Lecture Note for Solid Mechanics - contents.kocw.netcontents.kocw.net/KOCW/document/2013/kumoh/yoonsungho/1.pdf · 2016. 9. 9. · Advanced Materials & Smart Structures Lab. 금오공대기계공학과윤성호교수

Advanced Materials & Smart Structures Lab.금오공대기계공학과 윤성호교수

Lecture Note for Solid Mechanics- Analysis of Stress -

Prof. Sung Ho YoonDepartment of Mechanical EngineeringKumoh National Institute of Technology


Text book : Mechanics of Materials, 6th ed.,

W.F. Riley, L.D. Sturges, and D.H. Morris, 2007.

Prerequisite : Knowledge of Statics, Basic Physics, Mathematics, etc.


Analysis of stress

Application of equilibrium equations

Determine the forces on a structure by its support and internal forces on the parts of the structure

Objective of this chapter

Determine the internal effect of forces on the structure in order to understand the behavior of materials under the action of forces

Design considerations : safety and economy

Calculate the internal force intensity and deformation on the parts

By knowing the properties of the material, engineers can establish the most effective size and shape of the parts


Normal stress under axial loading Axial force is defined as the force collinear with the centroidal axis of member

Stress (force intensity) =

Normal stress : Normal force divided by the area over which the force is distributed

- Distribution of internal force- develops on exposed cross section- has a resultant F that is normal to the exposed

cross section and is equal in magnitude of P- has a line of action collinear with the line of

action of P

- Average intensity of internal force

AF

avg + : tensile- : compressive

- Normal stress at a point on the cross section

AF

avg

AF

A

0

lim

areaforce

Analysis of stress


- Shear stress at a point on the transverse cross section

AV

avg

AV

A

0

lim

Single shear

Double shear

Shear stress in connections The bolted and pinned connections are used to introduce the concept of shear

stress

Shear stress : shear force divided by the area over which the force is distributed

- Average shear stress on the transverse cross section

AV

avg

- The shear stress can not be uniformly distributed over the area. Actual shear stress or maximum shear stress is different from the average stress.

Analysis of stress


Bearing stress (compressive normal stress) Bearing stress occurs on the contact surface between two interacting members

and just names normal stress resulting from contact between two different bodies

AF

b

- For the case of connection on the surfaces of contact between the head of the bolts and the top plate

: annular area of the bolt or nut)(4

22io ddA

- For the case of connection where the shank of the bolt is pressed against the sides of the hole

: projected area (not actual contact area)dtA

: Average bearing stress

Analysis of stress


Unit of stress Force per unit area (FL-2) psi or ksi (=1000psi) in English unit Pa (N/m2) or MPa (MN/m2 or N/mm2) in SI unit

(Example) Determine the normal stress in the bar (a) On a section 20 in. to the right of point A, (b) On a section 20 in. to the right section of point B, and (c) On a section 20 in. to the right of point C. Assume that the cross sectional area is 3 in2.

(Sol)

(a) kipFFF

kipFFF

kipFFF

CDCD

BCBC

ABAB

450811854

2708154

81081

ksiA

F

ksiA

F

ksiA

F

CDCD

BCBC

ABAB

0.153

45

0.93

27

0.273

81

Analysis of stress


Stresses on an inclined plane in an axially loaded member

cossin

cos

AAPV

PN

n

- Normal stress is either max. or min. on the planes for which shear stress is zero. Also, shear stress is zero on the planes of max. or min. normal stress.

- Normal and shear stresses

2sin2

cossincossin

)2cos1(2

coscoscos 2

AP

AP

AP

AV

AP

AP

AP

AN

nn

nn

- Variations of normal and shear stresses

APAP

2

max

max

- Resultant F of internal force is equal to the applied load P

- Average total stress AF

Savg

at 0° and 180°

at 45° and 135°

Analysis of stress


(Example) The block has a 200x100-mm rectangular cross section. Normal stress on plane a-a is 12.0MPa (compression). If Φ is 36°, determine (a) load P (b) shear stress on plane a-a (c) maximum normal and shear stresses.

