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Engineering Mechanics–Statics
Engineering Mechanics –Statics
R. GaneshNarayanan
Department of Mechanical Engineering
IIT Guwahati
May 2008R. Ganesh NarayananBatch: Jan -1
-These lecture slides were prepared and used by me to conduct lectures for 1styear
B. Tech. students as part of ME 101 –Engineering Mechanics course at IITG.
-Theories, Figures, Problems, Concepts used in the slides to fulfill the course
requirements are taken from the following textbooks
-Kindly assume that the referencing of the following books have been done in this
slide
-I take responsibility for any mistakes in solving the problems.Readers are requested
to rectify when using the same
-I thank the following authors for making their books available for reference
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R. GaneshNarayanan
1.Vector
Mechanics
for
Engineers
–Statics&
Dynamics,
Beer
&
Johnston;
7thedition2.Engineering Mechanics Statics& Dynamics, Shames; 4thedition
3.Engineering Mechanics StaticsVol. 1, Engineering Mechanics Dynamics Vol. 2,
Meriam& Kraige; 5thedition4.Schaum’ssolved problems series Vol. 1: Statics; Vol.
2: Dynamics, Joseph F. Shelley
Batch: Jan -May 2008
R. Ganesh Narayanan
2
Engineering mechanics
-Deals with effect of forces on objects
Mechanics principles used in vibration, spacecraft design, fluid flow, electrical,
mechanical m/cdesign etc.
Statics: deals with effect of force on bodies which are not moving
Dynamics: deals with force effect on moving bodiesWe consider RIGID BODIES–Non
deformable
R. Ganesh Narayanan
3
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Scalar quantity: Only magnitude; time, volume, speed, density, mass…
Vector quantity: Both direction and magnitude; Force, displacement, velocity,
acceleration, moment…
V = IvIn, where IvI= magnitude, n = V / IvI
n
-dimensionless and in direction of vector ‘V’
y
j
In our course:
x
i
i, j, k –unit vectors
z
k
R. Ganesh Narayanan4
Dot product of vectors: A.B = AB cosθ; A.B = B.A (commutative)
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A.(B C) = A.B A.C (distributive operation)i . i = 1
A.B = (Axi Ayj Azk).(Bxi Byj Bzk) = AxBx AyBy AzBz
i . j = 0
Cross product of vectors: A x B = C; ICI = IAI IBI Sin θ; AxB= -(BxA)
C x (A B) = C x A
j
i
k
j
i
k
i
kAZBZ
j
5
AYBY
C x B
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k x j = -i; i x i = 0
AxBX
R. Ganesh Narayanan
AxB= (Axi Ayj Azk)x(Bxi Byj Bzk) = (AyBz-AzBy
)i (
)j (
)k
Force:
-action of one body on another
-required force can move a body in the direction of action, otherwise no effect
-some times plastic deformation, failure is possible-Magnitude, direction, point of
application; VECTOR
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Force, Direction of motion
Body movesForce
bulging
6
Force system:
Magnitude, direction and point of application is important
WIRE
External effect: Forces applied (applied force); Forces exerted by bracket, bolts,
foundation….. (reactive force)
Internal effect: Deformation, strain pattern –permanent strain; depends on material
properties of bracket, bolts…
R. Ganesh Narayanan7
Transmissibility principle:
A force may be applied at any point on a line of action without changing the resultant
effects of the force applied external to rigid body on which it actsMagnitude,
direction and line of action is important; not point of application
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Line of
P
P
R. Ganesh Narayanan8
Concurrent force:
Forces are said to be concurrent at a point if their lines of action intersect at
that point
Parallelogram law of forcesPolygon law of forces
F1, F2 are concurrent forces
R will be on same planeR = F1 F2
R does not
Two dimensional force system
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Rectangular components:
j
Fy
F
θFx
i
ve
-ve
ve
-ve
F = Fx
Fy; both are vector components in x, y directionFx= fxi ; Fy= fyj; fx, fyare
scalar quantities
Therefore, F = fx
i
fyj
Fx= F sin θ
2
fy2; θ= tan -1(fy/fx)
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R. Ganesh Narayanan10
Two concurrent forces F1, F2Rx = ΣFx; Ry= ΣFyF1
j
F2
DERIVATION
R. Ganesh Narayanani
11
Moment: Tendency to rotate; torque
Moment about a point: M = Fd
Magnitude of moment is
proportional to the force ‘F’andmoment arm ‘d’
to the LOA of force
UNIT : N-m
Moment is perpendicular to plane about axis O-OCounter CW =
ve; CW = -veR. Ganesh
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Narayanan
12
Cross product:
M = r x F; where ‘r’is the position vectorwhich runs from the moment reference point
‘A’to any point on the LOA of ‘F’M = Fr sin α; M = Fd
M = r x F = -(F x r): sense is important
Sin α= d / r
R. Ganesh Narayanan13
Varignon’stheorem:
The moment of a force about any point is equal to the sum of the moments of the
components of the forces about the same point
Concurrent forces –P, Q
Mo= r x R = r x (P Q) = Moment of ‘P’
Usefulness:
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Resultant ‘R’–moment arm ‘d’
Moment of ‘Q’
Force ‘P’–moment arm ‘p’; Force ‘Q’–moment arm ‘q’Mo= Rd = -pP
R. Ganesh Narayanan
14
Calculate the magnitude of the momentabout ‘O’of the force 600 N
1) Mo = 600 cos40 (4)
600 sin 40 (2)
= 2610 Nm (app.)
