Engineering Mechanics: Module 1 - Vidyarthiplus

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Engineering Mechanics: Module 1

1/8/2013

Engineering Mechanics, Module 1

Pre read

1/8/2013Basic Civil Engineering, Module 4

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1/8/2013Engineering Mechanics, Module 1

Learning Objectives

Force systems in Space

Introduction to Vector approach

Elements of vector algebra

Position Vector

Moment of a force about a point & axis

Resultant of forces

Equilibrium of forces in space using vector approach

Forces in Space

Fx = Fcos ƟxFy = Fcos ƟyFz = Fcos Ɵz

(OE)2 = (OD)2 + (DE)2

= (OA)2 + (OC)2+ (DE)2

(OE)2 = (OD)2 + (DE)2

= (OA)2 + (OC)2+ (DE)2

= (Fx)2 + (Fz)2+ (Fy)2

(OE)2 = (F)2

(F)2 = (Fx)2 + (Fz)2+ (Fy)2

Fx = Fcos Ɵx ; Fy = Fcos Ɵy; Fz = Fcos Ɵz

A force of magnitude 5 kN makes 30o, 50o &

100o with X, Y & Z axes respectively. Find

the magnitude of its components along X, Y

& Z axes

Ɵx = 30o ,Ɵy = 50o ,Ɵz = 100o

Fx = 4.33 kN, Fy = 3.21 kN, Fz = -0.868 kN

Direction cosines

The cosines of Ɵx, Ɵy & Ɵz are known as direction cosines of the force F

Denoted by l, m & n

l = cos Ɵx

m = cos Ɵy

n = cos Ɵz

l2 + m2 + n2 =1

A force acts at the origin in a direction defined by the

angles Ɵy = 60o & Ɵz= 45o. Taking the component of

the force along X direction as -800N, (1) find the

value of Ɵx. (2) find the other components and the

value of the force

Fcos Ɵx = -800 N

Ɵy = 60o & Ɵz= 45o

l = cos Ɵx , m = cos Ɵy & n = cosƟz

l2 + m2 +n2 =1 Ɵx =120

Fx = Fcos Ɵx F =1600 N

Fy= 800 N, Fz = 1131.37 N

F passes through the origin and the co-ordinates of a point along the line of action of F is known

d2 = (dx)2 + (dy)2+ (dz)2

Fx = (F X dx) /d ; Fy = (F X dy)/ d ; Fz= (F X dz)/d

Calculate the components of force 1000 Nshown in figure along X, Y & Z co-ordinates.

120 mm

Co-ordinate of A is (120, 40,50) dx =120, dy

=40, dz = 50

120 mm

Distance of A from the origin, d = sqrt ( 1202+ 402+ 502)

= 136.01 mm

Fx = 882.29 N, Fy = 294.1 N, Fz = 367.62 N

d2 = (dx)2 + (dy)2+ (dz)2

Fx = (F X dx) /d ; Fy = (F X dy)/ d ; Fz= (F X dz)/d

When A & B are 2 points on the line of actionof F, neither of which is at the origin

A (xa,ya,za)

B (xb,yb,zb)

dx = (xb- xa)dy = (yb-ya)dz = (zb- za)d2 = dx2 + dy2 +dz2

A force has a line of action that goes through the

coordinates (2,4,3) & (1, -5,2). If the magnitude of

the force is 100 N, find the components of the force

and its angles with the axes

dx = -1 , dy = -9 , dz = -1, d = 9.11

Fx = -10.976 N, Fy = -98.792 N, Fz = -10.976 N

Ɵx = 96.30, Ɵy = 171.085, Ɵz = 96.30

Unit Vector

Unit Vector Vector having unit length

Unit Vector = (Force Vector)/ Magnitude of force vector

= F/ |F|

Unit vectors along X, Y &Z i, j, k

Vector F= Fxi + Fyj + Fzk

and its angles with the axes. Find the force vector.

Fx = -10.976 N, Fy = -98.792 N, Fz = -10.976 N

Force Vector is F = Fxi + Fyj + Fzk

F = -10.976 i – 98.792 j -10.976k

and its angles with the axes. Find the force

vector in terms of unit vectors

The vector joining the points =

(1-2)i + (-5-4)j + (2-3)k

Unit vector along this vector = (-i -9j-k)/ 9.11

Unit vector = F/|F|Hence F = Unit Vector X |F|

= ((-i -9j-k)/ 9.11 ) X 100

= -10.976i – 98.79j – 10.976k

A force has a line of action that goes throughpoints A(4,2,5) & point B(12,4,6). If themagnitude of the force in the direction of AB is100N, find the force vector in terms of unit vector

Practice Problem

Vector joining A & B = 8i + 2 j + kUnit vector along AB = 8i + 2 j + k/ 8.30

Force vector = unit vector X magnitude of

the force

= (8i + 2 j + k/ 8.30) X 100

=96.38i + 24.096j + 12.04k

A force P acts as shown in fig. Express theforce P as a vector. Magnitude of force P is100N. Length of the side of the cube is 2 cm.

(0, 2, 2)

(2, 0,0)

Express the force P as a vector. Magnitude of forceP is 100N. Length of the side of the cube is 2 cm.

