Lecture Note 0

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Lecture Note 0 Review Second Semester, Academic Year 2012 Department of Mechanical Engineering Chulalongkorn University Contents Basic Statics Force system RigidbodiesinequilibriumRigid bodies in equilibrium Structures (trusses and frames/machines) in equilibrium Basic Mechanics of Materials Stress, strain and transformation Stress and deformation in axially loaded members, torsion, bdidl 2

bending and pressure vessels

Equilibrium

Definition An object is in equilibrium when it is stationary or in steady translation relative to an inertial reference frame. 



 0RFF 3





 0 OROMM When a body is in equilibrium, the resultant force and the resultant couple about any point are both zero.

R R

F MO

 

Example

Hibbeler Ex 5-6 #1

2D Equilibrium 4

Determine the horizontal and vertical components of reaction for the beam loaded as shown. Neglect the weight of the beam in the calculation.

(2)

Example

Hibbeler Ex 5-6 #2 Find:,,yxyABB

2D Equilibrium   

 

_Fi^nd^:

, , Equilibrium of 0(600 N)cos450 424.26 N424 NAns 0(100N)(2m)(600N)sin45(5m)

yxy xx x B

ABB ADB FB B M 5

   

 

0(100 N)(2 m)(600 N)sin45(5 m) (600 N)cos45(0.2 m)(7 m)0 319.50319 NAns 0(600 N)sin45(100

B y y yy

M A A FA 

N)(200 N)0 404.76 N405 NAnsy y

B B

2D Supports

Summary #1

2D Equilibrium 6

2D Supports

Summary #2

2D Equilibrium 7

2D Supports

Summary #3

2D Equilibrium 8

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Example

Hibbeler Ex 5-15 #2

3D Equilibrium FBD of plate 9

 

 

Equilibrium of plate 00 00

xx yy

ABC FB FB

Example

Hibbeler Ex 5-15 #3 



_E^q^u^ilib

rium of plate 0(300 N)(980.7 N)0(1)zzzC

ABC FABT

3D Equilibrium   

 

0(2 m)(980.7 N)(1 m)(2 m)0(2) 0(300 N)(1.5 m)(980.7 N)(1.5 m)(3 m) (3 m)(200 Nm)0(3)

xCz yz z

MTB MB A Solve (1), (2) & (3) 79035N21667NAB 10

   

790.35 N, 216.67 N, 707.02 N 790 N, 707 N, 0,217 N Ans

zz C zC xyz

AB T AT BBB

3D Supports

Summary #1

3D Equilibrium 11

3D Supports

Summary #2

3D Equilibrium 12

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3D Supports

Summary #3

3D Equilibrium 13

3D Supports

Summary #4

3D Equilibrium 14Comparison with 2D supports

Example

Disassembling the structure

Structure 15

Imaginary Section

Choice of Planes Imaginary section cuts through the considered region. The values of internal loads depends on the plane of imaginarysection

Loads imaginary section. 16

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Stress

Normal and Shear StressesLoads 17

avgN AavgV A

Strain

Normal and Shear StrainsLoads   ss

  2nt 18

 avg sL

Axially Loaded Members

A homogeneous and isentropic prismatic bar is subjected to axial load.

Axial Load  , , latP ALL 19

Changes in Lengths

Prismatic Bars A prismatic bars has straight longitudinal axis and constant cross section.

Axial Load secto PL AE , , P E AL 20

AE

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Example

Hibbeler 4-4 #1 The bronze C86100 shaft is subjected to the axial loads shown. If the diameters of each segment are dAB=0.75 in., dBC=2 in., and

Axial Load dCD=0.5 in., determine the displacement of end Awith respect to end D. 21

3 15.010 ksiE

Example

Hibbeler 4-4 #2Axial Load 22

Example

Hibbeler 4-4 #3 



_F^B^D^{of sec}

tion 0 2 kip0xAB

A FF

Axial Load    

 

2 kip (T) FBD of section 0 3(2 kip)0 6 kip (T)

xAB AB xBC BCF AB FF F 23





FBD of section 0 8 kip0

BC xCDAD FF 8 kip (T)CDF

Example

Hibbeler 4-4 #4Axial Load 24

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Example

Hibbeler 4-4 #5   23(2.0 kip)(48 in.) 0.014487 in.ABAB ABFL AE

Axial Load     

   

23 23 23

(0.75 in.)(15.010 ksi) 4 (6.0 kip)(120 in.) 0.015279 in. (2.0 in.)(15.010 ksi) 4 (8.0 kip)(36 in.) (05i)(15010k

AB BCBC BC BC CDCD CD CD

AE FL AE FL AE0.097785 in. i) 25

 23 (0.5 in.)(15.010 ks 4CDAE  

i) 0.127551 in.0.158 in.Ans

ADABBCCD

Torsion

Linearly Elastic Circular Bars  , maxmaxGGrG c

Torsion 26

Torsion Formula

Torsion 



ATdM Tr TL 27

 J GJ

Polar Moment of Inertia



 

23 00

Solid shaft (2)2 rr Jdd

 

Torsion 44 232rd J 



²²23 Tubular shaft or circular tube rr

 28





22 11

23 (2 )2rr rrJdd 4444 1212()() 232Jrrdd 

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Example

Gere 3.4-2 #1 A stepped shaft ABCD consisting of solid circular segments is subjected to three torques as shown. The material is steel with shear modulus of elasticityG=80 GPa.

