Engineering Mechanics

Chapter 6 : Centre of Gravity „„„„„ 89 We know that distance between centre of gravity of the body and bottom of hemisphere D, 3 ...

(^90) „„„„„ A Textbook of Engineering Mechanics We know that distance between centre of gravity of the combined body from Py(), ...

Chapter 6 : Centre of Gravity „„„„„ 91 6.11. CENTRE OF GRAVITY OF SECTIONS WITH CUT OUT HOLES The centre of gravity of such a se ...

(^92) „„„„„ A Textbook of Engineering Mechanics and 1 120 300 2 200 56 mm 3300200 y ⎛⎞+× =×⎜⎟= ⎝⎠+ (ii) Semicircle 2 222 11 (90) ...

Chapter 6 : Centre of Gravity „„„„„ 93 3 40 40 60 mm 2 x =+ = and 3 4420 8.5 mm 33 r y × == = ππ We know that distance between c ...

(^94) „„„„„ A Textbook of Engineering Mechanics 626 ––– 566 563 rr rr rr ⎡⎤⎛⎞⎡ ⎤ =+=⎢⎥⎜⎟⎢⎥ ⎣⎦⎝⎠⎣ ⎦ππ 65 2 –– 56 3 5 rr rr ⎛⎞ ==⎜ ...

Chapter 6 : Centre of Gravity „„„„„ 95 Example 6.17. A frustum of a solid right circular cone has an axial hole of 50 cm diamete ...

(^96) „„„„„ A Textbook of Engineering Mechanics Example 6.18. A solid hemisphere of 20 mm radius supports a solid cone of the sa ...

Chapter 6 : Centre of Gravity „„„„„ 97 and distance between centre of gravity of the cut out cone and its apex, 3 3 0.75 4 h yh= ...

(^98) „„„„„ A Textbook of Engineering Mechanics Now we find that the answer obtained is more than the right hand side of equatio ...

Chapter 6 : Centre of Gravity „„„„„ 99 How many centres of gravity a body has? Describe the various methods of finding out the ...

(^100) „„„„„ A Textbook of Engineering Mechanics 100 Moment of Inertia 7 CHAPTERCHAPTERCHAPTERCHAPTERCHAPTER 7.1. INTRODUCTION W ...

Chapter 7 : Moment of Inertia „„„„„ 101 or second moment of mass. But all such second moments are broadly termed as moment of in ...

(^102) „„„„„ A Textbook of Engineering Mechanics This may also be obtained by Routh’s rule as discussed below : 3 XX= AS I ...( ...

Chapter 7 : Moment of Inertia „„„„„ 103 22 22 –– 22 ... dd xx dd Ibydybydy ++ ==∫∫ (^33332) 2 ( / 2) (– / 2) 33312 d d yddbd b ...

(^104) „„„„„ A Textbook of Engineering Mechanics Fig. 7.5. Circular section. Fig. 7.4. Theorem of perpendicular axis. Solution. ...

Chapter 7 : Moment of Inertia „„„„„ 105 4 44 0 2()() 42 32 r ZZ x Ird ⎡⎤ ππ =π⎢⎥= = ⎣⎦ ... substituting 2 d r ⎛⎞ ⎜⎟= ⎝⎠ We know ...

(^106) „„„„„ A Textbook of Engineering Mechanics Fig. 7.7. Theorem of parallel axis. Example 7.4. A hollow circular section has ...

Chapter 7 : Moment of Inertia „„„„„ 107 7.13. MOMENT OF INERTIA OF A TRIANGULAR SECTION Consider a triangular section ABC whose ...

(^108) „„„„„ A Textbook of Engineering Mechanics Fig. 7.10. Semicircular section ABC. Example. 7.5. An isosceles triangular sect ...