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BTECH MECHANICAL PRINCIPLES AND APPLICATIONS Level 3 Unit 5 FORCES AS VECTORS Vectors have a magnitude (amount) and a direction. Forces are vectors FORCES AS VECTORS (2 FORCES) Forces F1 and F2 are in different directions They are NOT in equilibrium F1 F2 FORCES AS VECTORS (2 FORCES) 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) F2 F1 FORCES AS VECTORS (2 FORCES) If the two forces do not meet, the system is not in equilibrium F2 F1 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) This force is called the EQUILIBRANT F2 F1 FE FORCES AS VECTORS (2 FORCES) 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 F2 F1 FR FORCES AS VECTORS (3 FORCES) F3 F2 F3 FE F1 F2 F1 FORCES AS VECTORS (3 FORCES) F3 F2 F3 FR F1 F2 F1 F2 FORCES AS VECTORS ( 3 FORCES) ND 2 EXAMPLE 50o F1 F3 Equilibrant FE F1 Resultant F2 F3 FR F1 50o 50o F2 F3 Forces on a flat rectangular plate FORCES AS VECTORS Equilibrant F2 = 4N 40cm 60cm 50o F3 = 12N Resultant FE FR F1 10N 50o 50o 4N 12N F1 = 10N 4N 12N FINDING FE IDENTIFY THE DIRECTION AND MAGNITUDE OF THE FORCES THEN CONSTRUCT A VECTOR DIAGRAM F2 = 4N 60cm 40cm 50o F3 =12 N F1 = 10 N DRAW TO SCALE TO FIND THE MAGNITUDE AND DIRECTION OF FE (EQUILIBRANT) FE 10N 50o 4N 12N FE = 22N (Measured) DRAW IT IN THE OPPOSITE DIRECTION TO FIND THE MAGNITUDE AND DIRECTION OF RESULTANT FORCE FR 10N 50o 4N 12N FE = 22N (Measured) FINDING THE POSITION OF THE EQUILIBRANT (FE) 4N 60cm 22N 40cm F1 50o 4N 12N 12N 10N Put FE where you think it should be to balance the other forces Clockwise = 22N x X Anticlockwise = 4N x 40cm + 10N x 0 + 12N x 0 4N 60cm 10N 40cm x = 160Ncm X = 160 ÷ 22 = 7.27 cm 22N 12N The 10N and the 12N pass through the pivot A so the turning moment = 0 50o A 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 a) Calculate the magnitude and direction of the resultant force 2.6kN 1.4kN A 35o 4m 3m 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) 130o 1.4 kN Weight of plate = 200Kg x 9,81 = 1.96kN (acting from the centre of gravity of the uniform plate 2.6kN 1.4kN A 35o 4m 3m 130o 1.4 kN 1.96 kN VECTOR DIAGRAM WITH RESULTANT 2.2kN 1.4kN 1.96kN 2.6kN 1.4kN This shows a) the magnitude and direction of the resultant VECTOR DIAGRAM WITH EQUILIBRANT 2.2kN 1.4kN 1.96kN 2.6kN 1.4kN This shows b) the magnitude and direction of the equilibrant Resolve the diagonal forces 2.6kN and 1.4kN into vertical and horizontal components V1, H1 and V2 (H2 not needed) 2.2kN For explanation Click here 2.6kN x A 1.4kN V2 = 2.6xsin 35 = 1.49kN V1 = 1.4xsin40 = 0.90kN H1 1.4 x cos40 = 1.07kN V2 35o H2 not needed , it passes through A 3m 1.96 kN H1 40o 4m 1.4 kN V1 130 - 90 Resolve turning moments 1.49kN V2 2.2kN Clockwise 1.96 x 2 + V1 x 4 + 2.2 x X = Anticlockwise V2 x 4 + H1 x 3 1.4kN 2.6kN x A 35o H2 not needed , it passes through A 3m 1.96 kN H1 40o 4m 1.4 kN V1 0.9kN 1.07kN Resolve turning moments 1.49kN V2 2.2kN Clockwise = Anticlockwise 1.96 x 2 + 1.49 x 4 + 0.9 x 4 + 1.07 x 3 2.2 x X 7.52 + 2.2X = 9.17 2.2X = 9.17 – 7.52 2.2X = 1.65X = 1.65 ÷ 2.2 = 0.75m 1.4kN 2.6kN x A 35o H2 not needed , it passes through A 1.96 kN H1 40o 4m 1.4 kN V1 