Engineering Mechanics - Statics

Engineering Mechanics - Statics Instructor: R. Ganesh Narayanan, IITG

R.Ganesh Narayanan, IIT Guwahati

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Engineering mechanics - Deals with effect of forces on objects Mechanics principles used in vibration, spacecraft design, fluid flow, electrical, mechanical m/c design etc. Statics: deals with effect of force on bodies which are not moving Dynamics: deals with force effect on moving bodies We consider RIGID BODIES – Non deformable

Scalar quantity: Only magnitude; time, volume, speed, density, mass… Vector quantity: Both direction and magnitude; Force, displacement, velocity, acceleration, moment… V = IvI n, where IvI = magnitude, n = unit vector n = V / IvI n - dimensionless and in direction of vector ‘V’ y

In our course:

j x

z

i

i, j, k – unit vectors

k

R.Ganesh Narayanan, IIT Guwahati

3

Dot product of vectors: A.B = AB cos θ; A.B = B.A (commutative) A.(B+C) = A.B+A.C (distributive operation) A θ B

i.i=1 A.B = (Axi+Ayj+Azk).(Bxi+Byj+Bzk) = AxBx+AyBy+AzBz

i.j=0

Cross product of vectors: A x B = C; ICI = IAI IBI Sin θ; AxB = -(BxA) C x (A+B) = C x A + C x B j

i

j k

k i

k x j = -i; ixi=0

i

j

k

Ax

AY

AZ

BX

BY

BZ

R.Ganesh Narayanan, IIT Guwahati AxB = (Axi+Ayj+Azk)x(B xi+Byj+Bzk) = (AyBz- AzBy)i+( )j+( )k

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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 Direction of motion

Force, P kN

Body moves P, kN

Force < P kN

Body does not move bulging R.Ganesh Narayanan, IIT Guwahati

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Force system: Magnitude, direction and point of application is important

θ P Bracket

WIRE

External effect: Forces applied (applied force); Forces exerted by bracket, bolts, foundation….. (reactive force) Internal effect: Deformation, strain pattern – permanent strain; depends on material properties of bracket, bolts…

R.Ganesh Narayanan, IIT Guwahati

6

Transmissibility principle: A force may be applied at any point on a line of action without changing the resultant effects of the force applied external to rigid body on which it acts Magnitude, direction and line of action is important; not point of application Line of action P

P

R.Ganesh Narayanan, IIT Guwahati

7

Concurrent force: Forces are said to be concurrent at a point if their lines of action intersect at that point Parallelogram law of forces Polygon law of forces

F2

F1, F2 are concurrent forces R will be on same plane

R

R = F1+F2

A F1

Plane

R does not pass through ‘A’

Use triangle law F2

R = F1+F2

F2

R = F1+F2

R

F2

A

R A

F1

A F1

F1

R.Ganesh Narayanan, IIT Guwahati

F1

F2 R

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Two dimensional force system Rectangular components: j

F = Fx + Fy; both are vector components in x, y direction

Fy F

θ Fx

i

Fx = fx i ; Fy = fy j; fx, fy are scalar quantities Therefore, F

= fx i + fy j

Fx = F cos θ; Fy = F sin θ

+ ve

F=

fx2 + fy2

;θ

= tan -1 (fy/fx)

+ ve - ve - ve

R.Ganesh Narayanan, IIT Guwahati

9

Two concurrent forces F1, F2

F1

F2

j R

Rx = Σ Fx; Ry = Σ Fy i

DERIVATION

R.Ganesh Narayanan, IIT Guwahati

10

Moment: Tendency to rotate; torque O

Moment about a point: M = Fd M F

Magnitude of moment is A r proportional to the force ‘F’ and d B moment arm ‘d’ i.e, perpendicular distance from the axis of rotation to the LOA of force O UNIT : N-m Moment is perpendicular to plane about axis O-O Counter CW = + ve; CW = -ve R.Ganesh Narayanan, IIT Guwahati

α

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Cross product: M = r x F; where ‘r’ is the position vector which runs from the moment reference point ‘A’ to any point on the LOA of ‘F’ M = Fr sin α; M = Fd

A d

M = r x F = -(F x r): sense is important

R.Ganesh Narayanan, IIT Guwahati

α

r B

Sin α = d / r

12

Varignon’s theorem: The moment of a force about any point is equal to the sum of the moments of the components of the forces about the same point Concurrent forces – P, Q P

R

Mo = r x R = r x (P+Q) = r x P + r x Q

B o

r

Q Moment of ‘P’ Moment of ‘Q’

Usefulness: Resultant ‘R’ – moment arm ‘d’

Force ‘P’ – moment arm ‘p’; Force ‘Q’ – moment arm ‘q’ Mo= Rd = -pP + qQ

R.Ganesh Narayanan, IIT Guwahati

13

2

Pb:2/5 (Meriam / Kraige):

A

Calculate the magnitude of the moment about ‘O’ of the force 600 N

40 deg

4

1) Mo = 600 cos 40 (4) + 600 sin 40 (2) = 2610 Nm (app.)

