# BTECH Mechanical principles and applications

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BTECH Mechanical principles and applications
Level 3 Unit 5

Vectors have a magnitude (amount) and a direction.
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) F1 F2

Forces as vectors (2 forces)
If the two forces do not meet, the system is not in equilibrium F1 F2

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) F E This force is called the EQUILIBRANT

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 F1 F2 FR

Forces as vectors (3 forces)
FE F2 F1

Forces as vectors (3 forces)
FR F2 F1

Forces as vectors ( 3 forces) 2nd example
FE Equilibrant Resultant F1 F2 F3 FR 50o 50o

Forces on a flat rectangular plate
F1 = 10N F2 = 4N F3 = 12N 50o Forces as vectors 40cm 60cm Equilibrant Resultant F1 4N 12N FE 10N 4N 12N FR 50o 50o

F1 = 10 N F2 = 4N F3 =12 N 50o Finding FE Identify the direction and magnitude of the forces then construct a vector diagram 40cm 60cm

Draw to scale to find the magnitude and direction of FE (equilibrant)
FE = 22N (Measured)

10N 4N 12N FR Draw it in the opposite direction to find the magnitude and direction of resultant force 50o FE = 22N (Measured)

Finding the position of the equilibrant (FE)
40cm 60cm

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

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 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) 1.4kN 130o 35o A 2.6kN 1.4 kN 4m 3m

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

Vector diagram with RESULTANT
2.2kN 1.4kN 1.96kN This shows a) the magnitude and direction of the resultant 2.6kN 1.4kN

Vector diagram with equilibrant
2.2kN 1.4kN 1.96kN This shows b) the magnitude and direction of the equilibrant 2.6kN 1.4kN

Resolve the diagonal forces 2. 6kN and 1
Resolve the diagonal forces 2.6kN and 1.4kN into vertical and horizontal components V1, H1 and V2 (H2 not needed) V2 1.4kN 35o A 2.6kN 1.4 kN 2.2kN For explanation Click here x H2 not needed , it passes through A V2 = 2.6xsin 35 = 1.49kN V1 = 1.4xsin40 = 0.90kN H1 1.4 x cos40 = 1.07kN 3m 1.96 kN H1 40o 4m V1

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

Resolve turning moments
1.49kN V2 1.4kN 35o A 2.6kN 1.4 kN 2.2kN x Clockwise = Anticlockwise x x 4 + 0.9 x x 3 2.2 x X X = 2.2X = – 7.52 2.2X = X = 1.65 ÷ 2.2 = 0.75m H2 not needed , it passes through A 1.96 kN H1 1.07kN 40o 4m V1 0.9kN

Resolving forces 2000 Newtons

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

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) 20o 55o 105o a Angle C = 105o and side c represents 2000N 2000 N (c) b

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 708.17N 20o 55o 105o a b 2000 N (c) b/sin B = c/sin C therefore b/sin 55o = 2000/sin105o a = 2000 x sin 55o/sin105o 1696.1N

Using the cosine rule ( if you know one angle and two sides)
F2 = 60N 70o F1 = 30N F3

Using the cosine rule ( if you know one angle and two sides)
A2 = B2 + C2 -2BCcosA 70o F2 = 60N (C) F1 = 30N (B) F3 (A) A =110o (F3)2 = – 2x60x30x cos110o = 75.7N

Vertical and horizontal components of forces
Sketch the diagram Fv can be drawn at the other end of the sketch F Fv FH θ

Vertical and horizontal components of forces
Sketch the diagram sin θ = Fv/F F.sin θ = Fv F Fv FH 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 25o 70o F1(55N) F2 (25N) F3 25o 70o Can be drawn to scale 74.8N

Restoring force of two forces
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

Restoring force of two forces
F1(55N) F2 (25N) F3 F1v = F1.sin70o 55sin70o = 51.68N F1h = F1.cos70o 55cos70o = 18.81N

Restoring force of two forces
F1(55N) F2 (25N) F3 F2v = F2.sin25o 25sin25o = 10.57N F2h = F2.cos25o 25cos25o = 22.66N

