Purpose of Mohr’s Circle

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# Purpose of Mohr’s Circle - PowerPoint PPT Presentation

Purpose of Mohr’s Circle. Visual tool used to determine the stresses that exist at a given point in relation to the angle of orientation of the stress element. There are 4 possible variations in Mohr’s Circle depending on the positive directions are defined. . Sample Problem. y.

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## Purpose of Mohr’s Circle

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Presentation Transcript
Purpose of Mohr’s Circle
• Visual tool used to determine the stresses that exist at agiven point in relation to the angle of orientation of the stress element.
• There are 4 possible variations in Mohr’s Circle depending on the positive directions are defined.
Sample Problem

y

A particular point on the part

x

sy = -2 ksi

Some Part

sx = 6 ksi

txy = 3 ksi

x & y  orientation

x-axis

(6 ksi, 3 ksi)

2

6

3

Center of Mohr’s Circle

3

(-2 ksi, -3 ksi)

y-axis

Mohr’s Circle

t(CW)

sy = -2 ksi

sx = 6 ksi

txy = 3 ksi

s

(savg, tmax)

Mohr’s Circle

t(CW)

sy = -2 ksi

x-face

sx = 6 ksi

(6 ksi, 3ksi)

txy = 3 ksi

s

s2

s1

savg= 2ksi

(-2 ksi, -3ksi)

y-face

(savg, tmin)

(savg, tmax)

(2 ksi, 5 ksi)

Mohr’s Circle

t(CW)

sy = -2 ksi

x-face

sx = 6 ksi

(6 ksi, 3ksi)

R

3 ksi

txy = 3 ksi

s

s2

s1

4 ksi

y-face

(savg, tmin)

(2 ksi, -5 ksi)

s1 = savg + R = 7 ksi

s2 = savg – R = -2 ksi

(savg, tmax)

(2 ksi, 5 ksi)

Mohr’s Circle

t(CW)

sy = -2 ksi

x-face

sx = 6 ksi

(6 ksi, 3ksi)

3 ksi

2q

txy = 3 ksi

s

s2

s1

4 ksi

y-face

(savg, tmin)

(2 ksi, -5 ksi)

(savg, tmax)

(2 ksi, 5 ksi)

Principle Stress

t(CW)

s2 = -3 ksi

x-face

(6 ksi, 3ksi)

q = 18.435°

3 ksi

2q

s1 = 7 ksi

s

s2

s1

Principle Stress Element

4 ksi

Rotation on element is half of the rotation from the circle in same direction from x-axis

(savg, tmin)

(2 ksi, -5 ksi)

(savg, tmax)

(2 ksi, 5 ksi)

Shear Stress

t(CW)

x-face

savg = 2 ksi

f = 26.565°

(6 ksi, 3ksi)

2f

tmax = 5 ksi

3 ksi

2q

s

savg = 2 ksi

s2

s1

Maximum Shear Stress Element

4 ksi

y-face

(savg, tmin)

(2 ksi, -5 ksi)

Relationship Between Elements

savg = 2 ksi

tmax = 5 ksi

sy = -2 ksi

savg = 2 ksi

f = 26.565°

sx = 6 ksi

q = 18.435°

s2 = -3 ksi

txy = 3 ksi

s1 = 7 ksi

q + f = 18.435 ° + 26.565 ° = 45 °

What’s the stress at angle of 15° CCW from the x-axis?

y

A particular point on the part

V

x

s = ? ksi

Some Part

U

s = ? ksi

15°

x

t = ? ksi

U & V  new axes @ 15° from x-axis

(savg, tmax)

Rotation on Mohr’s Circle

t(CW)

(sU, tU)

x-face

30°

s

s2

s1

savg= 2ksi

y-face

15° on part and element is 30° on Mohr’s Circle

(sV, tV)

(savg, tmin)

(savg, tmax)

Rotation on Mohr’s Circle

t(CW)

(sU, tU)

x-face

R

sU = savg + R*cos(66.869°)

sU = 3.96 ksi

sV = savg– R*cos(66.869°)

sV = 0.036 ksi

tUV = R*sin(66.869°)

tUV = 4.60 ksi

66.869°

s

s2

s1

savg= 2ksi

y-face

(sV, tV)

(savg, tmin)

What’s the stress at angle of 15° CCW from the x-axis?

y

A particular point on the part

V

x

sV= .036 ksi

sU = 3.96 ksi

Some Part

U

15°

x

t = 4.60 ksi

### Questions?

Next: Special Cases

Mohr’s Circle

sy

t(CW)

sx

s

sy

sx

This isn’t the whole story however…

Mohr’s Circle for X-Y Planes

sx = s1 and sy = s2

sz = 0

Mohr’s Circle

sy

t(CW)

sx

tmax = txz

sx

sz

sz = 0 since it is perpendicular to the free face of the element.

sz = s3 and sx = s1

s

s1

s3

Mohr’s Circle for X-Z Planes

Mohr’s Circle

sy

t(CW)

tmax = txz

sx

sz

s

s2

s1

s3

sz = 0 since it is perpendicular to the free face of the element.

s1 > s2 > s3

Pure Uniaxial Tension

sy = 0

sx = P/A

s1= sx

s2 = 0

Note when sx = Sy, Sys = Sy/2

Ductile Materials Tend to Fail in SHEAR

Pure Uniaxial Compression

sy = 0

sx = P/A

s1 = 0

s2= sx

Pure Torsion

T

T

s2 = -txy

s1 = txy

Brittle materials tend to fail in TENSION.

s1

CHALK

Uniaxial Tension & Torsional Shear Stresses
• Rotating shaft with axial load.
• Basis for design of shafts.

(sx,txy)

sx = P/A

sx/2

s2 = sx/2-R

s1 = sx/2+R

txy = Tc/J

(0,tyx)