Chapter 9Combined Stresses

9-1 Introduction

• Basic types of loading: axial, torsional and flexural

• Stress formulas:

Axial loading -

Torsional loading -

Flexural loading -

A

Pa

J

T

I

Myf

9-2 Combined Axial & Flexural Loads

f

My

I

a

PA

af

PA

MyI

y

2

6( )

P Mc MA I bh

3 3 3

2

6 6

20 10 6(0.45 15 10 0.15 20 10 )(0.05)(0.150) (0.05)(0.150)

(2.67 10 ) (20.00 10 ) 22.67 MPa

A

3 3 3

2

6 6

20 10 6(0.45 15 10 0.15 20 10 )

(0.05)(0.150) (0.05)(0.150)

(2.67 10 ) (20.00 10 ) = 17.33 MPa

B

A

B 20

15

o2000sin(15 )

o2000cos(15 )C D

0 :DM O O3

(6) ( )2000cos(15 ) (4)2000sin(15 )12

264.598 lb.

y

y

C

C

264.598 lb.yC

o2000cos(15 )

0 :xF O2000cos(15 )xD

1931.852 lb.

0 :yF O2000sin(15 ) lb.y yD C

253.04 lb.yD

253.04 lb.yD

1931.852

517.638 lb

A

B264.598 lb.yC

1931.852 lb

517.638 lb

759.12 lb.ft

Section AB:

3(3) ( )1931.852 (1)517.638

12yM C

759.12 lb.ft

1931.852

517.638 lb

A

B264.598 lb.yC

1931.852 lb

517.638 lb

759.12 lb.ft

Normal Stresses

2

6( )

P Mc MA I bh

2

1931.852 6 759.12 12

2 6 2 6A

2920.1 lb/in

2

1931.852 6 759.12 12

2 6 2 6B

2598.1 lb/in

BMD

1012.16 lb.ft

529.20 lb.ft

min 2

1931.852 6 1012.16 12

2 6 2 6

21173.15 lb/in

max 2

1931.852 6 1012.16 12

2 6 2 6

2851.17 lb/in

2 24 0.0025 0.07854 mA D

4 4 6 464 (0.1) 1.5625 10 mI D

P

P

0.25P

P McA I

0.25 0.050.0025 1.5625

P P

840080 MPa

P

680 1029.92 kN

8400P

4.0 1.5

5.0

122400 [ 25 6 25 3]P

180,000 180,000 360,000 kg.

12

180,000 0.5 180,000 3.0

(1000 15) 15 5

M

90,000 540,000 562,500

112,500 kg-m.

2

6( )

P Mc MA I bh

2

min 2

360,000 6 112,50048,333.33 kg/m

1 9 1 9

For long slender members or columns, the effect of P- is significant

PA

MyI

y

For stiff members the formula is appropriate

P P

1in

21

in2

2 21 1( ) in2 4

A

4 4112

1 1( ) in2 192

I

Fig.(a)

Fig.(b)

1 12 4

max,( ) 1 14 192

( )( )28

( ) ( )a

P Mc P PP

A I

max. compressive stress in Fig.(a)

max. compressive stress in Fig.(b)

max,( ) 14

4( )b

P PP

A

max,( )

max,( ) 287 :1

4a

a

PP

Hw10

allow

B

D1D2

D1=(1+z1) in. D2 = D1(1+z2) in.

I1-1=1000(1+z3) in4 Area=10(1+z4) in2

B =10(1+z5) in. allow=10(1+z6) ksi.

ค่�า z1-z6 ได้�จากเลขประจ�าตั�วนิ�สิ�ตั ด้�งตั�อไปนิ��

46z1z2z3z4z5z6

Fig. P-908

หมายเหตุ� D2 = D1(1+z2) in.

เพื่��อให้�ห้นิ�าตั�ด้มี�ประสิ�ทธิ�ภาพื่ด้�ในิการร�บห้นิ�วยแรง

Hw11

L1

L2 L3 L4

b

h

L1= (1+z1) in. L2 = (1+z2) in.