600N

r o

in mm

j

2) Mo = r x F = (2i + 4j) x (600cos40i-600sin40j) i

= -771.34-1839 = 2609.85 Nm (CW); mag = 2610 Nm

R.Ganesh Narayanan, IIT Guwahati

14

Couple: Moment produced by two equal, opposite and non-collinear forces M o

-F

a d

=>-F and F produces rotation +F

=>Mo = F (a+d) – Fa = Fd; Perpendicular to plane ⇒Independent of distance from ‘o’, depends on ‘d’ only ⇒ moment is same for all moment centers R.Ganesh Narayanan, IIT Guwahati

15

Vector algebra method CCW Couple o

CW Couple rb

-F

M = ra x F + rb x (-F) = (ra-rb) x F = r x F

r ra

+F

Equivalent couples •Changing the F and d values does not change a given couple as long as the product (Fd) remains same •Changing the plane will not alter couple as long as it is parallel R.Ganesh Narayanan, IIT Guwahati

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EXAMPLE M -F

d

M

M +F

-F

d

d

+F

+F

-F

M

All four are equivalent couples

+2F

R.Ganesh Narayanan, IIT Guwahati

d/2

-2F

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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-couple syatem ⇒ a force can be replaced by a force and couple B

B F

A

F

B F

-F

F M = Fd

A

R.Ganesh Narayanan, IIT Guwahati

18

EXAMPLE 80N

80N

9m

9 60deg

o

80 N

o

60 deg

80 N

80 N

o

Mo = Y N m

Mo = 80 (9 sin 60) = 624 N m; CCW

R.Ganesh Narayanan, IIT Guwahati

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Resultants To describe the resultant action of a group or system of forces Resultant: simplest force combination which replace the original forces without altering the external effect on the body to which the forces are applied R

R = F1+F2+F3+….. = Σ F Rx = Σ Fx; Ry = Σ Fy; R = (Σ Fx)2 + (Σ Fy)2 -1 (Ry/Rx) Narayanan, IIT Guwahati Θ = tanR.Ganesh

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How to obtain resultant force ? F1

F2

Mo= ΣFd

F1

R= ΣF

F2 F3

F3 F1 – D1; F2 – D2; F3 – D3

R

M1 = F1d1; M2 = F2d2; M3 = F3d3

d

Mo=Rd

NON-CONCURRENT FORCES R.Ganesh Narayanan, IIT Guwahati

21

Principle of moments Summarize the above process:

R = ΣF Mo = ΣM = Σ(Fd) Mo = Rd

Mo= ΣFd

R

d R= ΣF

Mo=Rd

First two equations: reduce the system of forces to a force-couple system at some point ‘O’ Third equation: distance ‘d’ from point ‘O’ to the line of action ‘R’ => VARIGNON’S THEOREM IS EXTENDED HERE FOR NONCONCURENT FORCES R.Ganesh Narayanan, IIT Guwahati

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Text Books 1. Meriam/kraige 2. J. F. Shelley – Schaum’s series 3. Shames 4. Beer/Johnston 5. Lakshmana rao, Lakshmimarasimhan ….. STATICS – MID SEMESTER – DYNAMICS Tutorial: Monday 8 am to 8.55 am

R.Ganesh Narayanan, IIT Guwahati

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ENGINEERING MECHANICS TUTORIAL CLASS: Monday 8 AM TO 8.55 AM

Tutorial Groups

Roll Numbers From

Class Room

Tutors

To

TG1

07010101

07010141 (41 Students)

L2

Prof. R. Tiwari

TG2

07010142

07010149 (8 Students)

1G1

Dr. senthilvelan

07010201

07010233 (33 Students)

07010234

07010249 (16 Students)

1G2

R. Ganesh Narayanan

07010301

07010325 (25 Students)

07010326

07010353 (28 Students)

1202

Dr. M. Pandey

07010401

07010413 (13 Students)

07010414

07010449 (36 Students)

1205

Dr. Saravana Kumar

07010601

07010605 (5 Students)

TG3

TG4

TG5

LECTURE CLASSES: LT2 (one will be optional): Monday 3 pm to 3.55 pm Tuesday 2 pm to 2.55 pm Thursday 5 pm to 5.55 pm Friday 4 pm to 4.55 pm R.Ganesh Narayanan, IIT Guwahati

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Three dimensional force system Rectangular components Fx = F cos θx; Fy = F cos θy; Fz = F cos θz l, m, n are directional cosines of ‘F’

F = Fx i + Fy j + Fz k = F (i cos θx + j cos θy + k cos θz) = F (l i + m j + n k) F = F nf F

Fz k θy

θz o

θx Fx i

Fy j

Moment in 3D Mo

A

A - a plane in 3D structure

r α d

Mo = F d (TEDIOUS to find d) F

or Mo = r x F = – (F x r) (BETTER)

Evaluating the cross product Described in determinant form:

i

j

k

rx

rY

rZ

FX

FY

FZ

Expanding … Narayanan, IIT Guwahati R.Ganesh

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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 λ: Magnitude of the moment Mλ of F about λ λ Mo

= Mo . n

(scalar reprn.)

Similarly, Mλ = (r x F.n) n (vector reprn.)

n F

Scalar triple product o

r

rx

ry

rz

Fx

FY

FZ

β

γ

α Narayanan, IIT Guwahati α, β, γ – DCs of R.Ganesh n

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Varignon’s theorem in 3D F2

F3

Mo = rxF1 + rxF2 + rx F3 +…= Σ(r x F) = r x (F1+F2+F3+…)

B o

r

= r x (ΣF) = r x R

F1

M

Couples in 3D

d

+F

-F B

r

M = ra x F + rb x –F = (rarb) x F = rxF

A ra

rb

R.Ganesh Narayanan, IIT Guwahati

28

2D force system; equ. Force-couple; principle of moments

Beer-Johnston; 2.3 F2 = 80N

20

F1 = 150N

• Evaluate components of F1, F2, F3, F4

30

• Rx = ΣFx; Ry = ΣFy

15

F4 = 100N

• R = Rx i + Ry j • α = tan -1 (Ry/Rx)

