Force Systems 2D
Lecture Overview 1.) - Classification of Vectors - Vector addition, parallelogram law - Decomposition of vectors, force components 2.) - Moment of a vector - Force couples
Statics concerning rigid bodies - focus on sliding vectors - principle of transmissibility (force can be applied on any point along its line of action without changing its external effect on a rigid body)
Vector Addition 1. parallelogram law 2. triangle rule 3. analytic method
Vector Addition – graphical method the parallelogram law – resultant force
Vector Addition – graphical method the parallelogram law – resultant force 2 forces replaced by a single resultant force
scale
A
B
Vector Addition – graphical method the parallelogram law – resultant force 2 forces replaced by a single resultant force
scale
A
B
Vector Addition – graphical method the parallelogram law – resultant force 2 forces replaced by a single resultant force
A point of intersection
B
Vector Addition – graphical method the parallelogram law – resultant force 2 forces replaced by a single resultant force
A B
Vector Addition – graphical method the parallelogram law – resultant force Line of action of R through point of intersection of A and B scale
A
R B
Vector Addition – graphical method the triangle rule (from parallelogram law)
Vector Addition – graphical method the triangle rule (from parallelogram law) line of action of one force to be moved!
A
B
Vector Addition – graphical method the triangle rule (from parallelogram law) for two forces
r r r r r R = A+B = B+A B
R
A
A
R B
Vector Addition – graphical method the triangle rule (from parallelogram law) for three forces
r r r r r r r R = A + B + C = B + A + C = ... B C
A R
R B
A
C
Vector Addition – graphical method the triangle rule (from parallelogram law) location of line of action of resultant force R? Initial situation:
Result of addition:
A
B C
A B
R C
Vector Addition – graphical method the triangle rule (from parallelogram law) location of line of action of resultant force R? Initial situation:
A
A B
B C
C
Vector Addition – graphical method lines of action – point of intersection Initial situation:
A
A B
B C
C
Vector Addition – graphical method
Initial situation:
A A B
B C
C
Vector Addition – graphical method
Initial situation:
B
A
A B C
C
Vector Addition – graphical method result for 2 forces A and B : A + B = RAB location of resultant force RAB Initial situation:
B
A
A RAB C
B C
Vector Addition – graphical method location of RAB and C Initial situation:
A RAB C
B C
Vector Addition – graphical method lines of action Initial situation:
A RAB C
B C
Vector Addition – graphical method point of intersection Initial situation:
RAB
A B
C
C
Vector Addition – graphical method
Initial situation:
RAB
C
A B C
Vector Addition – graphical method resultant force of the 3 forces A, B and C Initial situation:
C
RAB RABC
A B C
Vector Addition – graphical method location of line of action of resultant force Initial situation:
C
RAB RABC
A B C
Vector Addition – graphical method location of line of action of resultant force
Initial situation:
B
A
C
A RABC
B C
R
Vector Addition – graphical method e.g.: equilibrium at particle: addition of all vectors = 0 situation at particle:
A
resultant reaction force
R
B C
B C
A R
Recalling equilibrium at particle
Recalling equilibrium at particle
Recalling equilibrium at particle
Vector Addition – analytic method
Vector Addition – analytic method trigonometric rules
B
R
β
θ
B
A 2
2
2
R = A + B − 2 A B ⋅ cosθ
sinβ sinθ = B R
magnitude
Inclination or R from A
Vector Decomposition 1. components along arbitrary axis a) graphical method b) analytical method 2. rectangular components vector addition by rectangular components
Vector Decomposition – graphical method
Vector Decomposition – graphical method vector components along 2 axis of arbitrary incline b F
φ
θ
a
given: Force F, 2 axis of arbitrary incline a and b
Vector Decomposition – graphical method vector components along 2 axis of arbitrary incline b F
φ
θ a
applying parallelogram law
Vector Decomposition – graphical method vector components along 2 axis of arbitrary incline b Fb
φ
F
θ Fa
a
Vector Decomposition – analytical method
Vector Decomposition – analytical method vector components along 2 axis of arbitrary incline b Fb
φ
F
θ
α Fa
by trigonometric rules:
components in a and b directions:
B
α = 180 o − (φ + α)
a
sinφ sinθ sinα = = Fa Fb F sinφ Fa = F sinα
and
sinθ Fb = F sinα
Rectangular Components y
F
θ x
Rectangular Components y
F
θ x
horizontal component of F:
Fx = F ⋅ cosθ
Rectangular Components y
F
Fy
θ Fx
x
horizontal component of F:
Fx = F ⋅ cosθ
vertical component of F:
Fy = F ⋅ sinθ
magnitude of F:
F = Fx2 + Fy2
Vector addition by rectangular components
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
R =R=
(F1x + F2 x ) + (F1 y + F2 y ) 2
2
example 1 given:
detail of a symmetric cable support θ = 30˚, P = 10 kN
determine: tensile force (horizontal) and shear force (vertical) exerted to the connection.
example 1 θ = 30˚, P = 10 kN equilibrium condition:
reaction force
R= -P
Py
support – position of bolt
Px
Px and Py are the rectangular components of P! fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 1 θ = 30˚, P = 10 kN
Py R= -P Px
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
║Px ║= cosθ·║P║ = cos30˚·10kN = 8.66kN ║Py ║= sinθ·║P║ = sin30˚·10kN = 5.00kN
example 2
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 2
T
R=?
