btech mechanical principles and applications level 3 unit 5
TRANSCRIPT
BTECH MECHANICAL PRINCIPLES AND
APPLICATIONSLevel 3 Unit 5
FORCES AS VECTORS
Vectors have a magnitude (amount) and a direction.
Forces are vectors
FORCES AS VECTORS (2 FORCES)
F1
F2
Forces F1 and F2 are in different directions They are NOT in equilibrium
FORCES AS VECTORS (2 FORCES)
F1
F2The 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)
FORCES AS VECTORS (2 FORCES)
F1
F2
If the two forces do not meet, the system is not in equilibrium
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
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 forcesThe forces area not in equilibrium and the resultant shows the direction and magnitude of the combination of the two forces
FORCES AS VECTORS (3 FORCES)
F1
F2F3
F1F2
F3FE
FORCES AS VECTORS (3 FORCES)
F1
F2F3
F1F2
F3FR
FORCES AS VECTORS ( 3 FORCES) 2ND EXAMPLE
F1
F2
F3
FE
F1
F2
F3
FR
Equilibrant
Resultant
F1
F2
F3
50o
50o50o
FORCES AS VECTORS
F1
4N
12N
FE10N
4N
12N
FR
Equilibrant Resultant
F1 = 10N
F2 = 4N
F3 = 12N50o
50o50o
40cm
60cm
Forces on a flat
rectangular plate
FINDING FE
IDENTIFY THE DIRECTION AND MAGNITUDE OF
THE FORCES THENCONSTRUCT A
VECTOR DIAGRAM
F1 = 10 N
F2 = 4N
F3 =12 N
50o
40cm
60cm
DRAW TO SCALE TO FIND THE MAGNITUDE AND DIRECTION OF FE (EQUILIBRANT)
10N
4N
12N
FE
50o
FE = 22N (Measured)
DRAW IT IN THE OPPOSITE DIRECTION TO FIND THE MAGNITUDE AND DIRECTION OF RESULTANT FORCE
10N
4N
12N
FR
50o
FE = 22N (Measured)
FINDING THE POSITION OF THE EQUILIBRANT (FE)
F1
4N
12N
22N 10N
4N
12N
50o
40cm
60cm
40cm
60cm10N
4N
12N
50o
Ax
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
1.4kN
130o
35oA
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
1.4kN
130o
35oA
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
VECTOR DIAGRAM WITH RESULTANT
2.2kN
1.4kN
1.4kN
1.96kN
2.6kN
This showsa) the magnitude and direction of the resultant
VECTOR DIAGRAM WITH EQUILIBRANT
2.6kN
1.4kN
1.4kN
1.96kN
2.2kN
This shows b) the magnitude and direction of the equilibrant
1.4kN 35oA
2.6kN
1.4 kN
V2 = 2.6xsin 35 = 1.49kNV1 = 1.4xsin40 =
0.90kNH1 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
1.4kN 35oA
2.6kN
1.4 kN
Clockwise = Anticlockwise 1.96 x 2 + V2 x 4 +V1 x 4 + H1 x 32.2 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
1.4kN 35oA
2.6kN
1.4 kN
Clockwise = Anticlockwise 1.96 x 2 + 1.49 x 4 +0.9 x 4 + 1.07 x 32.2 x X 7.52 + 2.2X = 9.172.2X = 9.17 – 7.522.2X = 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
RESOLVING FORCES
2000 Newtons
70o35o
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
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
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
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)
70o
F2 = 60N (C)
F1 = 30N (B)
F3 (A)
A =110o
A2 = B2 + C2 -2BCcosA
(F3)2 = 302 + 602 – 2x60x30x cos110o
= 75.7N
VERTICAL AND HORIZONTAL COMPONENTS OF FORCES
FFv
FH
θ
Sketch the diagram
Fv can be drawn at the other
end of the sketch
VERTICAL AND HORIZONTAL COMPONENTS OF FORCES
FFv
FH
θ
Sketch the diagram
Fv
sin θ = Fv/F
F.sin θ = Fv
cos θ = FH/F
F.cos θ = FHback
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
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
RESTORING FORCE OF TWO FORCES
25o
70o
F1(55N)
F2 (25N)
F3
F1v = F1.sin70o
55sin70o
= 51.68N
F1h = F1.cos70o
55cos70o
= 18.81N
RESTORING FORCE OF TWO FORCES
25o
70o
F1(55N)
F2 (25N)
F3F2v = F2.sin25o
25sin25o
= 10.57N
F2h = F2.cos25o
25cos25o
= 22.66N
RESTORING FORCE OF TWO FORCES
25o
70o
F1(55N)
F2 (25N)
F3F3v = F1v + F2v51.68 +10.57
= 62.25N
F3h = F1h +F2h18.81 + 22.66
= 41.47N
RESTORING FORCE OF TWO FORCES
F3
62.25N
41.47N
(F3)2 = 62.252 + 41.472
