Principal Stresses PDF

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PRINCIPAL STRESSES

THEORY What are principal stresses and principal planes? The maximum and minimum normal stresses acting on an element are known as principal stresses and the planes on which they are acting are known as principal planes. Fundamentals on principal stresses

Fig shows a element in plane stress condition Note: In analysis of problems we consider x-plane as reference plane i.e., zero degree plane ( = 0 ) and y-plane = 90

The plane which is perpendicular to x-axis is called x-plane. The plane which is perpendicular to y-axis is called y-plane. First stress axis means x-axis ( Second stress axis means y-axis (

axis) axis)

= Normal stress on x-plane in x direction

= Shear stress on x-plane in y direction

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PRINCIPAL STRESSES

= Shear stress on y-plane in the x-direction

Numerically,

=

is counterclockwise (ccw) direction considered as positive. This means rotates x-plane in counterclockwise (ccw) direction.

is clockwise (cw) direction considered as negative. This means rotates x-plane in clockwise (ccw) direction.

in counterclockwise (ccw) direction considered as positive.

in clockwise (cw) direction considered as negative.

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PRINCIPAL STRESSES

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PRINCIPAL STRESSES

DERIVATIONS 1. Derive expression for normal stress and shear stress on plane inclined at „ ‟ to the vertical axis in a biaxial stress system with a shear stress. From fig, „ „ is positive „ ‟ is positive is positive ( counter clockwise ) is positive. means normal stress on plane. = means shear stress on plane. Rotate x-plane by to match with inclined plane.

Fig: stress transformation for plane stress element

Fig: Force distribution diagram

4

PRINCIPAL STRESSES Original x-axis is perpendicular to original y-axis.

(-ve)

New x-axis is perpendicular to new y-axis

For equilibrium, ∑ = 0 , ∑ = 0 , [Resolving forces on and axis] Considering ∑ = 0 × A – ( A cos +

A cos) sin = 0

A sin ) cos - ( × A sin +

Divide throughout by A and transpose, = ( cos +

+ = =

cos ) sin

sin ) cos + ( sin + sin cos +

+2

+

sin cos

sin cos + )+

( + +

= =

(

2sin cos + (

cos 2 +

sin2 +

)

+

)

-

cos 2

sin 2

Considering, ∑ =0 × A + ( A cos +

A cos ) cos = 0

A sin ) sin - ( × A sin +

Divide throughout by A and transpose, = = - ( cos + = = = =

sin ) sin + ( sin +

sin cos -

cos ) cos

+ sin cos + +

+

(

-

)

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PRINCIPAL STRESSES

= =

+

= =

sin 2 +

=√

Also,

(

-

)

[

-

= cos 2 ]

cos 2

= Resultant stress

= Angle of obliquity tan = =

(

)

2. Derive the expression to find the orientation for direction of maximum and minimum normal stress as well as shear stress.

For max and min normal stress ( ) Since there are 2 planes oriented at and mutually perpendicular at each other. tan 2 = ( )

( )=0 [

( (

) (

)

]=0

( - sin 2 ) 2 +

)

2 sin 2 + (

tan 2 =

(

cos2 2 = 0 2 =

2 cos2 = 0

= tan 2

)

where implies

)

or

=

[

(

)

]=0

(

)

+

)

tan 2 =

2 cos 2 - 2 sin 2 = 0 Divide throughout by - 2cos2 (

)

Here, „ ‟ refers to angle of maximum shear stress. i.e. = &

( )=0 )

)

)

= 2 + 180° = + 90°

and

For max shear stress ( )

(

(

(

2

tan 2 = 0 6

(

)

)

PRINCIPAL STRESSES

tan 2 =

( (

)

= + 90°

)

tan 2 =

3. Derive an expression for finding principal stresses. Ans: Plane stress carry 4 stresses

, ,

,

These 4 components can be represented by square matrix =*

+

i.e

=

According to Cayley-Hamilton theorem, “Every square matrix satisfies its characteristics equation”. If A is a square matrix, then its characteristic equation is given by, |

|=0

|

|=0

where I = Identity matrix I=*

+

I =*

+

Characteristic roots are but principal stresses & =*

+

i.e., I = *

+ [

|

|=0

Expanding, (

)( -

-

-

+

)-

=0

+ -

=0

-

=0

7

= &

= ]