Restoring force of two forces
F1(55N) F2 (25N) F3 F3v = F1v + F2v = 62.25N F3h = F1h +F2h = 41.47N

Restoring force of two forces
F3 = 74.80N 41.47N

Resultant of two forces
Tan θ = opposite/adjacent Tan θ = 62.25/41.47 Tan θ = 1.5 θ = 56.33o F3 62.25N θ 41.47N Direction of F3 = = o

Moments of force Total Anticlockwise moments = Total Clockwise moments
8Nm 8Nm

Moments of force 8Nm + 4Nm = 12Nm = 12 Nm
Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = Nm

Moments of force Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = Nm

Moments of force Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = Nm

Force on A and B 10m 3m 1m 4m 2m A 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 = 44 ÷ 10 = 4.4N Total downward force = 8 N Force on A = 3.6 N Check this out using B as the pivot

Force on A and B Take A as the pivot
4 N/m uniformly distributed load 10m 1m 4m 2.5m 2.5m A B 20 N (UDL) 2N 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 UDL 4N/m x length 5m acting from centre of UDL

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

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

B TECH Question example solution
4kN/m (uniform distributed load) 4kN 6kN 2m 3m 5 m 40kN Clockwise moments 6 x 2 = 12kNm 4 x 5 = 20 kNm 40 x 5 = 200 kNm total = 232kNm Anticlockwise B x 10 kNm B = 232 ÷10 23.2kNm

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

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

Tensile Stress and strain
Necking Strain hardening Ultimate tensile strength Yield strength Fracture Y (Stress) X (Strain) Stress Strain

Force direction is perpendicular to cross sectional area
Tensile stress (σ) FORCE CROSS SECTIONAL AREA ( πr2) Stress = Force ÷ Cross sectional area FORCE Force direction is perpendicular to cross sectional area

Tensile strain Young’s Modulus Stress ÷ Strain Strain = ∆L ÷ Lo
Lo (original length) Increase in length ∆L Young’s Modulus Stress ÷ Strain Strain = ∆L ÷ Lo

Shear stress (τ) Force is parallel to cross sectional area of shear
Shear stress = Force ÷ cross sectional area of shear

Shear strain = Change in length ÷ original length
Force Shear strain = Change in length ÷ original length ∆l ÷ l

Shear Stress ÷ shear Strain
Force Shear Modulus Shear Stress ÷ shear Strain

a) Calculate the maximum direct stress in the connecting rods
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

Young’s modulus for steel = stress ÷ strain = 210GPa = 2.1 x10 11Pa
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 = πr2 = π x = 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

strain = elongation ÷ original length
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 = πr2 = π x = 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

Shear modulus for brass = 7 x 1010 Pa. Strain = stress ÷ modulus
20kN C B A Shear stress in pin Force ÷ area = 20kN ÷ cross sectional area of pin (π x ) = 20x103 ÷ 1.77 x10 -4m2 = x108 Pa Shear modulus for brass = 7 x 1010 Pa. Strain = stress ÷ modulus strain = 1.13 x 108 ÷ 7 x =

a) Determine the operational factor of safety in tension.
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.

F = 8kN F 70o Direct force F1 Shear force F2 8kN
Sin70o = F1 ÷ 8kN F1 = 8kN x Sin70o F1 = 7.5 kN 70o F2 F1 Cos70o = F1 ÷ 8kN F2 = 8kN x Cos70o F2 = 2.7 kN

Cross sectional area of the bolt = πr2
F = 8kN F Operational factor of safety = Tensile strength ÷ working stress Cross sectional area of the bolt = πr2 = π x (.006)2 = 1.13 x10-4 m2

Tensile stress = F ÷ area
70o F = 8kN F 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

Operational factor of safety = Tensile strength ÷ working stress
F = 8kN F 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

Operational factor of safety = Tensile strength ÷ working stress
F = 8kN F 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

Operational factor of safety = Tensile strength ÷ working stress
F = 8kN F 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

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