7/21/2019 Lecture Statics
1/213
Batch: Jan - May 2008 R. Ganesh Narayanan 1
Engineering Mechanics Statics
InstructorR. Ganesh Narayanan
Department of Mechanical EngineeringIIT Guwahati
7/21/2019 Lecture Statics
2/213
Batch: Jan - May 2008 R. Ganesh Narayanan 2
-These lecture slides were prepared and used by me to conduct lectures for 1st year 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 rectifywhen using the same
- I thank the following authors for making their books available for reference
R. Ganesh Narayanan
1. Vector Mechanics for Engineers Statics & Dynamics, Beer & Johnston; 7th edition
2. Engineering Mechanics Statics & Dynamics, Shames; 4th edition
3. Engineering Mechanics Statics Vol. 1, Engineering Mechanics Dynamics Vol. 2, Meriam &
Kraige; 5th edition
4. Schaums solved problems series Vol. 1: Statics; Vol. 2: Dynamics, Joseph F. Shelley
7/21/2019 Lecture Statics
3/213
R. Ganesh Narayanan 3
Engineering mechanics
- Deals with effect of forces on objects
Mechanics principles used in vibration, spacecraft
design, fluid flow, electrical, mechanical m/c designetc.
Statics: deals with effect of force on bodies whichare not moving
Dynamics: deals with force effect on moving bodies
We consider RIGID BODIES Non deformable
7/21/2019 Lecture Statics
4/213
R. Ganesh Narayanan 4
Scalar quantity: Only magnitude; time, volume, speed,
density, massVector quantity: Both direction and magnitude; Force,
displacement, velocity, acceleration, moment
V = IvI n, where IvI = magnitude, n = unit vectorn = V / IvI
n - dimensionless and in direction of vector V
In our course:y
x
z
j
i
k
i, j, k unit vectors
7/21/2019 Lecture Statics
5/213
R. Ganesh Narayanan 5
Dot product of vectors: A.B = AB cos ; A.B = B.A (commutative)
A.(B+C) = A.B+A.C (distributive operation)
A.B = (Axi+Ayj+Azk).(Bxi+Byj+Bzk) = AxBx+AyBy+AzBz
Cross product of vectors: A x B = C; ICI = IAI IBI Sin ; AxB = -(BxA)
C x (A+B) = C x A + C x B
i j k
A
B
i . i = 1i . j = 0
k i
j
k x j = -i;i x i = 0
AxB = (Axi+Ayj+Azk)x(Bxi+Byj+Bzk) = (AyBz- AzBy)i+( )j+( )k
i j k
Ax AY AZ
BX BY BZ
7/21/2019 Lecture Statics
6/213
R. Ganesh Narayanan 6
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
Force< P kN
Force,P kN
Direction of motion
Body moves
Body doesnot move
P, kN
bulging
7/21/2019 Lecture Statics
7/213
R. Ganesh Narayanan 7
Force system:
P WIREBracket
Magnitude, direction and point of applicationis important
External effect: Forces applied (applied force); Forces exerted bybracket, bolts, foundation.. (reactive force)
Internal effect: Deformation, strain pattern permanent strain;depends on material properties of bracket, bolts
7/21/2019 Lecture Statics
8/213
R. Ganesh Narayanan 8
Transmissibility principle:
A force may be applied at any point on a line of actionwithout changing the resultant effects of the forceapplied external to rigid body on which it acts
Magnitude, direction and line of action is important; notpoint of application
PP
Line ofaction
7/21/2019 Lecture Statics
9/213
R. Ganesh Narayanan 9
Concurrent force:
Forces are said to be concurrent at a point if their lines ofaction intersect at that point
A
F1
F2
R
F1, F2 are concurrent forces
R will be on same plane
R = F1+F2
Plane
Parallelogram law of forces
Polygon law of forces
AF1
F2
R
F2
F1A
F1
F2R
Use triangle law
A F1
R
F2
R does notpass through A
R = F1+F2 R = F1+F2
7/21/2019 Lecture Statics
10/213
R. Ganesh Narayanan 10
Two dimensional force system
Rectangular components:
Fx
Fy
j
i
F
F = Fx + Fy; both are vector components in x, y direction
Fx = fx i ; Fy = fy j; fx, fy are scalar quantities
Therefore, F = fx i + fy j
Fx = F cos ; Fy = F sin
F = fx2 + fy2 ; = tan -1 (fy/fx)+ ve
+ ve
- ve
- ve
7/21/2019 Lecture Statics
11/213
R. Ganesh Narayanan 11
Two concurrent forces F1, F2
Rx = Fx; Ry = Fy
DERIVATION
F2F1
R
i
j
7/21/2019 Lecture Statics
12/213
R. Ganesh Narayanan 12
Moment: Tendency to rotate; torque
Moment about a point: M = Fd
Magnitude of moment is
proportional to the force F and
moment arm d i.e, perpendiculardistance from the axis of rotation
to the LOA of force
UNIT : N-m
Moment is perpendicular to plane about axis O-O
Counter CW = + ve; CW = -ve
B
A
F
d
r
O
O
M
7/21/2019 Lecture Statics
13/213
R. Ganesh Narayanan 13
Cross product:
M = r x F; where r is the position vectorwhich runs fromthe moment reference point A to any point on theLOA of F
M = Fr sin ; M = Fd
M = r x F = -(F x r): sense is important
B
A
d
r
Sin = d / r
7/21/2019 Lecture Statics
14/213
R. Ganesh Narayanan 14
Varignons theorem:
The moment of a force about any point is equal to thesum of the moments of the components of the forcesabout the same point
oQ
P R
r
B
Mo = r x R = r x (P+Q) = r x P + r x Q
Moment of PMoment of Q
Resultant R moment arm d
Force P moment arm p; Force Q moment arm q
Mo= Rd = -pP + qQ
Concurrent forces P, Q
Usefulness:
7/21/2019 Lecture Statics
15/213
R. Ganesh Narayanan 15
Pb:2/5 (Meriam / Kraige):
Calculate the magnitude of the moment
about O of the force 600 N
1) Mo = 600 cos 40 (4) + 600 sin 40 (2)
= 2610 Nm (app.)
2) Mo = r x F = (2i + 4j) x (600cos40i-600sin40j)
= -771.34-1839 = 2609.85 Nm (CW);mag = 2610 Nm
o
600N4
2
A
in mm
40 deg
r
i
j
7/21/2019 Lecture Statics
16/213
R. Ganesh Narayanan 16
Couple: Moment produced by two equal, opposite and
non-collinear forces
-F
+Fa
d
o =>-F and F produces rotation
=>Mo = F (a+d) Fa = Fd;
Perpendicular to planeIndependent of distance from o,depends on d only
moment is same for all momentcenters
M
7/21/2019 Lecture Statics
17/213
R. Ganesh Narayanan 17
Vector algebra method
-F
+F
orb
rar M = ra x F + rb x (-F) = (ra-rb) x F = r x F
CCWCouple
CWCouple
Equivalent couples
Changing the F and d values does not change a given coupleas long as the product (Fd) remains same
Changing the plane will not alter couple as long as it is parallel
7/21/2019 Lecture Statics
18/213
R. Ganesh Narayanan 18
M
-F +Fd
M
-F+F
d
M
-F+F d
-2Fd/2+2F
M
EXAMPLE
All four are equivalent couples
7/21/2019 Lecture Statics
19/213
R. Ganesh Narayanan 19
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-couplesyatem
a force can be replaced by a force and couple
F
A
B
F
A
B F
-F
B F
M = Fd
7/21/2019 Lecture Statics
20/213
R. Ganesh Narayanan 20
o
80N
o
80N
80 N80 N o80 NMo = Y N m
60deg
9 m
Mo = 80 (9 sin 60) = 624 N m; CCW
EXAMPLE
9
60 deg
7/21/2019 Lecture Statics
21/213
R. Ganesh Narayanan 21
Resultants
To describe the resultant action of a group or system of forces
Resultant: simplest force combination which replace the originalforces without altering the external effect on the body to which
the forces are appliedR
R = F1+F2+F3+.. = F
Rx = Fx; Ry = Fy; R = ( Fx)2 + ( Fy)2
= tan -1 (Ry/Rx)
7/21/2019 Lecture Statics
22/213
R. Ganesh Narayanan 22
F1 F2
F3
F1 D1; F2 D2; F3 D3
F1 F2
F3
M1 = F1d1;
M2 = F2d2;
M3 = F3d3
R= F
Mo= Fd
NON-CONCURRENT FORCES
R
d
Mo=Rd
How to obtain resultant force ?
7/21/2019 Lecture Statics
23/213
R. Ganesh Narayanan 23
Principle of moments
Summarize the above process: R = F
Mo = M = (Fd)
Mo = Rd
First two equations: reduce the system of forces to a force-couplesystem at some point O
Third equation: distance d from point O to the line of action R
=> VARIGNONS THEOREM IS EXTENDED HERE FOR NON-CONCURENT FORCES
R= F
Mo= Fd R
d
Mo=Rd
7/21/2019 Lecture Statics
24/213
R. Ganesh Narayanan 24
STATICS MID SEMESTER DYNAMICS
Tutorial: Monday 8 am to 8.55 am
1. Vector Mechanics for Engineers Statics & Dynamics, Beer & Johnston; 7th edition
2. Engineering Mechanics Statics & Dynamics, Shames; 4th edition
3. Engineering Mechanics Statics Vol. 1, Engineering Mechanics Dynamics Vol. 2,Meriam & Kraige; 5th edition
4. Schaums solved problems series Vol. 1: Statics; Vol. 2: Dynamics, Joseph F. Shelley
Reference books
7/21/2019 Lecture Statics
25/213
R. Ganesh Narayanan 25
ENGINEERING MECHANICSTUTORIAL CLASS: Monday 8 AM TO 8.55 AM
07010605 (5 Students)07010601
Dr. Saravana Kumar120507010449 (36 Students)07010414TG5
07010413 (13 Students)07010401
Dr. M. Pandey120207010353 (28 Students)07010326TG4
07010325 (25 Students)07010301
R. Ganesh Narayanan1G207010249 (16 Students)07010234TG3
07010233 (33 Students)07010201Dr. senthilvelan1G107010149 (8 Students)07010142TG2
Prof. R. TiwariL207010141 (41 Students)07010101TG1
ToFrom
TutorsClass RoomRoll NumbersTutorial Groups
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 pm
Friday 4 pm to 4.55 pm
7/21/2019 Lecture Statics
26/213
R. Ganesh Narayanan 26
Three dimensional force system
Rectangular componentsFx = F cos x; Fy = F cos y; Fz = F cos z
F = Fx i + Fy j + Fz k= F (i cos x + j cos y + k cos z) = F (l i + m j + n k)
F = F nf
o
Fx i
Fy j
Fz kF
z
x
y
l, m, n are directional cosines of F
7/21/2019 Lecture Statics
27/213
R. Ganesh Narayanan 27
F
r
Mo
dA A - a plane in 3D structure
Mo = F d (TEDIOUS to find d)
or Mo = r x F = (F x r) (BETTER)
Evaluating the cross product
Described in determinant form: i j krx rY rZ
FX FY FZ
Moment in 3D
Expanding
7/21/2019 Lecture Statics
28/213
R. Ganesh Narayanan 28
Mo = (ryFz - rzFy) i + (rzFx rxFz)j + (rxFy ryFx) k
Mx = ryFz rzFy; My = rzFx rxFz; Mz = rxFy ryFx
Moment about any arbitrary axis :
F
r
Mo n
o
Magnitude of the moment M of F about
= Mo . n (scalar reprn.)
