4. distributed loads - chulalongkorn...

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4. Distributed Loads

2142111 S i 2011/22142111 Statics, 2011/2

© Department of Mechanical Engineering Chulalongkorn UniversityEngineering, Chulalongkorn University

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Objectives Students must be able to #1Objectives Students must be able to #1

Course ObjectiveI l d di t ib t d l d i t ilib i l Include distributed loads into equilibrium analyses

Chapter Objectives Describe the characteristics and determine the centroids, Describe the characteristics and determine the centroids,

centers of mass and centers of gravity by integration and composite body methods

Apply the Pappus Theorems for surface and volume of Apply the Pappus Theorems for surface and volume of revolution

Describe the characteristics and determine the first moment of f farea, second moment of area and polar moment of inertia by

integration, parallel-axis theorem and perpendicular-axis theorem

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Determine the resultant of loads (force/couple) with line, area and volume distribution by integration and area/volume analogy

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Analyze bodies/structures with distributed loads for unknown loads/reactions by appropriate FBDsloads/reactions by appropriate FBDs

For fluid statics Describe the characteristics and determine hydrostatic and

aerostatic pressures as distributed loads Determine the resultant of fluid statics by integration volume Determine the resultant of fluid statics by integration, volume

analogy and block-of-fluid methods Describe and determine the buoyancy and stability of floating

bodies Analyze bodies/structures with fluid statics for unknown

loads/reactions by appropriate FBDs

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y pp p

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For flexible cablesSt t th ti d t i l d fi iti f fl ibl State the assumptions and geometrical definitions of flexible cables

Appropriately approximate real-life cables into parabolic or pp p y pp pcatenary cables by load distribution

Prove and apply profile, length and tension formula for parabolic & catenary cablesparabolic & catenary cables

Identify and utilize techniques for obtaining numerical solutions of parabolic & catenary cables

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ContentsContents Centroid, Center of Mass and Center of Gravity

Pappus Theorems Pappus Theorems First Moment of Area, Moment of Inertia, Polar Moment of

Inertia Distributed Loads Fluid Statics

Flexible Cables Flexible Cables

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Software HelpsSoftware Helps 3M Software

Maple Computer Center Maple – Computer Center MatLab – Computer Center Mathematica

MapleS l t M l i th ‘St t M ’ Select Maple in the ‘Start Menu’

Type in commands, then ‘Enter’ Use ‘Help Menu’ for command templates Use Help Menu for command templates

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Centers of Gravity High School

Center

Centers of Gravity High School

F

x11x2

x3x4

F

G1 2 3 4x x x x+ + +

7

x 1 2 3 4

4x x x xx + + +

=

8

Centers of Gravity, CGCenter

Centers of Gravity, CG Weight of a body can be represented by an equivalent force

acting at its center of gravity G.

Assume a uniform and parallel force field due to gravitational attraction for most problemsattraction for most problems

Weight W = mg where m is the mass of the body and g is the magnitude of gravitational acceleration.

The center of gravity of is a unique point which is a function of weight distribution only.

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CG Principle of Moments

Center

CG Principle of Moments

N

x dW 1

1

i iNi

y i ii

x dWM x dW xW x

W=

=

= = → =

, x dW y dW

x yW W

= = If 0 :idW →

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, yW Wi

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CG Composite Bodies #1

Center


1 2W W W= +1 2

1 1 2 2

1 1 2 2

W W WWx W x W xWy W y W y

+= += +

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1 1 2 2y y y

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Center


1 2W W W= −1 2

1 1 2 2

1 1 2 2

Wx W x W xWy W y W y

= −= −

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1 1 2 2y y y

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Centers of Mass, CMCenter

An object’s distribution of mass can be represented by an

Centers of Mass, CM

equivalent mass acting at its center of mass.

The center of mass of is a unique point which is a function solely of mass distributionsolely of mass distribution.

Centers of mass coincides with G as long as the gravity field is treated as uniform and parallel.

x dm x dmx

my dm

=

12

y dmy

m=

13

CentroidsCenter

If the density ρ is constant and and gravity field is uniform and parallel G and center of mass coincide with the

Centroids

and parallel, G and center of mass coincide with the centroid of the body.

The centroid C is the geometrical center or the weighted average position of an object.

Locating the centroid by averaging the ‘moments’ of elements of objects about axeselements of objects about axes.

The centroid lies on the axis of symmetry.

Geometry of the body is the only factor that influence theGeometry of the body is the only factor that influence the position of the centroid.

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Centroids Formula

Center

Centroids Formula

For a line of length , , ...x dL y dL

L x yL L

→ = =

For a surface with area , , ...

L Lx dA y dA

A x yA A

→ = =

For a body of volume , , ...

A Ax dV y dV

V x yV V

→ = =

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Centroids Symmetry

Center

Centroids Symmetry

If a body has an axis of symmetry, its centroid lies on this axis.

