Why are bushes of softer material inserted in the eyes of levers?

the forces acting on the boss of lever & the pin are equal & opposite .There is a relative motion between the pin & the lever and bearing pressure becomes design criteria. The projected area of the pin is d1 x l1therefore Reaction R= P (d1 x l1 ). A softer material like phosphorous bronze bush with 3 mm thick is fitted in eyes to reduce the friction. & bear a bearing pressure upto5 to 10 N/mm2. Bushes are cheaper and can be easily replaceable.

Design a knuckle joint to transmit 150 KN. The design stresses may be taken as 75 MPa in tension, 60 MPa in shear and 150 MPa in compression.

Design of knuckle joint: Step 1) Diameter of Rod: d : =? Consider tensile failure of Rod 1. P =σt x A , 150 x 103 = 75 x π/4 xd2 , d = 50.4 mm 52 mm ( say)

Using Imperial relations Diameter of Knuckle pin Outside

Cotter Joint

COTTER JOINT

Cotter Joint: " A cotter joint is temporary joint and used to connect two coaxial rods or bars which are subjected to axial tensile and or compressive forces."

It consist of

1) spigot : It is the male part of the joint , it has a rectangular slot for passing the cotter through it. Spigot has a collar which rests against the socket end.

2) socket :It is the female part of the joint, it also has a rectangular slot for passing the cotter through it. It has a circular hole in which spigot fits.

Fig.1Show of hacksaw The belt is assembled with tensoion

Explain various failures to be considered in designing a cotter joint along with the necessary sketches and strength equations.

It consist of 3 elements: i. Socket ii. Spigot iii. Cotter Where, d= End diameter of rod d1= Diameter of spigot/Inside diameter of socket d2= Diameter of spigot collar D1= Outer diameter of socket D2= Diameter of socket collar C=Thickness of socket collar t1= Thickness of spigot collar t= thickness of cotter b= Mean width of cotter a= Distance of end of slot to the end of spigot P= Axial tensile/compressive force σt , σc , τ= Permissible tensile,

Give two applications of knuckle joint.

(i) A knuckle joint is used to connect two rods which are under the action of tensile loads. However, if the joint is guided, the rods may support a compressive load. (ii) Its use may be found in the link of a cycle chain, tie rod joint of roof truss, valve rod joint with eccentric rod, pump rod joint, tension link in bridge structure and lever and rod connections of various types.

The crank of a slider crank mechanism rotates clockwise at a constant speed of 300 rpm. The crank is 150 mm and the connecting rod is 600 mm long. Determine : (i) linear velocity and acceleration of the mid-point of the connecting rod, and (ii) angular ve

Note: Since D is the midpoint of AB, therefore d' is also midpoint of vector b' a'. Join o' d'. The vector o' d' represents the acceleration of midpoint D of the connecting rod i.e. aD.

By measurement, we find that aD = vector o' d' = 117 m/s2

Given: Lift = 40 mm Rise = ¼ x 360 = 900 Fall =1/6 x 360 = 600 Dwell = 1/10 x 360 =360

In a slider crank mechanism, the length of crank OB and connecting rod AB are 125 mm and 500 mm respectively. The centre of gravity G of the connecting rod is 275 mm from the slider. The crank speed is 600 rpm clockwise. When the crank has turned 45 from

In a four bar chain ABCD, AD is fixed and is 150 mm long. The crank AB is 40 mm long and rotates at 120 r.p.m. clockwise, while the link CD = 80 mm oscillates about D. BC and AB are of equal length. Find the angular velocity of link CD when angle BAD = 60

Explain the Klein’s construction to determine velocity and acceleration of single slider crank mechanism

If ωAO is the angular velocity of the crank, then Linear velocity’s of the links is given byVAO = ωAO x AO, VAP = ωAO x AM, VPO = ωAO x MO Acceleration of the links is given bya r AO = ω 2 AO x AO, a r AP = ω 2 AO x AC, a t AP = ω 2 AO x CN, aPO = ω 2 AO x NO

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