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Explain with the help of neat sketches three basic types of lever. State one application of each type.

In the first type of levers, the fulcrum is in between the load and effort. In this case, the effort arm is greater than load arm, therefore M.A. obtained is more than 1 Application: Bell crank levers used in railway signaling arrangement, rocker arm in I.C. Engines , handle of a hand pump, hand wheel of a punching press, beam of a balance, foot lever (any 1) In the second type of levers, the load is in between the fulcrum and effort. In this case, the effort arm is more than the load arm, therefore M.A. is more than 1.

A hollow shaft is required to transmit 50 kW power at 600 rpm. Calculate its inside and outside diameters if its ratio is 0.8. Consider yield strength of material as 380N/mm2 and factor of safety as 4.

Given : P= 50 KW = 50000W Speed = 600rpm k=Di/do = 0.8 σyt= 380 N/mm2 Factor of safety= 4 Design stress σt=σyt/fos =380/4 =95 Shear stress = τ =σt/2 = 95/2 =47.5N/mm2 Torque transmitted by hollow shaft T = P x 60/2πN T = 50000 x 60/2π x600 T = 795.67 N-m T= 795670 Nmm T= π/16 Xτ X do3 (1-k 4 ) 795670 =π/16 X 47.5 X do3 ( 1-0.84 ) Do3=144529.313 Do = 53 mm say 55 mm Di = 0.8X 55 = 44mm

State and explain main considerations in machine design.

Main considerations in machine design Type of loads and stresses caused by the load: the load on a machine component, may act in several ways, due to which, the internal stresses are set up. Mechanism: the successful operation of any machine depends largely upon the simplest arrangement of the parts, which will give desired motion Selection of material: designer should know the deep knowledge of properties of materials and behavior under working conditions

What are rolling contact bearings? State their advantages over sliding contact bearings.

Rolling contact bearing- contact between the surfaces is rolling ,it is antifriction bearing Advantages (any six) (1)low starting and running friction except at very high speed (2) ability to withstand momentary shock loads (3) accuracy of shaft alignment (4) low cost of maintenance (5) reliability of service (6) easy to mount and erect (7) cleanliness (8) small overall dimension

A bracket as shown in Figure No. 1 is fixed to a vertical steel column by means of five standard bolts.

Horizontal component of 45 KN, WH= 45Sin 600 =45x 0.866=38971N and vertical component of 45 KN, Wv,= 45xcos600 =45x0.5=22500N Direct tensile load in each bolt,Wt1= WH /5=38971/5=7794.20N Turning moment due to WH about G TH = WHx 25=38971x25=974275N (anticlockwise) direct shear load on each bolt =Ws=Wv/5 =22500/5=4500N Turning moment due to Wv about edge of the bracket, Tv= Wvx175=22500x175=3937500N-mm( clockwise( clockwise) Net turning moment =3937500-974275=2963225N---------(I) total moment of the load on the bolts @ th tilting edge = 2w x(L1) 2 + 2w x(L2) 2 =2xwx(50)2 + 2xwx(150)2 = 50000


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