F3 = 110N

Ry

R α Rx

R.Ganesh Narayanan, • R = 199iIIT+Guwahati 14.3j; α

= 4.1 deg

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F1 30 DEG

Boat

Find F1 and F2

15 DEG 45 DEG

F2

R =3000 N

R = F1 + 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.Ganesh Narayanan, IIT Guwahati

30

Meriam / kraige; 2/37 780 N Find the moment Mo of 780 N about the hinge point

A

10m

10

20 DEG

10 D

B

C

o

T = -780 COS20 i – 780 sin20 j = -732.9 i – 266.8 j

OC – FLAG POLE OAB – LIGHT FRAME D – POWER WINCH

r = OA = 10 cos 60 i + 10 sin 60 j = 5 i + 8.6 j Mo = r x F = 5014 k ; Mag = 5014 Nm

R.Ganesh Narayanan, IIT Guwahati

31

2

Meriam / kraige; 2/6 Replace couple 1 by eq. couple p, -p; find Θ

M 40

1

-P

M = 100 (0.1) = 10 Nm (CCW)

P

θ 100

θ

100

2

M = 400 (0.04) cos θ

60

1

100

10 = 400 (0.04) cos θ

100N

100N

=> Θ = 51.3 deg

R.Ganesh Narayanan, IIT Guwahati

32

60 N

Meriam / kraige; 2/8

2m

50 N

45

5m

140Nm

2m

Find the resultant of four forces and one couple which act on the plate

80N 2m

40 N

o

30 deg

1m

Rx = 40+80cos30-60cos45 = 66.9 N Ry = 50+80sin 30+60cos45 = 132.4 N

R = 148.3N 63.2 deg

237 Nm

o

R = 148.3 N; Θ = tan-1 (132.4/66.9) = 63.2 deg Mo = 140-50(5)+60cos45(4)-60sin45(7) = -237 Nm

R = 148.3N

Final LOA of R:

63.2 deg

148.3 d = 237; d = 1.6 m o

LOA of R with x-axis: (Xi + yj) x (66.9i+132.4j) = -237k

R = 148.3N

(132.4 x – 66.9 y)k = -237k o

132.4 x -66.9 y = -237 Y = 0 => x

R.Ganesh Narayanan, IIT Guwahati

= b = -1.792 m

b

y x 33

Couples in 3D M

d

+F

-F B

r

M = ra x F + rb x –F = (rarb) x F = rxF

A ra

rb

Equivalent couples

M = Fd

F

F B

A

B

r

F

F

A B

-F

R.Ganesh Narayanan, IIT Guwahati

34

How to find resultant ? R = ΣF = F1+F2+F3+… Mo = ΣM = M1+M2+M3+… = Σ(rxF) M = Mx2 + My2 + Mz2; R = ΣFx2 + ΣFy2 + ΣFz2 Mx = ; My = ; Mz =

R.Ganesh Narayanan, IIT Guwahati

35

Equilibrium Body in equilibrium - necessary & sufficient condition: R = ΣF = 0; M = ΣM = 0

Equilibrium in 2D Mechanical system: body or group of bodies which can be conceptually isolated from all other bodies System: single body, combination of bodies; rigid or non-rigid; combination of fluids and 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 R.Ganesh Narayanan, IIT Guwahati

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Modeling the action of forces

Imp

Imp

R.Ganesh Narayanan, IIT Guwahati

Meriam/Kraige

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FBD - Examples

Equilibrium equns. Can be solved, • Some forces can be zero • Assumed sign can be different

R.Ganesh Narayanan, IIT Guwahati

38

Meriam/Kraige

Types of 2D equilibrium Concurrent at a point: ΣFx = 0; ΣFy = 0 x

Collinear: ΣFx = 0

F2

F1

F3

Y F2

X F1

F3 F4 F1 F2

Y

F3

M

X

F4 Parallel: ΣFx = 0; ΣMz = 0 R.Ganesh Narayanan, IIT Guwahati

General: ΣFx = 0; ΣFy = 0; ΣMz = 0

39

General equilibrium conditions ΣFx = 0; ΣFy = 0; ΣFz = 0 ΣMx = 0; ΣMy = 0; ΣMz = 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 body ⇒Necessary and sufficient equilibrium conditions R.Ganesh Narayanan, IIT Guwahati

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P C

Q

PY

R

Px

D

QY

RY

Qx

Rx D

C W

A

B Pin

AX

A

Roller

AY

B BY

Written in three alternate ways,

ΣFx = 0; ΣFy = 0; ΣMA = 0

I

ΣMB = 0 => will not provide new information; used to check the solution; To find only three unknowns R.Ganesh Narayanan, IIT Guwahati

41

ΣFx = 0; ΣMA = 0; ΣMB = 0

II

Rigid body in equilibrium => • 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. (ΣMB = 0) says this force must be zero

ΣMA = 0; ΣMB = 0; ΣMc = 0;

III

Body is statically indeterminate: more unknown reactions than independent equilibrium equations R.Ganesh Narayanan, IIT Guwahati

42

Y

3D force system A

Meriam / Kraige; 2/10 15

Find the moment Mz of T about the z-axis passing thro the base O

x

O

9

z

R.Ganesh Narayanan, IIT Guwahati

T = 10kN

B

12 m

43

F = T = ITI nAB = 10 [12i-15j+9k/21.21] = 10(0.566i-0.707j+0.424k) k N Mo = 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

R.Ganesh Narayanan, IIT Guwahati

44

Merial / Kraige; 2/117 Replace the 750N tensile force which the cable exerts on point B by a forcecouple system at point O