= -771.34-1839 = 2609.85 Nm (CW);mag= 2610 Nm
R. Ganesh Narayananj
i
15
in mm
Couple:Moment produced by two equal, opposite and non-collinear forces
qQ
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=>-F and F produces rotation=>Mo = F (a d) –Fa= Fd;
Perpendicular to plane?Independent of distance from ‘o’, depends on ‘d’only
?moment is same for all moment centers
R. Ganesh Narayanan
16
Vector algebra method
CCW CW M = rax F
rbx (-F) = (ra-rb) x F = r x F
Equivalent couples
?Changing the F and d values does not change a given couple
as long as the product
(Fd) remains same
?Changing the plane will not alter couple as long as it is parallel
R. Ganesh Narayanan
17
EXAMPLE
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F
All four are equivalent couples
R. Ganesh Narayanan18
Force-couple system
=>Effect of force is two fold –1) to push or pull, 2) rotate the body about any axis
?Dual effect can be represented by a force-couple syatem
?a forcecan be by a force and couple
R. Ganesh Narayanan19
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EXAMPLE
o
Mo = Y
N m
Mo = 80 (9 sin 60) = 624
N m; CCW
R. Ganesh Narayanan20
Resultants
To describe the resultant action of a group or system of forcesResultant:simplest
force combination which replace the original forces without altering the external
effect on the body to which
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the forces are applied
R = F1 F2 F3 ….. = ΣF
Rx = ΣFx; Ry= ΣFy; R = (ΣFx)2
Θ= tan -1(Ry/Rx)
R. Ganesh Narayanan21
How to obtain resultant force ?
F3
D1; F2 –D2; F3 –D3
M1 = F1d1; M2 = F2d2;
M3 = F3d3
R. Ganesh Narayanan
(ΣFy)2
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22
F1 –
Principle of moments
Summarize the above process:
R = ΣF
Mo = ΣM = Σ(Fd)Mo = Rd
First two equations: reduce the system of forces to a force-couple system at some
point ‘O’
Third equation: distance ‘d’from point ‘O’to the line of action ‘R’=>
VARIGNON’S THEOREM IS EXTENDED HERE FOR NON-CONCURENT FORCES
R. Ganesh Narayanan
23
Reference books
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1.Vector
Mechanics
for
Engineers
–Statics&
Dynamics,
Beer
&
Johnston;
7thedition2.Engineering Mechanics Statics& Dynamics, Shames; 4thedition
3.Engineering Mechanics StaticsVol. 1, Engineering Mechanics Dynamics Vol. 2,
Meriam& Kraige; 5thedition4.Schaum’ssolved problems series Vol. 1: Statics; Vol.
2: Dynamics, Joseph F. Shelley
STATICS –MID SEMESTER –DYNAMICS
Tutorial: Monday 8 am to 8.55 am
R. Ganesh Narayanan24
ENGINEERING MECHANICS
TUTORIAL CLASS: Monday 8 AM TO 8.55 AM
Tutorial Groups
Roll Numbers
Class RoomTutors
From
To
TG10701010107010141 (41 Students)L2Prof. R. TiwariTG2
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0701014207010149 (8 Students)1G1
Dr. senthilvelan
07010201
07010233 (33 Students)TG3
0701023407010249 (16 Students)1G2R. GaneshNarayanan
07010301
07010325 (25 Students)TG4
0701032607010353 (28 Students)1202Dr. M. Pandey
07010401
07010413 (13 Students)TG5
0701041407010449 (36 Students)1205Dr. SaravanaKumar
07010601
07010605 (5 Students)
LECTURE CLASSES: LT2 (one will be optional):Monday 3 pm to 3.55 pmTuesday 2 pm to
2.55 pmThursday 5 pm to 5.55 pmFriday 4 pm to 4.55 pm
R. Ganesh Narayanan
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25
Three dimensional force system
Rectangular components
Fx= F cosθx; Fy= F cosθy; Fz= F cosθz
F = Fxi
Fyj
Fzk
l, m, n are directional cosines of ‘F’
= F (i cosθx
j cosθy
Fzk
F
θzθy
Fyj
o
θxFxi
R. Ganesh Narayanan
26
k cosθz) = F (l i
m j
n k)F = F nf
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Moment in 3D
A -a plane in 3D structureMo = F d (TEDIOUS to find d)
or Mo = r x F = –(F x r) (BETTER)
Described in determinant form:
i
j
krxrYrZFX
F
Y
FZ
Expanding …R. Ganesh Narayanan
27
Evaluating the cross product
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Mo = (ryFz-rzFy)i
(rzFx–rxFz)j
(rxFy–ryFx)kMx= ryFz–rzFy; My = rzFx–rxFz;
Mz= rxFy–ryFxMagnitude of the moment Mλof F about λ= Mo . n (scalar reprn.)
