BTECH MECHANICAL PRINCIPLES AND APPLICATIONS. Level 3 Unit 5

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1 BTECH MECHANICAL PRINCIPLES AND APPLICATIONS Level 3 Unit 5

2 FORCES AS VECTORS Vectors have a magnitude (amount) and a direction. Forces are vectors

3 FORCES AS VECTORS (2 FORCES) Forces F1 and F2 are in different F1 directions They are NOT in equilibrium F2

4 FORCES AS VECTORS (2 FORCES) The two forces can be drawn like this. (In the F2 correct direction and the lengths should be drawn to F1 scale to represent the magnitude of the forces)

5 FORCES AS VECTORS (2 FORCES) If the two forces do not F2 meet, the system is not in F1 equilibrium

6 FORCES AS VECTORS (2 FORCES) If a third force (FE) was added in the way shown the three would be in equilibrium (They are all joined up following each other, The force system is balanced) F1 F E F2 This force is called the EQUILIBRANT

7 FORCES AS VECTORS (2 FORCES) If the line joining the two forces is in the opposite direction to F2 the equilibrant it is the F1 RESULTANT of the two forces The forces area not in equilibrium and the resultant shows the direction and magnitude of the combination of the two forces FR

8 FORCES AS VECTORS (3 FORCES) F3 F2 FE F3 F1 F1 F2

9 FORCES AS VECTORS (3 FORCES) F3 F2 FR F3 F1 F1 F2

10 FORCES AS VECTORS ( 3 FORCES) 2 ND EXAMPLE F3 50 o F2 F1 Equilibrant FE F1 Resultant F3 F2 50 o FR 50 o F1 F3 F2

11 Forces on a flat F2 = 4N FORCES AS VECTORS rectangular plate 40cm 60cm 50 o F3 = 12N F1 = 10N Equilibrant Resultant FE F1 FR 10N 50 o 50 o 4N 4N 12N 12N

12 FINDING FE IDENTIFY THE DIRECTION AND MAGNITUDE OF THE FORCES THEN CONSTRUCT A VECTOR DIAGRAM F2 = 4N 60cm 40cm 50 o F1 = 10 N F3 =12 N

13 DRAW TO SCALE TO FIND THE MAGNITUDE AND DIRECTION OF FE (EQUILIBRANT) FE 50 o 10N 4N 12N FE = 22N (Measured)

14 DRAW IT IN THE OPPOSITE DIRECTION TO FIND THE MAGNITUDE AND DIRECTION OF RESULTANT FORCE FR 4N 50 o 10N 12N FE = 22N (Measured)

15 FINDING THE POSITION OF THE EQUILIBRANT (FE) 60cm 4N 22N F1 40cm 10N 12N 4N 12N 50 o

16 Put FE where you think it should be to balance the other forces Clockwise = Anticlockwise 4N 22N x X = 4N x 40cm + 10N x N x 0 60cm 40cm 10N = 160Ncm X = = 7.27 cm 22N 12N x 50 o A The 10N and the 12N pass through the pivot A so the turning moment = 0

17 B TECH Question example P1 The diagram shows a uniform rectangular plate supported in a vertical plane by forces acting at the three corners of the plate. The plate is 4m x 3m and has a mass of 200kg 2.6kN a) Calculate the magnitude and direction of the resultant force 1.4kN A 4m 35 o b) Show the magnitude and direction of the equilibrant force 3m 130 o c) Calculate the position of the resultant force with respect to the corner A (ie. Use A as a pivot) 1.4 kn

18 Weight of plate = 200Kg 1.4kN A 4m 35 o 2.6kN x 9,81 = 1.96kN 3m (acting from 130 o the centre of gravity of the 1.96 kn 1.4 kn uniform plate

19 VECTOR DIAGRAM WITH RESULTANT 2.2kN 1.4kN 1.96kN 2.6kN This shows a) the magnitude and direction of the resultant 1.4kN

20 VECTOR DIAGRAM WITH EQUILIBRANT 2.2kN 1.4kN 1.96kN 2.6kN This shows b) the magnitude and direction of the equilibrant 1.4kN

21 Resolve the diagonal forces 2.6kN and 1.4kN into vertical and horizontal components V1, H1 and V2 (H2 not needed) V2 = 2.6xsin 35 = 1.49kN V1 = 1.4xsin40 = 0.90kN H1 1.4 x cos40 = 1.07kN For explanation Click here 2.2kN A 1.4kN 3m x 1.96 kn 4m V2 V1 40 o 35 o 2.6kN 1.4 kn H2 not needed, it passes through A H

22 Resolve turning moments Clockwise = Anticlockwise 1.96 x 2 + V2 x 4 + V1 x 4 + H1 x 3 1.4kN 2.2kN 3m A x 1.49kN V2 35 o 2.6kN H2 not needed, it passes through A 2.2 x X 1.96 kn 4m 40 o H1 1.07kN 1.4 kn V1 0.9kN

