me a chapter 3

Click here to load reader

Download Me a Chapter 3

Post on 24-Oct-2014




1 download

Embed Size (px)




3.1 INTRODUCTIONA rigid body is one which does not suffer deformation.

It can be continuous

connected members.F2 F3

F1 P2 P1

F3 Continuous Member

F1 Connected Members



The forces acting on rigid bodies can be internal or external. F1, F2 and F3 which are applied by an external force on the rigid body are called external forces. P1 and P2 which are forces internal to the rigid body are called internal forces.


The external forces are completely responsible for the bulk motion of the rigid body. As far as this bulk motion is concerned, the internal forces are in equilibrium.


forces generally cause translation i.e. linear motion and/or rotation (motion about a pivot) of the rigid body.


Principle of transmissivity states that the condition of rest or motion of a rigid body is unaffected if a force, F acting on a point A is moved to act at a new point, B provided that the point B lies on the same line of action of that force.F A B F


The moment of a force will be formulated using Cartesian vectors in the next section. It is necessary to first expand our knowledge of vector algebra and introduce the cross-product method of vector multiplication.

Cross Products of ForcesP and Q

The cross product of two vectors , P and Q yields the vector, V which is written: V = P x Q ( i.e V = P cross Q). 3.3.1 Magnitude: The magnitude of V is the product of the magnitudes of P and Q and sine of the angle between their tails ( 0 < < 180o). Thus : V = P Q Sin



Vector, V has a direction that is perpendicular to the plane containing P and Q such that the direction of V is specified by the right hand rule i.e. curling the fingers of the right hand from vector P (cross) to vector Q, the thumb then points in the direction of V.

Direction of Cross ProductThe sense of V is such that a person located at the tip of V will observe as counterclockwise the rotation through that brings the vector P in line with vector Q. Knowing the magnitude and direction of V: V = P x Q = (P Q sin ) v Where: the scalar PQ sin defines the magnitude of V and the unit vector, v defines the direction of V. V = PxQ Q P V PQ sin = V Q P

3.3.3. Laws of Operation

1. The cummutative law does not apply i.e. P x Q Q x P Rather: P x Q = - Q x P 2. Multiplication by a scalar a ( P x Q) = (a P) x Q = P x ( a Q) = (P x Q)a 3. The distributive law: P x (Q + S ) = ( P x Q ) + ( P x S ) 4. The associative property does not apply to vector products (P x Q ) x S P ( Q x S )

Cartesian (Rectangular) Vector Formulationy j k = (i x j) z i x

To find i x j, the magnitude of the resultant is: i x j = i . j . sin 90o = 1 x 1 x 1 = 1. Using the right hand rule, the resultant vector points in the k direction. Thus i x j = 1 k. But: j x i = - k since the 90o rotation that brings j into i is observed as counterclockwise by a person located at the tip of - k.

Cartesian Vector Formulation

Contd.Likewise: i x j = k j xk=i kxi =j i x k = -j j x i = -k k x j =-i i xi =0 j xj =0 kx k =0 kk jjii -ve

Rules: If a circle is constructed as shown, then the vector product of two unit vectors in a counterclockwise fashion around the circle yields the positive third unit vector eg. k x i = j. Moving clockwise, a negative unit vector is obtained e.g. i x k = - j.

Cross Product of Two Vectors P and QConsider the cross product of two vectors P and Q expressed in Cartesian vector form: V = P x Q = ( Px i + Py j + Pz k) x ( Qx i + Qy j + Qz k) = Px Qx (i x i) + Px Qy (i x j) + Px Q z ( i x k) + Py Qx ( j x i) + Py Qy (j x j) + Py Q z (j x k) + Pz Q x (k x i) + Pz Qy (k x j) + Pz Qz (k x k) = ( Px Qy ) k + Px Qz (-j) + Py Qx (-k) + Py Qz (i) + Pz Qx (j) + Pz Qy (-i) V = (Py Q z - Pz Qy) i + (Pz Qx - Px Qz) j + (Px Qy - Py Qx) k

Equation For Product, V in Determinant Form

The equation for V may be written in a more compact determinant form as: V = P x Q= iPx Qx

j Py Qy

k Pz Qz


The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.

Moment of a Force About a Point

Consider a force, F acting at point A and a point O which lie on the same plane. The position of A is defined by the position vector, r, which joins the reference point O with A

Moment of a Force About a Point

Moment (Mo) of F about O is defined as the vector product of r and F: i.e. Mo = r x F

Direction of Mo

Using the right hand rule, the direction of Mo can be found. Curling the fingers of the right hand to follow the sense of rotation, the thumb points along the moment axis so the direction and sense of the moment vector is upward and perpendicular to the plane containing r and F.