Torsion (a) Calculate the maximum shear stress τmaxin the shaft. (b) Calculate the angle of twist φD(in degrees) at end D. 29

Example

Gere 3.4-2 #2Torsion 30

   



FBD of shaft 0 3000 Nm 2000 Nm800 Nm0 5800 Nm

x A A

M T T

Example

Gere 3.4-2 #3Torsion 



_F^B^D^{of sec}

tion in equilibrium 005800 NmxAABAB

A MTTT 31

  

  

_F^B^D^{of sec}

tion in equilibrium 03000 Nm02800 Nm FBD of section in equilibrium 02000 Nm0

xABBCBC xBCCDCD

B MTTT C MTTT800 Nm

Example

Gere 3.4-2 #4Torsion 32

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Example

Gere 3.4-2 #5  43421632 , Section

maxTrTrTTLTL JGJrdGd AB

Torsion    

    

6 33 4924

Section 1616(5800 Nm) ()57.69410 Pa (0.08 m) 3232(5800 Nm)(0.5 m) 0.0090146 rad (8010 N/m)(0.08 m) Section

AB maxAB AB ABAB AB AB

AB T d TL Gd BC 33

1 ()maxBC   

    

6 33 4924

616(2800 Nm) 66.02010 Pa (0.06 m) 3232(2800 Nm)(0.5 m) 0.013754 rad (8010 N/m)(0.06 m)

BC BC BCBC BC BC

T d TL Gd

Example

Gere 3.4-2 #6  6

Section 1616(800 Nm) ()6366210PaCD C

CD T

Torsion     

    

34 4924

()63.66210 Pa (0.04 m) 3232(800 Nm)(0.5 m) 0.019894 rad (8010 N/m)(0.04 m) 66.0 MPa betweenAns

maxCD CD CDCD CD CD max

d TL Gd BC 34

 

 

0.042663 rad 2.44

DABBCCD D CCWAns

Flexure Formula





The neutral axis NA does not change length. 0NA passes through the centroid.xF

Bending My



 

Load & stress relationshipzMM 35

 I

C & I:

Centroid and Moment of Inertia 33Area(,) (0,0) 1212

xyxyII bhhb bh

Bending 244

1212 (0,0) 46464DDD 36

2 1cxxIIAd

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Shear Formula

DerivationBending 0xF VQ It 37

It QyA

Example

Hibbeler 4th 6-68 #1 The T-beam is subjected to the loading shown, determine the absolute maximum bending stress in the T-beam and sketch the normal stress distribution on the cross section.

StressBending 38

Example

Hibbeler 4th 6-68 #2 FBD of whole T-beam 00xBFH 

 

StressBending 0261.813.5180 12.3/180.68333 kip 021.80 56.1/183.1167 kip

AB B yAB A

MR R FRR R

    

 

39

Example

Hibbeler 4th 6-68 #3  



_{FBD 1}

: 06 ft 020y

x FV

StressBending     

 

2 kip 020 2 kipft FBD 2: 6 ft15 ft

O

V MxM Mx x 40

   

 

023.11670 1.1167 kip 03.1167(6)20 1.116718.700 kipft

y O

FV V MxxM Mx

(11)

Example

Hibbeler 4th 6-68 #4Bending  



11 1

FBD 3: 09 ft (24 ft) 00.20.683330y

xxx FVx 41

    

 

1 1 111 2 11 2

0.20.683330.24.1167 kip 00.2(/2)0.683330 0.10.6833 0.14.116741.200 kipft

y F

Vxx MMxxx x MMx xx

Example:

Hibbeler 4th 6-68 #5Bending 42

Example

Hibbeler 4th 6-68 #6 Consider T-beam cross section

Bending 1122 12

NA locates on centroid(,). By symmetry about axis, 0 ii i

Cyz yz yAyAyA y AAA

   

 

43

(100.5)(61)5(110) (61)(110) 7.0625 in.yy   

Example

Hibbeler 4th 6-68 #7Bending 22 1122

Consider T-beam cross section NA locates on centroid(,). Find about neutral axis ()() 1

yy

Cyz I IIAdIAd 44

32 32 4

1 6(1)6(10.57.0625) 12 1 1(10)10(7.06255) 12 197.27 in.

I I I

  

(12)

Example

Hibbeler 4th 6-68 #8Bending Abs max bending moment 12 kipft at 6 ft at6ft

maxMx My x

   45

, at6 ft (1212)3.9375 2.8742 ksi 197.27 (1212)(7.0625) 5.1554 ksi 197.27

top bottom

x I  



 

   

Example:

Hibbeler 4th 6-68 #9 46Abs max bending stress5.16 ksi (C)Ansmax

Example

Hibbeler 4th 6-68 #10 02.9375 in. y

StressBending 2 3 2 4 3 4

61 in. (2.9375)1 in. 2.93750.5 in. (2.9375)/2 in.