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Example Centroids 1 #1

Center


Find centroid C of area A

2 1/ 2 ( )xx cy y= → =

22at , ,

cax a y b a cb cb

= = → = =

1/ 2( )xdA y dx dx 1/ 2

3 / 21/ 2

1/ 20

( )

2 2( )3 3

a

A

dA y dx dxc

x aA dA dx ab

= =

= = = =

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1/ 20 3 3A c c

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Center

p

Find by moment of area about axisy x

= =

0

( ) 2

1 1

a

A

a a

yAy y dA y dx

= =

2

0 0

2 2 22

1 12 2

1 1

a a

a a

y dx x dxc

x b abA

Ay

= = =

= = =

2

002 2

2 2 2 2 4

3 3 Ans

xc a

ab ab b

Ay

y = = = Ans4 4 2 8b

yA a

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Center

pFind by moment of area about axisx y

= =

1/2

0

3 / 2 5 / 2

( )

1 2

a

A

a a

xAx x dA x dxc

= =

= =

3 / 2 5 / 21/ 2 1/ 2 00

5 / 2 2

1 2 5

2 2

a axAx

A

dx xc ca a bx = =

= = =

1/ 2

2 2

552 2 3 3 Ans5 5 2 5

Ac

a b a b axA ab

x

5 5 2 5A ab

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Center


Find centroid C of line L, given a = b = 100 mm

2 2dxx cy cydy

= → =

2 2

dy

dL dx dy= +2

2

( / ) 1

1 ( / )b

dL dx dy

dx dy dy

L dL d d d

= +

= +

2

0

2

0

1 ( / )

1 (2 )

L

b

L dL dx dy dy

cy dy

= = +

= +

19

0

147.9 mm=

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Center


Find by taking moment of line about axisb

y x

= = +

= → =

2

02

1 (2 )

8483.6 mm 57.4 mm Ans

b

LLy y dL y cy dy

Ly y

= = + 2 2

0

Find by taking moment of line about axis

1 (2 ) b

L

x y

Lx x dL cy cy dy

20

= → = 0

26063.4 mm 41.0 mm AnsL

Lx x

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Center Formula


2 2 2 2 2 2 21 ln( )21

x a dx x x a a x x a ± = ± ± + ±

2 2 2 2 3

2 2 2 2 2 3 2 2 2

1 ( )31 1( )

x x a dx x a

x x a dx x x a a x x a

± = ±

± = ± ±

4 2 2

( )4 81 ln( )8

x x a dx x x a a x x a

a x x a

± = ± ± −

+ ±=

8

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Center


Find centroid C of arc L

By symmetry, lies on axis.C x

α=

By symmetry, lies on axis.

length of arc 2

C x

L rα

αα

θ θ

α α θ α

−

= =

= =

2 2

( ) cos ( )

2 2 cos 2 sin

LLx x dL r r d

r x r d rα

α α θ α

αα

−= =

=

2 2 cos 2 sin

sin Ans

r x r d r

rx

22

α

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Center


Built around 2560 BC, the Great Pyramid of Khufu (Cheops) is one of the Seven Wonders of the Ancient World. It was 481 ft high; g ;the horizontal cross section of the pyramid is square at any level, with each side measuring 751 ft at theeach side measuring 751 ft at the base. By discounting any irregularities, find the position of Pharaoh’s burial chamber which isPharaoh s burial chamber, which is located at the heart [centroid] of the pyramid.

23(http://ce.eng.usf.edu/pharos/wonders/pyramid.html)

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Center


By symmetry about and axes, x y lies on axis.

At 0, and , 00

C zx z h x a z

h h= = = =

00

( ) ( )

h hz mx C x h x ha a

a ax z h h z

−= + = + = − +−

= =

22

2

( ) ( )

4 ( )

x z h h zh h

adV dz h z dzh

= − − = −

= = −

22

2 0

4 ( )h

V

aVz z dV z h z dzh

= = −

2

22

0

4(2 ) 3

h

V

ha hV dV x dz= = =

2 2 24

24 1( )

12 3a a hh

hh

= =

24

120.25 ft Ans4hz = =

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Center


Find centroid C of the body.

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Center


6

3

B 8 08 10 0 0 0Comp (mm ) (mm) (mm) (mm)V x y z

×××

6

6

3

Box 8.08 10 0 0 0Cylinder 2.26 10 185 0 0Rod 17 67 10 0 175 0×

×× 6

6

Rod 17.67 10 0 175 0Sphere 0.524 10 0 275Total 1 10

00 88 ×Total 1 100.88

= =

38.5 mmi i

i

V xx x

V

= =

13.52 mmi i

i

V yy y

V

V

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= =

0 mm Ansi i

i

V zz z

V

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Pappus Theorems Formula #1Theorem IPappus

Pappus Theorems Formula #1

The area of a surface of revolution equals the product of the length of the generating curve and the distancethe length of the generating curve and the distance traveled by the centroid of the curve in generating the surface area.

For surface area generated bycomplete revolutioncomplete revolution

(2 )( )S x Lπ=

2S LFor incomplete revolutionS

π= 2S xL

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S xLθ=

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Pappus Theorem II


The volume of a body of revolution equals the product of the generating area and the distance traveled by thethe generating area and the distance traveled by the centroid of the area in generating the volume.

For volumn generated bycomplete revolution

(2 )( )V x Aπ=

For incomplete revolution

2V xAπ=

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V xAθ=

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Pappus Theorems Misc

Pappus

Pappus Theorems Misc

Also called the Pappus - Guldinus TheoremThe theorem require that the generating curves and areas The theorem require that the generating curves and areas do not cross the axis about which they rotates.

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Example Pappus Theorems 1 #1

Pappus

Example Pappus Theorems 1 #1

Find centroid C of a quarter circular area with radius r.

By symmetry x yπ

π

===

2

3

By symmetry, Quarter-circular area 4Volume of hemisphere 2 3

x yA rV rπ=Volume of hemisphere 2 3

From Pappus 2nd theorem

V r

[ ]π π π π= =

= =

3 22 2 3 2 ( 4)4 Ans

V xA r x rrx y

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π= = Ans

3x y

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SummarySummary Centroid, CM and CG are centers of geometry, mass and gravity.

Centroid and CM coincide if the density ρ is constant. CM and CG coincide if g is constant.