PRINCIPAL STRESSES –( - ) +

-

=0

This is in the form where a + bx + c = 0 = )

); c = (

Where a = 1; b = - (

=

) ] √, (

[ (

=

=

(

)

= (

)

=

(

)

√ (

)

√ (

)

-

)

-

√{(

)

√

)

=

=

)

) √,(

(

(

=

√

(

(

) )

√(

}

)

√ (

) This is the expression for principal stresses and

4. Prove that Principal planes carry zero shear stress Ans: The above statement means, principal planes carry minimum, maximum normal stresses. On principal planes, ( )=0 [

(

0+

(

) (

)

)

(-sin 2 ) 2 +

]=0 cos 2 × 2 = 0

Divide throughout by 2 8

PRINCIPAL STRESSES (

-

)

sin 2 +

cos 2 = 0

The above equation is the equation for shear stress. i.e. = 0 Therefore, on principal planes shear stress is equal to zero.

5. Show that sum of the normal stresses on any 2 perpendicular planes in general 2D stress system is constant. Solution: Meaning of above question is to prove, + = terms in each equation are mutually perpendicular. [i.e., i.e,

+ =

+

To prove WKT, =

and also

+ =

+

+ ;

= + ;

= + ]

= +

+

= =

) cos 2 +

+(

=

sin 2 ............................... (1)

) cos ( 2 ( + 90°)) +

+(

=

+(

) cos (180 + 2 ) +

=

-(

) cos 2 -

sin 2 ( + 90°)

sin (180 + 2 )

sin 2 .............................(2)

Add equation (1) and (2) +(

= = –(

=

Adding,

We get,

+

+

To prove +

) cos 2 + ) cos 2 –

=2(

sin 2 …............................ (1)

sin 2 …..........................(2)

)

= + …...........The sum of 2 normal stresses is a constant. +

9

.The

PRINCIPAL STRESSES

+

+√

=

+

…………………..(3)

–√

=

…………………………….(4)

Add equation (3) and (4) +

+√

=

+

–√

=

Adding, + +

=

+

= +

+

= +

…………………..(3)

=

=

…………………………….(4) +

+ ...........................Hence proved

Sum of two normal stresses on two mutually perpendicular planes is always a constant.

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PRINCIPAL STRESSES

PROBLEM 1

Data: = -30 N/m

;

= 120 N/m ;

= -40 N/m (clockwise [cw] direction;

=

-60° (clockwise [cw] direction) Analytical method The direction of shear stress is taken as,  

Positive [+ve] for counter clock wise [ccw] direction. Negative [-ve] for clock wise [cw] direction.

Results on inclined / oblique plane (a) Normal stress (b) Shear stress = (

)+(

=( - 40 sin2 (-60°)

)cos2 + )+(

= -(

sin2 )cos2 (-60°)

) sin2 +

) sin2 (-60°) - 40

= -( cos2 (-60°) =

= 117.14 N/m

cos2

44.95 N/m

(c) Resultant stress

(d) Angle of obliquity

=√ =√

11

=

(

=

(

) )

PRINCIPAL STRESSES

= 125.711 N/m

= -20.99°

Results on principal plane (e) Major Principal Stress (f) Minor principal stresses + √(

=

+ √(

=

)

= )

=

= 130 N/m

(

=

)

- √(

)

= -40 N/m

)

(g) Location of principal stress planes = + 90°

&

= 14.036° + 90°

)

(

=

- √(

= 104.036°

= 14.036° Results on maximum and minimum shear stress plane (h) Maximum and minimum shear stresses (i) Average normal stress on maximum shear stress planes = = = = = = = = = 85 MPa

= + 45°

= 45 MPa

(j) Location of maximum & minimum shear stress plane = + 45°

= 14.036° + 45°

= 104.036° + 45°

= 59.036°

= 149.036°

12

&

PRINCIPAL STRESSES

Graphical method Steps to construct Mohr’s circle Step 1: Draw x axis Normal stress ( ) and y – axis Shear stress ( ). „O‟ is the origin. As both Tensile stress [positive] and compressive stress [negative] present in the problem, therefore graph is plotted on both side of origin „O‟. „ ‟ is considered positive [+ve] in counter clock wise direction [ccw] and negative [ve] in clockwise direction [cw].

