CHAPTER 4 Stress Transformation · An element is subjected to the plane stresses shown in the figure. (a) Determine normal stress and shear stress acting on the plane that is inclined

CHAPTER 4 Stress Transformation

ANALYSIS OF STRESS

• For this topic, the stresses to be considered are not on the perpendicular and parallel planes only but also on other inclined planes.

x

z

y

b

A

a

b a

P P

(A body subjected to load P)

• On plane a-a, normal force, N produces normal stress.

• On plane b-b, normal force, N produces normal stress and shear force, V produces shearing stress.

ANALYSIS OF STRESS

P P

N = P cos

V = P sin

Area = A/cos

𝜎 =𝑁

𝐴=

𝑃

𝐴

ANALYSIS OF STRESS

𝜎𝑛 =𝑃 𝑐𝑜𝑠𝜃

𝐴𝑐𝑜𝑠𝜃

=𝑃

𝐴 𝑐𝑜𝑠2𝜃 = 𝜎𝑥 𝑐𝑜𝑠2𝜃

𝜏 =𝑃 𝑠𝑖𝑛𝜃

𝐴𝑐𝑜𝑠𝜃

= −𝑃

𝐴 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 = −𝜎𝑥 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃

• On an element, there are 6 components of stress: x , y , z , xy , yz , and zx

• If the z axis is not considered or the stresses are independent with the z axis, then there exist only stresses in the x and y directions.

x z

y

ANALYSIS OF STRESS

• This state of 2-dimensional stress is known as plane stress.

z = xz = yz = 0

All directions shown (axes and on element) are taken as positive.

x

y’ y x’

x

a

a

y

x

y

xy

ANALYSIS OF STRESS UNDER 2D

• Stresses on an element can be transformed using 2 methods:

(i) Equations Method

(ii) Mohr Circle

STRESS TRANSFORMATION

i) Equations Method

• Consider an element rotated an amount of about the z-axis.

• Stresses on an inclined plane will be yielded and can be expressed in terms of x , y , xy and .


x

y’

A sin

y

x’ x’y’

xy

x

x’

xy (A sin) cos

A A cos

Sign Convention:

Positive if counter-clockwise and usually taken from the vertical surface (x-plane) to the intended plane

Positive if tension or in the direction of positive axis

Positive if in the direction of positive shear (counter-clockwise) xy = yx


+xy

+y

+x

Normal Stress:

Shear Stress:

Note: If is clockwise, then put a negative sign in the equation(s)


𝜎𝑥′ =𝜎𝑥 + 𝜎𝑦

2+

𝜎𝑥 − 𝜎𝑦

2 𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃

𝜎𝑦′ =𝜎𝑥 + 𝜎𝑦

2−


2 𝑐𝑜𝑠2𝜃 − 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃

𝜏𝑥′𝑦′ = −𝜎𝑥 − 𝜎𝑦

2𝑠𝑖𝑛2𝜃 + 𝜏𝑥𝑦 𝑐𝑜𝑠2𝜃


x

x’

y’

x’

y’ x’y’

+ 90

In-Plane Principle Stress:


Taking 𝑑𝜎𝑥′

𝑑𝜃= 0;

𝑡𝑎𝑛2𝜃𝑝 =2𝜏𝑥𝑦


𝜎𝑚𝑎𝑥 & 𝜎𝑚𝑖𝑛 =𝜎𝑥+𝜎𝑦

2±

𝜎𝑥−𝜎𝑦

2

2+ 𝜏𝑥𝑦

2

Maximum In-Plane Shear Stress:


Taking 𝑑𝜏

𝑥′𝑦′

𝑑𝜃= 0;

𝑡𝑎𝑛2𝜃𝑠 = −𝜎𝑥 − 𝜎𝑦

2𝜏𝑥𝑦

𝜏𝑚𝑎𝑥 & 𝜏𝑚𝑖𝑛 = ±𝜎𝑥−𝜎𝑦

2

2+ 𝜏𝑥𝑦

2

Average Normal Stress:

When substituting the values for 2s into the equation for normal stress (σx’), there is also a normal stress on the plane of maximum in-plane shear stress, which can be determined by:


𝜎𝑎𝑣𝑔 =𝜎𝑥 + 𝜎𝑦

2

The state of plane stress at a point on a body is shown on the element in the Figure. Represent this stress state in terms of the principal stresses.