R.Ganesh Narayanan, IIT Guwahati

45

F = f λ, where λ 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

R.Ganesh Narayanan, IIT Guwahati

46

2D equilibrium Meriem / Kraige; 3/4 Find T and force at A; I-beam with mass of 95 kg/meter of length 95 kg/meter => 95(10-3)(5)(9.81) = 4.66kN ΣMA = (T cos 25) (0.25) + (T sin 25) (5-0.12) – 10(5-1.5-0.12) – 4.66 (2.5-0.12) = 0 y

T = 19.6 kN ΣFx = Ax – 19.6 cos 25 = 0

T 25 deg

Ax = 17.7 kN ΣFy = Ay+19.61 sin 25-4.66-10 = 0

Ax

0.5 m

0.12 m

Ay = 6.37 kN

Ay

1.5m

4.66 kN 10 kN

A = Ax2 + Ay2 = 18.88kN R.Ganesh Narayanan, IIT Guwahati

5m 47

B

Beer/Johnston; 4.5

mm, N

Find reactions at A, B if (a) a = 100 mm; (b) a=70 mm

A

60

a = 100 mm

60

40

a

80

50

30

ΣMa = 0 => (-40x60)+(-50x120)+(-30x220)+

10

By

(-10x300)+(-Byx120) = 0

Ay

By = 150 N

Bx

ΣFy = 0 => By-Ay-40-50-30-10 = 0 = 150-Ay-130 = 0 => Ay = 20 N 40

50

30

10

a = 70 mm By = 140 N

Ay = 10 N R.Ganesh Narayanan, IIT Guwahati

48

Beer/Johnston; 4.4

D

1.8

Find the reaction at the fixed end ‘E’

A

B

2.25 C

20

20

20

3.75

20

DF = 7.5 m

E

ΣFx = Ex + 150 (4.5/7.5) = 0 => Ex = - 90 kN (sign change)

F 4.5

ΣFy = Ey – 4(20)-150 (6/7.5) = 0 => Ey = 200 kN D

ΣME= 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

A

B

2.25 C

20

20

20

3.75

20 E

1.8

Ex

ME R.Ganesh Narayanan, IIT Guwahati

F

Ey

4.5 49 kN 150

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 R.Ganesh Narayanan, IIT Guwahati

50

3D equilibrium 3D equilibrium equns. can be written in scalar and vector form ΣF = 0 (or) ΣFX = 0; ΣFY = 0; ΣFZ = 0 ΣM = 0 (or) ΣMX = 0; ΣMY = 0; ΣMZ = 0

ΣF = 0 => Only if the coefficients of i, j, k are zero; ΣFX = 0 ΣM = 0 => Only if the coefficients of i, j, k are zero; ΣMX = 0

R.Ganesh Narayanan, IIT Guwahati

51

Modeling forces in 3D

R.Ganesh Narayanan, IIT Guwahati

52

Types of 3D equilibrium

R.Ganesh Narayanan, IIT Guwahati

53

Meriem / Kraige

B

7 = 22 + 62 + h2 => h = 3 m

z

7m

h

2m

rAG = -1i-3j+1.5k m; rAB = -2i-6j+3k m

A

6m x

ΣMA = 0 => rAB x (Bx+By) + rAG x W = 0

y

Bx By

3.5

(-2i-6j+3k) x (Bx i + By j) + (-i-3j+1.5k) x (-1962k) = 0

G 3.5

(-3By+5886)i + (3Bx-1962)j + (-2By+6Bx)k = 0

Ay

W=mg=200 x 9.81

=> By = 1962 N; Bx = 654 N

W = 1962 N

ΣF = 0 => (654-Ax) i + (1962-Ay) j + (-1962+Az)k = 0

Ax

Az

=> Ax = 654 N; Ay = 1963 N; Az = 1962 N; find A R.Ganesh Narayanan, IIT Guwahati

54

Meriem / Kraige; 3/64

R.Ganesh Narayanan, IIT Guwahati

55

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

R.Ganesh Narayanan, IIT Guwahati

56

F.B.D. - 1

F.B.D. - 2

F.B.D. - 2 ΣMc = 0 => (Dy) (15) – 200 (15) (15/2) – (1/2)(15)(300)[2/3 (15)] = 0 Dy = 3000 N R.Ganesh Narayanan, IIT Guwahati

57

F.B.D. - 1 ΣMB = 0 => -Ay (13) +(3000) (21) – 200 (34) (34/2-13) – ½ (300) (15) [6+2/3(15)] =0 Ay = -15.4 N ΣFy = 0 => Ay+By+3000-200(34)(1/2)(300)(15) = 0 Sub. ‘Ay’ here, => By = 6065 N

R.Ganesh Narayanan, IIT Guwahati

58

2D, 3D force system • Rectangular components • Moment • Varignon’s theorem • Couple • Force-couple system • Resultant

Equilibrium equations ΣFx = 0; ΣFy = 0; ΣMA = 0 ΣFx = 0; ΣMA = 0; ΣMB = 0

2D

ΣMA = 0; ΣMB = 0; ΣMc = 0 ΣF = 0 (or) ΣFX = 0; ΣFY = 0; ΣFZ = 0 ΣM = 0 (or) ΣMX = 0; ΣMY = 0; ΣMZ = 0

3D

• Principle of moment

R.Ganesh Narayanan, IIT Guwahati

59

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 R.Ganesh Narayanan, IIT Guwahati

60

Structures Truss: Framework composed of members joined at their ends to form a rigid structures Plane truss: Members of truss lie in same plane