Similarly, M = (r x F.n) n (vector reprn.)
Scalar triple product
rx ry rz
Fx FY FZ
, , DCs of n
7/21/2019 Lecture Statics
29/213
R. Ganesh Narayanan 29
Varignons theorem in 3D
o F1
F3F2
r
B
Mo = rxF1 + rxF2 + rx F3 += (r x F)
= r x (F1+F2+F3+)
= r x (F) = r x R
Couples in 3D
B
M
Ar
ra
rb
d
-F+F
M = ra x F + rb x F = (ra-rb) x F = rxF
7/21/2019 Lecture Statics
30/213
R. Ganesh Narayanan 30
Beer-Johnston; 2.3
F1 = 150N
30
F4 = 100N
15
F3 = 110N
F2 = 80N20
Evaluate components of F1, F2, F3, F4
Rx = Fx; Ry = Fy
R = Rx i + Ry j
= tan -1 (Ry/Rx)
Ry
Rx
R
R = 199i + 14.3j; = 4.1 deg
2D force system; equ. Force-couple; principle ofmoments
7/21/2019 Lecture Statics
31/213
R. Ganesh Narayanan 31
F1
F2 R =3000 N
30 DEG
45 DEG15 DEG
Find F1 and F2
3000 (cos15i sin 15j) = F1 (cos 30i Sin 30j)+ F2 (cos45i sin 45j)
EQUATING THE COMPONENTS OF VECTOR,
F1 = 2690 N; F2 = 804 N
R = F1 + F2
Boat
7/21/2019 Lecture Statics
32/213
R. Ganesh Narayanan 32
o
A
B
20 DEG
C
OC FLAG POLE
OAB LIGHT FRAME
D POWER WINCH
D
780 N
Find the moment Mo of 780 Nabout the hinge point
10m
10
10
T = -780 COS20 i 780 sin20 j
= -732.9 i 266.8 j
r = OA = 10 cos 60 i + 10 sin 60 j = 5 i + 8.6 j
Mo = r x F = 5014 k ; Mag = 5014 Nm
Meriam / kraige; 2/37
7/21/2019 Lecture Statics
33/213
R. Ganesh Narayanan 33
Meriam / kraige; 2/6
Replace couple 1 by eq. couple p, -p; find
M = 100 (0.1) = 10 Nm (CCW)
M = 400 (0.04) cos
10 = 400 (0.04) cos
=> = 51.3 deg
MP
-P
40100
100
100
100N 100N
60
1
2
1
2
60 N
7/21/2019 Lecture Statics
34/213
R. Ganesh Narayanan 34
80N30 deg40 N
50 N2m 5m45
2m
2m
1m
o
140Nm
Find the resultant of four forces and onecouple which act on the plate
Rx = 40+80cos30-60cos45 = 66.9 N
Ry = 50+80sin 30+60cos45 = 132.4 N
R = 148.3 N; = tan-1 (132.4/66.9) = 63.2 deg
Mo = 140-50(5)+60cos45(4)-60sin45(7) = -237 Nm
o
R = 148.3N63.2 deg237 Nm
o
R = 148.3N63.2 deg
148.3 d = 237; d = 1.6 mFinal LOA of R:
o
R = 148.3N
b
xy
(Xi + yj) x (66.9i+132.4j) = -237k
(132.4 x 66.9 y)k = -237k
132.4 x -66.9 y = -237
Y = 0 => x = b = -1.792 m
Meriam / kraige; 2/8
LOA of R with x-axis:
7/21/2019 Lecture Statics
35/213
R. Ganesh Narayanan 35
Couples in 3D
B
M
Ar
ra
rb
d
-F+F
M = ra x F + rb x F = (ra-rb) x F = rxF
F
AB
F
M = Fd
F
AB
F
-F
rB
Equivalent couples
7/21/2019 Lecture Statics
36/213
R. Ganesh Narayanan 36
How to find resultant ?
R = F = F1+F2+F3+
Mo = M = M1+M2+M3+ = (rxF)
M = Mx2 + My2 + Mz2; R = Fx2 + Fy2 + Fz2
Mx = ; My = ; Mz =
7/21/2019 Lecture Statics
37/213
R. Ganesh Narayanan 37
Equilibrium
Body in equilibrium - necessary & sufficient condition:R = F = 0; M = M = 0
Equilibrium in 2D
Mechanical system: body or group of bodies which can be conceptuallyisolated from all other bodies
System: single body, combination of bodies; rigid or non-rigid;combination of fluids and solids
Free 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.
=> After FBD, equilibrium equns. can be formed
7/21/2019 Lecture Statics
38/213
R. Ganesh Narayanan 38
Modeling the action of forces
Meriam/Kraige
Imp
Imp
FBD E l
7/21/2019 Lecture Statics
39/213
R. Ganesh Narayanan 39
FBD - Examples
Meriam/Kraige
Equilibrium equns. Can besolved,
Some forces can be zero
Assumed sign can bedifferent
T f 2D ilib i
7/21/2019 Lecture Statics
40/213
R. Ganesh Narayanan 40
Types of 2D equilibrium
x
F1
F2
F3
Collinear: Fx = 0 F1 F2
F3
F4
Parallel: Fx = 0; Mz = 0
F1
F2
F3F4
X
Y
Concurrent at a point: Fx = 0; Fy = 0
X
Y
M
General: Fx = 0; Fy = 0; Mz = 0
G l ilib i diti
7/21/2019 Lecture Statics
41/213
R. Ganesh Narayanan 41
General equilibrium conditions
Fx = 0; Fy = 0; Fz = 0
Mx = 0; My = 0; Mz = 0
These equations can be used to solve unknown forces,reactions applied to rigid body
For a rigid body in equilibrium, the system of external forces will
impart no translational, rotational motion to the bodyNecessary and sufficient equilibrium conditions
PY QY RYP Q R
7/21/2019 Lecture Statics
42/213
R. Ganesh Narayanan 42
Written in three alternate ways,
AB
C D
PY
Px
QY
Qx
RY
Rx
W
BY
AX
AY
MB = 0 => will not provide new information; used to check thesolution; To find only three unknowns
A B
C D
P Q R
RollerPin
Fx = 0; Fy = 0; MA = 0 I
7/21/2019 Lecture Statics
43/213
R. Ganesh Narayanan 43
Fx = 0; MA = 0; MB = 0 II
Point B can not lie on the line that passes through point A
First two equ. indicate that the ext. forces reduced to a single vertical force at A
Third eqn. (MB = 0) says this force must be zero
Rigid body in equilibrium =>
MA = 0; MB = 0; Mc = 0; III
Body is statically indeterminate: more unknown reactions thanindependent equilibrium equations
Y3D force system
7/21/2019 Lecture Statics
44/213
R. Ganesh Narayanan 44
Meriam / Kraige; 2/10
z
x
12 m
9
B
O
A
15 T = 10kN
Y
Find the moment Mz of T about the z-axis passingthro the base O
3D force system
7/21/2019 Lecture Statics
45/213
R. Ganesh Narayanan 45
F = T = ITI nAB = 10 [12i-15j+9k/21.21] = 10(0.566i-0.707j+0.424k) k NMo = rxF = 15j x 10(0.566i-0.707j+0.424k) = 150 (-0.566k+0.424i) k Nm
Mz = Mo.k= 150 (-0.566k+0.424i).k = -84.9 kN. m
Merial / Kraige; 2/117
7/21/2019 Lecture Statics
46/213
R. Ganesh Narayanan 46
Merial / Kraige; 2/117
Replace the 750N tensile force which the cable exerts on point B by a force-couple system at point O
7/21/2019 Lecture Statics
47/213
R. Ganesh Narayanan 47
F = f , where is unit vector along BC
= (750) BC/IBCI = 750 (-1.6i+1.1j+0.5k/2.005)
F = -599i+412j+188.5k
rob = OB = 1.6i-0.4j+0.8k
Mo = rob x F
= (1.6i-0.4j+0.7k) x (-599i+412j+188.5k)
Mo = - 363i-720j+419.2k
2D equilibrium
7/21/2019 Lecture Statics
48/213
R. Ganesh Narayanan 48
Meriem / Kraige; 3/4
MA = (T cos 25) (0.25) + (T sin 25) (5-0.12) 10(5-1.5-0.12) 4.66 (2.5-0.12) = 0
T = 19.6 kN T
25 deg
y
Ax
Ay
10 kN
0.5 m
4.66 kN
5m
1.5m
0.12 m
Fx = Ax 19.6 cos 25 = 0
Ax = 17.7 kN
Fy = Ay+19.61 sin 25-4.66-10 = 0
Ay = 6.37 kN
A = Ax2 + Ay2 = 18.88kN
Find T and force at A; I-beam with mass of 95kg/meter of length
95 kg/meter => 95(10-3)(5)(9.81) = 4.66kN
BBeer/Johnston; 4.5
7/21/2019 Lecture Statics
49/213
R. Ganesh Narayanan 49
A
40 50 30 10
6060 a 80
mm, N;
Find reactions at A, B if (a) a = 100 mm; (b)a=70 mm
4050 30 10
Bx
By
Ay
a = 100 mm
Ma = 0 => (-40x60)+(-50x120)+(-30x220)+
(-10x300)+(-Byx120) = 0
By = 150 N
Fy = 0 => By-Ay-40-50-30-10 = 0
= 150-Ay-130 = 0 => Ay = 20 N
a = 70 mm
By = 140 N Ay = 10 N
D1 8Beer/Johnston; 4.4
7/21/2019 Lecture Statics
50/213
R. Ganesh Narayanan 50
A B
20 20 20 20
C2.25
3.75
E
4.5
F
1.8
A B
20 20 20 20
C2.25
3.75
E
D
4.5
F1.8
150 kN
Ex
Ey
Find the reaction at the fixed endE
DF = 7.5 m
Fx = Ex + 150 (4.5/7.5) = 0 => Ex = - 90 kN (sign change)
Fy = Ey 4(20)-150 (6/7.5) = 0 => Ey = 200 kN
ME= 20 (7.2) + 20 (5.4) + 20 (3.6) +20 (1.8) (6/7.5) (150) (4.5) +ME= 0
ME= +180 kN.m => ccw
ME
ee /Jo sto ;
Instructions for TUTORIAL
7/21/2019 Lecture Statics
51/213
R. Ganesh Narayanan 51
Instructions for TUTORIAL
Bring pen, pencil, tagged A4 sheets, calculator, text books
Submitted in same tutorial class
Solve div II tutorial problems also
Solve more problems as home work
Tutorial : 10 % contribution in grading Do not miss any tutorial class
QUIZ 1 FEB, 11TH, 2008
3D equilibrium
7/21/2019 Lecture Statics
52/213
R. Ganesh Narayanan 52
q
3D equilibrium equns. can be written in scalar and vector form
F = 0 (or) FX = 0; FY = 0; FZ = 0
M = 0 (or) MX = 0; MY = 0; MZ = 0
F = 0 => Only if the coefficients of i, j, k are zero; FX = 0M = 0 => Only if the coefficients of i, j, k are zero; MX = 0
7/21/2019 Lecture Statics
53/213
Types of 3D equilibrium
7/21/2019 Lecture Statics
54/213
R. Ganesh Narayanan 54
Meriem / KraigeB
z
7/21/2019 Lecture Statics
55/213
R. Ganesh Narayanan 55
A
7 m
6 m
2 my
x
By
Bx
G
W=mg=200 x 9.81
W = 1962 N
Ay
AzAx
h
3.5
3.5
7 = 22 + 62 + h2 => h = 3 m
rAG = -1i-3j+1.5k m; rAB = -2i-6j+3k m
MA = 0 => rAB x (Bx+By) + rAG x W = 0
(-2i-6j+3k) x (Bx i + By j) + (-i-3j+1.5k) x (-1962k) = 0(-3By+5886)i + (3Bx-1962)j + (-2By+6Bx)k = 0
=> By = 1962 N; Bx = 654 N