Finally, denoting by the angle between the lines of action of the position vector, r an the force, F, the magnitude of the moment of F about O is: Mo = r F sin = F d Where: d is the perpendicular distance from O to the line of action of F. In two dimensions: F

F FMoment, M0 = + F d (Counter-clockwise , +ve) Moment, M0 = + F d clockwise , +ve)

3.5 VARIGNONS THEOREMIt states that the moment of a force about a point is equal to the sum of the moments of its components about the same point. Consider the Force, F with two components: F = F1 + F2. Using the distributive law of vectors ( see section 3.3.3) Mo = r x F1 + r x F2 = r x (F1 + F2) = r x F yF1 Moment of F about 0

F F2 r

Mo = r x F





Alternative Method

Fx = 100 cos 25o = 90.63 N Fy = 100 sin 25o = 42.26 NMB = - (0.204 x 90.63) + (42.26 x 0.0951) = - 14.47 N m = 14.47 N m Clockwise


Cross or vector product has a distinct advantage over the scalar formulation when solving problems in three dimensions. This is because, it is easier to establish the position vector, r to the force rather than determine the moment-arm distance, d perpendicular to the line of the force. For two dimensional problems, it is easier to use the Scalar Formulation

3.7 SCALAR OR DOT PRODUCT OF TWO VECTORS Scalar product of two vectors, P and Q P.Q = P Q cos ..... magnitude Where: 0 < < 180o The dot product is often referred to as scalar product of vectors, since the result is a scalar, not a vector.Q


3.7.1. Laws of Operation

(i) Commutative law: P. Q = Q. P

(ii)Multiplication by a scalar: a (P. Q) = (a P). Q = P. (a Q) = (P. Q) a (iii) Distributive law: P. ( Q1 + Q2) = P Q1 + P Q2

3.7.2 Cartesian Vector Formulation

3.7.3 Applications of Dot Product1. lines. The angle between two vectors P and Q is required. P = Px i + Py j + Pz k Recall that: P. Q = P Q cos ............. (1) P. Q = Px Qx + Py Qy + Pz Qz ...... (2) Equating (1) and (2): PQ cos = Px Qx + Py Qy + Pz QzPx Qx Py Qy Pz Qz i. e. cos PQ

Dot product is used to determine the angle formed by two vectors or inters

Q = Qx i + Qy j + Qz k

Solution Contd.

AC = - 100 mm i + 300 mm j - 600 mm k,

AC = 678.2 mm

BC = - 600 mm i + 300 mm j - 600 mm k, BC = 900 mm Let AC be P and BC be QPx Qx Py Qy Pz Qz P Q cos PQ ( 100 x 600) (300 x 300) ( 600 x 600) cos1 678.2 x 900 cos1 0.8355 33.3o

Solution Using Cosine Law


Using triangle ABC and cosine rule: C 678.2 A 500

900 B

5002 = 678.22 + 9002 - 2 x 678.2 x 900 cos 250000 = 1269955.2 - 1220760 cos cos = (12.7 - 2.5)/ 12.207 = 0.8356 = 33.3o.

Moment of a Force About A Specified AxisFirst, compute moment of force, F about any point O that lies on the aa axis. Moment of F about O, Mo = rA/O x F .................... (1) Where rA/O = OA. Mo acts along the moment axis bb which is perpendicular to the plane containing rA/O and F. The component or projection of Mo onto the aa axis is then represented by Ma. The magnitude of Ma is given by the dot product: Ma = Mo cos = Mo . a ....... (2) Where a defines the direction of the aa axis. Combining the two steps ie. Equations (1) and (2): Ma = (rA/O x F) . a

Moment of a Force About A Specified Axis ConcludedSince dot product is commutative, Ma = a .( rA/O x F) In vector algebra, this combination of dot and cross products yielding the scalar, Ma is called the triple scalar product. The triple scalar product can be written as: Ma = (ax i + ay j + az k) . i rA/Ox Fx or Ma = a .( rA/O x F) = . ax rA/Ox Fx j rA/Oy Fy ay j rA/Oy Fy k rA/Oz Fz az rA/Oz Fz

Solution ConcludedBG = - 400 mm i + 740 mm j - 320 mm k , BG = 900 mm

TBG = TBG. BG/BG = 1125 N x 1/900mm x BG = - 500 N i + 925 N j - 400 N k rB/A = AB = 400 mm i Unit vector, AD, along AD = AD/AD = (800 mm i - 600 mm k)/1000 mm = 0.8 i - 0.6 k MAD = AD . ( rB/A x TBG)


0.8 400 mm -500 N

0 0 925 N

-0.6 0 - 400

= - 0.6 ( 925 N x 400 mm) = - 222000 Nmm = - 222 Nm

3.9 MOMENT OF A COUPLE3.3.4 Introduction: A couple is defined as two parallel forces which have the same magnitude, opposite directions and are separated by a perpendicular distance, d


F Since the resultant force of the two forces of the couple is zero, the only effect of a couple is to produce a rotation or tendency of a rotation in a specified direction.

-F B




rB rA O To de

View more