(0, 2, 2)

(2, 0,0)

Vector joining A&B is 2i-2j-2k, |AB| = 3.46Unit vector along AB = 0.577i -0.577j-0.577k

Express the force P as a vector. Magnitude of forceP is 100N. Length of the side of the cube is 2 cm.

(0, 2, 2)

(2, 0,0)

Force Vector = 57.7i -57.7j – 57.7k

A vertical post guy wire is anchored by means of abolt at A as shown in fig. The tension in the wireis 2000N. Determine, (1) the components of theforce acting on the bolt (2) the angles definingthe direction of the force and (3) the force vector

40 m A

Coordinates of A (40, 0, -30)Coordinates of B (0,80,0)Vector joining A&B is -40i+80j+30k|AB| = 94.33Unit vector in the direction of AB = -.424i + 0.848 j +0.318 k

The force in the bolt is the same as the tension in the wire

Hence the Force vector is-848i + 1696j +636kDirection of forceƟx = 115.09o

Ɵy = 32o

Ɵz = 71.46o

Vector addition

The sum of the vectors can be obtained by adding the respective components of the vectors

If A= Axi + Ayj+ Azk &

B = Bxi + Byj+ Bzk, then,

A+B = B+A = (Ax+Bx)i + (Ay+By)j + (Az+Bz)k

Find the unit vector in the direction of theresultant of vectors A = (2i-j+k), B= (i+j+2k) ,C= (2i+2j+4k)

Resultant R = A+B+C

= 5i + 2j + 7k

Unit Vector = R/|R|

= 0.57i + 0.23j + 0.79k

Resultant of forces in space

The horizontal component of R= Rx = ∑Fx

Ry = ∑Fy

Rz= ∑Fz

R = sqrt ( Rx2+ Ry2 + Rz2)

Cos Ɵx = Rx/ RCos Ɵy = Ry/RCos Ɵz = Rz/R

4 forces act on a particle as shown in fig.Determine the resultant of the forces in vectorform

30o70o

120 N100 N

R= (140 cos 30-100cos70+100cos15)i + (140sin30+100sin70-120-100sin15)j=183.63i +18.09j

4 forces 30 kN, 20 kN, 25 kN & 100kN arerespectively directed through the points whose co-ordinates are A(2,1,5), B(3,-1,4) C(-3,-2,1) &D(4,1,-2). If these forces are concurrent at theorigin, calculate the resultant of the forces invector form

Practice

Problem

90.01i +10.01j + 6.68 k

2 cables ‘AB’ & ‘AC’ are attached at A as shownin fig. Determine the resultant of the forcesexerted at A by the 2 cables, if the tension is2000 N in cable AB & 1500 N in the cable AC.Also calculate the angles that this resultantmakes with the X, Y & Z axes

Co-ordinates of A (52,0,0)Coordinates of B (0, 50,40)Co-ordinates of C (0, 62,-50)

Unit vector along AB = -0.63i + 0.606j + 0.485k

FAB = -1260i + 1212 j +970k

Unit vector along AC = -0.546i +0.6518j -0.5256k

FAC= -819i +977.7j -788.4k

R = -2079i +2189.7j +181.6k

|R| = 3024.39

Ɵx = 133.42o

Ɵy = 43.61o

Ɵz = 86.56o

A post is held in vertical position by 3 cables AB,AC &AD as shown the fig. If the tension in thecable AB is 40 N, calculate the required tension inAC& AD so that the resultant of the 3 forcesapplied at A is vertical

A (0,48, 0)

B(16,0,12)

C(16,0,-24)

Vector along AB = 16i-48j+12k

Unit Vector along AB = 0.307i-0.923j+0.23k

Vector along AC = 16i-48j-24k

Unit Vector along AC = 0.2857i-0.857j-0.4257k

Vector along AD = -14i-48j

Unit Vector along AD= -0.28i-0.96j

FAB = 12.28i -36.92j +9.2k

Given , Rx = 0 & Rz= 0

Rx=0 12.28 +0.287 FAC – 0.28 FAD = 0

Rz= 0 9.2- 0.42587 FAC= 0

FAC = 21.47N

FAD = 65.86 N

Equilibrium of forces in space

For equilibrium, the components Rx, Ry & Rz must be zero

∑Fx= Rx = 0,

∑Fy = Ry= 0 &

∑Fz= Rz= 0

A horizontal force ‘P’ normal to the wallholds the cylinder in the position shown infig. Determine the magnitude of P and thetensions in the cables ‘AB’ & ‘AC’

Mass -250 kg

Mass =250 kgP

C (0,14,-12)

(0,14,9)

(1,2,0)

Unit vector along AB= -0.06i +0.798j +0.5988k

Unit vector along AC = -0.0588i+ 0.705j-0.705k

Since the cylinder is in equilibrium, ∑Fx= 0, ∑Fy=0 &

∑Fz = 0

∑Fy=0 0.798 FAB + 0.705 FAC -250X9.81= 0

∑Fz=0 0.598 FAB - 0.705 FAC = 0

∑Fx=0 P- 0.06 FAB - 0.058 FAC = 0

P = 193.02 N, FAB = 1755.47 N, FAC=1491.48 N

A tripod supports a load of 2 kN as shown in fig.