F = 0 => (654-Ax) i + (1962-Ay) j + (-1962+Az)k = 0
=> Ax = 654 N; Ay = 1963 N; Az = 1962 N; find A
Meriem / Kraige; 3/64
7/21/2019 Lecture Statics
56/213
R. Ganesh Narayanan 56
I.H. Shames
7/21/2019 Lecture Statics
57/213
R. Ganesh Narayanan 57
Find forces at A, B, D. Pin connection at C; E has welded connection
F.B.D. - 1
7/21/2019 Lecture Statics
58/213
R. Ganesh Narayanan 58
F.B.D. - 2
Mc = 0 => (Dy) (15) 200 (15) (15/2)(1/2)(15)(300)[2/3 (15)] = 0
Dy = 3000 N
F.B.D. - 2
F.B.D. - 1
7/21/2019 Lecture Statics
59/213
R. Ganesh Narayanan 59
MB = 0 => -Ay (13) +(3000) (21) 200(34) (34/2-13) (300) (15) [6+2/3(15)]= 0
Ay = -15.4 N
Fy = 0 => Ay+By+3000-200(34)-(1/2)(300)(15) = 0
Sub. Ay here,
=> By = 6065 N
2D, 3D force system Equilibrium equations
7/21/2019 Lecture Statics
60/213
R. Ganesh Narayanan 60
Rectangular components Moment
Varignons theorem
Couple Force-couple system
Resultant
Principle of moment
Fx = 0; Fy = 0; MA = 0
Fx = 0; MA = 0; MB = 0
MA = 0; MB = 0; Mc = 0
2D
F = 0 (or) FX = 0; FY = 0; FZ = 0
M = 0 (or) MX = 0; MY = 0; MZ = 03D
Structures
7/21/2019 Lecture Statics
61/213
R. Ganesh Narayanan 61
Truss: Framework composed of members joined at their ends to form a rigid
structuresPlane truss: Members of truss lie in same plane
Bridge truss
Roof truss
B
7/21/2019 Lecture Statics
62/213
R. Ganesh Narayanan 62
Three bars joined with pins at end Rigid bars and non-collapsible
Deformation due to induced internal strains is negligible
B D
CA
Non-rigid rigid
E
Non rigid body can be made rigid byadding BC, DE, CE elements
B
C
D
A
A
B
c
Instructions for TUTORIAL
7/21/2019 Lecture Statics
63/213
R. Ganesh Narayanan 63
Bring pen, pencil, tagged A4 sheets, calculator, text books
Submitted in same tutorial class
Solve div II tutorial problems also
Solve more problems as home work
Tutorial : 10 % contribution in grading Do not miss any tutorial class
QUIZ 1 FEB, 11TH, 2008
StructuresT F k d f b j i d t th i d t f i id
7/21/2019 Lecture Statics
64/213
R. Ganesh Narayanan 64
Truss: Framework composed of members joined at their ends to form a rigid
structuresPlane truss: Members of truss lie in same plane
Bridge truss
Roof truss
Three bars joined with pins at endB
7/21/2019 Lecture Statics
65/213
R. Ganesh Narayanan 65
Rigid bars and non-collapsible
Deformation due to induced internal strains is negligible
Non rigid body can be made rigid byadding BC, DE, CE elementsB D
CA
Non-rigid rigid
E
B
C
D
A
A c
Simple truss: structures built from basic triangleMore members are present to prevent collapsing => statically indeterminate truss;they can not be analyzed by equilibrium equations
Additional members not necessary for maintaining equilibrium - redundant
In designing simples truss or truss => assumptions are followed
1. Two force members equilibrium only in two forces; either tension or compression
7/21/2019 Lecture Statics
66/213
R. Ganesh Narayanan 66
q y p
2. Each member is a straight link joining two points of application of force3. Two forces are applied at the end; they are equal, opposite and collinear for
equilibrium
4. Newtons third law is followed for each joint
5. Weight can be included; effect of bending is not accepted
6. External forces are applied only in pin connections
7. Roller or rocker is also provided at joints to allow expansion and contraction due to
temperature changes and deformation for applied loads
T
T c
c weight
TWO FORCE MEMBERS
Two methods to analyze force in simple truss
7/21/2019 Lecture Statics
67/213
R. Ganesh Narayanan 67
Method of joints
This method consists of satisfying the conditions of equilibrium for theforces acting on the connecting pin of each joint
This method deals with equilibrium of concurrent forces and only twoindependent equilibrium equations are solved
Newtons third law is followed
Example
EF
7/21/2019 Lecture Statics
68/213
R. Ganesh Narayanan 68
A
B C
D
EF
L
Fy = 0; Fx = 0
Finally sign can be changed if
not applied correctly
Internal and external redundancy
external redundancy: If a plane truss has more supports than are necessary to
7/21/2019 Lecture Statics
69/213
R. Ganesh Narayanan 69
external redundancy: If a plane truss has more supports than are necessary to
ensure a stable equilibrium, the extra supports constitute external redundancy
Internal redundancy: More internal members than are necessary to prevent collapse,the extra members constitute internal redundancy
Condition for statically determinate truss: m + 3 = 2j
- Equilibrium of each joint can be specified by two scalar force equations, then 2jequations are present for a truss with j joints
-The entire truss composed of m two force members and having the maximum ofthree unknown support reactions, there are (m + 3) unknowns
j no. of joints; m no. of members
m + 3 > 2 j =>more members than independent equations; staticallyindeterminate
m + 3 < 2 j => deficiency of internal members; truss is unstable
DBI. H. Shames
Determine the force transmitted by each member;
7/21/2019 Lecture Statics
70/213
R. Ganesh Narayanan 70
AC E
F
1000
10
1010 10
1000
Determine the force transmitted by each member;
A, F = 1000 N
Pin A
FAB
FAC
1000
AFx = 0 =>FAC 0.707FAB = 0
Fy = 0 => -0.707FAB+1000 = 0
FAB = 1414 N; FAC = 1000 N
Pin B
B
FBC
1414
FBD
Fx = 0 => -FBD + 1414COS45 = 0 => FBD = 1000 N
Fy = 0 => -FBC+1414 COS45 = 0 => FBC = 1000 N
1000
FAC
FAB
1414
FBC
FBD
Pin C
1000 FFCE
7/21/2019 Lecture Statics
71/213
R. Ganesh Narayanan 71
B
1000
1000FCE
FDC1000
10001000
1000
FDC
Fx = 0 => -1000 + FCE + FDC COS 45 = 0 => FCE = 1000 N
Fy = 0 => -1000+1000+ FDC COS 45 = 0 => FDC = 0
SIMILARLY D, E, F pins are solved
B D5Find the force in each member of the loaded
il b h d f j i
Meriem / Kraige (similar pbm. 6.1 in Beer/Johnston)
7/21/2019 Lecture Statics
72/213
R. Ganesh Narayanan 72
B D
AC
E30 20
5 5
5
5 55 5
kN, m
cantilever truss by method of joints
ME = 0 => 5T-20(5)-30 (10) = 0; T = 80 kN
FBD of entire truss
7/21/2019 Lecture Statics
73/213
R. Ganesh Narayanan 73
Fx = 0 => 80 cos 30 Ex = 0; Ex = 69.28 kN
Fy = 0 => Ey +80sin30-20-30 = 0 => Ey = 10kN
Fx = 0; Fy = 0
Find AB, AC forcesFx = 0; Fy = 0
Find BC, BD forces
Fx = 0; Fy = 0
Find CD, CE forces
Fy = 0
Find DE forces
Fx = 0 can be checked
FBD of joints
Q = 100 N; smooth surfaces; Findreactions at A, B, C Q
Qroller
7/21/2019 Lecture Statics
74/213
R. Ganesh Narayanan 74
AB
30
c100100
RA
Rc
roller
RB
F = 0 => (-RA cos 60 - RB cos 60 + Rc) i + (-2 x 100 + RBsin 60 + RA sin 60)j = 0
RC = (RA + RB)/2RB + RA = 230.94
RC = 115.5 N
RB
100
Rc
RAB30
F = 0 => (-RAB cos 30 - RB cos 60 + Rc) i + (RB Sin 60 100 - RABsin 30)j = 0
0.866 RAB + 0.5 RB = 115.5; -0.5 RAB + 0.866 RB = 100RAB = 50 N (app.); RB = 144.4 N; RA = 230.94-144.4 = 86.5 N
7/21/2019 Lecture Statics
75/213
Methodology for method of joints
EF
7/21/2019 Lecture Statics
76/213
R. Ganesh Narayanan 76
A
B C
D
L
Fy = 0; Fx = 0
Finally sign can be changed if
not applied correctly
7/21/2019 Lecture Statics
77/213
Method of sections
7/21/2019 Lecture Statics
78/213
R. Ganesh Narayanan 78
In method of joints, we need only two equilibrium equations, as wedeal with concurrent force system
In method of sections, we will consider three equilibrium
equations, including one moment equilibrium eqn. force in almost any desired member can be obtained directly froman analysis of a section which has cut the member
Not necessary to proceed from joint to joint
Not more than three members whose forces are unknown shouldbe cut. Only three independent equilibrium eqns. are present
Efficiently find limited information
A
Methodology for method of sections
7/21/2019 Lecture Statics
79/213
R. Ganesh Narayanan 79
AB
C
D
EF
L
The external forces are obtained initially from method of joints, byconsidering truss as a whole
Assume we need to find force in BE, then entire truss has to be
sectioned across FE, BE, BC as shown in figure; we have only 3equilibrium equns.