Calculation by moment Integration Composite body

P th f b di t d b l ti Pappus theorems for bodies generated by revolutions

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Moment of …Moments

Moment of …

moment of force about axismoment of line about axis

F z rFL z rL

==moment of line about axis

moment of area about axismoment of volume about axis

L z rLA z rA

V z rV

==

=

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First Moment of Area Qx & Qy

Moments

First Moment of Area Qx & Qy The first moment of area Q

The first moment of area with The first moment of area with respect to x axes

Q dA A

The first moment of area with

xQ y dA yA= =

The first moment of area with respect to y axes

Q dA A yQ x dA xA= =

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Second Moment of Area Ixx & Iyy

Moments

Second Moment of Area Ixx & Iyy The second moment of area or

the area moment of inertia Ithe area moment of inertia I The second moment of area

with respect to x axes

2

xxI y dA=

The second moment of area with respect to y axes

2

yyI x dA=

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Polar Moment of Inertia JMoments

Polar Moment of Inertia J

2 2 2( )J r dA x y dA+ ( )A A

xx yy

J r dA x y dA

I I

= = +

= +

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Ix & Iy vs Iz Parallel-Axis Theorem

Moments

Ix & Iy vs Iz Parallel Axis Theorem

2( )I y d dA′= +2 2

2

( )

( ) 2

0

xx yA

y yA A A

I y d dA

y dA d y dA d dA

I Ad

= +

′ ′= + +

+ +

20x x yI Ad′ ′= + +

2xx x x yI I Ad′ ′= +

2yy y y x

xx yy

I I AdJ I I

′ ′= += +

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Example Second Moment of Area 1 #1

Moments

Example Second Moment of Area 1 #1Find the second moments of areaand polar moment of inertia aboutpaxes passing though the center

2 2

42 2 2

2

(2 )

xx A AI y dA y x dy

ay a y dyI π

= =

= − =

(2 )4

By symmetry

Axx y a y dyI = =

4

By symmetry

4yy xxaI I π= =

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4

2xx yyaJ I I π= + =

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Moments


Find the second moments of area and polar moment of inertia aboutaxes c cx y−

4 42 2 2 5

4 4c cx x xx ya aI I Ad a aπ ππ= + = + =

y4 4

2 2 2

4 45

4 4c cy y yy xa aI I Ad a aπ ππ= + = + =

xc

yc

452c c c cx x y yaJ I I π= + =

c

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Table Centroid and Moment of Inertia #1

Moments


( , )x y xxI yyIArea

bh (0,0)3

12bh 3

12hb

12 12

2

4Dπ (0,0)

4

64Dπ 4

64Dπ

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Moments


( , )x y xxI yyIArea

bh(0 )h 3bh 3hb

2(0, )

3 36 48

abπ (0,0)3

4abπ 3

4baπ

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Example Hibbeler 6-68 Mech of Mat #1

Moments

Example Hibbeler 6 68 Mech of Mat #1

Find the second moment of area with respect to the neutral axis (the horizontal axis which passes through the centroid ofaxis (the horizontal axis which passes through the centroid of the area)

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Moments


Find the centroid locationFind the centroid location( , )

By symmetry, 0C C y z

z=

=

1

1

y y y,(10 0.5) in.(6 in.)(1 in.)

yA

= +=

2

2

5 in.(1 in.)(10 in.)

yA

A

==

1 1 2 2

1 2

7 0625 in

i i

i

y A y A y Ay

yA A A

+= =+

=

42

7.0625 in.y =

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Moments


Find with respect to (wrt) horizontal I p ( )axis passing through C

2 21 1 2 2

3

( ) ( )

1 (6 in.)(1 in.)12c c

c c

z z

z z zz y zz yI I Ad I

I

Ad= + + +

= +

2 2

3

12(6 in. )(10.5 in. 7.0625 in.)

1

c c

c c

z z

z zI = − +

3

2

1 (1 in.)(10 in.)12(10 in. )(7.0

c c

c c

z z

z zI

I =

=

+

2625 in. 5 in.)−

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4 4197.2708 in. 197 in.c cz zI = =

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Loads Concentration vs. Distribution

Distributed Loads

Loads Concentration vs. Distribution

Concentrated Load acts at a point, does not exist in the exact sense,

t bl i ti h t t i ll acceptable approximation when contact area is small.

Distributed Loads distributed over line, area, volume.

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Loads Hand Tools

Distributed Loads

Loads Hand Tools

Blacksmith Hammer Wood ChiselBlacksmith Hammer Wood Chisel

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Loads Distribution Types

Distributed Loads

Loads Distribution Types

Line DistributionIntensity w (N/m) = force per unit length Intensity w (N/m) = force per unit length

Area Distribution Intensity (N/m2 or Pa) = force per unit area Action of fluid force pressure

I t l i t it f f i lid t Internal intensity of force in solid stress

Volume DistributionVolume Distribution Intensity (N/m3) = body force per unit volume Specific weight ρg is the intensity of gravitational

tt ti

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attraction.

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Loads Resultant Force System

Distributed Loads

Loads Resultant Force System

Distributed loads can be represented by an equivalent force-couple system consisting offorce-couple system, consisting of Resultant force FR

Magnitude Direction Line of action

Resultant couple M Resultant couple MR

Magnitude Direction

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Line Distribution Descriptions

Distributed Loads

Line Distribution Descriptions

The distributed loads may be described by functions of positionspositions.