Roof R.Ganesh truss Narayanan, IIT Guwahati

Bridge truss

61

B •Three bars joined with pins at end • Rigid bars and non-collapsible • Deformation due to induced internal strains is negligible A

c B B

D

Non rigid body can be made rigid by adding BC, DE, CE elements

D E

A

C

A C

Non-rigid

rigid

Simple truss: structures built from basic triangle More 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 R.Ganesh Narayanan, IIT Guwahati

62

In designing simples truss or truss => assumptions are followed 1. Two force members – equilibrium only in two forces; either tension or compression 2. Each member is a straight link joining two points of application of force 3. Two forces are applied at the end; they are equal, opposite and collinear for equilibrium 4. Newton’s 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

R.Ganesh Narayanan, IIT Guwahati

63

Two methods to analyze force in simple truss

Method of joints •This method consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint •This method deals with equilibrium of concurrent forces and only two independent equilibrium equations are solved • Newton’s third law is followed

R.Ganesh Narayanan, IIT Guwahati

64

Example F

E

A

D B

ΣFy = 0; ΣFx = 0

C

L

Finally sign can be changed if not applied correctly

R.Ganesh Narayanan, IIT Guwahati

65

Internal and external redundancy 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 ‘2j’ equations are present for a truss with ‘j’ joints -The entire truss composed of ‘m’ two force members and having the maximum of three 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; statically indeterminate m + 3 < 2 j => deficiency of internal members; truss is unstable R.Ganesh Narayanan, IIT Guwahati

66

I. H. Shames B

D

Determine the force transmitted by each member; A, F = 1000 N

10

Pin A

A

FAB FAB A

FAC

1000

10

F C 10

E

10

FAC ΣFx = 0 =>FAC – 0.707FAB = 0

1000

1000

ΣFy = 0 => -0.707FAB+1000 = 0 1000

FAB = 1414 N; FAC = 1000 N

Pin B ΣFx = 0 => -FBD + 1414COS45 = 0 => FBD = 1000 N

B

FBD ΣFy = 0 => -FBC+1414 COS45 = 0 => FBC = 1000 N

1414

FBD FBC

1414Narayanan, IIT Guwahati R.Ganesh

FBC

67

Pin C 1000 FDC

B

FCE

1000

FDC

FCE 1000

1000 1000

1000

ΣFx = 0 => -1000 + FCE + FDC COS 45 = 0 => FCE = 1000 N ΣFy = 0 => -1000+1000+ FDC COS 45 = 0 => FDC = 0

SIMILARLY D, E, F pins are solved

R.Ganesh Narayanan, IIT Guwahati

68

Meriem / Kraige (similar pbm. 6.1 in Beer/Johnston) Find the force in each member of the loaded cantilever truss by method of joints

B 5

5

D

5 5

5

A

5 C

30

R.Ganesh Narayanan, IIT Guwahati

5

20

E kN, m

69

FBD of entire truss ΣME = 0 => 5T-20(5)-30 (10) = 0; T = 80 kN ΣFx = 0 => 80 cos 30 – Ex = 0; Ex = 69.28 kN ΣFy = 0 => Ey +80sin30-20-30 = 0 => Ey = 10kN FBD of joints ΣFx = 0; ΣFy = 0 Find AB, AC forces

ΣFx = 0; ΣFy = 0 Find BC, BD forces

ΣFx = 0; ΣFy = 0

ΣFy = 0

Find CD, CE forces

Find DE forces ΣFx = 0 can be checked

R.Ganesh Narayanan, IIT Guwahati

70

Q

Q = 100 N; smooth surfaces; Find reactions at A, B, C

Q

roller

100 100

c

roller

A B

30°

Rc RA RB

ΣF = 0 => (-RA cos 60 - RB cos 60 + Rc) i + (-2 x 100 + RB sin 60 + RA sin 60) j = 0 RC = (RA + RB)/2 RB + RA = 230.94

RC = 115.5 N

100 RAB 30°

Rc

ΣF = 0 => (-RAB cos 30 - RB cos 60 + Rc) i + (RB Sin 60 – 100 - RAB sin 30) j = 0 0.866 RAB + 0.5 RB = 115.5; -0.5 RAB + 0.866 RB = 100

RB

RAB = 50 N (app.); RB = 144.4 N; RA = 230.94-144.4 = 86.5 N R.Ganesh Narayanan, IIT Guwahati

71

Two methods to analyze force in plane truss

Method of joints

Method of sections

•This method consists of satisfying the conditions of equilibrium for the forces acting on the connecting pin of each joint •This method deals with equilibrium of concurrent forces and only two independent equilibrium equations are solved • Newton’s third law is followed

R.Ganesh Narayanan, IIT Guwahati

72

Methodology for method of joints F

E

A

D B

ΣFy = 0; ΣFx = 0

C

L

Finally sign can be changed if not applied correctly

R.Ganesh Narayanan, IIT Guwahati

73

B

5 5

5

D 5

5

A

5 C

30

R.Ganesh Narayanan, IIT Guwahati

5

20

E kN, m

74

Method of sections • In method of joints, we need only two equilibrium equations, as we deal 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 from an 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 should be cut. Only three independent equilibrium eqns. are present •Efficiently find limited information R.Ganesh Narayanan, IIT Guwahati

75

Methodology for method of sections F

E

A

D B L

A

F

E

A

D B

C

C R1

L

A

R2

•The external forces are obtained initially from method of joints, by considering 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 3 equilibrium equns. • AA – section across FE, BE, BC; Forces in these members are R.Ganesh Narayanan, IIT Guwahati initially unknown