AA section across FE, BE, BC; Forces in these members are
initially unknown
A BC
D
EF
LR1 R2A
A
7/21/2019 Lecture Statics
80/213
R. Ganesh Narayanan 80
Now each section will apply opposite forces on each other
The LHS is in equilibrium with R1, L, three forces exerted on the cutmembers (EF, BE, BC) by the RHS which has been removed
IN this method the initial direction of forces is decided by moment aboutany point where known forces are present
For eg., take moment about point B for the LHS, this will give BE, BC tobe zero; Then moment by EF should be opposite to moment by R1;
Hence EF should be towards left hand side - compressive
Section 1 Section 2
Now take moment about F => BE should be opposite to R1moment; Hence BE must be up and to the right; So BE is tensile
7/21/2019 Lecture Statics
81/213
R. Ganesh Narayanan 81
Now depending on the magnitudes of known forces, BC directionhas to be decided, which in this case is outwards i.e., tensile
MB = 0 => FORCE IN EF; BE, BC = 0
Fy = 0 => FORCE IN BE; BC, EF = 0 ME = 0 => FORCE IN BC; EF, BE = 0
Section 1 Section 2
7/21/2019 Lecture Statics
82/213
R. Ganesh Narayanan 82
Section AA and BB arepossible
convenient
Important points
IN method of sections, an entire portion of the truss is considered a
7/21/2019 Lecture Statics
83/213
R. Ganesh Narayanan 83
single body in equilibrium Force in members internal to the section are not involved in theanalysis of the section as a whole
The cutting section is preferably passed through members and notthrough joints
Either portion of the truss can be used, but the one with smaller
number of forces will yield a simpler solution Method sections and method of joints can be combined
Moment center can be selected through which many unknown forcespass through
Positive force value will sense the initial assumption of force direction
Meriem/Kraige
Find the forces included in members KL,
7/21/2019 Lecture Statics
84/213
R. Ganesh Narayanan 84
CL, CB by the 20 ton load on the cantilevertruss
Section 1 Section 2
x
Moment abt. L => CB is compressive => creates CW moment
Moment abt. C => KL is tensile => creates CW moment
CL is assumed to be compressive
yKL
20 T
C CB
CL
G P
L
K
yKL
L
K
7/21/2019 Lecture Statics
85/213
R. Ganesh Narayanan 85
20 T
C CB
CL
GP
x
Section 1 Section 2
ML = 0 => 20 (5) (12)- CB (21) = 0 => CB = 57.1 t (C)
Mc = 0 => 20 (4)(12) 12/13 (KL) (16) = 0; KL = 65 t (T)
Mp = 0 => find PC distance and find CL; CL = 5.76 t (C)
BL = 16 + (26-16)/2 = 12 ft
= tan -1 (5/12) => cos = 12/13
Meriem/Kraige
Find the force in member DJ of the trussshown. Neglect the horizontal force in
7/21/2019 Lecture Statics
86/213
R. Ganesh Narayanan 86
supports
Section 2 cuts four members, but we have only3 equi. Equns
Hence consider section 1 which cuts only 3members CD, CJ, KJ
Consider FBD for whole truss and findreaction at A
MG = -Ay (24) +(10) (20) + 10(16) + 10(8) = 0
Ay = 18. 3 kN => creates CW moment
Force direction Moment abt. A => CD, JK Eliminated; CJ will be upwards creating CCW moment
Moment abt. C => JK must be towards right creating CCW moment
ASSUME CD TO HAVE TENSILE FORCE
7/21/2019 Lecture Statics
87/213
R. Ganesh Narayanan 87
MA = 0 => CJ (12) (0.707) 10 (4) -10( 8) =0; CJ = 14.14 KnMJ = 0 => 0.894 (CD) (6) +18.33 (12)-10(4)-10(8) = 0; CD = -18.7 kN
CD direction is changed
From section 1 FBD
From section 2 FBD
MG = 0 => 12 DJ +10(16)+10(20)-18.3 (24)-
14.14 (0.707)(12) = 0DJ = 16.7 kN
I.H. Shames FBD - 1
7/21/2019 Lecture Statics
88/213
R. Ganesh Narayanan 88
FBD - 2
From FBD-2
MB = 0 => -(10)(500)+30 (789)- FAC Sin 30 (30) = 0
FAC = 1244.67 N
From FBD -1Fx = 0 => FDA Cos 30 (1244.67) cos 30 1000 sin 30 = 0 ;
FDA = 1822 N
Fy = 0 => (1822)Sin 30 + (1244.67) sin 30 +FAB 1000 Cos 30 = 0; FAB = -667 N
Frames and machines
Multi force members: Members on which three or more forces acting
7/21/2019 Lecture Statics
89/213
R. Ganesh Narayanan 89
on it (or) one with two or more forces and one or more couples actingon it
Frame or machine: At least one of its member is multi force member
Frame: Structures which are designed to support applied loads andare fixed in position
Machine: Structure which contain moving parts and are designed to
transmit input forces or couples to output forces or couples
Frames and machines contain multi force members, the forces inthese members will not be in directions of members
Method of joints and sections are not applicable
Inter-connected rigid bodies with multi force members
Previously we have seen equilibrium of single rigid bodies
7/21/2019 Lecture Statics
90/213
R. Ganesh Narayanan 90
Now we have equilibrium of inter-connected members whichinvolves multi force members
Isolate members with FBD and applying the equilibrium equations Principle of action and reaction should be remembered
Statically determinate structures will be studied
Force representation and FBD
Representing force by rectangular components
7/21/2019 Lecture Statics
91/213
R. Ganesh Narayanan 91
Calculation of moment arms will be simplified
Proper sense of force is necessary; Some times arbitrary assignment
is done; Final force answer will yield correct force direction Force direction should be consistently followed
Frames and machines
Multi force members: Members on which three or more forces acting
it ( ) ith t f d l ti
7/21/2019 Lecture Statics
92/213
R. Ganesh Narayanan 92
on it (or) one with two or more forces and one or more couples actingon it
Frame or machine: At least one of its member is multi force member
Frame: Structures which are designed to support applied loads andare fixed in position
Machine: Structure which contain moving parts and are designed to
transmit input forces or couples to output forces or couples
Frames and machines contain multi force members, the forces inthese members will not be in directions of members
Method of joints and sections are not applicable
Inter-connected rigid bodies with multi force members
Previously we have seen equilibrium of single rigid bodies
7/21/2019 Lecture Statics
93/213
R. Ganesh Narayanan 93
Now we have equilibrium of inter-connected members whichinvolves multi force members
Isolate members with FBD and applying the equilibrium equations Principle of action and reaction should be remembered
Statically determinate structures will be studied
Force representation and FBD
Representing force by rectangular components
7/21/2019 Lecture Statics
94/213
R. Ganesh Narayanan 94
Calculation of moment arms will be simplified
Proper sense of force is necessary; Some times arbitrary assignment
is done; Final force answer will yield correct force direction Force direction should be consistently followed
Full truss
K, J are un-necessaryhere
7/21/2019 Lecture Statics
95/213
R. Ganesh Narayanan 95
AFAE
BD
Meriem/KraigeB
30 lb
1220 ft
7/21/2019 Lecture Statics
96/213
R. Ganesh Narayanan 96
A C
D
E
F50 lb
12
12
30 ft
20 ft
20 ftFind the forces in all the frames;neglect weight of each member
AxAy
50 lb
30 lb
Cx
Cy
Mc = 0 => 50 (12) +30(40)-30 (Ay) = 0; Ay = 60 lb
Fy = 0 => Cy 50 (4/5) 60 = 0 => Cy = 100 lb
FBD of full frame
ED:
EF: Two force member; E, F arecompressive
FBD of individual members
7/21/2019 Lecture Statics
97/213
R. Ganesh Narayanan 97
ED:MD = 0 => 50(12)-12E = 0 => E = 50 lb
F = 0 => D-50-50 = 0 => D= 100 lb
(components will be eliminated)
EF: F = 50 lb (opposite and equal to E)
AB:
MA = 0 => 50(3/5)(20)-Bx (40) = 0 => Bx = 15 lb
Fx = 0 => Ax+15-50(3/5) = 0 => Ax = 15 lb
Fy = 0 => 50 (4/5)-60-By = 0 =>By = -20 lb
BC: Fx = 0 => 30 +100 (3/5)-15-Cx = 0 => Cx = 75 lb
D
Fx = -50 (cos 53.1)+15+15 = -30+15+15 = 0
F
Fx
Fy
53.1 deg
E
A
B160
Find the force in link DE and components offorces exerted at C on member BCD
7/21/2019 Lecture Statics
98/213
R. Ganesh Narayanan 98
B
C
E
D
60100 150
480 N
60
80
A
B
C
E
D
100 150
480 N160
Ax
Ay
Bx
80
Fy = 0 => Ay-480 = 0 =>Ay = 480 N
MA = 0 => Bx (160)-480 (100) = 0 => Bx = 300 N
Fx = 0 => 300+Ax = 0 => Ax = -300 N
FBD of full frame
= tan -1 (80/150) = 28.07 deg
FBD of BCD
B
480 N
300
Cy
D
FD
E
DE: Two force member
Ay
FBD of AE
7/21/2019 Lecture Statics
99/213
R. Ganesh Narayanan 99
B
CD
300 Cx
FDE
Mc = 0 => -FDE sin 28.07 (250) 300(80)-480 (100) = 0; FDE = -561 N Fx = 0 => Cx (-561) cos 28.07 +300 = 0 => Cx = -795 N
Fy = 0 => Cy (-561) sin 28.07 480 = 0 => Cy = 216 N
E
D
FDE
D
FDE
A
E
Ax
Ay
FDE
Cy
Cx
FBD of DE
7/21/2019 Lecture Statics
100/213
7/21/2019 Lecture Statics
101/213
Machines Machines are structures designed to transmit and modify forces. Their main purpose
is to transform input forces into output forces.
Gi h i d f d i h
7/21/2019 Lecture Statics
102/213
R. Ganesh Narayanan 102
Given the magnitude ofP, determine the
magnitude of Q.