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2 2Let (0.01 ) MN/m ( in m)w x x= −

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Line Distribution Resultants

Distributed Loads

Line Distribution Resultants

RF w dx=

magnitude of F

RM wx dx F x= = location of F

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Example Line Resultant 1 #1

Distributed Loads

p

Consider right half the load= − ×

= = −

2 2 6

0.01 m6 2 2

Consider right half the load(0.01 ) 10 N/m ( in m)

10 (0 01 m)

w x x

F w dx x dx= = −

= − =

0

0.016 2 3

0

10 (0.01 m)

10 0.01 ( 3) 2 3 N

RF w dx x dx

x x

= = −

= −

0.016 2 2

00.016 2 2 4

10 (0.01 )

10 (0.01 2) ( 4)

M wx dx x x dx

x x−

= × ⋅=

03

0 (0 0 ) ( )

2.5 10 N mM xF → = 3.75 mmR x

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R

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Line Distribution Area Analogy

Distributed Loads

Line Distribution Area Analogy

If y w=

If

R

y w

F w dx

d F

=

area A

Rx xw dx F= centroid of A

The single equivalent force FR exerted by the line distributed load is equal toby the line distributed load is equal to the “area” A and acts through the centroid of A between the loading curve and the x axis

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and the x axis.

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Example Area Analogy 1 #1

Distributed Loads


Determine the reactions at A and C.

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Distributed Loads


(200 N/ )(5 ) 1000 NF1

2

(200 N/m)(5 m) 1000 N(100 N/m)(6 m) 600 N0 5(100 N/m)(6 m) 300 N

FFF

= == == =3 0.5(100 N/m)(6 m) 300 NF = =

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Distributed Loads


0F 1

0

01000 N Ans

x

x

F

F AA

= − + =

=

1000 N Ans

0

(2 5 m) (2 m)

x

A

A

M

F F

=

= + − −

1 3

2

(2.5 m) (2 m)(3 m) (6 m) 016.667 N Ans

y

y

F FF C

C

+ =

= −

2 3

0

0

y

y

y y

F

A C F F

= + − − =

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2 3 0916.67 N Ans

y y

y

C

A =

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Distributed Loads


If the cable can sustain tension of up to 600 N, determine the maximum wthe maximum w.

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Distributed Loads


= + + + ° =

0

(15 N)(7 5 m) (7 5 N)(20 m) (15 m) sin30 (30 m) 0CM

w w T T

− − + + ==

= =max max max

(15 N)(7.5 m) (7.5 N)(20 m) (15 m) sin30 (30 m) 030 262.5 N/m

Given 600 N, thus 30 262.5 N/m

w w T Tw T

T w T

56

=max max max

max 68.57 N/m Answ

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Distributed Loads


Find support reactions

1 m 1/ 2

01 m3 / 2

( )

2( ) 2

AF dA x dx= =

3 / 2

0

2( ) 2 kN3 3x

= =

1 m 1/ 2

01 m5/ 2

( )xF x x dx=

1 m5 / 2

0

2( ) 0.4 kN m5

0 6 m

x

x

= = ⋅

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0.6 mx =

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Distributed Loads


Consider FBD of beam

0xF = 0 Ans

0x

y

C

F

=

= 2 / 3 kN 0

0.667 kN Ansy

y

C

C

+ =

= −

0

(0.6 m)(2 / 3 kN) 0CM

M

= + + =

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0.4 kN m AnsM = − ⋅

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Area Distribution Resultants

Distributed Loads

Area Distribution Resultants

RF p dA

M xp dA F x

= RM xp dA F x= =

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Example Hibbeler 9-121 #1

Distributed Loads

pFind the equivalent concentrated loadconcentrated load

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Distributed Loads

pp

40 lb/ft2 General equation of a planeP1

FR 20 lb/ft230 lb/ft2

21

0 (1)(0 ft,0 ft,40 lb/ft )

40 0

ax by cp dP

c d

+ + + =

+

P2P4

y

x10 lb/ft210 ft

5 ft

xy 2

2

40 0(5 ft,0 ft,30 lb/ft )

5 30 0

c dP

a c d

+ =

+ + =

P3

10 lb/ft210 ft2

3

5 30 0(5 ft,10 ft,10 lb/ft )

5 10 10 0

a c dP

a b c d

+ + =

+ + + =2 , 2 , 40a c b c d c= = = −2

4(0 ft,10 ft,20 lb/ft )10 20 0

Pb c d+ + =

Subst into (1)2 2 40 02 2 40 0cx cy cp c+ + − =

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2

2 2 40 02 2 40 lb/ft

x y pp x y

+ + − == − − +

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Distributed Loads

pp

40 lb/ft2

FR

40 lb/ft2

20 lb/ft230 lb/ft2

y

30 lb/ft

5 ft

xy

x10 lb/ft210 ft

5 ft

= = − − +

5 10

( 2 2 40)RdF p dA x y dx dy

62

= − − + =

5 10

0 0( 2 2 40) 1250 lb AnsRF x y dydx

63


Distributed Loads

pp

40 lb/ft2

FR

40 lb/ft2

20 lb/ft230 lb/ft2 = =

2 33 ftRRF

x dFx

y

30 lb/ft

5 ft

xy

= =

= =

2.33 ft

4 33 ft AnsR

R

RF

xF

y dFyx

10 lb/ft210 ft

5 ft = = 4.33 ft AnsR

yF

5 10

0 0

5 10

( 2 2 40) 2916.67 lb ftR

RFx dF x y x dydx= − − + = ⋅

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5 10

0 0( 2 2 40) 5416.67 lb ft

RRF

y dF x y y dydx= − − + = ⋅

64

Area Distribution Volume Analogy

Distributed Loads

Area Distribution Volume Analogy

If z p

F p dA

=

= volume V

p

x xp dA F=

centroid of V

y yp dA F=

The single equivalent force F exerted by the area distributed load is equal to the

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q“volume” V and acts through the centroid of V

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Fluid StaticsFluid Statics

Fluid Statics Required in studies and designs of pressure vessels,

piping ships dams and off-shore structures etcpiping, ships, dams and off-shore structures, etc.