76

Section 1

Section 2

• Now each section will apply opposite forces on each other • The LHS is in equilibrium with R1, L, three forces exerted on the cut members (EF, BE, BC) by the RHS which has been removed • IN this method the initial direction of forces is decided by moment about any point where known forces are present • For eg., take moment about point B for the LHS, this will give BE, BC to be zero; Then moment by EF should be opposite to moment by R1; R.Ganesh Narayanan, IIT Guwahati 77 Hence EF should be towards left hand side - compressive

• Now take moment about ‘F’ => BE should be opposite to R1 moment; Hence BE must be up and to the right; So BE is tensile • Now depending on the magnitudes of known forces, BC direction has to be decided, which in this case is outwards i.e., tensile

Σ MB = 0 => FORCE IN EF; BE, BC = 0 Σ Fy = 0 => FORCE IN BE; BC, EF = 0

Section 1

Section 2

Σ ME = 0 => FORCE IN BC; EF, BE = 0

R.Ganesh Narayanan, IIT Guwahati

78

Section AA and BB are possible

convenient R.Ganesh Narayanan, IIT Guwahati

79

Important points • IN method of sections, an entire portion of the truss is considered a single body in equilibrium • Force in members internal to the section are not involved in the analysis of the section as a whole • The cutting section is preferably passed through members and not through 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 forces pass through • Positive force value will sense the initial assumption of force direction R.Ganesh Narayanan, IIT Guwahati

80

Meriem/Kraige Find the forces included in members KL, CL, CB by the 20 ton load on the cantilever truss y

L

KL K CL G P

C

CB

x

20 T

Section 1

Section 2

Σ Moment abt. ‘L’ => CB is compressive => creates CW moment Σ Moment abt. C => KL is tensile => creates CW moment R.Ganesh Narayanan, IIT Guwahati

CL is assumed to be compressive

81

y

L

KL K

θ CL x

G P 20 T

C

CB

Section 1

Section 2

BL = 16 + (26-16)/2 = 12 ft Θ = tan -1 (5/12) => cos Θ = 12/13

Σ ML = 0 => 20 (5) (12)- CB (21) = 0 => CB = 57.1 t (C) Σ Mc = 0 => 20 (4)(12) – 12/13 (KL) (16) = 0; KL = 65 t (T) Σ Mp = 0 => find PC distance and find CL; CL = 5.76 t (C)

R.Ganesh Narayanan, IIT Guwahati

82

Meriem/Kraige Find the force in member DJ of the truss shown. Neglect the horizontal force in supports Consider FBD for whole truss and find reaction at A ΣMG = -Ay (24) +(10) (20) + 10(16) + 10 (8) = 0 Ay = 18. 3 kN => creates CW moment Section 2 cuts four members, but we have only 3 equi. Equns Hence consider section 1 which cuts only 3 members – CD, CJ, KJ Force direction Σ Moment abt. A => CD, JK – Eliminated; CJ will be upwards creating CCW moment Σ Moment abt. C => JK must be towards right creating CCW moment R.Ganesh Narayanan, IIT Guwahati ASSUME CD TO HAVE TENSILE FORCE

83

From section 1 FBD

ΣMA = 0 => CJ (12) (0.707) – 10 (4) -10( 8) =0; CJ = 14.14 Kn ΣMJ = 0 => 0.894 (CD) (6) +18.33 (12)-10(4)-10(8) = 0; CD = -18.7 kN CD direction is changed

From section 2 FBD ΣMG = 0 => 12 DJ +10(16)+10(20)-18.3 (24)14.14 (0.707)(12) = 0 DJ = 16.7 kN

R.Ganesh Narayanan, IIT Guwahati

84

I.H. Shames

FBD - 1

FBD - 2 From FBD-2 ΣMB = 0 => -(10)(500)+30 (789)- FAC Sin 30 (30) = 0 FAC = 1244.67 N From FBD -1 ΣFx = 0 => FDA Cos 30 – (1244.67) cos 30 – 1000 sin 30 = 0 ; FDA = 1822 N R.Ganesh Narayanan, ΣFy = 0 => (1822)Sin 30 + (1244.67) sin 30 +FABIIT – Guwahati 1000 Cos 30 = 0; FAB = -667 N

85

Frames and machines Multi force members: Members on which three or more forces acting on it (or) one with two or more forces and one or more couples acting on it Frame or machine: At least one of its member is multi force member Frame: Structures which are designed to support applied loads and are 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 in these members will not be in directions of members Method of joints and sections are not applicable R.Ganesh Narayanan, IIT Guwahati

86

Inter-connected rigid bodies with multi force members • Previously we have seen equilibrium of single rigid bodies • Now we have equilibrium of inter-connected members which involves 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

R.Ganesh Narayanan, IIT Guwahati

87

Force representation and FBD • Representing force by rectangular components • 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

R.Ganesh Narayanan, IIT Guwahati

88

Full truss K, J are un-necessary here

AE

AF

BD

R.Ganesh Narayanan, IIT Guwahati

89

Meriem/Kraige B

30 lb 12

20 ft Find the forces in all the frames; neglect weight of each member

D

F

50 lb

12 E

20 ft A

C 30 ft

FBD of full frame

30 lb

Σ Mc = 0 => 50 (12) +30(40)-30 (Ay) = 0; Ay = 60 lb Σ Fy = 0 => Cy – 50 (4/5) – 60 = 0 => Cy = 100 lb 50 lb Cx R.Ganesh Narayanan, IIT Guwahati