EXAMPLE 1

x

y

20 MPa

90 MPa

60 MPa

xy

x = – 20 MPa (Compression)

y = 90 MPa (Tension)

xy = 60 MPa (Clockwise)

EXAMPLE 1 – Solution

x

y

20 MPa

90 MPa

60 MPa

xy

Principal Stresses


𝜎𝑚𝑎𝑥 & 𝜎𝑚𝑖𝑛 =𝜎𝑥 + 𝜎𝑦

2±


2

2

+ 𝜏𝑥𝑦2

𝜎𝑚𝑎𝑥 & 𝜎𝑚𝑖𝑛 =−20 + 90

2±

−20 − 90

2

2

+ 602

𝜎𝑚𝑎𝑥 & 𝜎𝑚𝑖𝑛 = 35 ± 81.4 max = 35 + 81.4 = 116 MPa min = 35 – 81.4 = 46.4 MPa

Orientation of Element


𝑡𝑎𝑛2𝜃𝑝 =2𝜏𝑥𝑦


𝑡𝑎𝑛2𝜃𝑝 =2 × 60

−20 − 90

2p2 = 47.49 p2 = 23.7 2p1 = 47.49 + 180 = 132.51 p1 = 66.3

The principal plane on which each normal stress acts can be determined by applying:

𝜎𝑥′ =𝜎𝑥 + 𝜎𝑦

2+


2 𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦 𝑠𝑖𝑛2𝜃

𝜎𝑥′ =−20 + 90

2+

−20 − 90

2 𝑐𝑜𝑠 −47.49° + 60𝑠𝑖𝑛 −47.49°

x’ = 46.4 MPa

Hence, σmin = 46.4 MPa acts on the plane defined by θp2 = 23.7°, whereas σmax = 116 MPa acts on the plane defined by θp1 = 66.3°.


By replacing the θp1 and θp2 into the equation for shear stress:

𝜏𝑥′𝑦′ = −𝜎𝑥 − 𝜎𝑦

2𝑠𝑖𝑛2𝜃 + 𝜏𝑥𝑦 𝑐𝑜𝑠2𝜃

𝜏𝑥′𝑦′ = −−20 − 90

2𝑠𝑖𝑛 −47.49° + 𝜏𝑥𝑦 𝑐𝑜𝑠 −47.49°

x’y’ = 0 MPa

No shear stress acts on this element.


The state of plane stress at a point on a body is represented on the element shown in the Figure. Represent this stress state in terms of the maximum in-plane shear stress and associated average normal stress.

EXAMPLE 2

x

y

20 MPa

90 MPa

60 MPa

xy

Maximum In-Plane Shear Stresses

MPa

xy

yx

4.81

602

9020

2

max

2

2

max

2

2

max


Orientation of Element

3.111

5.2225.421802

3.21

5.422

602

90202tan

22tan

1

1

2

2

s

s

s

s

s

xy

yx

s


The proper direction of max on the element can be determined by applying the equation:

MPayx

yx

xy

yx

yx

4.81

5.42cos605.42sin2

9020

2cos2sin2

''

''

''


Average Normal Stress Besides the maximum shear stress, as calculated above, the element is also subjected to an average normal stress determined from the equation:

MPaavg

avg

yx

avg

35

2

9020

2


ii) Mohr Circle

• In this method, stresses on a plane are drawn as one point on a Mohr circle.

