lecture statics

7/21/2019 Lecture Statics

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Batch: Jan - May 2008 R. Ganesh Narayanan 1

Engineering Mechanics Statics

InstructorR. Ganesh Narayanan

Department of Mechanical EngineeringIIT Guwahati


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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


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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


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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


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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


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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


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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


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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


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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


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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


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Two concurrent forces F1, F2

Rx = Fx; Ry = Fy

DERIVATION

F2F1

R

i

j


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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


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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


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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:


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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


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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


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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


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M

-F +Fd

M

-F+F

d

M

-F+F d

-2Fd/2+2F

M

EXAMPLE

All four are equivalent couples


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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


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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


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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)


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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 ?


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


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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


LOA of R with x-axis:


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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


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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 =


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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


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Modeling the action of forces

Meriam/Kraige

Imp

Imp

FBD E l


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FBD - Examples

Meriam/Kraige

Equilibrium equns. Can besolved,

Some forces can be zero

Assumed sign can bedifferent

T f 2D ilib i


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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


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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


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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


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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


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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


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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


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Merial / Kraige; 2/117

Replace the 750N tensile force which the cable exerts on point B by a force-couple system at point O


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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


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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


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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


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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


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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


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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


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Types of 3D equilibrium


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Meriem / KraigeB

z


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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


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I.H. Shames


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Find forces at A, B, D. Pin connection at C; E has welded connection

F.B.D. - 1


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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


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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


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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


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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


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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



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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


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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


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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


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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


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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


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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


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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;


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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


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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)


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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


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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


74/213


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


75/213

Methodology for method of joints

EF


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A

B C

D

L

Fy = 0; Fx = 0

Finally sign can be changed if

not applied correctly


77/213

Method of sections


78/213


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


79/213


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


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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


81/213


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


82/213


Section AA and BB arepossible

convenient

Important points

IN method of sections, an entire portion of the truss is considered a


83/213


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,


84/213


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


85/213


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


86/213


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


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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


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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


89/213


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


90/213


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


91/213


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


92/213


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


93/213


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


94/213


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


95/213


AFAE

BD

Meriem/KraigeB

30 lb

1220 ft


96/213


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


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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


98/213


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


99/213


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


100/213


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


102/213


Given the magnitude ofP, determine the

magnitude of Q.

Taking moments about A,

Pb

aQbQaPMA === 0

Center of mass & center of gravity


103/213


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


104/213


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


105/213


= 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


106/213


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


107/213


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


108/213


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


109/213


Internal effects in beams

Shear, bending, torsion of beams

v

v

M M

SHEARBENDING TORSION

compression

Tension


110/213


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


111/213


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


112/213


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


113/213


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


114/213


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


115/213


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


116/213


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


117/213


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


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


119/213


: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=


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


121/213


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


122/213


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


123/213


4 ft 4 2 2

Machine


124/213



125/213


FrictionEarlier we assumed action and reaction forces at contacting surfaces

are normal Seen as smooth surface not practically true


126/213


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


127/213


(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


128/213


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


129/213


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


130/213


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


131/213


F N

No motion,

(Px


132/213


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


133/213


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


134/213


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


135/213


: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


136/213


( )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


137/213


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


138/213


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)


139/213


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


140/213


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.


141/213


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


142/213


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


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


144/213


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


145/213


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


146/213


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


147/213

Screws

Used for fastening, transmitting power or motion, lifting body

Square threaded jack - screw jack V-thread is alsopossible


148/213


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 (+)


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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


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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


151/213


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


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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


153/213


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


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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


155/213

F i ti b t tFriction in full circular area


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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


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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


158/213


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


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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;


160/213


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.


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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


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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


163/213


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


164/213


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


165/213


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


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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


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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


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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


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


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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


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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


172/213


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


173/213


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


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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


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= 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


176/213


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


177/213


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)


178/213


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


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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


180/213


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


181/213


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


182/213


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)


183/213


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


184/213


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


185/213


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


186/213


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


187/213


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 )


188/213


0=d

dV

d2V / d2 < 0d2V / d2 > 0Must examine higher

order derivatives and are

zero

AB AB


189/213

( )( ) ( )( )( )

cos43.296.25

cosm3.0sm81.9kg10m08.0mkN4

cos

22

2

2

2

==

= mgbkad

Vd

2Vd


190/213


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)


191/213


AB

AB

(Sin varies from 0 to 1)


192/213


(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


193/213


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


194/213


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


195/213


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


196/213


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


197/213


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


198/213


( )

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


199/213


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)


200/213


Triangle: bh3/12 (about base)

Circular area: /4 r4 (about dia)


201/213


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


202/213


( )

++=

+==

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


203/213


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


204/213


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


205/213


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

lecture statics

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