Ay

Ax

Cy

90

FBD of individual members EF: Two force member; E, F are compressive ED: ΣMD = 0 => 50(12)-12E = 0 => E = 50 lb ΣF = 0 => D-50-50 = 0 => D= 100 lb (components will be eliminated) EF: F = 50 lb (opposite and equal to E) AB: ΣMA = 0 => 50(3/5)(20)-Bx (40) = 0 => Bx = 15 lb

D

ΣFx = 0 => Ax+15-50(3/5) = 0 => Ax = 15 lb

E

ΣFy = 0 => 50 (4/5)-60-By = 0 =>By = -20 lb Fx

Σ Fx = -50 (cos 53.1)+15+15 = -30+15+15 = 0 BC: ΣFx = 0 => 30 +100 (3/5)-15-Cx = 0 => Cx = 75 lb R.Ganesh Narayanan, IIT Guwahati

53.1 deg

Fy

F

91

Find the force in link DE and components of forces exerted at C on member BCD

A 160

B

480 N

60

FBD of full frame Ay

D

C

80

E A

Ax

60

100

150

160 480 N

B Bx

Σ Fy = 0 => Ay-480 = 0 =>Ay = 480 N

C

D θ

80

Σ MA = 0 => Bx (160)-480 (100) = 0 => Bx = 300 N Σ Fx = 0 => 300+Ax = 0 => Ax = -300 N

E 100

150

-1 (80/150) R.Ganesh IIT Guwahati Θ = tanNarayanan, = 28.07

deg

92

DE: Two force member FDE

FBD of BCD FBD of AE Cy

Ay

B

D

300 A

480 N

Cx

D

C

Ax

θ

FBD of DE

FDE FDE Cx

E

D

FDE

Cy E

FDE

Σ Mc = 0 => -FDE sin 28.07 (250) – 300(80)-480 (100) = 0; FDE = -561 N Σ Fx = 0 => Cx – (-561) cos 28.07 +300 = 0 => Cx = -795 N Σ Fy = 0 => Cy – (-561) sin 28.07 – 480 = 0 => Cy = 216 N R.Ganesh Narayanan, IIT Guwahati

93

Meriem/Kraige

3m

Find the horizontal and vertical components of all the forces; neglect weight of each member

2m

A 1.5m R =0.5 m 0.5m B

F

E

FBD of full frame

1.5m

Ay

C

Ax

1.5m D 400 kg

0.4 x 9.81 = 3.92

Dx

Σ MA = 0 => 5.5 (-0.4) (9.81) + 5Dx = 0 => Dx = 4.32 kN Σ Fx = 0 => -Ax + 4.32 = 0 => Ax = 4.32 kN Σ Fy = 0 => Ay – 3.92 = 0 => Ay = 3.92 kN

R.Ganesh Narayanan, IIT Guwahati

94

FBD of individual members 3.92

3.92 4.32

F

3.92

A

Bx B

3.92

3.92

E 3.92 Ex

By

Bx

Ey

3.92 3.92

Cy

Cx

Ey

By

Ex

E

4.32 D

3m

Cx

2m

A

C 1.5m

Cy

R =0.5 m 0.5m B E

Apply equilibrium equn. And solve for forces

F

1.5m C 1.5m

R.Ganesh Narayanan, IIT Guwahati

D

95 400 kg

Machines • Machines are structures designed to transmit and modify forces. Their main purpose is to transform input forces into output forces. • Given the magnitude of P, determine the magnitude of Q.

Taking moments about A,

∑ M A = 0 = aP − bQ

a P R.GaneshbNarayanan, IIT Guwahati Q=

96

Center of mass & center of gravity

C

B

B

B

A

A

C

G

G

W

W

A

C

G

BODY

G

W

•Body of mass ‘m’ •Body at equilibrium w.r.t. forces in the cord and resultant of gravitational forces at all particles ‘W’ •W is collinear with point A •Changing the point of hanging to B, C – Same effect •All practical purposes, LOA coincides with G; G – center of gravity R.Ganesh Narayanan, IIT Guwahati

97

Y

z

Moment abt. Y axis = dw (x) Sum of moments for small regions through out the body: ∫ x dw

dw

Moment of ‘w’ force with Y axis = w x

G

r r

w

X

∫ x dw = w x Sum of moments

X = (∫ x dw) / w

Y = (∫ y dw) / w W = mg

X = (∫ x dm) / m

Y = (∫ y dm) / m R.Ganesh Narayanan, IIT Guwahati

Moment of the sum

Z = (∫ z dw) / w 1

Z = (∫ z dm) / m 98

2

In vector form,

r = (∫ r dm) / m

3

ρ = m/V; dm = ρ dv

X = (∫ x ρ dv) / ∫ ρ dv 4

Y = (∫ y ρ dv) / ∫ ρ dv

ρ = not constant through out body

Z = (∫ z ρ dv) / ∫ ρ dv 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 R.Ganesh Narayanan, IIT Guwahati

99

Centroids of lines, areas, volumes Suppose if density is constant, then the expression define a purely geometrical property of the body; It is called as centroid Centroid of volume

X = (∫ xc dv) / v

Y = (∫ yc dv) / v

Z = (∫ zc dv) / v

Y = (∫ y dA) / A

Z = (∫ z dA) / A

Y = (∫ y dL) / L

Z = (∫ z dL) / L

Centroid of area

X = (∫ x dA) / A Centroid of line

X = (∫ x dL) / L

R.Ganesh Narayanan, IIT Guwahati

100

y

Find the y-coordinate of centroid of the triangular area

dy

X / (h-y) = b/h y

AY = ∫ y dA

h x x

b h

h

½ b h (y) = ∫ y (x dy) = ∫ y [b (h-y) / h] dy = b h2 / 6 0

0

Y=h/3

R.Ganesh Narayanan, IIT Guwahati

101

Beams Structural members which offer resistance to bending due to applied loads

• Reactions at beam supports are determinate if they involve only three unknowns. Otherwise, they are statically indeterminate R.Ganesh Narayanan, IIT Guwahati