Step 2: From origin O, and are plotted on both side of the plane. Plot to suitable scale [Scale 1 cm = 10 N/mm2]. OA = = -30 N/mm2 OB = = 120 N/mm2

Step 3: In Mohr‟s circle,  Clockwise [cw] shear stress taken upward.  Counter clockwise [ccw] shear stress taken downward. So draw AE perpendicular to OA i.e. = -40 N/mm2 in clockwise direction [ccw] upward and BF perpendicular to OA in counter clockwise [cw] direction downward.

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PRINCIPAL STRESSES

Step 4:Join EF cutting x-axis at C which is the center of the Mohr‟s circle. „C‟ as center and radius equal to CE or EF draw a circle. This is called Mohr‟s circle and it is cut x-axis at P and Q. Here, CE is x-plane [Reference plane] CF is y-plane Always rotate only x-plane.

Step 5: From „C‟ draw a line CJ at angle 2 = -120° with CE in CW direction. Draw JK perpendicular to x-axis and join O and J. Join OJ such that  give the angle of obliquity. In JOK,

Fig shows measurement of angle of obliquity in Mohr‟s circle,

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PRINCIPAL STRESSES

Step 6: Fig represent 2 and 2 [angles of principal plane] from the reference plane.

Step 7: Represent 2 and 2 [angles of maximum shear stress plane] from the reference plane a shown.

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PRINCIPAL STRESSES

Mohr’s circle full view

Conversion: Scale 1 cm = 10 N/m 1.

= 3 cm

3 10 = 30 N/m

2.

= 12cm

12 10 = 120 N/m

3.

= 4 cm

4 10 = 40 N/m

4.

= 11.8 cm

5.

= 4.4 cm

6.

= 12.6 cm

7.

= 4 cm

8.

= 13 cm

13 10 = 130 N/m

9.

= 8.6 cm

8.6 10 = 86 N/m

10.

= 4.4 cm

4.4 10 = 44 N/m

11.8 10 = 118 N/m 4.4 10 = 44 N/m 12.6 10 = 126 N/m 4 10 = 40 N/m

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PRINCIPAL STRESSES

Results on inclined plane / oblique plane 1.

117.14 N/m

2.

-44.95 N/m

3.

125.711 N/m

4.

Formula used = (

)cos2 +

)+(

) sin2 +

= -(

cos2

=√

-20.99°

=

(

)

(

)

Results on principal planes 5.

14.036°

6.

°

= =

+ 90°

7. =

130 N/m

=

+ √(

)

=

- √(

)

8. =

40 N/m

Results on maximum and minimum shear stress planes 59.036° = + 45°

9. 10.

= + 45°

149.036°

11.

= =

85 N/m

=

12. 45 N/m

=

17

=

sin2

PRINCIPAL STRESSES

Results on the element

Principal planes with Principal stress

Maximum and minimum shear stress planes with shear stress and normal stress.

18

PRINCIPAL STRESSES

PROBLEM 2

Data: = -120 N/m ; = -80 N/m ; -45° (clockwise [ccw] direction)

= - 60 N/m (clockwise [ccw] direction;

=

Analytical method The direction of shear stress is taken as,  

Positive [+ve] for counter clock wise [ccw] direction. Negative [-ve] for clock wise [cw] direction. Normal stresses

= ( sin2 =( sin2 (-45°)

)+( )+(

Shear stresses ) sin2 + cos2

= -(

)cos2 +

) sin2 (-45°) - 60cos2 (-

= -( 45°)