0.9kN 1.07kN RESOLVING FORCES 2000 Newtons Weight suspended by two ropes Draw the perpendicular Identify the angles between the forces A and B and the perpendicular A B 55o Draw the triangle using the angles 20o 35o 55o 70o A 105o 2000 N B 20o 2000 Newtons The length of the sides of the triangle represent the magnitude of the forces NOT the length of rope USING THE SINE RULE (IF YOU KNOW THE ANGLES) a/sin A = b/Sin B = c/sin C angle A = 20o angle B = 55o (opposites to sides a & b) 55o Angle C = 105o and side c represents 2000N a 105o 2000 N (c) b 20o USING THE SINE RULE (IF YOU KNOW THE ANGLES) a/sin A = c/sin C therefore a/sin 20o = 2000/sin105o a = 2000 x sin 20o/sin105o 55o 708.17N b/sin B = c/sin C therefore b/sin 55o = 2000/sin105o a 105o 2000 N (c) b a = 2000 x sin 55o/sin105o 1696.1N 20o USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES) F2 = 60N F3 70o F1 = 30N USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES) A2 = B2 + C2 -2BCcosA (F3)2 = 302 + 602 – 2x60x30x cos110o F2 = 60N (C) = 75.7N F3 (A) A =110o F1 = 30N (B) 70o VERTICAL AND HORIZONTAL COMPONENTS OF FORCES Sketch the diagram Fv can be drawn at the other end of the sketch Fv F θ FH VERTICAL AND HORIZONTAL COMPONENTS OF FORCES Sketch the diagram sin θ = Fv/F Fv F θ FH Fv F.sin θ = Fv cos θ = FH/F F.cos θ = FH back RESTORING FORCE OF TWO FORCES F3 is the restoring force of F1 and F2 F1(55N) F3 70o 25o Can be drawn to scale 74.8N F2 (25N) 25o 70o RESTORING FORCE OF TWO FORCES F1(55N) F3 is the restoring force of F1 and F2 F3 70o 25o F2 (25N) Can be solved by resolving the horizontal and vertical components of F1 and F2 RESTORING FORCE OF TWO FORCES F1(55N) F3 70o 25o F1v = F1.sin70o 55sin70o = 51.68N F2 (25N) F1h = F1.cos70o 55cos70o = 18.81N RESTORING FORCE OF TWO FORCES F1(55N) F3 70o 25o F2v = F2.sin25o 25sin25o = 10.57N F2 (25N) F2h = F2.cos25o 25cos25o = 22.66N RESTORING FORCE OF TWO FORCES F1(55N) F3 70o 25o F3v = F1v + F2v 51.68 +10.57 = 62.25N F2 (25N) F3h = F1h +F2h 18.81 + 22.66 = 41.47N RESTORING FORCE OF TWO FORCES F3 (F3)2 = 62.252 + 41.472 62.25N (F3)2 = 5594.82 F3 = 74.80N 41.47N RESULTANT OF TWO FORCES Tan θ = opposite/adjacent Tan θ = 62.25/41.47 F3 62.25N Tan θ = 1.5 θ = 56.33o θ 41.47N Direction of F3 = 180 + 56.33 = 236.33o MOMENTS OF FORCE 4m 2N 2m 4N Total Anticlockwise moments = Total Clockwise moments 8Nm 8Nm MOMENTS OF FORCE 3m 4m 2m 2N 2N 4N Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = 12 Nm MOMENTS OF FORCE 3m 4m 2m 2N 2N 4N Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = 12 Nm MOMENTS OF FORCE 3m 4m 2m 2N 2N 4N Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = 12 Nm FORCE ON A AND B 10m 1m 3m 4m 2m A 2N 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 4N B FORCE ON A AND B 4 N/m uniformly distributed load 10m 1m 4m A 2N 2.5m 2.5m 2N B 20 N (UDL) 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 UDL 4N/m x length 5m acting from centre of UDL B TECH Question example 6kN A 2m 4kN/m (uniform distributed load) 4kN 3m 5m B P2 Calculate the support reactions A and B for the simply supported beam in the diagram B TECH Question example solution A 6kN 2m 4kN/m (uniform distributed load) 