A, B & C are in XZ plane. Find the forces in the 3

legs of the tripod

A (1.2, 0,0) ; B (0, 0,1.2) ; C(-1, 0, -0.8);

D(0,1.8,0)

A (1.2, 0,0) ; B (0, 0,1.2) ; C(-1, 0, -0.8);

D(0,1.8,0)

FAD= FAD( -0.55i + 0.833j)

FBD= FBD(+0.833j – 0.55k)

FCD = FCD(+0.452i + 0.815j + 0.362k)

∑Fx= 0 , ∑Fy= 0 & ∑Fz= 0-0.556 FAD + 0 FBD + 0.453 FCD = 0

0.333 FAD + 0.833 FBD + 0.815 FCD = 0

0 FAD – 0.556 FBD + 0.362 FCD = 0

FAD = 0.801 kN

FBD = 0.641 kN

FCD = 0.983 kN

Dot Product

A . B = |A||B| cos Ɵ, where Ɵ is the angle between

the vectors

Dot product Scalar quantity

A = Axi + Ayj + Azk

B = Bxi + Byj + Bzk

A . B = AxBx + AyBy + AzBz

Angle between A&B , cosƟ = A.B/ (|A||B|)

Dot Product

Application

Work done by a force F in causing a displacement ‘s’ is given by

W = Fs cos Ɵ

W = F . s

The point of application of a force F(5i+10j-15k)is displaced from point A(i+oj+3k) to the point(3i-j-6k). Find the work done by the force

W = F.s

s = 2i-j-9k

W = 135 Nm

Dot Product

Another Application to find the component or projection of a force along a line

F = Fxi+ Fyj +Fzk

Projection of F along a line AB = F. Unit vector along AB

A force F= 3i-4j+12k acts at a point A(1,-2,3).

Compute the component of F along line AB if B

has co-ordinates (2,1,2)

Vector along AB = i+3j-k

Unit vector along AB = 0.301i+0.904j-0.301k

Component of F along AB = (3i-4j+12k) .

(.301i+0.904j-0.301k)

= -6.325.

To convert this in vector form, -6.325(0.301i+0.904j-

0.301k)

= -1.895i-5.717j+1.903k

Cross Product

A x B = |A| |B| sin Ɵ

A x B =

A x B = i(AyBz – AzBy) – j(AxBz – AzBx) + k( AxBy –AyBx)

Ax Ay Az

Bx By Bz

Moment of a force in space

Moment of a force F about a point is given by

M = r x F where r = xi + yj + zk; r is the position vector

M = r x F =

M = i(yFz-zFy) – j(xFz- zFx) + k (xFy- yFx)

Mx = yFz-zFy ; My= -xFz+zFx ; Mz = xFy- yFx

Mx,My& Mz are the moments about the X,Y & Z axes

Fx Fy Fz

Position Vector

Position Vector of a point passing through the origin

A (x, y, z)

r = xi + yj + zkMagnitude of position vector r = sqrt (x2+ y2+z2)

Position Vector of a point wrt another point

Position vector of point A, wrt point B

rAB is given by the line BA

If A (xa,ya,za) & B(xb,yb,zb) then

rAB is given by line BA (xa-xb)i + (ya-yb)j+ (za-zb)k

A force F= 2i+ 3j-4k is applied at the pointA(1, -1,2).Find the moment of the force abouta point B(2,-1,3).

A(1, -1, 2) B(2, -1,3)

M = r x F

r = Position vector of the point A wrt point B = (1-2)i + (-1- -1)j + (2-3)k

= -i + 0j –k

Moment of force at A about B = r x F

M = 3i -6j -3k

A force of magnitude 44 N acts through the point A(4,-1,7) in the direction of vector 9i+6j-2k. Find the moment of the force about the point O(1,-3,2)

Unit vector in the direction of 9i+6j-2k is 0.818i + 0.545j + 0.181k

Force vector, F = 36i +24j -8k

Position vector of A wrt point O is 3i+2j+5k

M = -136i + 204j

Moment of a force about an axis

(1) Calculate the Moment of F

about O= Mo= r x F

(2) Calculate the component (projection) of Mo on the line OLMOL = Mo .Unit vector along OL

Find the moment of a force F(20i+30j) kNacting at a point with position vector (6i+10j)about the line passing through the pointsB(6i+10j-4k) and C(-4i+5j-2k)

(1) Calculate the Moment of F about point B.

(2) Find the projection of this moment MB about the line BC

(1) Moment of F about B

=Position vector of A wrt B x F

= 4k x (20i+30j) = -120i +80j

(1) Moment of F about point B = -120i +80j

(2) Find the projection of this moment MB about the line BC

Vector along BC = -10i-5j+2k

Unit vector along BC = -0.88i-0.44j+0.176k

MBC = (-120i +80j). (-0.88i-0.44j+0.176k)

= 70.4 kNm

Engineering Mechanics: Module 1 - Vidyarthiplus

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