Taking moments about A,
Pb
aQbQaPMA === 0
Center of mass & center of gravity
7/21/2019 Lecture Statics
103/213
R. Ganesh Narayanan 103
A B
C
W
G
A
B
C
W
G
W
GA
B
C
G
Body of mass mBody at equilibrium w.r.t. forces in the cord and resultant of gravitationalforces at all particles W
W is collinear with point AChanging the point of hanging to B, C Same effect
All practical purposes, LOA coincides with G; G center of gravity
BODY
dw
z YMoment abt. Y axis = dw (x)
Sum of moments for small regions through out thebody: x dw
7/21/2019 Lecture Statics
104/213
R. Ganesh Narayanan 104
G
w
X
Moment of w force with Y axis = w x
x dw = w x
Sum of moments Moment of the sum
W = mg
r
r
X = ( x dm) / m
X = ( x dw) / w Y = ( y dw) / w Z = ( z dw) / w
Y = ( y dm) / m Z = ( z dm) / m
1
2
r = ( r dm) / mIn vector form,
= m/V; dm = dv
3
7/21/2019 Lecture Statics
105/213
R. Ganesh Narayanan 105
= m/V; dm = dv
X = ( x dv) / dv
Y = ( y dv) / dv
Z = ( z dv) / dv
= not constant through outbody
4
Equns 2, 3, 4 are independent of g; They depend only on mass distribution;This define a co-ordinate point center of mass
This is same as center of gravity as long as gravitational field is uniform and parallel
Centroids of lines, areas, volumes
Suppose if density is constant, then the expression define a purelygeometrical property of the body; It is called as centroid
7/21/2019 Lecture Statics
106/213
R. Ganesh Narayanan 106
X = ( xc dv) / v Y = ( yc dv) / v Z = ( zc dv) / v
g p p y y;
Centroid of volume
X = ( x dA) / A Y = ( y dA) / A Z = ( z dA) / A
Centroid of area
X = ( x dL) / L Y = ( y dL) / L Z = ( z dL) / LCentroid of line
y
h
xydy
Find the y-coordinate of centroid of the triangular area
X / (h-y) = b/h
7/21/2019 Lecture Statics
107/213
R. Ganesh Narayanan 107
x
yAY = y dA
b h (y) = y (x dy) = y [b (h-y) / h] dy = b h2 / 6
b
0
h
0
h
Y = h / 3
Beams
Structural members which offer resistance to bending due toapplied loads
7/21/2019 Lecture Statics
108/213
R. Ganesh Narayanan 108
pp
Reactions at beam supports are determinate if they involve only three
unknowns. Otherwise, they are statically indeterminate
External effects in beams
Reaction due to supports, distributed load, concentrated loads
7/21/2019 Lecture Statics
109/213
R. Ganesh Narayanan 109
Internal effects in beams
Shear, bending, torsion of beams
v
v
M M
SHEARBENDING TORSION
compression
Tension
7/21/2019 Lecture Statics
110/213
R. Ganesh Narayanan 110
Cx
W
D
E
B
CF
G A
SECTION - J
F
D
J
TV
M
V SHEAR FORCE
F AXIAL FORCE
M BENDING MOMENT AT J
TD
CyFBE
AX
AY
J
A
A
J
F
VM
Internal forces in beam
Shear force and bending moment in beam
7/21/2019 Lecture Statics
111/213
R. Ganesh Narayanan 111
To determine bending moment and shearing
force at any point in a beam subjected to
concentrated and distributed loads.
1. Determine reactions at supports by
treating whole beam as free-body
FINDING REACTION FORCES AT A AND B
2. SECTION beam at C and draw free-body
DIRECTION OF V AND M
7/21/2019 Lecture Statics
112/213
R. Ganesh Narayanan 112
2. SECTION beam at Cand draw free body
diagrams forACand CB. By definition,
positive sense for internal force-couple
systems are as shown.
SECTION C
SECTION CSECTION C
+ VE SHEAR FORCE
+VE BENDING MOMENTV
M
V
M
EVALUATING V AND M
7/21/2019 Lecture Statics
113/213
R. Ganesh Narayanan 113
Apply vertical force equilibrium eqn. to AC, shear force at C, i.e.,V can be determined
Apply moment equilibrium eqn. at C, bending moment at C, i.e.,M can be determined; Couple if any should be included
+ ve value of V => assigned shear force direction is correct
+ ve value of M => assigned bending moment is correct
Evaluate the Variation of shear and bendingmoment along beam
Beer/Johnston
7/21/2019 Lecture Statics
114/213
R. Ganesh Narayanan 114
MB= 0 =>RA (-L)+P (L/2) = 0; RA= +P/2RB = +P/2
SECTION AT C
Between A & D
SECTION AT E
Between D & B
SECTION AT C; C is at x distance from A
Member AC: Fy = 0 => P/2-V = 0; V = +P/2Mc = 0 => ( P/2) (X) + M = 0; M = +PX/2
7/21/2019 Lecture Statics
115/213
R. Ganesh Narayanan 115
Mc = 0 => (- P/2) (X) + M = 0; M = +PX/2
Any section between A and D willyield same result
V = +P/2 is valid from A to D
V = +P/2 yields straight line from Ato D (or beam length : 0 to L/2)
M = +PX/2 yield a linear straight line
fit for beam length from 0 to L/2
SECTION AT E; E is at x distance from A
7/21/2019 Lecture Statics
116/213
R. Ganesh Narayanan 116
CONSIDER AE:
Fy = 0 => P/2-P-V = 0; V = -P/2
ME = 0 => (- P/2) (X) +P(X-L/2)+ M = 0; M = +P(L-X)/2
EB CAN ALSO BE CONSIDERED
7/21/2019 Lecture Statics
117/213
R. Ganesh Narayanan 117
V = V0 + (NEGATIVE OF THE AREA UNDER THE LOADING
CURVE FROM X0 TO X) = V0 - w dxM = M0 + (AREA UNDER SHEAR DIAGRAM FROM X0 TO X) = M0+ V dx
c1
Slide 117
c1 cclab9, 1/24/2008
7/21/2019 Lecture Statics
118/213
Beer/Johnston Taking entire beam as free-body, calculatereactions atA andB.
Determine equivalent internal force-couplesystems at sections cut within segmentsAC,
CD and DB
7/21/2019 Lecture Statics
119/213
R. Ganesh Narayanan 118
:0=M
( ) ( )( ) ( )( ) 0cm22N400cm6N480cm32 =yB
N365=yB:0= BM
( )( ) ( )( ) ( ) 0cm32cm10N400cm26N480 =+ A
N515=:0= xF 0=xB
The 400 N load atEmay be replaced by a 400 N force and 1600 N-cm couple at
D.
CD, andDB.
:01=M 040515 21 =+ Mxxx
2
From A to C:
= :0yF 040515 = VxV 40515=
7/21/2019 Lecture Statics
120/213
R. Ganesh Narayanan 119:02 =M ( ) 06480515 =++ Mxx
( ) cmN352880 += xM
From CtoD:
= :0yF 0480515 = VN35=V
220515 xx=
V = 515 + (-40 X) = 515-40X = 515 - 40 dx
M = 515-40x dx = 515x-20 x2
0
x
0
x
Evaluate equivalent internal force-couple systems
7/21/2019 Lecture Statics
121/213
R. Ganesh Narayanan 120
q p y
at sections cut within segmentsAC, CD, andDB.
FromD toB:
= :0yF 0400480515 = V
N365=V:02 =M
( ) ( ) 01840016006480515 =+++ Mxxx
( ) cmN365680,11 = xM
Shear force & Bending moment plot
7/21/2019 Lecture Statics
122/213
R. Ganesh Narayanan 121
AC: (35X12) + (1/2 x 12 x 480) = 33000 to 3300
CD: 3300 +(35X6) = 3510
3300 to 3510DB: 365 x 14 = 5110
5110 to 0
AREA UNDER SHEAR FORCE DIAGRAM GIVES BM DIAGRAM
300 lb100 lb/ftFind the shear force andbending moment for theloaded beam
7/21/2019 Lecture Statics
123/213
R. Ganesh Narayanan 122
4 ft 4 2 2
Machine
7/21/2019 Lecture Statics
124/213
R. Ganesh Narayanan 123
7/21/2019 Lecture Statics
125/213
R. Ganesh Narayanan 124
FrictionEarlier we assumed action and reaction forces at contacting surfaces
are normal Seen as smooth surface not practically true
7/21/2019 Lecture Statics
126/213
R. Ganesh Narayanan 125
p y
Normal & tangential forces are important
Tangential forces generated near contacting surfaces areFRICTIONAL FORCES
Sliding of one contact surface to other friction occurs and it isopposite to the applied force
Reduce friction in bearings, power screws, gears, aircraft propulsion,missiles through the atmosphere, fluid flow etc.
Maximize friction in brakes, clutches, belt drives etc.
Friction dissipated as heat loss of energy, wear of parts etc.
Friction
Dry friction
( l b f i ti )
Fluid friction
7/21/2019 Lecture Statics
127/213
R. Ganesh Narayanan 126
(coulomb friction)
Occurs when un-lubricated surfaces arein contact during sliding
friction force always oppose the sliding
motion
Occurs when the adjacent layers in a
fluid (liquid, gas) are moving at differentvelocities
This motion causes friction betweenfluid elements
Depends on the relative velocitybetween layers
No relative velocity no fluid friction
depends on the viscosity of fluidmeasure of resistance to shearing actionbetween the fluid layers
Dry friction: Laws of dry frictionW
W weight; N Reaction of the surface Only vertical componentA
7/21/2019 Lecture Statics
128/213
R. Ganesh Narayanan 127
N
P applied load
F static friction force : resultant of many forces acting overthe entire contact area
Because of irregularities in surface & molecular attraction
W
P
N
F
A
P
W
F
A B
Fm
Fk
F
p
Equilibrium Motion
7/21/2019 Lecture Statics
129/213
R. Ganesh Narayanan 128
N
P is increased; F is also increased and continue to oppose P This happens till maximum Fm is reached Body tend to move till Fm is reached
After this point, block is in motion
Block in motion: Fm reduced to Fk lower value kinetic friction force and itremains same related to irregularities interaction
N reaches B from A Then tipping occurs abt. B
p
More irregularitiesinteraction
Less irregularitiesinteraction
EXPERIMENTAL EVIDENCE:
Fm proportional to NFm = s N; s static friction co-efficient
7/21/2019 Lecture Statics
130/213
R. Ganesh Narayanan 129
Similarly, Fk = k N; k kinetic friction co-efficient
s and k depends on the nature of
surface; not on contact area ofsurface
k = 0.75 s
Four situations can occur when a rigid body is in contact with a
horizontal surface:
We have horizontal and vertical force equilibrium equns. and
F = N F
7/21/2019 Lecture Statics
131/213
R. Ganesh Narayanan 130
F N
No motion,
(Px
7/21/2019 Lecture Statics
132/213
R. Ganesh Narayanan 131
No motion Motion No friction Motion impending
ss
sms N
N
N
F
===
tan
tan
kk
kkk N
N
N
F
===
tan
tan
s angle of staticfriction maximum angle(like Fm)
k angle of kineticfriction; k < s
Consider block of weight Wresting on board with variable inclination
angle .