(Sol)

)(695)10(7.694

)54(2cos1)10)(100)(200(2

)10(12

2cos12

3

6

6

CkNNP

P

A

P

A

N

nn

(a)

MPa

A

P

A

V

nn

52.16)54(2sin)10)(100)(200(2

)10()695(

2sin2

6

3

(b)

MPam

N

A

P

MPam

N

A

P

4.17)10(37.17)10)(100)(200(2

)10(695

2

7.34)10(74.34)10)(100)(200(

)10(695

2

6

6

3

max

2

6

6

3

max

(c)

2sin2

)2cos1(2

A

P

A

P

n

n

Analysis of stress


Stress at a general point in an arbitrarily loaded member

AFS n

An

0

lim

- For any other plane through O the values of F and M could be different- General concept of stress state at a point is needed

- Stress vector or resultant stress

AFA

F

nt

An

nn

An

0

0

lim

lim

- Normal stress and shear stress are defined as

- Consider plane having an outward normal in the x-direction

x

x

xz

xy

x

Analysis of stress


Stress vector itself depends on the orientation of the plane with which it is associated

Infinite number of planes can be passed through the point, resulting in an infinite number of stress vectors being associated with the point

Specification of stresses on three mutually perpendicular planes is sufficient to completely describe the state of stress at the point

- Normal stress : single subscript indicates the plane on which the stress acts

z

y

x

- Shear stress : first subscript indicates the plane on which the stress acts and second subscript indicates the coordinate axis to which the stress is parallel

zx

yz

xy

zz

yy

xx

Analysis of stress


Two dimensional or plane stress

0 yzxzzyzxz

- Two parallel faces to be perpendicular to z-axis of the small element are assumed to be free of stress

- The only components of stress in two dimensional problem

- Two dimensional or plane stress problem

yxxyyx ,,

Thin plates where z-dimension of the body is small and the z-components of force are zero

Analysis of stress


Stress transformation equation for plane stress

- Equation relating the normal and shear stresses n and nt on an arbitrary plane () and the known stresses x, y and xy on the reference planes

:0 nF

0sin)cos(cos)sin(

sin)sin(cos)cos(

dAdA

dAdAdA

xyyx

yxn

2222

22

212

21

222

sincos

sincoscoscossinsincos

xyyxyx

xyyx

xyyxn

:0 tF

0

sin)sin(cos)cos(cos)sin(sin)cos(

dAdAdAdAdA

yxxy

yxnt

222

22

cossin

)sin(coscossin)(

xyyx

xyyxnt

Analysis of stress


sign convention for stress transformation equation

Tensile normal stresses are positive; compressive normal stresses are negative. The sign of a normal stress is independent of the coordinate system.

Shear stress is positive if it points in the positive direction of the coordinate axis of the second subscript when it is acting on a surface whose outward normal is in a positive direction of the coordinate axis of the first subscript. The sign of a shear stress depends on the coordinate system.

Angle measured counterclockwise from the reference positive x-axis is positive.

(n, t, z) axes have the same order and form a right-hand coordinate system as (x, y, z) axes.

Analysis of stress


(Example) At a point on the outside surface of a thin walled pressure vessel, stresses are 8000 psi(T) and zero shear on a horizontal and 4000 psi(C) and zero shear on a vertical plane. Determine the stresses at this point on plane b-b having a slope of 3 vertical to 4 horizontal.

(Sol)

)(368036800)13.53(sin)8000()13.53(cos)4000(

cossin2sincos22

22

Tpsipsi

xyyxn

)(57605760

0)13.53cos()13.53sin()8000()4000(

)sincos(cossin)( 22

Cpsipsi

xyyxnt

Analysis of stress


(Example) Determine (a) normal and shear stresses on plane a-b, (b) normal and shear stresses on plane c-d, which is perpendicular to plane a-b, (c) show the stresses on planes a-b and c-d using a small element.

(Sol)

(a)

(b)

MPaMPaMPa yxxyyx 60,100,80

MPaMPa

CMPaMPa

xyyxnt

xyyxn

8.9578.95)]42(sin)42()[cos60()42cos()42sin()]100(80[

)sin(coscossin)()(3.6026.60

)42cos()42sin()60(2)42(sin)100()42(cos80

cossin2sincos

22

22

22

22

MPaMPa

TMPaMPa

xyyxnt

xyyxn

8.9578.95)]48(sin)48()[cos60()48cos()48sin()]100(80[

)sin(coscossin)()(3.4026.40

)48cos()48sin()60(2)48(sin)100()48(cos80

cossin2sincos

22

22

22

22

(c) Since planes a-b and c-d through the point are perpendicular, the stresses on these planes may be drawn on an element as figure (g).

