A round flange bearing, as shown in Fig. 7.26(b), is fastened to the machine frame by means of four cap screws spaced equally on a 300 mm pitch circle diameter. The diameter of the flange is 400 mm. The external force P is 25 kN, which is located at a distance of 150 mm from the machine frame. There are two dowel pins to take shear load. The cap screws are relieved of all shear force. Determine the size of the cap screws, if the maximum permissible tensile stress in the cap screw is limited to 50 N/mm².

A round flange bearing, as shown in Fig. 7.26(b), is fastened to the machine frame by means of four cap screws spaced equally on a 300 mm pitch circle diameter. The diameter of the flange is 400 mm. The external force P is 25 kN, which is located at a distance of 150 mm from the machine frame. There are two dowel pins to take shear load. The cap screws are relieved of all shear force. Determine the size of the cap screws, if the maximum permissible tensile stress in the cap screw is limited to 50 N/mm².

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Gentlemen, please help to solve a,b,c as I faced confusing questions about screw jack. Conditions are: screw jack is lifting 3.2 tonnes, screw pitch is 12 mm, diameter of the screw is 5 mm. Coefficient of friction was not given but I think we will need it for total force calculation, lets take it 0.2. a) Calculate horizontal component to move the load - what is ment here by horizontal component? b) Calculate vertical component to move the load - what is ment here by vertical component? c) Total force needed to move the load with a lever of 0.82m d) Velocity ratio (VR=distance by effort/distance by load) Distance by effort (lever)=2 x 3.14 x 0.82 = 5.15m Distance by load = pitch= 12mm= 0.012 m VR=5.15/0.012= approx 429. e) Calculate efficiency. For this we need to have MA (mechanical advantage). It can be calculated when we know total force. Efficiency=Load (3200 x 9.8) / effort
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