23 Resolve turning moments Clockwise = Anticlockwise 1.96 x x x x x X X = X = X = 1.65X = = 0.75m 1.4kN 2.2kN A x 1.96 kn 4m 1.49kN V2 V1 0.9kN 35 o 40 o 2.6kN 1.4 kn H2 not needed, it passes through A H1 1.07kN

24 RESOLVING FORCES 2000 Newtons

25 Weight suspended by two ropes Draw the perpendicular Identify the angles between the forces A and B and the perpendicular A B Draw the triangle using the angles 35 o 55 o 20 o 70 o 55 o A 2000 N 105 o B The length of the sides of the triangle represent the magnitude of the forces NOT the length of rope 20 o 2000 Newtons

26 USING THE SINE RULE (IF YOU KNOW THE ANGLES) a/sin A = b/sin B = c/sin C angle A = 20 o angle B = 55 o (opposites to sides a & b) Angle C = 105 o and side c represents 2000N 2000 N (c) 55 o 20 o a 105 o b

27 USING THE SINE RULE (IF YOU KNOW THE ANGLES) a/sin A = c/sin C therefore a/sin 20 o = 2000/sin105 o a = 2000 x sin 20 o /sin105 o 55 o a N 105 o b/sin B = c/sin C therefore b/sin 55 o = 2000/sin105 o a = 2000 x sin 55 o /sin105 o 2000 N (c) 20 o b N

28 USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES) F3 F2 = 60N 70 o F1 = 30N

29 USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES) A 2 = B 2 + C 2-2BCcosA (F3) 2 = x60x30x cos110 o = 75.7N F3 (A) F2 = 60N (C) A =110 o 70 o F1 = 30N (B)

30 VERTICAL AND HORIZONTAL COMPONENTS OF FORCES Sketch the diagram F v F F v can be drawn at the other end of the sketch θ F H

31 VERTICAL AND HORIZONTAL COMPONENTS OF FORCES Sketch the diagram sin θ = F v /F F v F F v F.sin θ = F v θ F H cos θ = F H /F back F.cos θ = F H

32 RESTORING FORCE OF TWO FORCES F1(55N) F3 is the restoring force of F1 and F2 F3 Can be drawn to scale 25 o 25 o 70 o F2 (25N) 74.8N 70 o

33 RESTORING FORCE OF TWO FORCES F1(55N) F3 is the restoring force of F1 and F2 70 o F2 (25N) F3 Can be solved by resolving the horizontal and vertical components of F1 and F2 25 o

34 RESTORING FORCE OF TWO FORCES F1(55N) F3 F1v = F1.sin70 o 55sin70 o = 51.68N 25 o 70 o F2 (25N) F1h = F1.cos70 o 55cos70 o = 18.81N

35 RESTORING FORCE OF TWO FORCES F1(55N) F3 F2v = F2.sin25 o 25sin25 o = 10.57N 25 o 70 o F2 (25N) F2h = F2.cos25 o 25cos25 o = 22.66N

36 RESTORING FORCE OF TWO FORCES F1(55N) F3 F3v = F1v + F2v = 62.25N 25 o 70 o F2 (25N) F3h = F1h +F2h = 41.47N

37 RESTORING FORCE OF TWO FORCES F3 (F3) 2 = N (F3) 2 = N F3 = 74.80N

38 RESULTANT OF TWO FORCES Tan θ = opposite/adjacent F3 Tan θ = 62.25/41.47 θ 62.25N Tan θ = 1.5 θ = o 41.47N Direction of F3 = = o

39 MOMENTS OF FORCE 4m 2m 2N 4N Total Anticlockwise moments = Total Clockwise moments 8Nm 8Nm

40 4m MOMENTS OF FORCE 3m 2m 2N 2N Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = 12 Nm 4N

41 4m MOMENTS OF FORCE 3m 2m 2N 2N Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = 12 Nm 4N

42 4m MOMENTS OF FORCE 3m 2m 2N 2N Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = 12 Nm 4N

43 FORCE ON A AND B 10m A 1m 4m 3m 2m B 2N 2N Take A as the pivot 4N Anticlockwise ( B x 10) = (2 x1) + (2 x 5) + (4 x 8) = 44 Nm Force on B = = 4.4N Total downward force = 8 N Force on A = 3.6 N Check this out using B as the pivot

44 Check this out using B as the pivot FORCE ON A AND B 10m 4 N/m uniformly distributed load A 1m 4m 2.5m 2.5m B 2N Take A as the pivot 2N 20 N (UDL) Anticlockwise ( B x 10) = clockwise (2 x1) + (2 x 5) + (20 x 7.5) = 162 Nm Force on B = = 16.2 N Total downward force = 24 N Force on A = 7.8 N UDL 4N/m x length 5m acting from centre of UDL