Analysis of stress


Principal stresses and maximum shear stress - plane stress Transformation equations for plane stress provide a means for determining the

normal stress and the shear stress on different planes through a point.

For design purposes, critical stresses at the point are the maximum normal and shear stresses

yx

xy

p

pp

xyyxn

xyyxyx

n

dd

22cos2sin

2tan

2cos22sin)(0

2sin2cos22

- The above relation gives the orientations of two principal planes.- Principal planes are normal to each other. (Two values of p differ by 180o)- A third principal plane for the plane stress state has outward normal in z-direction.

xyyxnt

yno

xynt

xno

at

at

90

0

Shear stress is zero on planes experiencing maximum and minimum values of normal stress.

Planes free of shear stress are the principal planes

Normal stresses on principal planes are principal stresses

Analysis of stress


yx

xy

p

pp

2

2cos2sin

2tan

0

22

3

2

2

2,1

zp

xyyxyx

pp

If xy and (x - y) have the same sign (tan2p is positive),

2p is between 0o and 90o with the other value 180o greaterp is between 0o and 45o with the other value 90o greater (p and p+90o)

Both sin2p and cos2p are positive

For a given set of stress values, there are two values of p differing by 180o

The above relation gives the orientations of two principal planes which are normal to each other. (Two values of p differ by 90o)

A third principal plane for the plane stress state has outward normal in z-direction.

2

2

2

2

2

2sin

2

22cos

xyyx

xyp

xyyx

yx

p

Both sin 2p and cos 2p are negative

2

2

2

2

2

2sin

2

22cos

xyyx

xyp

xyyx

yx

p

Analysis of stress


xy

yx

xyyxnt

xyyx

nt

dd

22cos

2sin2tan

2sin22cos)(0

2cos22sin2

2

2

21

2

2

pp

xyyx

p

There are two perpendicular planes of maximum in-plane shear stress, having the same magnitude but opposite signs

Maximum shear stress is equal in magnitude to one-half the difference between the two principal stresses

(a) Two tangents are negative reciprocals(b) Two angles 2p and 2 differ 90o, and p and differ 45o apart(c) Planes on which maximum in-plane shear stress occur are 45o from the principal planes

yx

xy

p

pp

2

2cos2sin

2tan

Maximum in-plane shear stress occurs on planes by the specified values of

Analysis of stress


2

22

2

2222

21

2

2

2

2

2

2

21

ppp

pxyyx

xyyxyx

xyyxyx

pp

There are two perpendicular planes of maximum in-plane shear stress, having the same magnitude but opposite signs

Maximum shear stress is equal in magnitude to one-half the difference between the two principal stresses

Relation between the principal stresses and the maximum in-plane shear stress

yxyx

xyyxyx

xyyxyx

pp

22

22222

2

2

2

21

Relation between the principal stresses and the normal stresses on the orthogonal planes

Analysis of stress


(Example) (a) Determine principal stresses and maximum shearing stress (b) locate planes on which these stresses act and show the stresses on a complete sketch.

(Sol) (a) psipsipsi xyyx 4000,8000,10000

psi

Cpsipsi

Tpsipsi

zp

p

p

xyxxyx

pp

0

)(88508850

)(1085010850

98491000

)4000(2

)8000(10000

2

)8000(10000

22

3

2

1

2

2

2

2

2,1

psi98502

)8850(10850

2minmax

max

Maximum in-plane shear stress

psi

xyyx

p

9850

)4000(2

)8000()10000(

2

2

2

2

2

Maximum shear stress

Analysis of stress


(b)