L3= (1+z3) in. L4 = (1+z4) in.

b = 0.2(1+z5) in. h = b(1+z6) in.

P = (1+z5) kips. F = (1+z6) kips.

ค่�า z1-z6 ได้�จากเลขประจ�าตั�วนิ�สิ�ตั ด้�งตั�อไปนิ��

46z1z2z3z4z5z6หมายเหตุ� h = b(1+z6) in.

เพื่��อให้�ค่านิมี�ค่วามีล'กไมี�นิ�อยกว�าค่วามีกว�างเสิมีอ

9-3 Kern of Section: Loads Applied off Axes of Symmetry

( )P My Pe a

A I I

Ia

Ae

for b h section

3( /12)2h bh

bh e

6h

e

That is in designing of masonry or other structures weak in tension, the resultant load should fall in the middle third of the section.

6h

e

The maximum eccentricity to avoid tension

The general case:

( )( ) yx

y x

Pe yPe xPA I I

2 2

( )( )0 yx

y x

Pe yPe xP

A Ar Ar

The position of neutral axis (line of zero stress)

2 20 1 yx

y x

eex y

r r

2

2

x x

y y

I Ar

I Ar

2y

x

ru

e

2x

y

re

( , )2 2

b h

3 3

( )( )Rectangular section: 0

/12 /12yx

Pe yPe xPbh bh hb

3 3

( )( / 2)( )( / 2)0

/12 /12yx

Pe bPe hP

bh bh hb

1

/ 6 / 6yx

ee

h b

A compressive load P= 12 kips is applied, as in Fig. 9-8a, at a point 1 in. to the right and 2 in. above the centroid of a rectangular section for which h=10 in. and b=6 in. Compute the stress at each corner and the location of the neutral axis. Illustrate the answers with a sketch similar to Fig. 9-8b.

918

12 kips

2

1

10

6

( )( ) yx

y x

Pe yPe xPA I I

3 3

Rectangular section:

( )( )

/12 /12yx

Pe yPe xPbh bh hb

3 3

12 (12 1) (12 2)0.08 ksi

6 10 6 10 /12 10 6 /12( 5) (3)

A

3 3

12 (12 1) (12 2)0.72 ksi

6 10 6 10 /12

( 5

10 6 /12

) ( 3)B

3 3

12 (12 1) (12 2)0.48 ksi

6 10 6 10 /12 1(5) ( 3)

0 6 /12C

0.32 ksiD

12 kips

2

1

10

63 3

Position of Neutral Axis:

( )( )0

/12 /12yx

Pe yPe xPbh bh hb

3 3

12 (12 1) (12 2)0

6 10 6 10 /12 10 6 /1( ) (

2)x y

3 21

25 3

x y

on x axis (y=0) 25/ 3 8.33x

on y axis (x=0 3/ 2 1.5) y

N.A.

921 Calcualte and sketch the kern of a W360 X 122 section.

2 2Position of Neutral Axis: 0 1 yx

y x

eex y

r r

257 363A( , )

2 2

2 2

257 363At corner A: 0 1

63 52 21 3yx

ee

22 63on x-axis ( =0): mm

230.89

57y xe e

22 153on y-axis ( =0): mm

3629 0

31 .x ye e

9-4 Variation of Stress with Inclination of Element

Mc

I

Tc

J

9-5 Stress at A Point

Stress at a point really defines the uniform stress distributed over a differential area.

• The most general state of stress at a point may be represented by 6 components,

),, :(Note

stresses shearing,,

stresses normal,,

xzzxzyyzyxxy

zxyzxy

zyx

state of stress เม�อแสดงด�วยระบบโคออร�ด�เนตุ (xyz)

xx xy xz x xy xz

yx yy yz yx y yz

zx zy zz zx zy z

σsymmetry

state of stress เม�อแสดงด�วยระบบโคออร�ด�เนตุ (xyz)

xx xy xz x xy xz

yx yy yz yx y yz

zx zy zz zx zy z

σsymmetry

• Plane Stress - state of stress in which two faces of the cubic element are free of stress. For the illustrated example, the state of stress is defined by

.0,, and xy zyzxzyx

• State of plane stress occurs in a thin plate subjected to forces acting in the midplane of the plate.