• From the equations, it can be shown that:

Centre of Circle R2 = max2


𝜎𝑥′ −𝜎𝑥 + 𝜎𝑦

2

2

+ 𝜏𝑥′𝑦′2 =


2

2

+ 𝜏𝑥𝑦2

Steps for constructing Mohr Circle: 1. Determine centre of circle, C (x, y)

x = avg , y = 0

2. Determine point A (x, xy) : coordinate when = 0

x = +ve (tension) or –ve (comp)

xy = +ve (counter clockwise) or –ve (clockwise)

3. Determine point A’ (y, yx) : coordinate when = 90

4. Draw circle connecting A and A’ with C as the centre

5. The datum (reference line) in Mohr circle is line AC

6. All angles must be determined from line AC ( = 0)


max

0

min (1)

(+ve)

(x+y)/2

(+ve)

max

(2)

2p(max)

2p(min)

y

y

yx

yx

Y(+y, -yx)

x x

xy

xy

X(+x, +xy) 2s(max)


X(+x, +xy)

Y(+y, -yx)

The state of plane stress at a point on a body is represented on the element shown in the Figure. Determine the principal stresses and the orientation acting at this point.

EXAMPLE 3

x

y

20 MPa

90 MPa

60 MPa

xy

Construction of the Circle

x = –20 MPa (Compression)

y = 90 MPa (Tension)

τxy = 60 MPa (Clockwise)


The Centre of Circle is at:

The Radius of the Circle is:

𝜎𝑎𝑣𝑔 =𝜎𝑥 + 𝜎𝑦

2=

−20 + 90

2= 35 𝑀𝑃𝑎

𝑅 =𝜎𝑥 − 𝜎𝑦

2

2

+ 𝜏𝑥𝑦2 =

−20 − 90

2

2

+ 602 = 81.4 𝑀𝑃𝑎

tan 2θp2 = 60 / 55 2θp2 = tan-1 1.09 2θp2 = 47.49° θp2 = 23.74 °


35 MPa

20 MPa

60 MPa

σmax = 116.4 MPa σmin = 46.4 MPa

2θp, max = 132.6°

A (-20, 60)

A’ (90, - 60)

C = (35, 0) 2θp2

The state of plane stress at a point on a body is represented on the element shown in the Figure. Determine the maximum in-plane shear stresses and the orientation of the element upon which they act.

EXAMPLE 4

x

y

20 MPa

90 MPa

60 MPa

xy

tan 2θs1 = 55 / 60 2θs1 = tan-1 1.09 2θs1 = 42.5° θs1 = 21.3 °


35 MPa

20 MPa

τmax = 81.4MPa

2θs1 = 42.5°

τmin = 81.4MPa

A’ (90,- 60)

A (-20, 60)


81.4 MPa x’

y’

x

35 MPa 21.3

The state of plane stress at a point on a body is represented on the element shown in the Figure. Represent this state of stress on an element oriented 30° counterclockwise from the position shown.

EXAMPLE 5

x

y

20 MPa

90 MPa

60 MPa

xy

tan 2θ = 55 / 60 2θ = tan-1 1.09 2θ = 42.5° θ = 21.3 °

= 60° – 42.5° = 17.5 °

At Point B:

x’ = 35 + 81.4 sin17.5°

x’ = 59.48 MPa

x’y’ = 81.4 cos 17.5°

x’y’ = 77.63 MPa


35 MPa 20 MPa

42.5°

A (-20, 60)

= 17.5°

x’

x’y’

B’

B

C


35 MPa 20 MPa

42.5°

A (-20, 60)

= 17.5°

x’

x’y’

B’

B

C

x’y’

x’

= 17.5°

At Point B’:

y’ = 81.4 sin 17.5° 35

y’ = 10.52 MPa

x’y’ = 81.4 cos 17.5°

x’y’ = 77.63 MPa


x’

y’

x 30

An element is subjected to the plane stresses shown in the figure. (a) Determine normal stress and shear stress acting

on the plane that is inclined at 20o as shown in the figure.

(b) Determine the maximum normal stress and its orientation.

(c) Sketch the plane of the maximum normal stress by showing its values and orientation.

(d) Determine the maximum shear stress.

30 N/mm2

30 N/mm2

50 N/mm2

50 N/mm2

20o

CLASS EXERCISE

Documents

CHAPTER 4 Stress Transformation · An element is subjected to the plane stresses shown in the figure. (a) Determine normal stress and shear stress acting on the plane that is inclined