ANGLE OF INCLINATION IS INCREASING
7/21/2019 Lecture Statics
133/213
R. Ganesh Narayanan 132
Angle of inclination =
angle of repose; = s
R Not vertical
Three categories of problems
All applied forces are given, co-effts. of friction are known
Find whether the body will remain at rest or slide
First category: to know a body slips or not
7/21/2019 Lecture Statics
134/213
R. Ganesh Narayanan 133
y
Friction force F required to maintain equilibrium is unknown(magnitude not equal to sN)
Determine F required for equilibrium, by solving equilibrium equns; Also findN
Compare F obtained with maximum value Fm i.e., from Fm = sN
F is smaller or equal to Fm, then body is at rest
Otherwise body starts moving
Actual friction force magnitude = Fk= kN
Solution
A 100 N force acts as shown on a 300 N block
placed on an inclined plane. The coefficients of
friction between the block and plane are s = 0.25and k= 0.20. Determine whether the block is in
equilibrium and find the value of the friction force.
Beer/Johnston
7/21/2019 Lecture Statics
135/213
R. Ganesh Narayanan 134
:0= xF ( ) 0N300-N100 53 =F
N80=F:0= yF ( ) 0N300- 5
4 =N
N240=
The block will slide down the plane.
Fm < FFm = s N = 0.25 (240) = 60 N
= 36.9 DEG
= 36.9DEG
If maximum friction force is less than friction force
required for equilibrium, block will slide. Calculatekinetic-friction force.
NFF
7/21/2019 Lecture Statics
136/213
R. Ganesh Narayanan 135
( )N240200
N
.
FF kkactual
=
==
N48=actualF
FmFk
F
p
Equilibrium Motion
Meriam/Kraige; 6/8
M
Cylinder weight: 30 kg; Dia: 400 mm
Static friction co-efft: 0.30 between cylinder and surface
Calculate the applied CW couple M which cause the cylinderto slip
7/21/2019 Lecture Statics
137/213
R. Ganesh Narayanan 136
30
30 x 9.81
NA
FA = 0.3 NANB
FB = 0.3 NB
M
C
Fx = 0 = -NA+0.3NB Cos 30-NB Sin 30 = 0Fy = 0 =>-294.3+0.3NA+NBCos 30-0.3NB Sin 30 = 0
Find NA & NB by solving these two equns.
MC = 0 = > 0.3 NA (0.2)+0.3 NB (0.2) - M = 0
Put NA & NB; Find M
NA = 237 N & NB = 312 N; M = 33 Nm
Meriam/Kraige; 6/5
Wooden block: 1.2 kg; Paint: 9 kg
Paint
Woodenblock
12
4
Roof
Second category: Impending relative motion when two orthree bodies in contact with each other
7/21/2019 Lecture Statics
138/213
R. Ganesh Narayanan 137
g; gDetermine the magnitude and direction of (1) the friction
force exerted by roof surface on the wooden block, (2)total force exerted by roof surface on the wooden block
= tan-1 (4/12) = 18.43
Roofsurface
(2) Total force = 10.2 x 9.81 = 100.06 N UP
N
F
10.2x 9.81
X
Y
(1)Fx = 0 => -F+100.06 sin 18.43 => F = 31.6 N
Fy = 0 => N = 95 N
Beer/Johnston
For 20 kg block For 30 kg block(a)
7/21/2019 Lecture Statics
139/213
R. Ganesh Narayanan 138
20 x 9.81 = 196.2 N
N1
F1
T
30 x 9.81 = 294.3 N
F2
For 20 kg block For 30 kg block
F1P
N1
N2
(a)
(B)490.5 N
7/21/2019 Lecture Statics
140/213
R. Ganesh Narayanan 139
N
P
Beer/JohnstonB
6 m
A 6.5-m ladder AB of mass 10 kg leans against a wall as shown.Assuming that the coefficient of static friction on s is the same at
both surfaces of contact, determine the smallest value of s for whichequilibrium can be maintained.
7/21/2019 Lecture Statics
141/213
R. Ganesh Narayanan 140
A
2.5 m
A
B
FB
NB
FANA
W
1.25 1.25
O
Slip impends at both A and B, FA= sNA, FB= sNB
Fx=0=> FANB=0, NB=FA=sNA
Fy=0=> NAW+FB=0, NA+FB=W
NA+sNB=W; W = NA(1+s2)
Mo = 0 => (6) NB
- (2.5) (NA
) +(W) (1.25) = 0
6sNA - 2.5 NA + NA(1+s2) 1.25 = 0
s = -2.4 2.6 = > Min s = 0.2
Wedges
Wedges - simple machines used to raise heavyloads like wooden block, stone etc.
Loads can be raised by applying force P to
7/21/2019 Lecture Statics
142/213
R. Ganesh Narayanan 141
wedge
Force required to lift block is significantly lessthan block weight
Friction at AC & CD prevents wedge from slidingout
Want to find minimum force P to raise block
A wooden block
C, D Wedges
7/21/2019 Lecture Statics
143/213
Two 8 wedges of negligible weight are used to moveand position a 530-N block. Knowing that the
coefficient of static friction is 0.40 at all surfaces ofcontact, determine the magnitude of the force P forwhich motion of the block is impending
Beer/Johnston
7/21/2019 Lecture Statics
144/213
R. Ganesh Narayanan 143
s = tan1
s = tan1
(0.4) = 21.801
21.8R1
FBD of block
20
21.8
530
R2530
R2
R141.8
91.8 46.4 (R2/Sin 41.8) = (530/sin 46.4)
R2 = 487.84 N
Using sine law,
slip impends at wedge/blockwedge/wedge and block/incline
7/21/2019 Lecture Statics
145/213
R. Ganesh Narayanan 144
P = 440.6 N
Beer/Johnston
A 6 steel wedge is driven into the end of an ax handleto lock the handle to the ax head. The coefficient of
static friction between the wedge and the handle is0.35. Knowing that a force P of magnitude 60 N wasrequired to insert the wedge to the equilibrium positionshown determine the magnitude of the forces exerted
7/21/2019 Lecture Statics
146/213
R. Ganesh Narayanan 145
shown, determine the magnitude of the forces exerted
on the handle by the wedge after force P is removed.
P = 60 N s = tan 1 s= tan 1 (0.35 ) = 19.29
19.293
19.29
36
By symmetry R1= R2; in EQUILIBRIUM
Fy = 0: 2R1 sin 22.29 60 N =0
R1 = R2 = 79.094 N
WHAT WILL HAPPEN IF P IS REMOVED ?
R1R2
7/21/2019 Lecture Statics
147/213
Screws
Used for fastening, transmitting power or motion, lifting body
Square threaded jack - screw jack V-thread is alsopossible
7/21/2019 Lecture Statics
148/213
R. Ganesh Narayanan 147
W- AXIAL LOAD
M APPLIED MOMENT ABOUT AXIS OF SCREW
M = P X r
L LEAD DISTANCE Advancement per revolution
HELIX ANGLE
M
Upwardmotion
L
W
RP = M/r
One full threadof screw
To raise load
M
F
R
P
w +
tan (+) = P/W = M/rW
=> M = rW tan (+)
7/21/2019 Lecture Statics
149/213
R. Ganesh Narayanan 148
2r
angle of friction
> M rW tan ( )
= tan-1 (L/2r)
To lower load unwinding condition
P = M/r
W
R
< Screw will remain in place self locking
=> M = rW tan (-)
= In verge of un-winding
Moment required tolower the screw
W > Screw will unwind itself
7/21/2019 Lecture Statics
150/213
R. Ganesh Narayanan 149
P = M/r
R
=> M = rW tan (-)Moment required toprevent unwinding
Beer/Johnston A clamp is used to hold two pieces of wood togetheras shown. The clamp has a double square thread of
mean diameter equal to 10 mm with a pitch of 2 mm.
The coefficient of friction between threads is s =0.30.
If a maximum torque of 40 Nm is applied in
tightening the clamp, determine (a) the force exerted
7/21/2019 Lecture Statics
151/213
R. Ganesh Narayanan 150
tightening the clamp, determine (a) the force exerted
on the pieces of wood, and (b) the torque required toloosen the clamp.
Lead distance = 2 x pitch = 2 x 2 = 4 mm
r = 5 mm
( )
30.0tan
1273.0
mm10
mm22
2
tan
==
===
ss
r
L
= 3.7
= 7.16s
(double square thread)
a) Forces exerted on the wooded pieces
M/r tan (+) = WW = 40 / (0.005) tan (24) = 17.96 kN
b) the torque required to loosen the clamp
7/21/2019 Lecture Statics
152/213
R. Ganesh Narayanan 151
b) the torque required to loosen the clamp
M = rW tan (-) = 0.005 (17.96) tan (9.4)M = 14.87 Nm
The position of the automobile jack shown iscontrolled by a screw ABC that is single-
threaded at each end (right-handed thread at A,left-handed thread at C). Each thread has a pitchof 2 mm and a mean diameter of 7.5 mm. If thecoefficient of static friction is 0.15, determine themagnitude of the couple M that must be applied
Beer/Johnston
7/21/2019 Lecture Statics
153/213
R. Ganesh Narayanan 152
magnitude of the couple M that must be applied
to raise the automobile.
FBD joint D:
Fy = 0 => 2FADsin254 kN=0
FAD = FCD = 4.73 kN
By symmetry:
4 kN
FAD FCD
25 25D
FBD joint A:
4
.73 kN
FAC
FAE = 4.73
25
25A
Fx = 0 => FAC2(4.73) cos25=0
FAC = 8.57 kN
Joint A
7/21/2019 Lecture Statics
154/213
R. Ganesh Narayanan 153
L = Pitch = 2 mm
W = FAC = 8.57
R
Joint A
P = M/r
(7.5)
Here is used instead of used earlier
MA = rW tan (+) = (7.5/2) (8.57) tan (13.38) = 7.63 Nm
Similarly, at C, Mc = 7.63 Nm (by symmetry); Total moment = 7.63 (2) = 15.27 Nm
7/21/2019 Lecture Statics
155/213
F i ti b t tFriction in full circular area
7/21/2019 Lecture Statics
156/213
R. Ganesh Narayanan 155
Friction between tworing shaped areas
Friction in full circular area
- DISK FRICTION (Eg., Disc clutch)
Consider Hollow shaft (R1, R2)
M Moment required for shaft
rotation at constant speedP axial force which maintainsshaft in contact with bearing
Couple moment required to overcome frictionresistance, M
7/21/2019 Lecture Statics
157/213
R. Ganesh Narayanan 156
Equilibrium conditions and moment equations arenecessary to solve problems
A .178 m-diameter buffer weighs 10.1 N. Thecoefficient of kinetic friction between the buffing padand the surface being polished is 0.60. Assumingthat the normal force per unit area between the padand the surface is uniformly distributed, determinethe magnitude Q of the horizontal forces required toprevent motion of the buffer.