A Ay y yyy

   

   47

3344 22 3

(2.9375) (3.43756) in. 2 , 1 in.

I I I I

IQyAyA y VQ t It

Q 

   



 

Example

Hibbeler 4th 6-68 #11 7.0625 in.0y

StressBending 2 223

(7.0625)1 in. 0.5(7.0625) in. 1 (70625)in

II II IIIIII

Ay yyy QyA yQ

   

   48

223 (7.0625) in. 2 , 1 in.

II II II II

yQ VQ t It

 

 

(13)

Example

Hibbeler 4th 6-68 #12   Between 06 ft Absmaxisveupwardstressdirectionx V

 

StressBending  

  

Abs max isve upward stress direction 02.9375 in., 7.0625 in.0,

maxmaxI I maxII II

max

V VQVQ y ItIt VQ y ItVQ It

 

 49



  max

at NA 0 20.5(7.

max maxII II

y VQ It

   ²² 06250) 0.253 ksiAns 197.271

Example

Hibbeler 4th 6-68 #13 1 2.9375 in.3.9375 in.y

StressBending    

2 223

6(3.9375) in. (3.9375)/2 in. 3(39375)in

III III IIIIIIIII

Ay yyy QyA yQ

    50



 

3(3.9375) in. , t6 in.III

III III III

y VQ It

Q 

Example

Hibbeler 4th 6-68 #14StressBending 51

Integration

Slopes & Deflections Successive integrations Find constants of integration through Bdditi

 

Bending Boundary conditions Find constants of integration through Continuity condition Symmetry condition

 



  52

2 2dvM EIdx

(14)

Example

Hibbeler 12-6 #1Bending Determine the equations of the elastic curve for the beam and specifythebeam’smaximumdeflection.EIisconstant. 53

specify the beams maximum deflection.EIis constant.

Example

Hibbeler 12-6 #2Bending FBD of whole beam 00 33

xAFH LP



54

33 0()0 22 00 2

ABB yABA

LP MRLPR P FRRPR

  

 

Example

Hibbeler 12-6 #3Bending Section 1: 0 00 22y

xL PP FVV

 



55

00 2APx MMVxM

 Example

Hibbeler 12-6 #4Bending 3 Section2:L Lx 56

Section 2: 2 3 00 22 33 00 22

y A

Lx PP FVVP P MLVxMMPxPL

   

 

(15)

Example

Hibbeler 12-6 #5Bending 57

Example

Hibbeler 12-6 #6Bending 2 2 12 3 12 2

Section 1: 0 (1) 24 (2) 12 At , 0,0 with (2)0

xL dvPdvP EIMxEIxC dxdx P EIvxCxC AxvC

    58

2 1 222

At , , 0 with (2) 12 At , , 4126BB

PL BxLvC dvPLPLPL BxLvEIv dx

 

Example

Hibbeler 12-6 #7Bending 2 2 32 32 34

3 Section 2: 2 33 (3) 222 3 x(4) 64

L Lx dvdvP EIMPxPLEIxPLxC dxdx P EIvxPLxCC

   59

22 3 3 4

64 5 At , , with (3) 66 At , , 0 with (4) 4

dvPLPL BxLC dxEI PL BxLvC

 

Example

Hibbeler 12-6 #8Bending

 

³² 22

Section 1: 0, 12 3Ans 12

P xLvxLx EI dvP xL dxEI

  60

 

3223 22

3 Section 2: ,2910x3 212 395Ans 6

LP LxvxLxLL EI dvP xLxPL dxEI

 

(16)

Example

Hibbeler 12-6 #9Bending 61

Example

Hibbeler 12-6 #10

 

22Max ve deflection in section 1 (0) 1 30

xL dvP LL



Bending

    

22 3 32 3

30 123 0.0321 12 33 Max ve deflection in section 2 () at 22

+ve max

xLxL dxEI PPL vxxL EIEI LL Lxx PPL

   62

 

3223 2910x3 128ve max maxve ma

PPL vxLxLL EIEI vv

 

 

3 in the downward directionAns 8xPL EI

Statically Indeterminate Members

The reactions of members cannot be determined by equilibrium equations alone. The degree of static indeterminacyis the number of reactions in excess of the number of equilibrium equations. 63

SI Members

Propped cantilever beams Statically indeterminate to the first degree Choose static redundants by remove support to obtain statically determinate released or primary structure 3 equations 4 unknowns 64

Add compatibility condition

(17)

Pressure Vessels

Sphere Consider half section 65

2 2 0(2)0 2 /2

xm m m

pr Frtpr rt rrtr

 



 2pr t

Pressure Vessels

Closed Cylinder 



1

Consider half section 0(2)20xFbtpbr 



2 2

Consider cross sectional section 0(2)0zmFrtpr 66

1 2

circumferencial or hoop stress longitudinal or axial stress  

1pr t2 2pr tσ2