,02.78,98.11

,04.156,96.232

444.0)8000(10000

)4000(222tan

p

p

yx

xyp

)(1085010850

)98.11cos()98.11sin()4000(2)98.11(sin)8000()98.11(cos10000

cossin2sincos

98.11

1

22

22

Tpsipsi

when

p

xyyxn

p

)(88508850

)02.78cos()02.78sin()4000(2)02.78(sin)8000()02.78(cos10000

cossin2sincos

02.78

2

22

22

Cpsipsi

when

p

xyyxn

p

psi

Tpsipsi

when

xyyxnt

xyyxn

9850

)]02.33(sin)02.33()[cos4000(

)02.33cos()02.33sin()]8000(10000[

)sin(coscossin)(

)(10001000

)02.33cos()02.33sin()4000(2)02.33(sin)8000()02.33(cos10000

cossin2sincos

02.334598.11

22

22

22

22

Analysis of stress


psi

Tpsipsi

when

xyyxnt

xyyxn

9850

)]02.123(sin)02.123()[cos4000(

)02.123cos()02.123sin()]8000(10000[

)sin(coscossin)(

)(10001000

)02.123cos()02.123sin()4000(2)02.123(sin)8000()02.123(cos10000

cossin2sincos

02.1239002.33

22

22

22

22

Normal stresses on the perpendicular planes of maximum shear stresses are equal both in magnitude and sign

Shear stress have equal magnitude but opposite sign

Analysis of stress


Mohr’s circle for plane stressGraphic interpretation of transformation equations for plane stress

22

2

2

22

2cos2sin2

2sin2cos22

xyyx

ntyx

n

xyyx

nt

xyyxyx

n

0,

2yxisCenter

22

2xy

yxRisRadius

How to draw a Mohr’s circle

Coordinates of each point on the circle represent the normal and shear stresses on one plane through the stresses point

Angular position of the radius to the point gives the orientation of the plane

Normal stresses are plotted as horizontal coordinates (tensile stresses plotted to the right of the origin)

Shear stresses are plotted as vertical coordinates (clockwise rotation of stress element plotted above -axis)

Analysis of stress


In the case of the x greater than y

1. Choose a set of x-y reference axes

2. Identify stresses and list them with proper sign3. Draw a set of coordinate axes4. Plot and label it point V (vertical plane)5. Plot and label it point H (horizontal plane)6. Draw line between V and H7. Establish center C and radius R8. Draw a circle

),( xyx ),( yxx

Procedures to draw a Mohr’s circle

Analysis of stress


Mohr’s Circle for plane stress

2sin2sin2cos2cos

)22cos('

pp

p

CVCVOC

CFOCOF

V : stresses on the vertical planeH : stresses on the horizontal planeC : center of circleLine CV : vertical plane from which angle is

measured

Using

avgyx

xyp

yxp

OC

VVCV

CVCV

2)(

2sin

2)(2cos

'

'

2sin2cos22

2sin'2cos''

xyxxxx

VVCVOCOF

2sin2

2cos

2sin'2cos'

2sin2cos2cos2sin

)22sin('

xxxy

pp

p

CVVV

CVCV

CFFF

Analysis of stress


22

22

1

2

22

xyxx

p

xyxxxx

p

CBCA

CVOCCDOCOD

22tan yxxy

p

• If the two nonzero principal stresses have the same sign, the maximum shear stress at the point will not be in the plane of the applied stresses

• The angle between the vertical plane and one of the principal planes is p

• All angle on Mohr’s circle are twice the corresponding angles for the actual stressed body

Analysis of stress


(Example 2-11) Determine (a) principal stresses and maximum shearing stress (b) normal and shearing stresses on a-a plane.

(Sol)

(a)

ksiCV

CHV

06.847

)0,1(:)4,6(:)4,8(:

22

0

)(06.706.706.81

)(06.906.906.81

3

2

1

zp

p

p

CksiksiOE

TksiksiOD

Since two principal stresses have opposite signs

ksiCBCAp 06.8max

The principal planes are represented by lines CD and CE

87.1474.292

5714.07

42tan

pp

p

or

Analysis of stress


Two orthogonal surfaces are sufficient to completely specify the principal stresses One of the surfaces is required to completely specify the maximum shear stress

Analysis of stress


(Homework)

(2-8), (2-18), (2-37), (2-58), (2-74), (2-93)

Documents

Lecture Note for Solid Mechanics - contents.kocw.netcontents.kocw.net/KOCW/document/2013/kumoh/yoonsungho/1.pdf · 2016. 9. 9. · Advanced Materials & Smart Structures Lab. 금오공대기계공학과윤성호교수