Beer/Johnston
7/21/2019 Lecture Statics
158/213
R. Ganesh Narayanan 157
prevent motion of the buffer.
O
M
Q - Q0.2 m
Mo = 0 => (0.2) Q M = 0; Q = M / 0.2
M = 2/3 (0.6) (10.1) (0.178/2) = 0.36 Nm
Q = M / 0.2 = 0.36/0.2 = 1.8 N
Belt frictionDraw free-body diagram for PP element of belt
( ) 02
cos2
cos:0 =
+= NTTTFsx
( ) 02
sin2
sin:0 =
+=
TTTNFy
Consider flat belt, cylindricaldrum
7/21/2019 Lecture Statics
159/213
R. Ganesh Narayanan 158
dT / T = S d
dT / T = S dT1
T2
0
ln (T2/T1) = S ; T2/T1 = e S
angle ofcontact
ln (T2/T1) = S ; T2/T1 = e S
Applicable to belts passing over fixed drums; ropes wrapped around a post; beltdrives
T2 > T1
This formula can be used only if belt, rope are about to slip;
7/21/2019 Lecture Statics
160/213
R. Ganesh Narayanan 159
V- Belt
T2/T1 = e S /sin (/2)
Angle of contact is radians; rope is wrapped n times - 2n radIn belt drives, pulley with lesser value slips first, with S remaining same
Beer/Johnston
A flat belt connects pulleyA to pulleyB. The
coefficients of friction are s = 0.25 and k= 0.20
between both pulleys and the belt.Knowing that the maximum allowable tension in the
belt is 600 N, determine the largest torque which can
be exerted by the belt on pulleyA.
7/21/2019 Lecture Statics
161/213
R. Ganesh Narayanan 160
Since angle of contact is smaller, slippage will occur on pulleyB first. Determinebelt tensions based on pulleyB; = 120 deg = 2/3 rad
( )
N4.3551.688
N600
688.1600
1
3225.0
11
2 s
==
===
T
e
T
e
T
T
( )( ) 0N600N4.355mc8:0 =+= MM
mcN8.1956 =M
7/21/2019 Lecture Statics
162/213
R. Ganesh Narayanan 161
Check for belt not sliping at pulley A:
ln (600/355.4) = x 4/3 => = 0.125 < 0.25
A 120-kg block is supported by a rope which iswrapped 1.5 - times around a horizontal rod. Knowingthat the coefficient of static friction between the ropeand the rod is 0.15, determine the range of values of Pfor which equilibrium is maintained.
Beer/Johnston
7/21/2019 Lecture Statics
163/213
R. Ganesh Narayanan 162
P W = 9.81 X 120 =1177.2 N
= 1.5 turns = 3 rad For impending motion of W up
P = W e s = (1177.2 N) e (0.15)3= 4839.7 N
For impending motion of W downP
=W es
= (1177.2 N)e(0.15)3
= 286.3 N
For equilibrium: 286 N P 4.84 kN
In the pivoted motor mount shown, the weight W of the175-N motor is used to maintain tension in the drivebelt. Knowing that the coefficient of static frictionbetween the flat belt and drumsA and B is 0.40, andneglecting the weight of platform CD, determine thelargest couple which can be transmitted to drum B whenthe drive drumA is rotating clockwise.
Beer/Johnston
7/21/2019 Lecture Statics
164/213
R. Ganesh Narayanan 163
For impending belt slip: CW rotation = radians
Obtain FBD of motor and mount; MD = 0 => find T1 and T2
Obtain FBD of drum at B; MB in CCW; MB = 0; Find MB
T1 = 54.5 N, T2 = 191.5 N
MB=10.27 N.m
Virtual work
We have analyzed equilibrium of a body by isolating it with a FBDand equilibrium equations
7/21/2019 Lecture Statics
165/213
R. Ganesh Narayanan 164
Class of problems where interconnected members move relative toeach other; equilibrium equations are not the direct andconventional method
Concept of work done by force is more direct => Method of virtualwork
Work of a force
U = +(F cos ) S (+ ve)
F
A AS
Work done U by the force F on the body duringdisplacement is the compt. Of force in thedisplacement direction times the displacement
7/21/2019 Lecture Statics
166/213
R. Ganesh Narayanan 165
F A
ASU = +F (cos S)
Work is a scalar quantity as we get same result regardlessof direction in which we resolve vectors
F
A AS U = -(F cos ) S
U = 0 if S = 0 and = 90 deg
FA
A
drA1
A2Work done by force F duringdisplacement dr is given by, dU = F.dr
dU = (Fx i + Fy j + Fz k).(dx i + dy j + dz k)
= Fx dx + Fy dy + Fz dz
7/21/2019 Lecture Statics
167/213
R. Ganesh Narayanan 166
U = F.dr = Fx dx + Fy dy + Fz dzWe should know relation between the force and their coordinates
Work of a couple
d
MdU = M d
U = M d
F
-FMoment can be takeninstead of forces
Forces which do no work
ds = 0; cos = 0
reaction at a frictionless pin due to rotation of a body around thepin
reaction at a frictionless surface due to motion of a body along the
surface
7/21/2019 Lecture Statics
168/213
R. Ganesh Narayanan 167
weight of a body with cg moving horizontally
friction force on a wheel moving without slipping
Only work done by applied forces, loads, friction forces need tobe considered
7/21/2019 Lecture Statics
169/213
Principle of virtual work
Imagine the small virtual displacement of particle which is
acted upon by several forces F1, F2, .. Fn
Imagine the small displacement A to A
This is possible displacement, but will not occur
7/21/2019 Lecture Statics
170/213
R. Ganesh Narayanan 169
AA ---- VIRTUAL DISPLACEMENT, r(not dr)
Work done by these forces F1, F2, .Fn during virtual
displacement r is called VIRTUAL WORK, U
U = F1. r + F2. r + ..+ Fn. r = R . r
Total virtual work of theforces
Virtual work ofthe resultant
Principle of virtual work for particle
Principle of virtual work for rigid body
7/21/2019 Lecture Statics
171/213
R. Ganesh Narayanan 170
Principle of virtual work for system of interconnected rigid bodies
Work of internal forces is zero (proved earlier)
Applications of Principle of virtual work
Mainly applicable to the solutions of problems involving machines or mechanismsconsisting of several interconnected rigid bodies
TOGGLE VISE
7/21/2019 Lecture Statics
172/213
R. Ganesh Narayanan 171
Wish to determine the force of the vice on the block for a given force
Passuming no frictionVirtual displacement is given; This results in xB and yc.Here no work is done by Ax, Ay at A and N at B
UQ = -Q xB ; UP = -P yc
7/21/2019 Lecture Statics
173/213
R. Ganesh Narayanan 172
In this problem, we have eliminated all un-known reactions, whileMA = 0 would have eliminated only TWO unknowns
The same problem can be used to find for which thelinkage is in equilibrium under two forces P and Q
Output work = Input work
Real machines
7/21/2019 Lecture Statics
174/213
R. Ganesh Narayanan 173
For an ideal machine without friction, the output work is equal to the input
work; 2Ql cos = Pl sin
In real machine, output work < input work => because of presence offriction forces
( )
=+=
==
tan
cossincos20
0
21 PQ
PlPlQl
xyPxQU BCBOutput work = Input work
friction force work
Q = 0 if tan = => = , angle of friction
Mechanical efficiency
Mechanical efficiency of m/c, = Output work / Input work
For toggle vise, = 2Ql cos / Pl sin
Substituting Q = P (tan ) here
1 cot
7/21/2019 Lecture Statics
175/213
R. Ganesh Narayanan 174
= 1 cot In the absence of friction forces, = 0 and hence = 1 => Ideal m/c
For real m/c, < 1
Beer/JohnstonDetermine the magnitude of the couple M
required to maintain the equilibrium of the
mechanism.
Virtual displacement = , xD, Work done by Ex,Ey, A is zero
B i t l k i i l
7/21/2019 Lecture Statics
176/213
R. Ganesh Narayanan 175
By virtual work principle,U = UM + Up = 0
M + P xD = 0
xD = -3l sin can be obtained fromgeometry
M = 3Pl sin
Beer/Johnston
A hydraulic lift table consisting of two identicallinkages and hydraulic cylinders is used to raise a1000-kg crate. Members EDB and CG are each
of length 2a and member AD is pinned to themidpoint of EDB.Determine the force exerted by each cylinder inraising the crate for = 60o, a = 0.70 m, and L =
3 20 m
7/21/2019 Lecture Statics
177/213
R. Ganesh Narayanan 176
3.20 m.
Work done is zero for Ex, Ey, Fcg; Work done by W, FDH will be considered
W, y are in opposite
direction ( )sign will come
FDH, s are in same direction,(+) sign will come
1)
---- (1)
7/21/2019 Lecture Statics
178/213
R. Ganesh Narayanan 177
direction, (-)sign will come2) Express y, s in terms of
y = 2a cos ; s = (aL sin/s)
Substituting y & s in (1) gives,-(1/2) W (2a cos ) + (FDH) (aL sin/s) = 0
FDH = W (s/L) cot
3) Apply numerical data
FDH = (1000 X 9.81) (2.91/3.2) cot 60 = 5.15 kN S is obtained from this triangle
The mechanism shown is acted upon by the force P.Derive an expression for the magnitude of the force Q
required for equilibrium.