102

External effects in beams Reaction due to supports, distributed load, concentrated loads

Internal effects in beams Shear, bending, torsion of beams v M v

M R.Ganesh Narayanan, IIT Guwahati

SHEAR

BENDING

TORSION

103

compression Tension

D T

D E

C

Cx

F

J

D T

V J

Cy FBE

V

J

M

M

F

F

B SECTION - J W G

AX

A

A

A AY V – SHEAR FORCE

Internal forces in beam

F – AXIAL FORCE

R.Ganesh Narayanan, IIT Guwahati

M – BENDING MOMENT AT J

104

Shear force and bending moment in beam To determine bending moment and shearing force at any point in a beam subjected to concentrated and distributed loads.

FINDING REACTION FORCES AT A AND B

1. Determine reactions at supports by treating whole beam as free-body

R.Ganesh Narayanan, IIT Guwahati

105

DIRECTION OF V AND M 2. SECTION beam at C and draw free-body diagrams for AC and CB. By definition, positive sense for internal force-couple systems are as shown.

M

M

+ VE SHEAR FORCE +VE BENDING MOMENT

V

V SECTION C

R.Ganesh Narayanan, IIT Guwahati

SECTION C

106

SECTION C

EVALUATING V AND M

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 R.Ganesh Narayanan, IIT Guwahati

107

Beer/Johnston

Evaluate the Variation of shear and bending moment along beam

ΣMB= 0 =>RA (-L)+P (L/2) = 0; RA= +P/2 RB = +P/2

SECTION AT C Between A & D

SECTION AT E Between D & B

R.Ganesh Narayanan, IIT Guwahati

108

SECTION AT C; C is at ‘x’ distance from A Member AC:

ΣFy = 0 => P/2-V = 0; V = +P/2 ΣMc = 0 => (- P/2) (X) + M = 0; M = +PX/2 Any section between A and D will yield same result V = +P/2 is valid from A to D V = +P/2 yields straight line from A to 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 R.Ganesh Narayanan, IIT Guwahati

109

SECTION AT E; E is at ‘x’ distance from A

CONSIDER AE:

ΣFy = 0 => P/2-P-V = 0; V = -P/2 ΣME = 0 => (- P/2) (X) +P(X-L/2)+ M = 0; M = +P(L-X)/2 EB CAN ALSO BE CONSIDERED R.Ganesh Narayanan, IIT Guwahati

110

V = V0 + (NEGATIVE OF THE AREA UNDER THE LOADING CURVE FROM X0 TO X) = V0 - ∫w dx

c1

M = M0 + (AREA UNDER SHEAR DIAGRAM FROM X0 TO X) = M0 + ∫V dx

R.Ganesh Narayanan, IIT Guwahati

111

Slide 111 c1

cclab9, 1/24/2008

Beer/Johnston

• Taking entire beam as free-body, calculate reactions at A and B. • Determine equivalent internal force-couple systems at sections cut within segments AC, CD, and DB.

∑M A = 0: B y (32 cm ) − (480 N )(6 cm ) − (400 N )(22 cm ) = 0 ∑MB = 0:

B y = 365 N

(480 N )(26 cm ) + (400 N )(10 cm ) − A(32 cm ) = 0 A = 515 N

∑ Fx = 0 :

Bx = 0

• The 400 N load at E may be replaced by a 400 N force and 1600 N-cm couple at R.Ganesh Narayanan, IIT Guwahati D.

112

From A to C: ∑ F y = 0 : 515 − 40 x − V = 0 V = 515 − 40 x

∑ M1 = 0 :

( )

− 515 x − 40 x 12 x + M = 0 M = 515 x − 20 x 2 x

V = 515 + (-40 X) = 515-40X = 515 - ∫40 dx 0 x

M = ∫515-40x dx = 515x-20 x2 0

From C to D:

∑ Fy = 0 :

515 − 480 − V = 0 V = 35 N

∑ M 2 = 0 : − 515 x + 480 ( x − 6 ) + M = 0 M = (2880 + 35 x ) N ⋅ cm R.Ganesh Narayanan, IIT Guwahati

113

• Evaluate equivalent internal force-couple systems at sections cut within segments AC, CD, and DB. From D to B:

∑ Fy = 0 :

515 − 480 − 400 − V = 0 V = −365 N

∑M2 = 0: − 515 x + 480 ( x − 6 ) − 1600 + 400 ( x − 18 ) + M = 0 M = (11,680 − 365 x ) N ⋅ cm

R.Ganesh Narayanan, IIT Guwahati

114

Shear force & Bending moment plot

AREA UNDER SHEAR FORCE DIAGRAM GIVES BM DIAGRAM AC: (35X12) + (1/2 x 12 x 480) = 3300 0 to 3300 CD: 3300 +(35X6) = 3510 3300 to 3510 DB: 365 x 14 = 5110 5110 to 0 R.Ganesh Narayanan, IIT Guwahati

115

Find the shear force and bending moment for the loaded beam

300 lb

1000 lb/ft

4 ft

R.Ganesh Narayanan, IIT Guwahati

4

2

2

116

Machine

R.Ganesh Narayanan, IIT Guwahati

117

R.Ganesh Narayanan, IIT Guwahati

118