Mechanical Engineering Principles

72 MECHANICAL ENGINEERING PRINCIPLES given by: M=RA×x− 6 (x− 3 )= 2. 4 x− 6 (x− 3 ) = 2. 4 x− 6 x+ 18 i.e. M= 18 − 3. 6 x(clockw ...

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 73 M M= 2.4x M=^18 − 3.6x 00 0 x A x= 0 x= 5 m F x F= 2.4 kN F=−3.6 kN + 6 kN − C B (a) ...

74 MECHANICAL ENGINEERING PRINCIPLES It is negative, because as the left of the section tends to slide downwards the right of th ...

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 75 0 0 (a) BM diagram (kN m) (b) SF diagram (kN) 0 x x M F A AD B B 0 E E D − 10 − 10 + ...

76 MECHANICAL ENGINEERING PRINCIPLES Bending moment (BM) Atx, M=15 kN m+1kNm×(x− 2 ) −30 kN m = 15 +x− 2 − 30 i.e. M=x− 17 (a st ...

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 77 Problem 5. Determine the bending moment and shearing force diagram for the cantilever ...

78 MECHANICAL ENGINEERING PRINCIPLES 1 m 1 m 3 kN C AB Figure 6.28 Figure 6.29 [see Figure 6.43(b) on page 82)] 2 m 1 m 4 kN C ...

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 79 downwards, and spread over the entire length of the cantilever. w= 10 kN/m AB x 5 m F ...

80 MECHANICAL ENGINEERING PRINCIPLES it is evident that: RA=RB ( 6. 30 ) Taking moments aboutBgives: Clockwise moments about B = ...

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 81 of the following simply supported beams; hence, plot the bending moment and shearing ...

82 MECHANICAL ENGINEERING PRINCIPLES Answers to Exercise 31 (page 77) A 00 00 CD 2 kN 2 kN −3 kN −3 kN 1 kN 3 kN B BMD SFD (c) C ...

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 83 Answers to Exercise 32 (pages 81) (a) A 0 BMD SFD 21 kN −21 kN 3.5 m 3.5 m 36.75 kN m ...

7 First and second moment of areas At the end of this chapter you should be able to: define a centroid define first moment of a ...

FIRST AND SECOND MOMENT OF AREAS 85 (ii) First moment of area of shaded strip about axisOY=(yδx)(x)=xyδx Total first moment of a ...

86 MECHANICAL ENGINEERING PRINCIPLES 7.5 Worked problems on centroids of simple shapes Problem 1. Show, by integration, that the ...

FIRST AND SECOND MOMENT OF AREAS 87 = 1 2 ∫ 3 0 16 x^2 dx 18 = 1 2 [ 16 x^3 3 ] 3 0 18 = 72 18 = 4 Hence the centroid lies at (2 ...

88 MECHANICAL ENGINEERING PRINCIPLES = 1 2 [ 25 x^3 3 − 10 x^4 4 + x^5 5 ] 5 0 125 6 = 1 2 ( 25 ( 125 ) 3 − 6250 4 + 625 ) 125 6 ...

FIRST AND SECOND MOMENT OF AREAS 89 theradius of gyrationof areaAabout the given axis. SinceAk^2 = ∑ ay^2 =I then the radius of ...

90 MECHANICAL ENGINEERING PRINCIPLES P G C x b P d G 2 d 2 dx Figure 7.9 From the parallel axis theorem IPP=IGG+(bd) ( d 2 ) 2 i ...

FIRST AND SECOND MOMENT OF AREAS 91 Table 7.1 Summary of standard results of the second moments of areas of regular sections Sha ...