Question 2: F Hole (d₁) 1.5d 1T Cylinder flange Figure Q1 [40 marks] It is required to design a helical compression/spring for the valve mechanism. The axial force acting on the spring is 300N when the valve is open and 150N when the valve is closed. The length of the spring is ° 30mm when the valve is open and 35mm when the valve is closed as shown in FigureQ2. The spring is squared and ground ends and the ultimate tensile strength is 1370 N/mm². The permissible shear stress (1) for spring wire should be taken as 30% of the ultimate tensile strength. The modulus of rigidity (G) is 81370N/mm². The spring is to be fitted over a valve rod of 20mm diameter and if 150 N 35 30 the wire diameter (d) of the spring is 4mm.) Closed position 30 2-1370 x 100 300 N (b) Open position Figure Q2 12 6= 1.6 MPa 1) Tmux: 6. A Jaa (100×10) ⅱ) = THE 1600N A x P Motor pulley Diameter 500 mm Speed 700 rpm Coefficient of friction=0.3 Angle of contact = 143.2 Using the data in Table Q-2, ité x100x10 = 1 Kg/m - (10×15³× Calculate, (a) Mean coil diameter (D). (b) Outside diameter (Do). (c) Number of active coils (N). (d) The spring rate (k). (e) Free length of the spring (L). (f) Maximum deflection of the spring (8). Question 3: Area D= Do- Ds Do An open belt 100 mm wide and 10 mm thick is transmitting powe 17.5 +444 N Spring 1444 8 - 8FD'N d'G 8FD'N G 1+ Formulae sheet 20 ? 8FC N G. d Dicxd 16666 C N P xassump Page 14 - 16 M.-[M + √M² +T]-x0,xd SFD xd 8EC K, C= K, - 4C-1 0.615 4C-4 C 5 M. [KM+ √(x, x M)² + (K, xT)²] P=X. Fr+ Y. Fa Dp Pe w= Do-20 FdLfad d-fl N₂ N₁ d₂ 16666 Z T₁ = m² k T1 = ене T Pring rate Spring Stifness d'G T₂ Ks 8D N Fd (fad-L d+fL 8DN J"G D<3 L T=√(KM)²+(K, ×T)² Solid length (Ls) T₂ Plain, ground ends. Ls = (Na+ 1) d Squared or closed Ls = (Na+3) d ends