Transcript Document

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