4kN 5m 3m B 40kN Uniform load is 4kN/m. total UDL = 4 x 10 = 40kN ( acting from centre of beam) B TECH Question example solution A 6kN 2m Anticlockwise B x 10 kNm B = 232 ÷10 23.2kNm 4kN/m (uniform distributed load) 4kN 3m 40kN 5m Clockwise moments 6 x 2 = 12kNm 4 x 5 = 20 kNm 40 x 5 = 200 kNm total = 232kNm B B TECH Question example solution A 6kN 2m Total downward force 6 + 4 + 40 = 50kN 4kN/m (uniform distributed load) 4kN 40kN 5m B Total upward force A + B = 50kN A + 23.2 = 50kN A = 50 – 23.2 = 26.8kN B TECH Question example solution A 6kN 2m Check using B as the pivot 4kN/m (uniform distributed load) 4kN 3m 40kN 5m CW = A x10 ACW = 4 x 5 = 20kNm 40 x 5 = 200kNm 6 x 8 = 48kNm = 268kNm A = 26.8kN B TENSILE STRESS AND STRAIN Strain hardening Necking Stress Ultimate tensile strength Fracture Yield strength Y (Stress) X (Strain) Strain TENSILE STRESS (σ) CROSS SECTIONAL AREA ( πr2) FORCE FORCE Stress = Force ÷ Cross sectional area Force direction is perpendicular to cross sectional area TENSILE STRAIN Lo (original length) Increase in length ∆L Strain = ∆L ÷ Lo Young’s Modulus Stress ÷ Strain SHEAR STRESS (τ) Force is parallel to cross sectional area of shear Force Shear stress = Force ÷ cross sectional area of shear SHEAR STRAIN Force Shear strain = Change in length ÷ original length ∆l ÷ l SHEAR STRESS Force Shear Shear Modulus Stress ÷ Strain shear C B 20kN A 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 C B 20kN A 20kN Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr2 = π x .012 = 3.14 x 10-4 m2 . Maximum stress = 20x103 ÷ 3.14 x 10-4 = 6.4 x 107 N/m2 (Pa) Young’s modulus for steel = stress ÷ strain = 210GPa = 2.1 x10 11Pa strain = stress ÷ Young’s modulus Strain = 6.4 x 107 ÷ 2.1 x10 11 =3 x 10-4 m/m C B 20kN A 20kN Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr2 = π x .012 = 3.14 x 10-4 m2 . strain = elongation ÷ original length elongation =strain x original length = 3 x 10-4 m/m x 0.5 = 1.5x10-4 m = 0.15mm C B 20kN A 20kN Shear stress in pin Force ÷ area = 20kN ÷ cross sectional area of pin (π x .00752) = 20x103 ÷ 1.77 x10 -4m2 = 1.13 x108 Pa Shear modulus for brass = 7 x 1010 Pa. Strain = stress ÷ modulus strain = 1.13 x 108 ÷ 7 x 1010 = 0.0016 F = 8kN F 70o 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. Direct force F1 F = 8kN F 70o Shear force F2 8kN Sin70o = F1 ÷ 8kN 70o F2 F1 = 8kN x Sin70o F1 = 7.5 kN F1 Cos70o = F1 ÷ 8kN Cos70o F2 = 8kN x F2 = 2.7 kN F = 8kN F 70o Operational factor of safety = Tensile strength ÷ working stress Cross sectional area of the bolt = πr2 = π x (.006)2 = 1.13 x10-4 m2 F = 8kN F 70o Operational factor of safety = Tensile strength ÷ working stress Tensile stress = F ÷ area 7.5 x 103 ÷ 1.14x 10-4 6.6 x 107 Pa F = 8kN F 70o Operational factor of safety = Tensile strength ÷ working stress Shear stress = F ÷ area 2.7 x 103 ÷ 1.14x 10-4 2.4 x 107 Pa F = 8kN F 70o Operational factor of safety = Tensile strength ÷ working stress operational factor of safety in tension. 500 x 106 Pa ÷ 6.6 x 107 Pa = 7.6 F = 8kN F 70o Operational factor of safety = Tensile strength ÷ working stress operational factor of safety in shear. 300 x 106 Pa ÷ 2.4 x 107 Pa = 12.5