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Lecture Note 0Lecture Note 0

Review

Second Semester, Academic Year 2012,Department of Mechanical Engineering

Chulalongkorn University

Contents

Basic Statics Force system Force system Rigid bodies in equilibrium Structures (trusses and frames/machines) in equilibrium

Basic Mechanics of MaterialsStress strain and transformation Stress, strain and transformation

Stress and deformation in axially loaded members, torsion, bending and pressure vessels

2

Equilibrium Definition

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

0

0O

R

R O

F F

M MOR O

When a body is in equilibrium the resultant force andF

3

When a body is in equilibrium, the resultant force and

the resultant couple about any point are both zero.R

R

F

M O

Example Hibbeler Ex 5-6 #1

2D Equilibrium

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

4

weight of the beam in the calculation.


2D Equilibrium

Find: , , y x yA B B

Equilibrium of

0 (600 N)cos 45 0x x

ADB

F B

424.26 N 424 N Ans

0 (100 N)(2 m) (600 N)sin45 (5 m)

(600 N) 45 (0 2 ) (7 ) 0

x

B

B

M

A

(600 N)cos 45 (0.2 m) (7 m) 0

319.50 319 N Ans

0 (600 N) i 45 (100

y

y

A

A

F A N) (200 N) 0B

5

0 (600 N)sin45 (100y yF A

N) (200 N) 0

404.76 N 405 N Ansy

y

B

B

2D Supports Summary #1

2D Equilibrium

6


2D Equilibrium

7


2D Equilibrium

8


3D Equilibrium

FBD of plate

E ilib i f l t ABC

Equilibrium of plate

0 0

0 0x x

ABC

F B

F B

9

0 0y yF B


3D Equilibrium

Equilibrium of plate

0 (300 N) (980 7 N) 0 (1)

ABC

F A B T

0 (300 N) (980.7 N) 0 (1)

0 (2 m) (980.7 N)(1 m) (2 m) 0 (2)z z z C

x C z

F A B T

M T B

0 (300 N)(1.5 m) (980.7 N)(1.5 m) (3 m)

(3 m) (200 N m) 0 (3)y z

z

M B

A

Solve (1), (2) & (3)

790.35 N, 216.67 N, z zA B 707.02 N

790 N 707 N

CT

A T

10

790 N, 707 N, 0, 217 N Ans

z C

x y z

A TB B B


3D Equilibrium

11


3D Equilibrium

12


3D Equilibrium

13


3D Equilibrium

14Comparison with 2D supports

Example Structure

Disassembling the structure

15

Imaginary Section Choice of PlanesLoads

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

imaginary section.

16

Stress Normal and Shear StressesLoads

17

avgNA

avgVA

Strain Normal and Shear StrainsLoads

2nt

avg

s ss L

18

Axially Loaded MembersAxial Load

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

, , latPA L L

, , latA L L

19

Changes in Lengths Prismatic BarsAxial Load

A prismatic bars has straight longitudinal axis and constant cross gsection.

P , , P EA L

PLAE

20

Example Hibbeler 4-4 #1Axial Load

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 dCD = 0.5 in., determine the displacement of end A with respect to end D.

3

21

315.0 10 ksiE


22


FBD of section A

0 2 kip 0

2 kip (T)x AB

AB

F F

F

FBD of section

0 3(2 kip) 0

AB

F F

0 3(2 kip) 0

6 kip (T)x BC

BC

F F

F

FBD of section

0 8 kip 0x CD

AD

F F

23

p

x CD

8 kip (T)CDF


24


(2.0 kip)(48 in.) 0 01448 iAB ABF L

2 3

(2.0 kip)(48 in.) 0.014487 in.(0.75 in.) (15.0 10 ksi)

4(6 0 ki )(120 i )

AB ABAB

AB

F LA E

F L

2 3

(6.0 kip)(120 in.) 0.015279 in.(2.0 in.) (15.0 10 ksi)

4

BC BCBC

BC

F LA E

2 3

(8.0 kip)(36 in.)

(0.5 in.) (15.0 10 ks4

CD CDCD

CD

F LA E

0.097785 in.i)

4

AD AB BC CD

25

0.127551 in. 0.158 in. Ans

Torsion Linearly Elastic Circular BarsTorsion

, max maxG Gr Gcc

26

Torsion FormulaTorsion

AT dM

TrJ

TLGJ

27

Polar Moment of InertiaTorsion

2 3

Solid shaft

(2 ) 2r r

J d d

2 3

0 0 (2 ) 2 J d d

4 4r d 2 32r dJ

2 2

1 1

2 3

Tubular shaft or circular tube

(2 ) 2 r r

r rJ d d

1 1r r

4 4 4 41 2 1 2( ) ( )J r r d d

28

1 2 1 2( ) ( )2 32

Example Gere 3.4-2 #1Torsion

A stepped shaft ABCD consisting of solid circular segments is subjected to three torques as shown. The material is steel with h d l f l ti it G 80 GPshear modulus of elasticity G = 80 GPa.

