2. A load of magnitude Pis applied at point B. } a. Using the method of virtual work, find the value of that keeps the system in equilibrium with the given geometry shown in the picture. Assume the beams are rigid for this part (3 marks) b. If P = 1000 N and h = 1 m, determine the spring constant, k (state units of k). (1 mark) c. Consider the beam A-B: determine the component of the load perpendicular to the beam due to the spring (1 mark) d. Now consider the beams have a stiffness, El. Calculate the deflection of A-B at the spring joint (assume the deflection is negligible compared to extension of the spring such that the answer to part a is not changed due to the flexibility of the beam). (3 marks) e. If beams are rods with circular cross-section, determine the minimum diameter if the deflection must be less than 1 x 103 m. The beams are made of Steel with E = 200 GPa. I = TD/64. (1 mark) f. Determine the maximum axial stress in A-B using the flexure formula: o My/1 where M is the bending moment in the beam. (1 mark)

Answer 1

Unloaded length of the spring = 1/2 32 f8 = KAL k [ 1 case - 1/2 virtual displacement K . А A - L Since L cosu 3 2413 un кр = & cz LCase + 22 Caso 3 It case 10 ts Ar Differentiating: - SB² 10 z & Lease sol Apy SAD Lsme se 3 free-body diagram sa 2 2 -St sinoso 22 Cas dit 3 at initial position hat sindi 2 casei 41 ( 15 - equation & divided by equation il L sindi 3 tand; = 2 + casa 4 hr tandi 8 3 69.44 395 ha Lsmei h=0.936329 L

Virtual - work equation. Suzo Is Sko - Es ste PS IR zo (ason's k [ 4 core - 0.941) * (-7 sinoso) - (4 con 0 ) 6-st. a PC e Caseso) 20 771 s Qu's Caso -0.47 a) 6-4 sino) -Pcasero (46*** -0.477 m)-( (KD ] P= 1 4 case-oda sino 3 + ssino smo) ] Cance P= (IL ( 2/3 Cond- 0.97 ) Cose Claus (to-pro ) ) de (ero 8807 ) grup 12 . 4 KL tance heim (b) af P1000N ha 0.936 329 L w 8 9 0.5 Put the all the value in equation 1000 a 4 & KK 1068 & tance 3 ARKKI [c Cance -0.07

than In this question equilibrium position not given assume that equilibrium angle 0 -60° 1000 = 4 KK & 1.068 & tanto a 을 tlasbo -0.47 42 ] K= 2061.57 N/m spring sone Es z k[ 41 case - 0.5 = 2061.52 & 1.068 & (as 6o - us] fs A fo 437.05284 N for singo Perpenticulary force F = 378.498 N.

(d) If suppose beam AB as a simple supported than deflection of beam B В So=Pa (34² – 492) 48EL D So- P (3) ( 31-412 F =P 16 4862 So = P13 (23) Iso= 23 PL3 3 as 12 1296 I e so-14163m E = 200 G Pa Ie 1 D4 64 23 x 378. 498 x (1.0683 200 x 10 к тарч 2 64 1296 04 8.33 489796 157 m D= 0.03021sm. Dz 30.215 mm (minimum diameter 6= my 243 t I I {fo & 2 RA TRB RA F RB EMA Bean AB RBAL farb B = FRA 2 3

2 28L M 2F 3 & z 2x 378. 1984 1.068 289.83 N.m م ابرا g g 6 6= my M D/2 32M N = 32K 89,83 9103 IT 23 И 69 DI na (30.11523 6= 33.171 MPa (marimum stress)