Topics Definitions Fluid pressure

Hydrostatic pressure on submerged surfaces Hydrostatic pressure on submerged surfaces Buoyancy Air pressurep

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Fluid Statics Definitions

Fluid Statics

Fluid Statics Definitions

Fluid is any continuous substance which, when at rest, is unable to support shear forceunable to support shear force.

Fluid Statics studies pressure of fluid at rest. Hydrostatics – stationary liquid Aerostatics – stationary gas

Pascal’s Law: the pressure at any given point in a fluid is the same in all directionsthe same in all directions.

Pressure p in fluid at rest is a function of vertical dimension and its density ρ.

Resultant force on a body from pressure acts at the center of pressure P.

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Fluid Statics Fluid Pressure #1

Fluid Statics


ρ = + − + = Consider forces on control volume in direction

0 ( ) ( ) ( ) 0x w

x

F p dA g dAdx p dp dA

67

ρ

=

( )x w

wdp g dx

68


Fluid Statics

In direction onlyxd d

0 0

wp x

wp

dp g dx

dp g dx

ρ

ρ

=

=

t 0

ρ= +0 wp p gx

0 pressure at 0p x= =

F p dA= A

F p dA=

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Pressure Submerged Surfaces #1

Fluid Statics


Find the resultant force andFind the resultant force and its position on the submerged face AB. It is given that the width of the plate is b and thewidth of the plate is b and the fluid has a constant density ρ.

MethodsMethods Integration Volume analogy

69

Equilibrium of block of fluid *new*

70


Fluid StaticsFluid Block


Consider equilibrium of

the block of fluid above the plate:

= = = 0, 0, 0x y OF F M

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Example Fluid Statics 1 #1

Fluid Statics


Find the resultant force and its position on the dam with 1 m width The density of water is 1000 kg/m3m width. The density of water is 1000 kg/m .

1 m

y

p1

2.5 m 2.5 m

x

y

F

2.5 m

p2

LP

711 m

72


Fluid StaticsIntegration


ρ= =

1(1 )x yF p dA gx dy1

2.5xy −= + ρ

ρ=−

0

0

2.5

( )

( )2.5

x yAp g y

gx dxy

p1

2.5

ρ = − =

02

2.5

12250 N2.5 2

g x

2.5 m

x

y

F Fx

ρ= =

2.5

02.52

(1 )

3062 N

y xAF p dA gx dx

x

2.5 m

p2

LPFy

ρ

= =

= +0

2 2 2

30625 N2

x y

xg

F F F1 m

72

= =32984 N 33.0 kN Ansx y

F

73


Fluid StaticsIntegration


ρ= =

1

0(1 )x yA

F y py dA gxy dy12.5

xy −= +

ρ= − +−

0

2.5

2 53

( 1)2.5 2.5gx x dx

y

p1

2.5

ρ = = ⋅

2.53

20

4083.3 N m32.5

0 33333 m

g x

y2.5 m

x

y

F Fx

ρ

=

= =

2.5 2

02 5

0.33333 m

(1 )y xA

y

F x px dA gx dx

2.5 m

p2

LPFy

ρ

= = ⋅

2.53

0

51042 N m3

1 6667

xg1 m

73

==1.6667 m

( , ) (1.67,0.333) m AnsxP x y

74


Fluid StaticsVolume Analogy


= =1At 0, 0x pρ= =

= +

1

2

At 0, 02.5 m, 2.5 N

1 ( )

x px p g

F p p LWy

p1

= +

= × × × + ×

1 2

2 2

( )21 2.5 1000 9.8 2.5 1 12

F p p LW

2.5 m

x

y

F

=

=

232.98 kN Ans

2 1( ) ( 2 5 1)

F

P x y

2.5 m

p2

LP

=

( , ) ( 2.5, 1)3 3

( , ) (1.67,0.333) m Ans

P x y

P x y1 m

74

75




2.5 mp12.5 m1 m

x

y

Fx

y2.5/3 m

G

pLP

FF1

p2

W2.5/3 m

11000 9.8 2.5 1 12250 N2

W gVρ= = × × × × =

1 m

751

21 1 1000 9.8 2.5 2.5 30625 N2 2

F ghHρ= = × × × × =

76




2.5 m

1 m

2/3 m =

− = 0

012250 N

x

x

F

W FF

x

y2.5/3 m

G=

= 12250 N

0

0

x

y

F

F

F F

F1

P

Fy

− =

=

+

1

2 2 2

0

30625 Ny

y

F F

F

F F F

FW

2.5/3 mFx

= +

= =32984 N 33.0 kN Ansx yF F F

F

76

77




+ 0M 2.5 m

1 m

2/3 mO = + ×− + − + =

1

0

1 2.5 2 03 3

O

y x

M

F x F y W F

x

y2.5/3 m

G L

× − − + −2

3 32 2.5 1 2.5

3 3y x

F1

P

Fy

− −= + = +From 1, 12.5 2.5

x xy yFW

2.5/3 mFx

= =5 1, 3 3

( ) (5 / 3 1/ 3) A

x y

P

77

=( , ) (5 / 3,1/ 3) m AnsP x y

78


Fluid Statics


Cross section of a long channel is shown Each ofchannel is shown. Each of the bottom plates, hinged at B, has a mass of 250 kg per meter of channelper meter of channel length. Find force per meter of channel length acting on each plate at B The0 9 m each plate at B. The density of water is 1000 kg/m3 .1.2 m 1.2 m

