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BTECH MECHANICAL PRINCIPLES AND APPLICATIONS Level 3 Unit 5.

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Presentation on theme: "BTECH MECHANICAL PRINCIPLES AND APPLICATIONS Level 3 Unit 5."— Presentation transcript:

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) F1 F2 Forces F1 and F2 are in different directions They are NOT in equilibrium

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

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

6 FORCES AS VECTORS (2 FORCES) F1 F2 F E 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) This force is called the EQUILIBRANT

7 FORCES AS VECTORS (2 FORCES) F1 F2 FR If the line joining the two forces is in the opposite direction to the equilibrant it is the 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

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

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

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

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

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

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

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

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

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

17 1.4kN 130 o 35 o A 2.6kN 1.4 kN 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 a) Calculate the magnitude and direction of the resultant force b) Show the magnitude and direction of the equilibrant force c) Calculate the position of the resultant force with respect to the corner A (ie. Use A as a pivot) B TECH Question example 4m 3m

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

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

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

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

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

23 1.4kN 35 o A 2.6kN 1.4 kN Clockwise = Anticlockwise 1.96 x x x x x X X = X = 9.17 – X = 1.65X = 1.65 ÷ 2.2 = 0.75m 1.96 kN x V2 H1 V1 H2 not needed, it passes through A 2.2kN 4m 40 o Resolve turning moments 0.9kN 1.49kN 1.07kN

24 RESOLVING FORCES 2000 Newtons

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

26 USING THE SINE RULE (IF YOU KNOW THE ANGLES) 20 o 55 o 105 o a b 2000 N (c) 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

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

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

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

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

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

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

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

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

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

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

37 RESTORING FORCE OF TWO FORCES F N 41.47N (F3) 2 = (F3) 2 = F3 = 74.80N

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

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

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

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

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

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

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

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

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

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

48 B A 4kN/m (uniform distributed load) 4kN 6kN Total upward force A + B = 50kN A = 50kN A = 50 – 23.2 = 26.8kN 5 m 40kN 2m B TECH Question example solution Total downward force = 50kN

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

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

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

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

53 SHEAR STRESS (τ) Force Shear stress = Force ÷ cross sectional area of shear Force is parallel to 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 20kN C B A 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 20kN C B A 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 m 2. Maximum stress = 20x10 3 ÷ 3.14 x = 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 10 7 ÷ 2.1 x10 11 =3 x m/m

58 20kN C B A 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 m 2. strain = elongation ÷ original length elongation =strain x original length = 3 x m/m x 0.5 = 1.5x10 -4 m = 0.15mm

59 20kN C B A Shear stress in pin Force ÷ area = 20kN ÷ cross sectional area of pin (π x ) = 20x10 3 ÷ 1.77 x10 -4 m 2 = 1.13 x10 8 Pa Shear modulus for brass = 7 x Pa. Strain = stress ÷ modulus strain = 1.13 x 10 8 ÷ 7 x =

60 70 o F = 8kN F 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 70 o F = 8kN F Direct force F1 Shear force F2 8kN F1 F2 70 o Sin70 o = F1 ÷ 8kN F1 = 8kN x Sin70 o F1 = 7.5 kN Cos70 o = F1 ÷ 8kN F2 = 8kN x Cos70 o F2 = 2.7 kN

62 70 o F = 8kN F 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 70 o F = 8kN F Operational factor of safety = Tensile strength ÷ working stress Tensile stress = F ÷ area 7.5 x 10 3 ÷ 1.14x x 10 7 Pa

64 70 o F = 8kN F Operational factor of safety = Tensile strength ÷ working stress Shear stress = F ÷ area 2.7 x 10 3 ÷ 1.14x x 10 7 Pa

65 70 o F = 8kN F 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 70 o F = 8kN F 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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