45 B TECH Question example 6kN 4kN 4kN/m (uniform distributed load) A 2m 3m 5 m B P2 Calculate the support reactions A and B for the simply supported beam in the diagram

46 B TECH Question example solution 6kN 4kN 4kN/m (uniform distributed load) A 2m 3m 40kN 5 m B Uniform load is 4kN/m. total UDL = 4 x 10 = 40kN ( acting from centre of beam)

47 B TECH Question example solution 6kN 4kN 4kN/m (uniform distributed load) A 2m Anticlockwise B x 10 knm B = kNm 3m 40kN 5 m Clockwise moments 6 x 2 = 12kNm 4 x 5 = 20 knm 40 x 5 = 200 knm total = 232kNm B

48 B TECH Question example solution 6kN 4kN 4kN/m (uniform distributed load) A 2m Total downward force = 50kN 40kN 5 m Total upward force A + B = 50kN A = 50kN B A = = 26.8kN

49 B TECH Question example solution 6kN 4kN 4kN/m (uniform distributed load) A 2m Check using B as the pivot 3m 40kN 5 m CW = A x10 ACW = 4 x 5 = 20kNm 40 x 5 = 200kNm 6 x 8 = 48kNm = 268kNm A = 26.8kN B

50 TENSILE STRESS AND STRAIN Stress Strain hardening Necking Ultimate tensile strength Yield strength Fracture Y X (Stress) (Strain) Strain

51 TENSILE STRESS (σ) CROSS SECTIONAL AREA ( πr 2 ) FORCE FORCE Stress = Force Cross sectional area Force direction is perpendicular to cross sectional area

52 TENSILE STRAIN Lo (original length) Increase in length L Young s Modulus Strain = L Lo Stress Strain

53 SHEAR STRESS (τ) Force Force is parallel to cross sectional area of shear Shear stress = Force cross sectional area of shear

54 SHEAR STRAIN Force Shear strain = Change in length original length l l

55 SHEAR STRESS Force Shear Modulus Shear Stress shear Strain

56 C 20kN A B 20kN P3 The diagram shows a shackle joint subjected to a tensile load. The connecting rods A and B are made from steel and the pin c is made from brass. Young s modulus is 210 GPa for steel and 100GPa foe brass. The shear modulus for steel is 140GPa and the shear modulus for brass is 70GPa. The smallest diameter for the connecting rods A and B is 20mm and the diameter of the pin C is 15 mm. a) Calculate the maximum direct stress in the connecting rods b) Calculate the maximum direct strain in the connecting rods c) Calculate the change in length of a 500mm length of connecting rod. d) Calculate the shear stress in the pin e) Calculate the shear strain in the pin

57 C 20kN A B 20kN Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr 2 = π x.01 2 = 3.14 x 10-4 m 2. Maximum stress = 20x x 10-4 = 6.4 x 10 7 N/m 2 (Pa) Young s modulus for steel = stress strain = 210GPa = 2.1 x10 11 Pa strain = stress Young s modulus Strain = 6.4 x x10 11 =3 x 10-4 m/m

58 C 20kN A B 20kN Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr 2 = π x.01 2 = 3.14 x 10-4 m 2. strain = elongation original length elongation =strain x original length = 1.5x10-4 m = 0.15mm = 3 x 10-4 m/m x 0.5

59 C 20kN A B 20kN Shear stress in pin Force area = 20kN cross sectional area of pin (π x ) = 20x x10-4 m 2 = 1.13 x10 8 Pa Shear modulus for brass = 7 x Pa. Strain = stress modulus strain = 1.13 x x =

60 F F = 8kN 70 o In the diagram the diameter in the of the bolt shown for the angle is 12mm. It is made from a material with a tensile strength of 500MPa and a shear strength of 300 MPa a) Determine the operational factor of safety in tension. b) Determine the operational factor of safety in shear.

61 Direct force F1 F F = 8kN 70 o Shear force F2 F2 8kN Sin70 o = F1 8kN F1 = 8kN x Sin70 o 70 o F1 = 7.5 kn F1 Cos70 o = F1 8kN Cos70 o F2 = 8kN x F2 = 2.7 kn

62 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress Cross sectional area of the bolt = πr 2 = π x (.006) 2 = 1.13 x10-4 m 2

63 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress Tensile stress = F area 7.5 x x x 10 7 Pa

64 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress Shear stress = F area 2.7 x x x 10 7 Pa

65 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress operational factor of safety in tension. 500 x 10 6 Pa 6.6 x 10 7 Pa = 7.6

66 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress operational factor of safety in shear. 300 x 10 6 Pa 2.4 x 10 7 Pa = 12.5

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