0.9 m

0 9 mA C

B

78

0.9 mA C

79


Fluid Statics


1At , 0.9 N A 1 8 N

B p gA

ρ=

2At , 1.8 NWidth of channelSymmetry of forces at : 0

A p gw

B B

ρ==

=Symmetry of forces at : 0yB B =

p1

B

By p1

B B

By

p2

0.9 m

BxBp2

0.9 m

Bx B

x

y1.2 m

A

0.6 m

1.2 mC

0.6 m

79

x

Ay

AxA mg = 250gw

Cy

CxCmg = 250gw

80


Fluid StaticsVolume Analogy


p2

p1

F1BxB

y

p2 F1F2

0.9 mx

1.2 mAx

A mg = 250gwρ= =1 1

Volume Analogy1.5 1.35 NF p w gw

Ay

g g

ρ= −=

2 2 10.5( )1.50.675 N

F p p wgw

ρ ρ

= + − − − =

0

0.9 250 0.6 1.35 0.75 0.675 0.5 00 9 1 00 1666

A

x

M

B gw gw gw

80

= → == =

0.9 1500 1666.7 kNForce per meter acting at 1.67 16.3 kN/m Ans

x xB gw B gwB g

81

Buoyancy Definition

Fluid Statics

Buoyancy Definition

The resultant force exerted on the surface of an object immersed in a fluid:immersed in a fluid: is equal and opposite to the weight of displaced fluid, pass through the center of buoyancy B (center of mass of the

displaced fluid).

ρ= = ( )( )

B wF pA gh AhA

81

ρρ

==

( )w

w f

hA gV g

82

Buoyancy Stability

Fluid Statics

Buoyancy Stability

82

stable unstable

83

Flexible CablesCables

Flexible Cables Found in suspension bridges, transmission lines, etc.

Topics Topics Assumptions of flexible cables Types of cable loadingsyp g Geometrical definition Parabolic cables

C t bl Catenary cables

83

84

Flexible Cables Assumptions

Cables

Flexible Cables Assumptions

Bear load only in tensionNegligible displacement due to stretching Negligible displacement due to stretching Inextensible

Perfectly flexibley Negligible bending resistance Tangential tension along the cable

84

85

Flexible Cables Concentrated Loads

Cables

Flexible Cables Concentrated Loads

Concentrated / discrete loadWeight of the cable is negligible Weight of the cable is negligible.

85

86

Cables Geometrical Definitions

Cables

Cables Geometrical Definitions

L

S = length (m)L = span (m)

h

h = sag, dip (m)S

w = load intensity (N/m)y T

θ = tangential angle at x (rad)T = tension at x (N)T = tension at lowest point (N)

x

T0

w

θ

86

T0 = tension at lowest point (N) x

87

Parabolic CablesCables

Parabolic Cables

Longest Suspension BridgeAkashi Kaikyo Bridge, Japan

3910 m total span(http://www.hsba.go.jp/bridge/)

main cablemain cable

suspender

87towerdeck

88

Parabolic Cables Analysis

Cables

Parabolic Cables Analysisy T

uniform vertical load w =

x

T0

w

θ per horizon unit talof length

0 00 cos 0 cos (1)xF T T T Tθ θ = − = → =

x

0 sin 0 sin

(2) tan

(2)y

y x

F T wx T wx

w dy wx dy x dx

θ θ

θ

= − = → =

→

0 00 0

2 2

tan (1)

(2) (1)

x dy x dxT dx T

θ = = → =

+

1 w

88

2 2 2 2 20

0

1 , 2

wy x T T w xT

= = +

89

Parabolic Cables Formula

Cables

Parabolic Cables Formula

ly

h x

s

20

2 1 , /1 ax a w Twy x = == 00

20

, /2

at , 1 2

2ax a w T

x l y h

y xTwT lh

= ==

( )2 2 2 2

0 max 0

2 2

2 2 20

2 2 2 2 2

1 1 at

1 1 1 (

2

)

T a x T Th

T T a l x lw x = += + → = + =

89

( )2 2 2 2 2 2 21 1 11 ln( 1 )22

ds dx dy s x a x ax a xa

= + → = + + + +

90

Example Parabolic Cables 1 #1

Cables


Find tension in the cable at A and the angle θ made by the cable with the horizontal at Bcable with the horizontal at B.

B70 m

40 m

30 kg/mA

90

91


Cables


2 2

At point Awx wlT= =70 m 40 ml h = =

=

0

2

0

2 2(294.21 N/m)(70 m)

Ty h

T

= == =70 m, 40 m30 N/m 294.21 N/m

l hw g

= =

0

02

2(40 m)18020.4 N 18.02 kN Ans

T

T

wx dy wxl B

T

T0

sy

θ= → = =0 0

tan2

At point

wx dy wxyT dx T

B

h

A

T0

w = 30g

x

θ = =0

p(294.21 N/m)(70 m)tan

18020.4 NwxT

g

91

θθ

== °

tan 1.1428648.8 Ans

92

Flexible Cables Catenary Cables

Cables

Flexible Cables Catenary Cables

Vertical load is the cable’s own weight.

