Mechanical Engineering Principles

52 MECHANICAL ENGINEERING PRINCIPLES 4.5 The method of sections (a mathematical method) In this method, an imaginary cut is made ...

FORCES IN STRUCTURES 53 i.e. F 2 = 9 .96 kN, F 5 =−4kN and F 6 =− 7 .5kN The above answers can be seen to be the same as those o ...

54 MECHANICAL ENGINEERING PRINCIPLES If the Young’s modulus is doubled in the members of a pin-jointed truss, and the external ...

FORCES IN STRUCTURES 55 Assignment 1 This assignment covers the material contained in Chapters 1 to 4. The marks for each questi ...

56 MECHANICAL ENGINEERING PRINCIPLES Determine the unknown internal forces in the pin-jointed truss of Figure A1.2. (12) R 1 R ...

5 Simply supported beams At the end of this chapter you should be able to: define a ‘moment’ of a force and state its unit calc ...

58 MECHANICAL ENGINEERING PRINCIPLES (b) Turning moment,M is 2100 N mm and the effective length d becomes 100 mm (see Figure 5.2 ...

SIMPLY SUPPORTED BEAMS 59 Hence, taking moments aboutPin Figure 5.3, F 2 ×b=the clockwise moment, and F 1 ×a=the anticlockwise, ...

60 MECHANICAL ENGINEERING PRINCIPLES i.e. force,F= 1340 20 =67 N (b) The clockwise moment is now due to a force of 23 N acting a ...

SIMPLY SUPPORTED BEAMS 61 40 N 15 mm 35 mm 25 N 60 N d RA Figure 5.9 5.3 Simply supported beams having point loads Asimply suppo ...

62 MECHANICAL ENGINEERING PRINCIPLES (b) For the beam to be in equilibrium, the forces acting upwards must be equal to the force ...

SIMPLY SUPPORTED BEAMS 63 1.2 m 12 kN 400 N 20 kN 1.3 m 1.5 m PQ RP RQ Figure 5.14 (a) At equilibrium, RP+RQ= 12 + 0. 4 + 20 = 3 ...

64 MECHANICAL ENGINEERING PRINCIPLES 12 mm 0.2 kN RA d 2.7 kN 0.4 kN 1.3 kN 10 mm 15 mm Figure 5.16 For the force system shown ...

SIMPLY SUPPORTED BEAMS 65 of a couple are in: N m, N mm, kN m, etc, unlike that of a force. The method of calculating reactions ...

66 MECHANICAL ENGINEERING PRINCIPLES Resolving forces vertically gives: RA+RB= 0 from which, RB=−RA=−2kN Problem 14. Determine t ...

SIMPLY SUPPORTED BEAMS 67 Figure 5.30 [RA= 0 ,RB=0] (e) 10 kN m 10 kN m 1 m 3 m 4 m RA RB Figure 5.30 Figure 5.31 [RA=7kN, R ...

68 MECHANICAL ENGINEERING PRINCIPLES 10 N 1 m 3 m 5 m B RA RB A 20 N Figure 5.34 The force acting at B (i.e. RB)in Figure 5.34 ...

6 Bending moment and shear force diagrams At the end of this chapter you should be able to: define a rigid-jointed framework de ...

70 MECHANICAL ENGINEERING PRINCIPLES (a) Positive shearing force (b) Negative shearing force FF FF Figure 6.2 Shearing forces sh ...

BENDING MOMENT AND SHEAR FORCE DIAGRAMS 71 For the present problem, to demonstrate the nature of bending moment and shearing for ...