(a) Calculate the maximum shear stress τmax in the shaft.(b) Calculate the angle of twist φD (in degrees) at end D. (b) Ca cu ate t e a g e o t st φD ( deg ees) at e d

29


FBD of shaft 0

3000 N mx

A

M

T

30

2000 N m 800 N m 05800 N mAT


FBD of section in equilibriumA FBD of section in equilibrium

0 0 5800 N m

FBD of section in equilibriumx A AB AB

A

M T T T

B FBD of section in equilibrium

0 3000 N m 0 2800 N m

FBD of section in equilibriumx AB BC BC

B

M T T T

C

31

FBD of section in equilibrium

0 2000 N m 0x BC CD CD

C

M T T T 800 N m


32


4 3 4

2 16 32, maxTr Tr T TL TLJ GJr d G d

6

Section 16 16(5800 N m)( ) 57 694 10 PaAB

J GJr d G dAB

T

3 3

4 9 2 4

( ) 57.694 10 Pa(0.08 m)

32 32(5800 N m)(0.5 m) 0.0090146 rad(80 10 N/ ) (0 08 )

max ABAB

AB ABAB

dT L

G d

4 9 2 4(80 10 N/m ) (0.08 m)Section

1

ABABG d

BC 66 16(2800 N m)BCT

1( )max BC

63 3

6 16(2800 N m) 66.020 10 Pa(0.06 m)

32 32(2800 N m)(0.5 m) 0 013754 rad

BC

BC

BC BC

Td

T L

33

4 9 2 4 0.013754 rad

(80 10 N/m ) (0.06 m)BCBCG d


Section 16 16(800 N )

CDT

63 4

16 16(800 N m)( ) 63.662 10 Pa(0.04 m)

32 32(800 N m)(0 5 m)

CDmax CD

CD

Td

T L

4 9 2 4

32 32(800 N m)(0.5 m) 0.019894 rad(80 10 N/m ) (0.04 m)

CD CDCD

CD

T LG d

66.0 MPa between Ans

0 042663 d

max BC

0.042663 rad

2 44

D AB BC CD

CCW Ans

34

2.44D CCW Ans

Flexure FormulaBending

The neutral axis NA does not change length.

0 NA passes through the centroid.

Load & stress relationshipx

z

F

M M

MyI

35

C & I: Centroid and Moment of InertiaBending

3 3

Area ( , )

(0 0)

x yx y I I

bh hbbh (0,0)12 12

bh

2 4 4

(0,0)4 64 64D D D

21cx xI I Ad

36

Shear Formula DerivationBending

0F 0xF

VQIt

Q y A

Q y A

37

Example Hibbeler 4th 6-68 #1Stress Bending

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 sectionsketch the normal stress distribution on the cross section.

38


FBD of whole T-beam

0 0BF H 0 0

0 2 6 1.8 13.5 18 0

12 3 /18 0 68333 kip

x B

A B

F H

M R

R

12.3 /18 0.68333 kip

0 2 1.8 0

56 1/18 3 1167 kip

B

y A B

R

F R R

R

56.1/18 3.1167 kipAR

39


FBD 1: 0 6 ft

0 2 0

x

F V

0 2 0

2 kip

0 2 0

yF V

V

M x M

0 2 0

2 kip ftOM x M

M x

FBD 2: 6 ft 15 ft

0 2 3.1167 0y

x

F V

1.1167 kip

0 3.1167( 6) 2 0

y

O

V

M x x M

40

1.1167 18.700 kip ftM x

Example Hibbeler 4th 6-68 #4Bending

FBD 3 0 9 ft ( 24 ft)

1 1

1

FBD 3: 0 9 ft ( 24 ft)

0 0.2 0.68333 0

0 2 0 68333 0 2 4 1167 kipy

x x x

F V x

V

1

1 1 1

2

0.2 0.68333 0.2 4.1167 kip

0 0.2 ( / 2) 0.68333 0

0 1 0 6833F

V x x

M M x x x

M

41

21 12

0.1 0.6833

0.1 4.1167 41.200 kip ft

x

M

M x

x x

Example: Hibbeler 4th 6-68 #5Bending

42


Consider T-beam cross sectionNA locates on centroid ( , ).C y z

1 1 2 2

By symmetry about axis, 0

i i

y zy A y A y A

1 1 2 2

1 2

(10 0.5) (6 1) 5 (1 10)

i i

i

y A y A y Ay

A A A

y

(6 1) (1 10)7.0625 in.y

y

43


Consider T-beam cross sectionNA locates on centroid ( )C y zNA locates on centroid ( , ).

Find about neutral axis

C y z

I2 2

1 1 2 2

3 2

( ) ( )

1 6(1) 6(10.5 7.0625)

y yI I

I

Ad I Ad

3 2

6(1) 6(10.5 7.0625)121 1(10) 10(7.0625 5)

12

I

I

44

4197.27 in.I


Abs max bending moment12 kip ft at 6 ftmaxM x

, at 6 ftMy xI

( 12 12) 3.9375 2.8742 ksi

197.27( 12 12) ( 7 0625)

top

45

( 12 12) ( 7.0625) 5.1554 ksi197.27bottom

Example: Hibbeler 4th 6-68 #9

46Abs max bending stress 5.16 ksi (C) Ansmax


0 2.9375 in. y

23

2

6 1 in.

