BTECH MECHANICAL PRINCIPLES AND - Physics Championphysicschampion.co.uk/Forces as vectors 1.pdf ·...
Transcript of BTECH MECHANICAL PRINCIPLES AND - Physics Championphysicschampion.co.uk/Forces as vectors 1.pdf ·...
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BTECH MECHANICAL PRINCIPLES AND APPLICATIONS
Level 3 Unit 5
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FORCES AS VECTORS
Vectors have a magnitude (amount) and a direction.
Forces are vectors
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FORCES AS VECTORS (2 FORCES)
F1
F2
Forces F1 and F2
are in different
directions They
are NOT in
equilibrium
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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)
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FORCES AS VECTORS (2 FORCES)
F1
F2
If the two
forces do not
meet, the
system is not in
equilibrium
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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
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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
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FORCES AS VECTORS (3 FORCES)
F1
F2
F3
F1
F2
F3 FE
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FORCES AS VECTORS (3 FORCES)
F1
F2
F3
F1
F2
F3 FR
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FORCES AS VECTORS ( 3 FORCES) 2ND EXAMPLE
F1
F2
F3
FE
F1
F2
F3
FR
Equilibrant
Resultant
F1
F2
F3
50o
50o 50o
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FORCES AS VECTORS
F1
4N
12N
FE 10N
4N
12N
FR
Equilibrant Resultant
F1 = 10N
F2 = 4N
F3 = 12N
50o
50o 50o
40cm
60cm
Forces on a flat
rectangular plate
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FINDING FE
IDENTIFY THE DIRECTION AND MAGNITUDE OF THE FORCES
THEN CONSTRUCT A VECTOR DIAGRAM
F1 = 10 N
F2 = 4N
F3 =12 N
50o
40cm
60cm
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DRAW TO SCALE TO FIND THE MAGNITUDE AND DIRECTION OF FE (EQUILIBRANT)
10N
4N
12N
FE
50o
FE = 22N (Measured)
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DRAW IT IN THE OPPOSITE DIRECTION TO FIND THE MAGNITUDE AND DIRECTION OF RESULTANT FORCE
10N
4N
12N
FR
50o
FE = 22N (Measured)
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FINDING THE POSITION OF THE EQUILIBRANT (FE)
F1
4N
12N
22N 10N
4N
12N
50o
40cm
60cm
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40cm
60cm
10N
4N
12N
50o
A x
Clockwise = Anticlockwise
22N
22N x X = 4N x 40cm + 10N x 0 + 12N 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
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1.4kN
130o
35o 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
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1.4kN
130o
35o 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
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VECTOR DIAGRAM WITH RESULTANT
2.2kN
1.4kN
1.4kN
1.96kN
2.6kN
This shows a) the magnitude and direction of the resultant
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VECTOR DIAGRAM WITH EQUILIBRANT
2.6kN
1.4kN
1.4kN
1.96kN
2.2kN
This shows b) the magnitude and direction of the equilibrant
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1.4kN 35o 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 40o
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)
130 - 90
3m
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1.4kN 35o A
2.6kN
1.4 kN
Clockwise = Anticlockwise
1.96 x 2 + V2 x 4 + V1 x 4 + H1 x 3 2.2 x X 1.96 kN
x
V2
H1
V1
H2 not needed , it passes through A
2.2kN
4m 40o
Resolve turning moments
0.9kN
1.49kN
1.07kN
3m
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1.4kN 35o A
2.6kN
1.4 kN
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
1.96 kN
x
V2
H1
V1
H2 not needed , it passes through A
2.2kN
4m 40o
Resolve turning moments
0.9kN
1.49kN
1.07kN
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RESOLVING FORCES
2000 Newtons
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70o 35o
20o 55o
20o
55o
105o
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
A B
A
B
The length of the sides of the triangle represent the magnitude of the forces NOT the length of rope
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USING THE SINE RULE (IF YOU KNOW THE ANGLES)
20o
55o
105o
a
b
2000 N (c)
a/sin A = b/Sin B = c/sin C
angle A = 20o angle B = 55o (opposites to sides a & b)
Angle C = 105o and side c represents 2000N
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USING THE SINE RULE (IF YOU KNOW THE ANGLES)
20o
55o
105o
a
b
2000 N (c)
a/sin A = c/sin C therefore a/sin 20o = 2000/sin105o
a = 2000 x sin 20o/sin105o
708.17N
b/sin B = c/sin C therefore b/sin 55o = 2000/sin105o
a = 2000 x sin 55o/sin105o
1696.1N
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USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES)
F2 = 60N
70o
F1 = 30N
F3
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USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES)
70o
F2 = 60N (C)
F1 = 30N (B)
F3 (A)
A =110o
A2 = B2 + C2 -2BCcosA
(F3)2 = 302 + 602 – 2x60x30x cos110o
= 75.7N
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VERTICAL AND HORIZONTAL COMPONENTS OF FORCES
F Fv
FH
θ
Sketch the diagram
Fv can be drawn at the other
end of the sketch
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VERTICAL AND HORIZONTAL COMPONENTS OF FORCES