( , )n n • State of plane stress also occurs on the free surface

of a structural element or machine component, i.e., at any point of the surface not subjected to an external force.

Two methods to compute the maximum stresses i.e.,

(1) Analytical approach

(2) Using of Mohr’s circle

Plane Stress

x

y

xy

xy x

y yx

yx

x

y

z x

y x

y x

y

z

9-6 Variation of Stress at A Point: Analytical Derivation

A

cosA

sinA

0nF ( sin

( s

)( cos ) ( cocos sin sin

cn o

s

i s

)

)

x xy

yx

yA A

A

A A

0tF ( sin )

( s

sin cos co( s

s

c

in )

os ) c s )

in

( ox xyy

yx

A

A

A A A

0nF ( sin

( s

)( cos ) ( cocos sin sin

cn o

s

i s

)

)

x xy

yx

yA A

A

A A

0tF ( sin )

( s

sin cos co( s

s

c

in )

os ) c s )

in

( ox xyy

yx

A

A

A A A

22cos 2 cossin sinyx xy

2 2sin cosinco sicss os nx yyy xx

2 2sisin 2

No11 cos2

cos ,2

te: , cos sinc 2

,2

n2

osxy yx

cos2 sin 22 2

x y x yxy

sin 2 cos22

x yxy

cos2 sin 22 2

x y x yxy

sin 2 cos22

x yxy

A

cosA

sinA

x y

xy

xy

x y yx

yx

cos2 sin 22 2

x y x yx xy

sin 2 cos22

x yxy xy

cos2 sin 22 2

x y x yy xy

2

2

cos2( ) cos( 2 ) cos2

sin 2( ) sin( 2 ) sin 2

cos2 sin 22 2

x y x yxy

sin 2 cos22

x yxy

A

cosA

sinA

d2 sin 2 2 cos2 0

d 2x y

xy

Find maximum or minimum differentiating Eq.(9-5) w.r.t. and setting the derivative equal to zero

Eq.(9-5)

2tan 2 xy

x y

Eq.(9-6)

Find maximum or minimum differentiating Eq.(9-6) w.r.t. and setting the derivative equal to zero

d2 cos2 2 sin 2 0

d 2x y

xy

tan 2

2x y

sxy

A

cosA

sinA

2tan 2 xy

x y

cos2 sin 22 2

x y x yxy

sin 2 cos22

x yxy

Eq.(9-5)

Eq.(9-6)

At zero shearing stress

0 sin 2 cos22

x yxy

ซึ่'�งเป)นิมี*มีเด้�ยวก�บสิมีการ Eq.(9-7) ด้�งนิ��นิ ค่�า maximum or minimum จะเก�ด้ข'�นิเมี��อ = 0

2tan 2 xy

x y

1 1

2 2 2 2

sin 2 , cos2

( ) 2 ( )2 2

xy x y

x y x yxy xy

2 2

2 2 2 2

sin 2 , cos2

( ) 2 ( )2 2

xy y x

x y x yxy xy

1 2 2

2

( )2 2

x y x yxy

Maximum or minimum (Principal stresses)

2tan 2 xy

x y

1

2

2

1

Maximum or minimum

tan 22x y

sxy

2 2 1 1max ( )

2 2x y

xy

1

22

1

s

มี*มี และ s ตั�างก�นิ 45O

22000.04 kN/mm 40 MPa, 0, 0

50 100x y xy

P

A

O Ocos2( ) sin 2-40 -( )2 2

40x y x yxy

O O40 0 40 0cos2( ) 0 sin 2(-40 -40 ) 16.5 MPa

2 2

sin 2 cos22

x yxy

O O20 0sin 2( ) 0 c-40 -4os2( ) 9.85 MPa

20

4,000 psi

8,000 psi

6,000 psi

x

y

xy

6,000 psi 4,000 psi

8,000 psi

1 2 2 2 2

2

4000 ( 8000) 4000 ( 8000)( ) ( ) ( 6000)