(2 9375 ) 1 i

A

A

24

3

(2.9375 ) 1 in. 2.9375 0.5 in.

(2 9375 ) / 2 in

A yyy y y

4

3 3 4 4

(2.9375 ) / 2 in.

I

y y y

Q y A y A

3 3 4 42 2

3(2.9375 ) (3.4375 6) in.2

I

I

y y

y

VQ

Q

47

, 1 in.II

I

VQt

It


7.0625 in. 0y

2 (7.0625 ) 1 in.II

y

A y

0.5(7.0625 ) in.IIy y y

Q A

2 2 3

1 (7.0625 ) in.2II

II II IIQ y A

yQ

2

, 1 in.IIII

VQt

48

, 1 in.IIII

tIt


Between 0 6 ftx

Abs max is ve upward stress direction V

V Q V Q

0 2.9375 in., max max I

I

V Q V Qy

It ItV QV Q

7.0625 in. 0, max II

II

max V Qy

ItV Q

It

max

at NA 0

2 0.5( 7.

max

max II

yV Q

2 20625 0 ) 0.253 ksi Ans

49

maxIIIt

0 53 s s197.27 1


2 9375 in 3 9375 iny

1

2

2.9375 in. 3.9375 in.

6 (3.9375 ) in.III

y

A y

( ) (3.9375 ) / 2 in.

III

III

yy y y

2 2 3

3(3.9375 ) in.III III III

III

Q y A

yQ

, t 6 in.IIIIII

III

VQIt

50

IIIIt


51

Integration Slopes & DeflectionsBending

Successive integrationsFind constants of integration through

Find constants of integration through Boundary conditions Find constants of integration through

Continuity condition Symmetry condition

2d v M2 EIdx

52

Example Hibbeler 12-6 #1Bending

Determine the equations of the elastic curve for the beam and specify the beam’s maximum deflection. EI is constant.

53


FBD of whole beam

0 0

3 30 ( ) 02 2

x A

A B B

F H

L PM R L P R

( )2 2

0 02

A B B

y A B APF R R P R

54


Section 1: 0 x L

0 02 2yP PF V V

Px

0 02A

PxM M Vx M

55


3Section 2: 2LL x

30 02 2

3 3

yP PF V V P

P

56

3 30 02 2APM L Vx M M Px PL


57


22

Section 1: 0 x Ld v P dv P

212

31 2

(1)2 4

(2)

d v P dv PEI M x EI x Cdxdx

PEIv x C x C

1 2

22

( )12

At , 0, 0 with (2) 0A x v C

PL

1

2 2 2

At , , 0 with (2) 12

At

PLB x L v C

dv PL PL PLB x L v EIv

58

At , , 4 12 6B BB x L v EIv

dx


3Section 2: 2LL x

22

32

23 3 (3)2 2 2

d v dv PEI M Px PL EI x PLx Cdxdx

3 23 4

2 2

2 2 23 x (4)

6 45

dxdxPEIv x PLx C C

d PL PL

2 2

3

3

5At , , with (3) 6 6

At 0 ith (4)

dv PL PLB x L Cdx EI

PLB L C

59

4At , , 0 with (4) 4

PLB x L v C


3 2Section 1: 0 ,12

Px L v x L xEI

2 2

3 2 2 3

3 Ans12

3S ti 2 2 9 10 3

dv P x Ldx EI

L PL L L L

3 2 2 3

2 2

3Section 2: , 2 9 10 x 32 12

3 9 5 Ans6

L PL x v x Lx L LEI

dv P x Lx PLd EI

60

6dx EI


61


Max ve deflection in section 1 (0 )x L

2 2

3

13 012 3

dv P x L x Ldx EI

P PL

3

3 2 0.032112

3 3Max ve deflection in section 2 ( ) at

+ve maxP PLv x xLEI EI

L LL x x

3

3 2 2 3

Max ve deflection in section 2 ( ) at 2 2

2 9 10 x 312 8ve max

L x x

P PLv x Lx L LEI EI

12 8ve max EI EI

3

i th d d di ti APL

62

max ve mav v in the downward direction Ans8xPLEI

Statically Indeterminate Members The reactions of members cannot be determined by

equilibrium equations alone.

The degree of static indeterminacy is 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 ppstatically determinate released or primary structure

3 equations4 unknowns

Add compatibility condition

64

Pressure Vessels Sphere

Consider half section

220 (2 ) 0

2x mprF r t p rr t

pr

65

2/ 2

m

m

r tr r t r

2p

t

Pressure Vessels Closed Cylinder

Consider half section

0 (2 ) 2 0F bt pbr 10 (2 ) 2 0

Consider cross sectional section

xF bt pbr

22

Consider cross sectional section

0 (2 ) 0z mF r t p r

pr

pr

1 circumferencial or hoop stress

1 t 2 2tσ2

66

1

2

plongitudinal or axial stress

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