R
T = 400 N
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 2
T
y
R=?
R
x
T = 400 N
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N o
R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N 0
R = R 2x + R 2y = 692.8N
example 2
T
y x
R=?
R
60˚ T
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N o
R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N 0
R = R 2x + R 2y = 692.8N
example 2
T
y x
R=?
R
60˚ T
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N o
R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N 0
R = R 2x + R 2y = 692.8N
example 2
T
y x
R=?
R
60˚ T fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N o
R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N 0
R = R 2x + R 2y = 692.8N
example 2
T
y x
R=?
R
60˚ T fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N o
R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N 0
R = R 2x + R 2y = 692.8N
example 2
T
y
R=?
R
60˚
x
T fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
R x = T + T ⋅ cos60 = 400N(1 + 0.5) = 600.0N o
R y = T ⋅ sin60 = 400N ⋅ 3 2 = 346.4N 0
R = R + R = 692.8N 2 x
2 y
Lecture Overview 1.) - Classification of Vectors - Vector addition, parallelogram law - Decomposition of vectors, force components 2.) - Moment of a vector - Force couples
Moment 1. moment of a force about a point - scalar product - cross product 2. force couples
Moment 1. moment of a force about a point
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
moment of a force force acting in a plane
F
moment of a force line of action and support
F
moment of a force line of action of reaction force distance
F
F
hinge support: no equilibrium - rotation!
moment of a force axis of rotation d
F
moment of a force Moment of Force F
M = F·d [kNm] d = moment arm perpendicular to line of action!
d
F
moment of a force sign convention – right hand rule (+)
M (+)
F
moment of a force, scalar product Moment of Force F
M = F·d [kNm] d = moment arm d
F
examples - moment of a force
M = F·d
F
d
Equilibrium!
left side
M = -F·d
M = +F·d
F
+
d
right side
M
M
F
-
d
Equilibrium!
d
d
d
d
2F F
d
d
left side
right side M = +F·d
M = -2F·d 2F
F
d
d
Rotation!
d
d/2
Equilibrium!
left side
right side
M = +F·d
M = -2F·d/2 2F
F
d
d/2
5m
1m
F = ?? 5m 500 N 1m
600 N ???
1m
3m
Moment 1. moment of a force about a point - scalar product - cross product 2. force couples
Moment - Cross Product (vector product) Moment of a vector V about any point 0 r is a position vector from reference point 0 to any point on the line of action of vector V.
r r r r M0 = r × V = F r r i j r M 0 = rx ry Vx Vy
r ⋅ r ⋅ sinθ r k rz Vz
0 ║r║·sinθ
r θ
F
Definition of Cross Product (vector product) Determination of resulting vector by three by three matrix with unit vectors i, j and k.
r i ax bx
r j ay by
r k az bz
r r r r r A × B = (a y b z - a z b y )i + (a z b x - a x b z ) j + (a x b y - a y b x )k for 2D system
Moment – Varignon’s Theorem Moment of a force F about any point equals the sum of the moments of the components of the force about the same point.
by cross product:
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
Moment – Varignon’s Theorem Moment of a force F about any point equals the sum of the moments of the components of the force about the same point.
by rectangular moment arms:
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways.
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways. 1.) product of force by moment arm
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways.
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways.
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways. 2.) replace force by rectangular components y x fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways.
y x
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways.
y x
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways.
y x
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways. 3.) by cross product
y x fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways.
y x fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 3 Calculate the magnitude of the moment about the base point 0 in different ways.
y
r i 2 600 ⋅ cos40o
r j 4 - 600 ⋅ sin40 o
r k 0 0
x fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
r r r r r =0 =0 A × B = (a y b z - a z b y )i + (a z b x - a x b z ) j + (a x b y - a y b x )k
Moment 1. moment of a force about a point - scalar product - cross product 2. force couples
Moment 2. force couples
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
moment of a couple moment produced by 2 equal and opposite forces is known as a couple.
F
d
F
moment of a couple moment of couple about any point of reference 0. 0 F
a d
F
moment of a couple moment of couple about any point of reference 0. 0 F
a
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
d
F
R=F–F=0 M = F(a+d) - Fa M = Fd
moment of a couple moment of couple about any point of reference 0. 0 F
a
r d
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
F
R=F–F=0 M = F(a+d) - Fa
cross product: M = r x F
M = Fd
moment of a couple
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
Force-couple systems resolution of a force into a force and a couple – equal external effect.
example 4
example 4 M = -F·y y = -M/F = - 0.375 kNm / 5 kN y = -75 mm
fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
example 4 M = -F·y y = -M/F = - 0.375 kNm / 5 kN 75 mm
y = -75 mm
initial
= fig by J.L. Meriam, L.G. Kraige, Engineering Mechanics I