)cos2 (-45°) - 60

= - 20 N/m

= - 40 N/m Resultant stress

Angle of obliquity

=√ =√ = 44.72 N/m

=

(

)

=

(

)

=26.56°

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PRINCIPAL STRESSES

Major Principal Stress =

+ √(

=

+ √(

Minor principal stresses

)

- √(

= )

- √(

=

= -36.75 N/m

) )

= -163.2 N/m

Maximum and minimum shear stresses

Average normal stress on maximum shear stress planes

= = = = = 63.24 N/m

= 100 N/m

Location of principal stress planes =

(

=

(

Location of maximum and minimum shear stress planes

)

= + 45°

)

= 35.78° + 45° = 80.78°

= 35.78° =

+ 90°

= + 45°

= 35.78° + 90°

=

=

= 170.75°

°

20

° + 45°

PRINCIPAL STRESSES

Graphical method Steps to construct Mohr’s circle Step 1: Draw x axis Normal stress ( ) and y – axis Shear stress ( ). „O‟ is the origin. As both are compressive stress [ negative ], graph is plotted on the left side of origin „O‟. „ ‟ is considered positive [+ve] in counter clock wise direction [ccw] and negative [ve] in clockwise direction [cw].

Step 2: From origin O, and are plotted on left side of the plane. Plot to suitable scale [Scale 1 cm = 10 N/mm2]. OA = = -120 N/mm2 OB = = -80 N/mm2

Step 3: In Mohr‟s circle,  Clockwise [cw] shear stress taken upward.  Counter clockwise [ccw] shear stress taken downward. So draw AE perpendicular to OA i.e. = -60 N/mm2 in clockwise direction [ccw] upward and BF perpendicular to OA in counter clockwise [cw] direction downward.

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PRINCIPAL STRESSES

Step 4: Join EF cutting x-axis at C which is the center of the Mohr‟s circle. „C‟ as center and radius equal to CE or EF draw a circle. This is called Mohr‟s circle and it is cut xaxis at P and Q.

Step 5: From „C‟ draw a line CJ at angle 2 = -90° with CE in CW direction. Draw JK perpendicular to x-axis and join O and J. In JOK,

Fig shows measurement of angle of obliquity in Mohr‟s circle,

22

PRINCIPAL STRESSES

Step 6: Fig represent 2 and 2 [angles of principal plane] from the reference plane.

Step 7: Represent 2 and 2 [angles of maximum shear stress plane] from the reference plane a shown.

23

PRINCIPAL STRESSES

Conversion: Scale 1 cm = 10 N/m 1.

= 12 cm

12 10 = 120 N/m

2.

= 8 cm

8 10 = 80 N/m

3.

= 6 cm

6 10 = 60 N/m

4.

= 4 cm

4 10 = 40 N/m

5.

= 2 cm

2 10 = 20 N/m

6.

= 4.4 cm

4.4 10 = 44 N/m

7.

= 16.3 cm

16.3 10 = 163 N/m

8.

= 3.6 cm

3.6 10 = 36 N/m

9.

= 6.3 cm

6.3 10 = 63 N/m

10.

= 10 cm

10 10 = 100 N/m

24

PRINCIPAL STRESSES

Mohr’s circle full view

Results on inclined plane / oblique plane 1.

-40 N/m

2.

-20 N/m

3.

44.72 N/m

4.

26.56°

Formula used = (

)+(

)cos2 + ) sin2 +

= -( =√ =

(

)

(

)

Results on principal plane 5.

35.78°

=

25

cos2

sin2

PRINCIPAL STRESSES

6.

°

= + 90°

Results on maximum and minimum shear stress plane 80.78° = + 45°

7. 8.

170.75°

= + 45°

Results of Stress parameters 1.

=

-36.75 N/m

2.

=

-163.24 N/m

Formula used =

+ √(

)

=

- √(

)

= = 3.

63.24 N/m

4.

100 N/m

=

=

Fig of value on element

Principal Planes with principal stress 26

=

PRINCIPAL STRESSES

Maximum and minimum she...