2 2 2 2x y x y

xy

2 22000 (6000) ( 6000) 10485. 64, psi85.33

O Ocos2 30( ) sin 2( )2 2

30x y x yxy

O O304000 ( 8000) 4000 ( 8000)

cos2( ) ( 6000) sin 2( ) 6,19630 .15 psi2 2

sin 2 cos22

x yxy

O O4000 ( 8000)sin 2( ) ( 6000) cos2( )30 30 2196.15 psi

2

4,000 psi

8,000 psi 6,000 psi

6,196.15 psi

2,196.15 psi

o30

cos2 sin 22 2

x y x yxy

sin 2 cos22

x yxy

Eq.(9-5)

Eq.(9-6)

9-7 Variation of Stress at A Point: Mohr’s Circle

Otto Mohr (1882)

Eq.(a)2 + Eq.(b)2

Rule for Applying Mohr Circle to Combined Stresses

( , )x xy

( , )y xy

x-ax

is

y-ax

is

(0,0)

( , )x xy

( , )y xy

(0,0)

x-ax

is

y-ax

isC

(0,0)

( , )x xy

x-ax

is

y-ax

isC

( , )y xy

n-axis

R

( , )n n

n

n

(0,0)

( , )x xy

( , )y xy

n-axis

R

( , )n n

n

n

x-ax

is

y-ax

isC

( , )x xy

( , )y xy

x-ax

is

y-ax

is

C

( ,0) ( ,0)2

x yC

C

R

2 2( )2

x yxyR

1( ,0)2( ,0)

1

2

C R

C R

max( , )C

max R

1

1

sin 2 or

2tan 2 =

xy

xy

x y

R

o2 12 180 2

( , )x xy

( , )y xy

x-axis

y-axisC

R 1( ,0)2( ,0)

(4000, 6000)

( 8000,6000)

( ,0) ( ,0)2

8000 4000( ,0) ( 2000,0)

2

x yC

C

2 2 2 24000 8000( ) ( ) 6000 6000 2 psi

2 2x y

xyR

1 2, 2000 6000 2 4485.3, 10485.3 psiC R

1

6000sin 2

6000 2xy

R

O1 22.5

12

1

22.5

( 2000,0)

x-axis

y-axis

CR 1( ,0)2( ,0)

(4000, 6000)

( 8000,6000)

o

o

o

30

o

o o

30

cos(15 )

2000 6000 2 cos(15 ) 6196.15 psi

sin(15 ) 6000 2 sin(15 ) 2196.15 psi

C R

R

( 2000,0)

o o30 30( , )

o o120 120( , )

o

o

o

120

o

o o

120

cos(15 )

2000 6000 2 cos(15 ) 10196.15 psi

sin(15 ) 6000 2 sin(15 ) 2196.15 psi

C R

R

30

6196.15

2196.15

10196.15

2196.15

9-8 Absolute Maximum Shearing Stress

Mohr’s circle: Rotation around z-axis

x1

2

1 2

2zR

zR 12

1

2

1

2

2

2xR

Mohr’s circle: Rotation around x-axis

xR

Mohr’s circle: Rotation around y-axis

1

2yR

yR

1

2

x1

2

1 2

2zR

zR

1

2yR

yR

2

2xR

xR

Mohr’s circles for plane stress

zR

yR

xR

Absolute maximum shearing stress for plane stress is equal to the largest of the following three values

1

2

1 2 1 2, ,2 2 2z z xR R R

Mohr’s circles for general state of stress

zR

yR

xR

1

2

z 3

Absolute maximum shearing stress for general state of stress is equal to the largest of the following three values

1 2 1 3 2 3, ,2 2 2z z xR R R

1 2

2

1 5

201

025 ksi,

2

50 2015 ksi,

2

0 ksi,

2

2

2

2

Maximum in-plane shearing stress =

1 2 50 2015 ksi

2 2

Absolute maximum shearing stress is the largest of

50x

Maximum in-plane shearing stress =

1 2 50 2035 ksi

2 2

Absolute maximum shearing stress is the largest of

1 2

2

1

50 2035 ksi,

5025 ksi,

2 2

2010 ksi,

2

2

2

2

(ksi)

(ksi)

1=-50 2 =20zRyR

xR

Ex.