92

93

Catenary Cables Analysis #1

Cables


uniform vertical load per unit of cable lengthμ =Ty

per unit of cable lengthT0

μx

θs

xμ

0 00 cos 0 cos

0 sin 0 sin

(1)

(2)x

y

F T T T T

F T s T s

θ θ

θ μ θ μ

= − = → = = − = → =

22

20 0 0

(2) tan 1(1)

y

dy d y ds dysT dx T dx T dxdxμ μ μθ

= = → = = +

93

0 0 0( )

94


Cables


0

Let tan , dy adx T

dy

μσ θ= = =

22

At 0 : 0, 0

1

dyx ydx

d y dy

σ

μ

= = = =

2

0

2

1

1

y yT dxdx

d da a dx

μ

σ σσ

= +

= + → =2

0 02

1 1

1

x

a a dxdx

d a dxσ

σσ

σ

+ →+

=

94

0 021 σ+

95

Catenary Cables Formula

Cables

Catenary Cables Formula

ly

( )0Let /

1a Tμ=

h x

( )

tan

1 cosh( ) 1

dy as

y axa

θ

= −

= =

s 1 sinh( )

dx

s axa

=

0

0

As cos , coscosh( )

TT T

T dx dsxa

θ θ=

= =

00

0

coscosh( ) at

T dsT TdxlT T a x l

θ= =

= =

hyp sin sinh( )

hyp cos cosh( )

x

x

+ →

+ →

95

ax 0m cosh( ) at lT T a x lhyp cos cosh( )x+ →

http://mathworld.wolfram.com/topics/HyperbolicFunctions.html

96

Catenary Cables Hyperbolic Functions

Cables

Catenary Cables Hyperbolic Functions

x xe e−−sinh( )2

cosh( )x x

e ex

e ex−

−=

+=cosh( )2

sinh( )tanh( )cosh( )

x

xxx

=

=cosh( )

sinh( ) cosh( )

xd x xdx

=

cosh( ) sinh( )d x xdx

=

96

http://en.wikipedia.org/wiki/Hyperbolic_function

97

Example Catenary Cables 1 #1

Cables


Find the length of the cable.

400 m

100AB

100 mA

y 100 m, 200 mh l= =

AB h

97

x

l

98


Cables


= −1From (cosh 1)y ax

=

From (cosh 1)

At : 200 m 1

y axa

A x

( )= −

= −

1100 cosh(200 ) 1

100 cosh(200 ) 1 (1)

aa

a a−= ×

=

3 -1Solve (1), 4.6541 10 m1From sinh

a

s axa

How to solve100 cosh(200 ) 1a a= −

−

−

× ×=×

3

3

sinh(4.6541 10 200) 4.6541 10

a

s

98

== =

230.158 mTotal length 2 460.3 m Ans

sS s

99

Example Catenary Cables 1 #3GraphicalCables = −100 cosh(200 ) 1a a


3.0

ion 2.0

2.5 f = 100xg = cosh(200x) − 1

func

t

0 5

1.0

1.5

= −100 cosh(200 ) 1

( ) 100

a a

f x x x (m-1)0.000 0.002 0.004 0.006 0.008 0.010

0.0

0.5

== −

( ) 100( ) cosh(200 ) 1

f x xg x x

x (m )

99

−= = 1 at 0.004654 m Ansf g x

100

Example Catenary Cables 1 #4Newton-RaphsonCables = −100 cosh(200 ) 1a a


1( )For an equation ( ) 0: ( )

kk k

f xf x x xf+= = −′1q ( )( )k k

kf x+ ′

′= − + = −( ) 100 cosh(200 ) 1, ( ) 100 200sinh(200 )f x x x f x x′= = − = −

′= − =0 0 0

1 0 0 0

( ) ( ) , ( ) ( )Guess 0.01, ( ) 1.7622, ( ) 625.37

( ) / ( ) 0.0071822x f x f x

x x f x f x′= − =′= − =′

2 1 1 1

3 2 2 2

( ) / ( ) 0.0054857( ) / ( ) 0.0047868( ) / ( ) 0 0046

x x f x f xx x f x f x

f f 584

See filechap4_catenary_1.xls

′= − =4 3 3 3( ) / ( ) 0.0046x x f x f x′= − =′= − =

5 4 4 4

584( ) / ( ) 0.0046541( ) / ( ) 0 0046541 Ans

x x f x f xx x f x f x

p _ y_

100

= =6 5 5 5( ) / ( ) 0.0046541 Ansx x f x f x

101


Cables = −100 cosh(200 ) 1a a Newton-Raphson


0 0

0.5

x0x1

f(x0)/f'(x0)

ctio

n f

-0.5

0.0 01

func

-1.5

-1.0 f = 100x

x (m-1)0.000 0.002 0.004 0.006 0.008 0.010

-2.0

101Numerical Methods in Engineering by Prof. Pramote Dechaumphai

102


Cables


Find the length of the cable.

40 m

14A 8 m14 m

B

6 mh

y

Al2

1

2

1 2

6 m14 m

40 m

hhl l

==

+ =x1x2

AB

l1h1h2

102

1 2 40 ml l+s2

s1

103


Cables

Example Catenary Cables 2 #2y

Al2

1

2

6 m14 m

40 m

hhl l

==

+ =x1x2

AB

l1h1h2

1 2 40 ml l+ =1x2

s2s1

1From (cosh 1)

1

y axa

= −

( )

11At : 6 (cosh 1) (1)

1 1: 14 (cosh 1) cosh (40 ) 1 (2)At

B ala

A al a l

= −

= =

103

( )2 1

-11

: 14 (cosh 1) cosh (40 ) 1 (2)

Solve (1) and (2): 0.04453 m , 16.07 m

At A al a la a

a l

= − = − −

= =

104


Cables

Example Catenary Cables 2 #3y

Al2

1

2

6 m14 m

40 m

hhl l

==

+ =x1x2

AB

l1h1h2

1 i h

1 2 40 ml l+ =1x2

s2s1

=

= × =1

1 sinh

1 sinh(0.04453 16.07) 17.477 m

s axa

s

( )

