MACHINE ELEMENT DESIGN Knuckle Joint Knuckle joint is used to: 1- connect two rods whose axes either coincide or intersect and lie in one plane. 2- The knuckle joint is used to transmit axial tensile force. 3- The construction of this joint permits limited angular movement between rods, about the axis of the pin. This type of joint is popular in machines and structures. Typical applications of knuckle joints are as follows: • Joints between the tie bars in roof trusses. • Joints between the links of a suspension bridge. • Joints in valve mechanism of a reciprocating engine. • Fulcrum for the levers. • Joints between the links of a bicycle chain. Note A knuckle joint is unsuitable to connect two rotating shafts, which transmit torque D = diameter of each rod (mm) D1 = enlarged diameter of each rod (mm) d = diameter of knuckle pin (mm) d0 = outside diameter of eye or fork (mm) a = thickness of each eye of fork (mm) b = thickness of eye end of rod-B (mm) d1 = diameter of pin head (mm) x = distance of the center of fork radius R from the eye (mm) Figure 1 Knuckle Joint Prepared by: Engineer Mohamed Hassan Page | 1 MACHINE ELEMENT DESIGN Stresses on Knuckle Joint 1- Tensile Failure of Rods Each rod is subjected to a tensile force P. The tensile stress in the rod is given by, where 𝝈𝒕 is the permissible tensile stress for the rods. The enlarged diameter D1 of the rod near the joint is determined by the following empirical relationship, D1 = 1.1 D Fig. 1.2 Free Body Diagram of Forces 2- Shear Failure of Pin The pin is subjected to double shear as shown in Fig. 1.3. The area of each of the two planes that 𝜋 resist shear failure is ( 𝑑 2 ). Therefore, shear stress in the pin 4 is given by, where 𝜏 is the permissible shear stress for the pin. The standard proportion for the diameter of the pin is as follows, d=D Figure 1.3 Shear Failure of Pin 3-Failure of Pin in Eye When a cylindrical surface such as a pin is subjected to a force along its periphery, its projected area is taken into consideration to find out the stress. As shown in Fig. 1.4, the projected area of the cylindrical surface is (l * d) and the compressive stress is given by, 𝜎𝑐 = 𝑓𝑜𝑟𝑐𝑒 𝑃 = 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑒𝑑 𝑎𝑟𝑒𝑎 𝐿 ∗ 𝑑 As shown in Fig.1.1, the projected area of the pin in the eye is (bd) and the compressive stress between the pin and the eye is given by, 𝝈𝒄 = Figure 1.4 Projected Area of Cylindrical Surface 𝑷 𝒃𝒅 4-Crushing Failure of Pin in Fork As shown in Fig. 1.1, the total projected area of the pin in the fork is (2ad) and the compressive stress between the pin and the fork is given by, 𝝈𝒄 = Prepared by: Engineer 𝑷 𝟐𝒂𝒅 Mohamed Hassan Page | 2 MACHINE ELEMENT DESIGN 5-Tensile Failure of Eye Section XX shown in Fig. 1.5(a) is the weakest section of the eye. The area of this section is given by, area = 𝒃 (𝒅𝒐 – 𝒅) The tensile stress at section XX is given by, 𝝈𝒕 = 𝑷 𝑷 = 𝒂𝒓𝒆𝒂 𝒃(𝒅𝒐 − 𝒅) 6-Shear Failure of Eye The eye is subjected to double shear as shown in Fig. 4.23(b). The area of each of the two planes resisting the 𝒃(𝒅 −𝒅) 𝒐 shear failure is approximately. 𝟐 Therefore, shear stress is given by, 𝝉= 𝑷 = 𝒂𝒓𝒆𝒂 𝑷 𝒃( 𝒅𝒐 − 𝒅) 𝟐( ) 𝟐 𝑷 𝝉= 𝒃(𝒅𝒐 − 𝒅) Figure 1.5 (a) Tensile Failure of Eye (b) Shear Failure of Eye Standard proportion for outside diameter of the eye or the fork is given by the following relationship 𝒅𝒐 = 𝟐𝒅 7-Tensile Failure of Fork: Fork is a double eye and as such, Fig. 1.5 is applicable to a fork except for dimension b which can be modified as 2a in case of a fork. The area of the weakest section resisting tensile failure is given by area = 2𝑎 (𝑑𝑜 – 𝑑) Tensile stress in the fork is given by 𝝈𝒕 = 𝑷 𝑷 = 𝒂𝒓𝒆𝒂 𝟐𝒂(𝒅𝒐 − 𝒅) 8-Shear Failure of Fork Each of the two parts of the fork is subjected to double shear. 𝝉= 𝑷 𝟐𝒂(𝒅𝒐 − 𝒅) Standard proportions for the dimensions a and b are as follows, a = 0.75 D b = 1.25 D The diameter of the pinhead is taken as, d1 = 1.5 d The gap x shown in Fig. 1.1 is usually taken as 10 mm. then x = 10 mm Prepared by: Engineer Mohamed Hassan Page | 3
