Mechanical Engineering Principles

32 MECHANICAL ENGINEERING PRINCIPLES 70 N 60 N 80 N 80 ° 30 ° 60 ° Figure 3.17 60 N 80 N 70 N 30 ° a b c 0 Scale 0 20 40 60 N φ ...

FORCES ACTING AT A POINT 33 The space diagram is shown in Figure 3.19. Since the system is in equilibrium, the vector diagram mu ...

34 MECHANICAL ENGINEERING PRINCIPLES 15 kN 37 ° 28 ° Figure 3.22 Four coplanar forces acting on a body are such that it is in e ...

FORCES ACTING AT A POINT 35 (ii) Determine the horizontal and vertical compo- nents of the 20 N force, i.e. horizontal component ...

36 MECHANICAL ENGINEERING PRINCIPLES By Pythagoras’ theorem, r= √ 1. 5352 + 9. 5302 = 9. 653 , and by trigonometry, angle φ=tan− ...

FORCES ACTING AT A POINT 37 The following coplanar forces act at a point: forceA, 15 N acting horizontally to the right, force ...

38 MECHANICAL ENGINEERING PRINCIPLES Which of the following is not a vector quantity? (a) displacement (b) density (c) velocit ...

FORCES ACTING AT A POINT 39 A space diagram of a force system is shown in Figure 3.31. Which of the vec- tor diagrams in Figure ...

4 Forces in structures At the end of this chapter you should be able to: recognise a pin-jointed truss recognise a mechanism de ...

FORCES IN STRUCTURES 41 indeterminatepin-jointed trusses, nor can they be used to determine forces inmechanisms. Statically inde ...

42 MECHANICAL ENGINEERING PRINCIPLES 4.3 Graphical method In this case, the method described in Chapter 3 will be used to analys ...

FORCES IN STRUCTURES 43 that the membersBDandDAare in compression and are defined as struts. It should also be noted from Figure ...

44 MECHANICAL ENGINEERING PRINCIPLES A B E D C R 1 R 2 H 2 Joint 1 Joint^2 4 kN 30 ° 60 ° Figure 4.15 In this case, the spaces b ...

FORCES IN STRUCTURES 45 Problem 7. Determine the internal forces in the pin-jointed truss of Figure 4.20. 2 m 4 kN 3 kN 5 kN Joi ...

46 MECHANICAL ENGINEERING PRINCIPLES Now try the following exercise Exercise 20 Further problems on a graph- ical method Determi ...

FORCES IN STRUCTURES 47 Problem 8. Solve Problem 5, Figure 4.8 on page 42, by the method of joints. Firstly, assume all unknowns ...

48 MECHANICAL ENGINEERING PRINCIPLES 4 kN 30 ° 60 ° Joint 1 Joint 3 Joint 2 R 1 R 2 H 2 Figure 4.30 Next, we will isolate each j ...

FORCES IN STRUCTURES 49 Resolving forces vertically: F 2 sin 60°+R 2 = 0 i.e. R 2 =−F 2 sin 60° ( 4. 11 ) Substituting equation ...

50 MECHANICAL ENGINEERING PRINCIPLES F 3 F 4 R 2 = 6.25 kN 30 ° Figure 4.36 Resolving forces vertically: R 2 +F 4 sin 30°= 0 i.e ...

FORCES IN STRUCTURES 51 Resolving forces vertically: 0 = 3 +F 6 sin 30°+F 7 +F 8 sin 30° i.e. 0= 3 + 0. 5 F 6 +F 7 + 0. 5 F 8 ( ...