(vector reprn.)
Scalar triple product
rxryrzFx
FYFZα, β, γ–DCsof n
αβ
γ
28
Varignon’stheorem in 3D
Mo = rxF1
rxF2
rxF3
…= Σ(r x F)
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= r x (F1 F2 F3 …)= r x (ΣF) = r x R
M
M = rax F
rbx –F = (ra-rb) x F = rxF
29
2D force system; equ. Force-couple; principle of
moments
?Evaluate components of F1, F2, F3, F4?Rx = ΣFx; Ry= ΣFy
?R = Rx i
Ryj?α= tan -1(Ry/Rx)
Ry
R
Rx
R. Ganesh Narayanan?R = 199i
30
F1
14.3j; α= 4.1 deg
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Find F1 and F2
R = F1
F2
3000(cos15i–sin 15j) = F1(cos30i–Sin 30j)
F2(cos45i–sin 45j)EQUATING THE
COMPONENTS OF VECTOR,F1 = 2690 N; F2 = 804 N
R. Ganesh Narayanan31
Find the moment Mo of 780 N about the hinge point
T = -780 COS20 i –780 sin20 j= -732.9 i –266.8 j
r = OA = 10 cos60 i
10 sin 60 j = 5 i
8.6 jMo = r x F = 5014 k ; Mag= 5014 Nm
R. Ganesh NarayananOC –FLAG POLEOAB –LIGHT FRAMED –POWER WINCH
32
Replace couple 1 by eq. couple p, -p; find
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M = 100 (0.1) = 10 Nm (CCW)M = 400 (0.04) cosθ10 = 400 (0.04) cosθ=> Θ= 51.3 deg
R. Ganesh Narayanan100N
33
100N
Find the resultant of four forces and one couple which act on the plateRx = 40
80cos30-60cos45 = 66.9 NRy= 50 80sin 30 60cos45 = 132.4 NR = 148.3 N; Θ
= tan-1
(132.4/66.9) = 63.2 deg
60 N
Mo = 140-50(5) 60cos45(4)-60sin45(7) = -237 Nm
Final LOA of R:148.3 d = 237; d = 1.6 m
LOA of R with x-axis:
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(Xi
yj) x (66.9i 132.4j) = -237k(132.4 x –66.9 y)k= -237k132.4 x -66.9 y = -237
Y = 0 => x = b = -1.792 m
R. Ganesh Narayanan
Couples in 3D
M
M = rax F
rbx –F = (ra-rb) x F = rxF
Equivalent couples
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M = Fd
35
How to find resultant ?
R = ΣF = F1 F2 F3 …
Mo = ΣM = M1 M2 M3 …= Σ(rxF)
M =
Mx2
My2
Mx=
; My =
Mz2;
;
R = ΣFx2
ΣFy2
ΣFz2
Mz=
R. Ganesh Narayanan36
Equilibrium
Body in equilibrium
-necessary & sufficient condition:R = ΣF = 0; M = ΣM = 0
Equilibrium in 2D
Mechanical system: body or group of bodies which can be conceptually isolated from
all other bodies
System: single body, combination of bodies; rigid or non-rigid; combination of fluids
and solidsFree body diagram -FBD:
=> Body to be analyzed is isolated; Forces acting on the body are represented –action
of one body on other, gravity attraction, magnetic force etc.
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=> After FBD, equilibrium equns. can be formed
R. Ganesh Narayanan
37
Modeling the action of forces
Imp
Imp
R. Ganesh Narayanan38
Meriam/Kraige
FBD -Examples
R. Ganesh NarayananEquilibrium equns. Can be solved,
?Some forces can be zero?Assumed sign can be different
39
Meriam/Kraige
Types of 2D equilibrium
Concurrent at a point: ΣFx= 0; ΣFy= 0
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x
F3
= 0Y
X
F1
F4
Parallel: ΣFx= 0; ΣMz
= 0
R. Ganesh Narayanan
Y
X
General: ΣFx= 0; ΣFy= 0; ΣMz= 0
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40
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