Beer/Johnston
W.D. by Ay, Bx, By will be zero
U = 0 => +Q (XA) P (YF) B
7/21/2019 Lecture Statics
179/213
R. Ganesh Narayanan 178
YF
U = 0 => +Q (XA) - P (YF)Find XA and YF in terms of
(Check calculation of XA and YF)
U = Q(2l cos ) - P(3l sin ) = 0
Q Bx
By
P
Ay
Yf
XAXA
x
y
Work of force using finite displacement
Work of force F corresponding to infinitesimal displacement,dr = dU = F. dr
Work of F corresponding to a finite displacement of particlefrom A1 to A2 and covering distances S1, S2,
U1-2 = F dr or U1-2 = (F cos) ds = F (S2-S1)
A2 S2
7/21/2019 Lecture Statics
180/213
R. Ganesh Narayanan 179
U = F . dr or U = (F cos) ds = F (S2-S1)A1 S1
S1, S2 distance along the path traveled by the particle
Area under curve = U1-2
Similarly, work of a couple of moment M, dU = M d
U1-2 = M d = M (2-1)2
1
Work of a weight
WdydU =
Work of a spring
F = k xk springconstant, N/m
7/21/2019 Lecture Statics
181/213
R. Ganesh Narayanan 180
yW
WyWy
WdyU
WdydU
y
y
=
=
=
=
21
21
2
1
Work is equal to product of W and
vertical displacement of CG of body;Body moves upwards; Body movingdownwards will have +ve work done
( )
222
1212
1
212
1
kxkx
dxkxU
dxkxdxdU
x
x
=
=
==
+ve work done is expected if x2 < x1, i.e.,when spring is returning to its un-deformedposition
( ) xFFU 2121
21 +=
Potential EnergyWork of a weight: 2121 WyWyU =
The work is independent of path and depends only on
positions (A1, A2) or Wy
potential energy of the body with
respect to theforce of gravity Wr
== gVWy
VVU
7/21/2019 Lecture Statics
182/213
R. Ganesh Narayanan 181
2121 gg VVU =
Vg1 < Vg2 => PE is increasing with displacement in this
case, work done is negativeWork is positive, if PE decreases
Unit of PE Joule (J)
Work of a spring
( ) ( )
=
=
=
e
ee
V
VV
kxkxU
21
2
22
12
12
1
21
potential energy of the body with
respect to the elastic force Fr
Here PE increases work done is (-ve)
7/21/2019 Lecture Statics
183/213
R. Ganesh Narayanan 182
Here PE increases, work done is (-ve)
Now in general, it is possible to find a function V, called potential energy, such
that, dU = -dVU1-2 = V1 V2 => A force which satisfies this eqn. is conservative force
Work is independent of path & negative of change in PE for the
cases seen
Potential energy & equilibrium
Considering virtual displacement, U = -V = 0
=> (dV / d) = 0 => position of the variable defined by single independentvariable,
In terms of potential energy, the virtual work principle states that if a system is in
equilibrium, the derivative of its total potential energy is zero
(V/) = 0
1
7/21/2019 Lecture Statics
184/213
R. Ganesh Narayanan 183
equilibrium, the derivative of its total potential energy is zero
Example:
Initial spring length = AD
Work is done only by W, F
For equilibrium, U = (Ve + Vg)2
Total potential energy of the system, V = Vg + Ve
For W For F
7/21/2019 Lecture Statics
185/213
R. Ganesh Narayanan 184
dv/d = 4kl2sin cos Wl sin = 0
= 0 and = cos-1 (W/4kl)
POC
k
BA
b
b
b
b
Two uniform links of mass, m are connected asshown. As the angle increases with P applied inthe direction shown, the light rod connected at A,
passes thro pivoted collar B, compresses thespring (k). If the uncompressed position of thespring is at = 0, find the force which willproduce equilibrium at the angle
Vg = 0
4bsin/2
7/21/2019 Lecture Statics
186/213
R. Ganesh Narayanan 185
Compression distance of spring, x = movement of A away from B; X = 2b sin /2
U = (Ve + Vg)
Find Ve, Ve; Vg, Vg; U (of P)
Ve = k x2; Vg = mgh
U = P (4b sin /2)
2Pb cos /2 = 2kb2sin /2 cos /2 + mgb sin /2 P = kb sin /2 + mg tan /2
Meriam/Kraige, 7/39
For the device shown the spring would be un-stretchedin the position =0. Specify the stiffness k of the springwhich will establish an equilibrium position in the
vertical plane. The mass of links are negligible.
kb b
b
m
y
Vg = 0
7/21/2019 Lecture Statics
187/213
R. Ganesh Narayanan 186
Spring stretch distance, x = 2b-2b cos
Ve = k [(2b)(1-cos )]2 = 2kb2 (1-cos )2
Vg = -mgy = -mg (2bsin) = -2mgbsin V = 2kb2 (1-cos )2 - 2mgbsin
For equilibrium, dv / d = 4kb2(1-cos ) sin - 2mgb cos = 0
=> K = (mg/2b) (cot /1-cos )
0=d
dV
Stability of equilibrium (one DOF )
7/21/2019 Lecture Statics
188/213
R. Ganesh Narayanan 187
0=d
dV
d2V / d2 < 0d2V / d2 > 0Must examine higher
order derivatives and are
zero
AB AB
7/21/2019 Lecture Statics
189/213
( )( ) ( )( )( )
cos43.296.25
cosm3.0sm81.9kg10m08.0mkN4
cos
22
2
2
2
==
= mgbkad
Vd
2Vd
7/21/2019 Lecture Statics
190/213
R. Ganesh Narayanan 189
at = 0: 083.32
+=d
Vd stable
SpringAB of constant 2 kN/m is attached to twoidentical drums as shown.Knowing that the spring is un-stretched when =
0, determine (a
) the range of values of the massm of the block for which a position of equilibriumexists, (b) the range of values of for which theequilibrium is stable.
Beer/Johnston (10.81)
7/21/2019 Lecture Statics
191/213
R. Ganesh Narayanan 190
AB
AB
(Sin varies from 0 to 1)
7/21/2019 Lecture Statics
192/213
R. Ganesh Narayanan 191
(Cos varies from 1 to 0)
XV = ( xc dv) YV = ( yc dv) ZV = ( zc dv)Centroid of volume:
Centroid of area:
Moments of inertia : The moment of inertia of an object about a given axisdescribes how difficult it is to change its angular motion about that axis.
First moment of volume
w.r.t. yz plane
7/21/2019 Lecture Statics
193/213
R. Ganesh Narayanan 192
XA = ( xc dA) YA = ( yc dA) ZA = ( zc dA)
Symmetry plane
Centroid ofvolume xc dv = 0
Consider a beam subjected to pure bending.
Internal forces vary linearly with distance from
the neutral axis which passes through the section
centroid.
X-axis => neutral axis => centroid of section
passes
F = k y A vary linearly with distance y
7/21/2019 Lecture Statics
194/213
R. Ganesh Narayanan 193
momentsecond
momentfirst022 ==
====
=
dAydAykM
QdAydAykR
AkyF
x
r
MX = y F = k y2 A;
Moment of inertia of beamsection w.r.t x-axis, IX (+VE)
Second moments or moments of inertia of an area with
respect to thex andy axes,
== dAxIdAyI yx22
For a rectangular area
Rectangular moment of inertia
7/21/2019 Lecture Statics
195/213
R. Ganesh Narayanan 194
For a rectangular area,
3
3
1
0
22
bhbdyydAyI
h
x ===
IY = x2 dA = x2 h dx = 1/3 b3h
0
b
Thepolar moment of inertia is an important parameter in problems involving
torsion of cylindrical shafts and rotations of slabs.
= dArJ2
0
Polar moment of inertia
7/21/2019 Lecture Statics
196/213
R. Ganesh Narayanan 195
The polar moment of inertia is related to the rectangular moments of
inertia,
xy II
dAydAxdAyxdArJ
+=
+=+== 22222
0
Consider areaA with moment of inertiaIx. Imagine that
the area is concentrated in a thin strip parallel to thex axis
with equivalentIx.
IkAkI xxxx ==
2
kx = radius of gyration with respect to thex axis
Radius of gyration
7/21/2019 Lecture Statics
197/213
R. Ganesh Narayanan 196
x f gy p
JkAkJ
A
IkAkI
OOOO
y
yyy
==
==
2
2
222yxO kkk +=
Beer/Johnston
( )hh
dhb
dyh
bdAI 3222
Determine the moment of inertia of a
triangle with respect to its base.
For similar triangles,
dyh
hbdA
h
yhbl
h
yh
b
l =
=
=
dA = l dy
Determination of MI by area of integration
7/21/2019 Lecture Statics
198/213
R. Ganesh Narayanan 197
( )
h
x
yyhh
b
dyyhyh
dyh
ybydAyI
0
43
0
32
0
22
43
=
===
12
3bhIx=
yMI of rectangular area:
dA = bdy
xb
h
y
dy
Ix = y2 dA = y2 bdy = 1/3 bh3; Iy = 1/3 hb3
0
h
MI - Ix and Iy for elemental strip:
About centroidal axis (X, Y): Ix = 1/12 bh3; Iy = 1/12 b3hX
Y
7/21/2019 Lecture Statics
199/213
R. Ganesh Narayanan 198
y dIx = 1/3 dx (y3) = 1/3 y3 dx
dIy = x2dA = x2y dx or 1/3 x3dy
x
YX
dA = Ydx From this, MI of whole area can becalculated by integration
dx
dy
y
x
a
b y = k x5/2
Find MI w.r.t Y axis
Beer/Johnston (9.1)
7/21/2019 Lecture Statics
200/213
R. Ganesh Narayanan 199
Triangle: bh3/12 (about base)
Circular area: /4 r4 (about dia)
7/21/2019 Lecture Statics
201/213
R. Ganesh Narayanan 200
Rectangular area: bh3/3 (about base)
Parallel axis theorem
Consider moment of inertiaIof an areaA with respect
to the axisAA
= dAyI2
The axisBBpasses through the area centroid and
is called a centroidal axis.
dA
A A
y
CB B
d
y
7/21/2019 Lecture Statics
202/213
R. Ganesh Narayanan 201
( )
++=
+==
dAddAyddAy
dAdydAyI
22
22
2
2AdI+=
C Centroid
BB Centroidal axis
MI of area withcentroidal axis
0
Parallel axis theorem
Jo
= Jc
+ Ad2First moment ofarea
Moments of Inertia of Composite Areas
The moment of inertia of a composite area A about a given axis isobtained by adding the moments of inertia of the component areas A1,
A2, A3, ... , with respect to the same axis.
y
7/21/2019 Lecture Statics
203/213
R. Ganesh Narayanan 202
x
It should be noted that the radius of gyration of a composite area is not
equal to sum of radii of gyration of the component areas
MI of some common geometric shapes
7/21/2019 Lecture Statics
204/213
R. Ganesh Narayanan 203
Moment of inertiaITof a circular area with respect to a
tangent to the circle T,
445
224412
r
rrrAdIIT
=
+=+=
Application 1:
Application 2: Moment of inertia of a triangle with respect to a
7/21/2019 Lecture Statics
205/213
R. Ganesh Narayanan 204
Application 2: Moment of inertia of a triangle with respect to acentroidal axis,
( )3
361
2
31
213
1212
2
bh
hbhbhAdII
AdII
AABB
BBAA
=
==
+=
IDD = IBB + ad2 = 1/36 bh3 + 1/2
Top Related