(F3)2 = 5594.82
F3 = 74.80N
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
MOMENTS OF FORCE2m
4N
4m
2N
Total Anticlockwise moments = Total Clockwise moments
8Nm 8Nm
MOMENTS OF FORCE3m4m
2m
4N
2N 2N
Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = 12 Nm
MOMENTS OF FORCE3m4m
2m
4N
2N 2N
Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = 12 Nm
MOMENTS OF FORCE3m4m
2m
4N
2N 2N
Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = 12 Nm
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 = 4.4NTotal downward force = 8 N Force on A = 3.6 N
Check this out using B as the pivot
2m1mA B
FORCE ON A AND B
4m
10m
2.5m
20 N (UDL)2N 2N
Take A as the pivotAnticlockwise ( B x 10) = clockwise (2 x1) + (2 x 5) + (20 x 7.5) = 162 Nm
Force on B = 162 ÷ 10 = 16.2 NTotal downward force = 24 N Force on A = 7.8 N
Check this out using B as the pivot
2.5m1mA B
4 N/m uniformly distributed load
UDL 4N/m x length 5m acting from centre of UDL
BA
4kN/m (uniform distributed load)
4kN6kN
P2 Calculate the support reactions A and B for the simply supported beam in the diagram
5 m3m2m
B TECH Question example
BA
4kN/m (uniform distributed load)
4kN6kN
Uniform load is 4kN/m. total UDL = 4 x 10 = 40kN ( acting from centre of beam)
5 m3m2m
B TECH Question example
solution
40kN
BA
4kN/m (uniform distributed load)
4kN6kN
Clockwise moments6 x 2 = 12kNm4 x 5 = 20 kNm 40 x 5 = 200 kNm
total = 232kNm
5 m3m2m
B TECH Question example
solution
40kN
AnticlockwiseB x 10 kNm
B = 232 ÷10 23.2kNm
BA
4kN/m (uniform distributed load)
4kN6kN
Total upward force A + B = 50kNA + 23.2 = 50kN
A = 50 – 23.2 = 26.8kN
5 m40kN
2m
B TECH Question example
solution
Total downward force6 + 4 + 40 = 50kN
BA
4kN/m (uniform distributed load)
4kN6kN
CW = A x10ACW = 4 x 5 = 20kNm 40 x 5 = 200kNm 6 x 8 = 48kNm = 268kNmA = 26.8kN
5 m3m2m
B TECH Question example
solution
Check using B as the pivot
40kN
TENSILE STRESS AND STRAIN
NeckingStrain hardening
Ultimate tensile strength
Yield strengthFracture
Y (Stress)
X (Strain)
Stress
Strain
TENSILE STRESS (σ)
FORCEFORCE
CROSS SECTIONAL AREA ( πr2)
Stress = Force ÷ Cross sectional area
Force direction is
perpendicular to cross
sectional area
TENSILE STRAINLo (original length)
Increase in length ∆L
Strain = ∆L ÷ Lo
Young’s Modulus Stress ÷ Strain
SHEAR STRESS (τ)
Force
Shear stress = Force ÷ cross sectional area of
shear
Force is parallel to cross sectional area
of shear
SHEAR STRAIN
Force
Shear strain = Change in length ÷ original length
∆l ÷ l
SHEAR STRESS
Force
Shear Modulus Shear Stress ÷ shear Strain
20kN
C
B
A20kN
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 rodsb) Calculate the maximum direct strain in the connecting rodsc) Calculate the change in length of a 500mm length of connecting rod.d) Calculate the shear stress in the pine) Calculate the shear strain in the pin
20kN
C
B
A20kN
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
20kN
C
B
A20kN
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
20kN
C
B
A20kN
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
70o
F = 8kNF
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.
70o
F = 8kNF
Direct force F1
Shear force F2
8kN
F1F2
70oSin70o = F1 ÷ 8kN F1 = 8kN x Sin70o
F1 = 7.5 kN
Cos70o = F1 ÷ 8kN F2 = 8kN x Cos70o
F2 = 2.7 kN
70o
F = 8kNF
Operational factor of safety = Tensile strength ÷ working stress
Cross sectional area of the bolt = πr2
= π x (.006)2 = 1.13 x10-4 m2
70o
F = 8kNF
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
70o
F = 8kNF
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
70o
F = 8kNF
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
70o
F = 8kNF
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