×

= − =

1

2

sinh(0.04453 16.07) 17.477 m0.04453

1 sinh 0.04453(40 16.07) 28.723 m0.04453

s

s

104= + =1 2

0.04453

Total cable length 46.2 m AnsS s s

105


Cables Newton-Raphson


( )f x1

( )For an equation ( ) 0: ( )

kk k

k

f xf x x xf x+= = −′

For equations ( , ) 0 and ( , ) 0,f x y g x y= =

1

1

( , ) ( , ) ( , )( , ) ( , ) ( , )

and

k k k k k k k

k k k k k k k

f x y x f x y y x f x yg x y x g x y y y g x y

x x x y y y

+

+

∂ ∂ ∂ ∂ Δ ⋅ = − ∂ ∂ ∂ ∂ Δ

+ Δ + Δ1 1 1 1and k k k k k kx x x y y y+ + + += + Δ = + Δ

105Numerical Methods in Engineering by Prof. Pramote Dechaumphai

106




1let , : ( , ) cosh 6 1 0x a y l f x y xy x= = = − − =

( )1

( , ) cosh (40 ) 14 1 0( , ) sinh 6

g x y x y xf x y x y xy

= − − − =

∂ ∂ = −

( )( )

( , ) sinh( , ) (40 )sinh (40 ) 14

f x y y x xyg x y x y x y

∂ ∂ =∂ ∂ = − − −

( )( , ) sinh (40 )g x y y x x y∂ ∂ = − −

106

107




= =0 0Guess 1.0, 20 :x y

= × = ×

∂ ∂ = × ∂ ∂ = ×

8 80 0 0 0

9 8

( , ) 2.4258 10 , g( , ) 2.4258 10 ,

/ 4.8517 10 , / 2.4258 10 ,

f x y x y

f x f y

+

∂ ∂ = × ∂ ∂ = − ×∂ ∂ ∂ ∂ Δ

⋅ = ∂ ∂ ∂ ∂ Δ

9 8

1

/ 4.8517 10 , / 2.4258 10 .( , ) ( , )( ) ( )

k k k k k

g x g yf x y x f x y y xg x y x g x y y y

−

( , )( )

k kf x yg x y+

∂ ∂ ∂ ∂ Δ 1( , ) ( , )k k k k kg x y x g x y y y

Δ × × − ×⋅ = Δ× × ×

9 8 82

9 8 8

( , )

4.8517 10 2.4258 10 2.4258 104 8517 10 2 4258 10 2 4258 10

k kg x y

xy

− −

Δ× × − × Δ = − × Δ = − ×

= + Δ = −

2

2 81 1

1 0 1

4.8517 10 2.4258 10 2.4258 10

5.0000 10 , 1.5665 101.0 0.5,

y

x yx x x

107

−

+ Δ

= + Δ = − × =1 0 1

81 0 1

1.0 0.5,

20.0 1.566 10 20.0

x x x

y y y

108


Cables Newton-RaphsonSee file chap4_catenary_2.xls

Example Catenary Cables 2 #8k x y Dx Dy0 1.00000 20.000 -5.0000E-02 -1.5665E-081 0.95000 20.000 -5.0000E-02 -4.3288E-082 0.90000 20.000 -5.0000E-02 -1.1835E-073 0.85000 20.000 -5.0000E-02 -3.2214E-074 0 80000 20 000 -5 0000E-02 -8 7523E-074 0.80000 20.000 5.0000E 02 8.7523E 075 0.75000 20.000 -5.0000E-02 -2.3757E-066 0.70000 20.000 -4.9999E-02 -6.4449E-067 0.65000 20.000 -4.9998E-02 -1.7475E-05

19 0.08365 19.000 -2.2676E-02 -8.8940E-0120 0.06097 18.110 -1.1950E-02 -1.0926E+0021 0 04902 17 018 3 9691E 03 7 6840E 0121 0.04902 17.018 -3.9691E-03 -7.6840E-0122 0.04505 16.249 -5.1462E-04 -1.7321E-0123 0.04454 16.076 -1.0609E-05 -4.8138E-0324 0.04453 16.071 -5.6995E-09 -2.7864E-06

108

25 0.04453 16.071 -1.7810E-15 -8.7017E-13

109

Concepts #1

Review

Concepts #1

Centroid, center of mass and center of gravityrespectively represent the centers of geometry massrespectively represent the centers of geometry, massand weight.

Pappus theorems can be used to find the surface area and volume of revolutions.

Apart from moments of forces, there are several kinds of moments. The first moment of area is a property of cross section that is used to predict its resistance to shearsection that is used to predict its resistance to shear stress. The moment of inertia quantifies the rotational inertia of a body. The polar moment of inertia is a property of cross section that is used to predict its

109

property of cross section that is used to predict its resistance to torsion.

110

Concepts #2

Review

p The equivalent resultant of the distributed loads can be used

in the analyses instead of the distributed loads The area andin the analyses instead of the distributed loads. The area and volume analogies help visualized the line and surface distributed loads on bodies.Th fl id t ti d l ith th ff t f fl id t t The fluid statics deal with the effects of fluid at rest. The pressure, exerted by the fluid in the perpendicular

direction with respect to the surface in contact with the fluid, varies linearly with depth.

The block of fluid is an addition method to determine resultants.resultants.

Flexible cables can support tension only. Parabolic cables are loaded with uniform force per unit

l th

110

span length. Catenary cables are loaded with uniform force per unit of

cable length.

4. distributed loads - chulalongkorn...

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