Hw17 the figure

( ส�าหร�บข้�อน��ให�ค�านวณ ค!า absolute maximum shearing

stress ด�วยโดยกำ�าหนดให� z = 0 )

210( 1) MPaz

110( 1) MPaz

310( 1) MPaz

ค่�า z1-z3 ได้�จากเลขประจ�าตั�วนิ�สิ�ตั ด้�งตั�อไปนิ�� 46xxxz1z2z3

9-9 Application of Mohr’s Circle to Combined Loadings

Combined Loadings (axial, torsional, flexural)

Combined stresses

Mohr’s Circlex-axis

( ),

y-axis

(0, )

12

max

Principal stresses and, Maximum shearing stress

1

2

2

1

1

22

1

s

maxmax

Design Criteria, ,allow allow

Stress Trajectories

1

2

max

Tc

J

12

1

Tc

J

Torsional Failure Modes

• A ductile specimen breaks along a plane of maximum shear

• A brittle specimen breaks along planes perpendicular to 1

• Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear.

max

Tc

J

1

Tc

J

45o

max

Tc

J

1

Tc

J

Stress Trajectories for Torsion

Stress Trajectories: lines of principal stress direction but of variable stress intensity

Stress Trajectories for Beam

Mohr’s Circle x-axis

( ),

y-axis(0, )

12

max

My

I

VQ

Ib

7 26

(2500 )(0.05)8 10 N/m 80 MPa

1.5625 10McI

6

26

(0.05) 1.6 10 1.6T( ) N/m MPa

3.125 10Tc T TJ

2500 N.mM

100 mm

80 MPa

100 MPa

D

4 46 4(0.1)

1.5625 10 m64 64D

I

4 46 4(0.1)

3.125 10 m32 32D

J

Mohr’s Circle

1.68( )0, T

12

max80 MPa

1.6 MPa

T

1.60,( )T

(40,0)

C40 MPaC

2 21

1.640 40 ( )

TC R

2 2max

1.640 ( )

TR

80 MPa

100 MPa

2 (30)(87.81)P

16,551.8 wattP 87.81 N.mT

2 21.640 ( ) 60 MPa

T

2P f T

4 4

,4 2

r rI J

3

4M

McI r

3

2

Tc TI r

Mohr’s Circle

3 34 2,( )

rTr

M

12

max3

4M

r

3

2T

r

320,( )Tr

(40,0)

C

32 /( )C M r

2 2 2 2max 3 3 3

2 2 2( ) ( )