F Fv
FH
θ
Sketch the diagram
Fv
sin θ = Fv/F
F.sin θ = Fv
cos θ = FH/F
F.cos θ = FH back
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RESTORING FORCE OF TWO FORCES
25o
70o
F1(55N)
F2 (25N)
F3
F3 is the restoring force of F1 and F2
25o
70o
Can be drawn to scale
74.8N
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RESTORING FORCE OF TWO FORCES
25o
70o
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
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RESTORING FORCE OF TWO FORCES
25o
70o
F1(55N)
F2 (25N)
F3
F1v = F1.sin70o
55sin70o
= 51.68N
F1h = F1.cos70o
55cos70o
= 18.81N
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RESTORING FORCE OF TWO FORCES
25o
70o
F1(55N)
F2 (25N)
F3 F2v = F2.sin25o
25sin25o
= 10.57N
F2h = F2.cos25o
25cos25o
= 22.66N
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RESTORING FORCE OF TWO FORCES
25o
70o
F1(55N)
F2 (25N)
F3 F3v = F1v + F2v
51.68 +10.57
= 62.25N
F3h = F1h +F2h
18.81 + 22.66
= 41.47N
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RESTORING FORCE OF TWO FORCES
F3
62.25N
41.47N
(F3)2 = 62.252 + 41.472
(F3)2 = 5594.82
F3 = 74.80N
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RESULTANT OF TWO FORCES
F3
62.25N
41.47N
θ
Tan θ = opposite/adjacent
Tan θ = 62.25/41.47
Tan θ = 1.5
θ = 56.33o
Direction of F3 = 180 + 56.33 = 236.33o
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MOMENTS OF FORCE 2m
4N
4m
2N
Total Anticlockwise moments = Total Clockwise moments
8Nm 8Nm
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MOMENTS OF FORCE 3m 4m
2m
4N
2N 2N
Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = 12 Nm
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MOMENTS OF FORCE 3m 4m
2m
4N
2N 2N
Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = 12 Nm
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MOMENTS OF FORCE 3m 4m
2m
4N
2N 2N
Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = 12 Nm
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FORCE ON A AND B
4m
10m
3m
4N
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
2m 1m A B
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FORCE ON A AND B
4m
10m
2.5m
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
2.5m 1m A B
4 N/m uniformly distributed load
UDL 4N/m x length 5m acting from centre of UDL
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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
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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
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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 ÷10 23.2kNm
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B A
4kN/m
(uniform
distributed
load)
4kN 6kN
Total upward force A + B = 50kN A + 23.2 = 50kN A = 50 – 23.2 = 26.8kN
5 m 40kN
2m
B TECH Question example solution
Total downward force 6 + 4 + 40 = 50kN
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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
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TENSILE STRESS AND STRAIN
Necking Strain hardening
Ultimate tensile strength
Yield strength
Fracture
Y (Stress)
X (Strain)
Stress
Strain
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TENSILE STRESS (σ)
FORCE FORCE
CROSS SECTIONAL AREA ( πr2)
Stress = Force ÷ Cross sectional area
Force direction is
perpendicular to cross
sectional area
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TENSILE STRAIN Lo (original length)
Increase in length ∆L
Strain = ∆L ÷ Lo
Young’s Modulus Stress ÷ Strain
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SHEAR STRESS (τ)
Force
Shear stress = Force ÷ cross sectional area of
shear
Force is parallel to cross sectional area
of shear
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SHEAR STRAIN
Force
Shear strain = Change in length ÷ original length
∆l ÷ l
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SHEAR STRESS
Force
Shear Modulus Shear Stress ÷ shear Strain
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20kN
C
B
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
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20kN
C
B
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
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20kN
C
B
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
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20kN
C
B
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
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70o
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.
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70o
F = 8kN F
Direct force F1
Shear force F2
8kN
F1 F2
70o Sin70o = F1 ÷ 8kN F1 = 8kN x Sin70o
F1 = 7.5 kN
Cos70o = F1 ÷ 8kN F2 = 8kN x Cos70o
F2 = 2.7 kN
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70o
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
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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
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70o
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
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70o
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
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70o
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