M TR M T

r r r

2 21 3

2C R M M T

r

10 ksi

If900 12

900 lb-ft 10.8 kips-in1000600 12

600 lb-ft 7.2 kips-in1000

T

M

2 2max 3

2 23 3

2

2 8.2637.2 10.8 ksi

M Tr

r r

2 21 3

2 23 3

2

2 12.8477.2 7.2 10.8 ksi

M M Tr

r r

max 16 ksi

max 10 ksi

16 ksi

0.938 in.r

0.929 in.r

2500 N

1250 N

3750 N

4000 N

2500 N

2875 N

3625 N

1500 N.m

750 N.m

750 N.m

2500 N

1250 N

3750 N

4000 N

2500 N

2875 N

3625 N

1500 N.m

750 N.m

750 N.m

750 N.m 750 N.m

1500 N.m

2500 N

4 m 2 m

750 N.m 750 N.m

4000 N 2500 N

1500 N.m

1 m 2 m 1 m 2 m3625 N 2875 N

3750 N1250 N

2500 N

1250 N

3750 N

4000 N

2500 N

2875 N

3625 N

1500 N.m

750 N.m

750 N.m

BMzD

3625 N.m

2875 N.m

TMD

1500 N.m

750 N.m

BMyD

5000 N.m3750 N.m

1250 N.m

2500 N

1250 N

3750 N

4000 N

2500 N

2875 N

3625 N

1500 N.m

750 N.m

750 N.m

BMzD

3625 N.m

2875 N.m

TMD

1500 N.m

750 N.m

BMyD

5000 N.m3750 N.m

1250 N.m

Cross section of solid shaft

and the resultant moment

zM

yM

2 2| | z yMM M

3834.5 N.m

4725.2 N.m 5000 N.m

|M|A B C D E

A

B

C

D

E

BMzD

3625 N.m

2875 N.m

TMD

1500 N.m

750 N.m

BMyD

5000 N.m3750 N.m

1250 N.m

3834.5 N.m

4725.2 N.m 5000 N.m

|M|A B C D E

2 2max 3

2M T

r

2 21 3

2M M T

r

From Prob. 951 and this problem.

70 MPa

120 MPa

Mohr’s Circle

x-axis

( ), y-axis

(0, )

12

max3

4 M

r

3

2Tr

2 2max 3

24725.2 1500 1000 mm

r

70 MPa

35.6 mmr

37.2 mmr

2 21 3

24725.2 4725.2 1500 1000

r

120 MPa

2 2max 3

25000 750 1000 mm

r

70 MPa

At section D

35.8 mmr

37.7 mmr

2 21 3

25000 5000 750 1000

r

120 MPa

37.7 mm≥r

At section C

state of stress on the element on the surface of vessel

1

2

67.5

67.5

R

R

Absolute maximum shearing stress

1 2

1

2

| |50 MPa

2| | 67.5

50 MPa2 2

| | 67.550 MPa

2 2

R

R

R

50 MPa

32.5 MPaR

2

2 2

2x y

xyR

2 2 2 222.5 32.5xyR 2 2 232.5 22.5 550xy

23.45 MPaxy

23.45 MPaTc

J

4 4

(455 mm)23.45 MPa

920 90032

T

301.8 kN.mT

20 mm

120 mm

36 420 120

=2.88 10 mm12

I

A

20 mm

40 mm

4 3

(20 40) 40

=3.2 10 mm

Q

N.A.

250 mm

40 kNP

30 kNV

7500 kN.mmM

6

40 7500 20

20 120 2.88 10 68.75 MPa

P My

A I

4

6

30 3.2 1016.67 MPa

2.88 10 20VQI b

250 mm

40 kNP

30 kNV

7500 kN.mmM

6

40 7500 20

20 120 2.88 10 68.75 MPa

P Mc

A I

4

6

30 3.2 1016.67 MPa

2.88 10 20VQI b

2 2

2 2

( )2

68.75( ) 16.67 38.20 MPa

2

x yxyR

1 2, 34.375 38.20

72.578, 3.825 MPa

C R

O

16.67sin 2

38.20

12.94

xy

R

12.94

72.58

72.58 3.83

3.83

( ,0) ( ,0)2

68.75 0( ,0) (34.375,0)

2

x yC

C

y-ax

is

Mohr’s Circle at point A

,1(6 68.75 .67)

12

max

160,( .67)

(34.375,0)C

x-ax

is

2

20 mm

120 mm

36 420 120

=2.88 10 mm12

I

B

20 mm

40 mm

4 3

(20 40) 40

=3.2 10 mm

Q

N.A.

300 mm

40 kNP

30 kNV

9000 kN.mmM

6

40 9000 ( 20)

20 120 2.88 10 45.83 MPa

P My

A I

4

6

30 3.2 1016.67 MPa

2.88 10 20VQI b

2 2

2 2

( )2

45.83( ) 16.67 28.34 MPa

2

x yxyR

300 mm

40 kNP

30 kNV

9000 kN.mmM

45.83 MPa

16.67 MPa

O16.67sin 2 2 36.03

28.34xy

R

Mohr’s Circle at point B

45.83,( )16.67

x-axis

160,( .67)C

y-axis

o60 o36.06

22.9( )15,0

48.81, 11 1( ).5

( ,0) ( ,0)2

45.83 0( ,0) ( 22.915,0)

2

x yC

C

0

o

3028.34sin (23.97 ) 11.51 MPa

0

o o

30

o

cos(60 36.03 )

22.915 28.34cos(23.97 )

48.81 MPa

C R

45.83 MPa

16.67 MPa

48.81 MPa

11.51 MPa

Hw18

1L

2L

3L

4L

D

1.2D

1.2D

L1= 4(1+z1) in. L2 = 4(1+z2) in.

L3= 4(1+z3) in. L4 = 4(1+z4) in.

D = 4(1+z5) in.

ค่�า z1-z5 ได้�จากเลขประจ�าตั�วนิ�สิ�ตั ด้�งตั�อไปนิ�� 46xz1z2z3z4z5

Hw19

L= 0.4(1+z1) m. P = 4(1+z2) kN

H= 40(1+z3) mm. W = 40(1+z4) mm

ค่�า z1-z4 ได้�จากเลขประจ�าตั�วนิ�สิ�ตั ด้�งตั�อไปนิ�� 46xxz1z2z3z4

Also find the maximum shearing stress at point A. Show your results on a complete sketch of a differential element.

LP

H

W

2(1 )

EG

http://www.kyowa-ei.co.jp/english/products.htm

Strain and deformation of line element

0, 0, 0x y xy 0, 0, 0x y xy 0, 0, 0x y xy

O

( )IIA

Aydy

O

A

( )IIIAxydy

Oxdx

( )IAA

O

A

A

0, 0, 0x y xy

O

A

dx

dy

ds

cos2 sin 22 2

x y x yxy

sin 2 cos22

x yxy

Eq.(9-5)

Eq.(9-6)

A

cosA

sinA

12

6300 2 10 radR

6300 10 rad2xy

6800 10 radx

6200 10 rady

6500 10 radC

(800,300)

(200, 300)

If we use the stress-strain relation directly the same answer can be obtained

จงพื่�สิ+จนิ, สิมีการ (9-19) (9-20) ด้�วยภาษาของตั�วเองHw20a

Hw20b

a= 100(1+z1) b= -100(1+z2)

c= 100(1+z3)

ค่�า z1-z3 ได้�จากเลขประจ�าตั�วนิ�สิ�ตั ด้�งตั�อไปนิ�� 46xxxz1z2z3

Hw21

ปร�มาณทาง Physics สามารถแทนด�วย Tensor

Order 0 = zero order Tensor (Scalar) – Magnitude (มีวล, ค่วามีห้นิาแนิ�นิ)

Order 1 = first order Tensor (Vector) – Magnitude, Direction (ค่วามีเร.ว, แรง)

Order 2 = second order Tensor – Magnitudes, Directions (stress, strain)

… Higher order ….

ปร�มาณทาง Physics ไม!เปลี่�ยนแปลี่งไปตุามระบบโคออร�ด�เนตุท�ใช้�ในกำารว�ด

mass

length

2 kg.= ?? lb.mass temperature

5 in. = 12.7 cm.length O O50 C = 122 Ftemperature

1

1

0

1 0.5 0.2

0.5 3 1

0.2 1 4

σ

x

y

z

xy

z

0.6

0.8

1

2 2manitude 1 1 2 2 2 2manitude 0.6 0.8 1 2

P

P

ปร�มาณทาง Physics ไม!เปลี่�ยนแปลี่งไปตุามระบบโคออร�ด�เนตุท�ใช้�ในกำารว�ด

แรง ย�งคงม�ข้นาดแลี่ะท�ศทางเท!าเด�ม ไม!ว!าจะแสดง component ข้องเวคเตุอร�ด�วยระบบโคออร�ด�เนตุอ�น

P

สถานะข้องหน!วยแรง (state of stress) ย�งคงม�ค�ณสมบ�ตุ�เหม�อนเด�ม ไม!ว!าจะแสดงด�วยระบบโคออร�ด�เนตุอ�น

O

A

A

0, 0, 0x y xy

O

B

A

0, 0, 0x y xy