Purpose of Mohr’s Circle Visual tool used to determine the stresses that exist at a given point in...
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Transcript of Purpose of Mohr’s Circle Visual tool used to determine the stresses that exist at a given point in...
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
x = 6 ksi
y = -2 ksi
xy = 3 ksi
Some Part
A particular point on the part
x
y
x & y orientation
Mohr’s Circle (CW)
x-axis
y-axis
x = 6 ksi
y = -2 ksi
xy = 3 ksi
(6 ksi, 3 ksi)
63
(-2 ksi, -3 ksi)
2
3 Center of of Mohr’s CircleMohr’s Circle
Mohr’s Circle (CW)
avg=
2ksi
x-face
y-face
(6 ksi, 3ksi)
(-2 ksi, -3ksi)
(avg, max)
x = 6 ksi
y = -2 ksi
xy = 3 ksi
(avg, min)ksiyxavg 2
2
Mohr’s Circle (CW)
x-face
y-face
(6 ksi, 3ksi)
x = 6 ksi
y = -2 ksi
xy = 3 ksi
4 ksi
avg + R7 ksiavg – R ksi
(avg, max)(2 ksi, 5 ksi)
(avg, min)(2 ksi, -5 ksi)
3 ksi
maxR
ksi
)ksi()ksi(R
5
43 22
R
Mohr’s Circle (CW)
x-face
y-face
(6 ksi, 3ksi)
x = 6 ksi
y = -2 ksi
xy = 3 ksi 2
4 ksi
(avg, max)(2 ksi, 5 ksi)
(avg, min)(2 ksi, -5 ksi)
3 ksi
43518
869362
4
32 1
.
.
ksi
ksiTan
Principle Stress (CW)
x-face
(6 ksi, 3ksi)
1 = 7 ksi
2 = -3 ksi
2
4 ksi
(avg, max)(2 ksi, 5 ksi)
(avg, min)(2 ksi, -5 ksi)
3 ksi = 18.435°
Principle Stress Element
Rotation on element is half of the rotation from the circle in same direction from x-axis
Shear Stress (CW)
x-face
y-face
(6 ksi, 3ksi)
avg = 2 ksi
avg = 2 ksi
max = 5 ksi
2
4 ksi
(avg, max)(2 ksi, 5 ksi)
(avg, min)(2 ksi, -5 ksi)
3 ksi
56526
130532
86936902
2902
.
.
.
2
Maximum Shear Stress Element
= 26.565°
Relationship Between Elements
avg = 2 ksi
avg = 2 ksi
max = 5 ksi
= 26.565°
1 = 7 ksi
2 = -3 ksi
x = 6 ksi
y = -2 ksi
xy = 3 ksi = 18.435°
+ = 18.435 ° + 26.565 ° = 45 °
What’s the stress at angle of 15° CCW from the x-axis?
= ? ksi
= ? ksi
= ? ksi
Some Part
A particular point on the part
x
y
U & V new axes @ 15° from x-axis
15°
U
x
V
Rotation on Mohr’s Circle
(CW)
avg=
2ksi
x-face
y-face
(avg, max)
(avg, min)
30°
15° on part and element is 30° on Mohr’s Circle
(U, U)
(V, V)
U = avg + R*cos(66.869°)
U = 3.96 ksi
V = avg – R*cos(66.869°)
V = 0.036 ksi
UV = R*sin(66.869°)
UV = 4.60 ksi
Rotation on Mohr’s Circle
(CW)
avg=
2ksi
x-face
y-face
(avg, max)
(avg, min)
66.869°
R
(U, U)
(V, V)
What’s the stress at angle of 15° CCW from the x-axis?
U = 3.96 ksiV= .036 ksi
= 4.60 ksi
Some Part
A particular point on the part
x
y
15°
U
x
V
Questions?
Next: Special Cases
Special Case – Both Principle Stresses Have the Same Sign
Cylindrical Pressure Vessel
X
Y
Z t
Dpx 2
t
Dpy 4
Mohr’s Circle
x
(CW)
y
x
y
2yx
Mohr’s Circle for X-Y PlanesThis isn’t the whole story however…
x = 1 and y = 2
Mohr’s Circle
1
(CW)
y
x
z
z = 0 since it is perpendicular to the free face of the element.
z = 3 and x = 1
3
Mohr’s Circle for X-Z Planes
x
z = 0
maxxz
231
max
Mohr’s Circle
1
(CW)
2
y
x
z
z = 0 since it is perpendicular to the free face of the element.
3
maxxz
1 > 2 > 3
Pure Uniaxial Tension
y = 0
x = P/A
Ductile Materials Tend to Fail in SHEAR
1= x
2x
max
2 = 0
Note when x = Sy, Sys = Sy/2
Pure Uniaxial Compression
y = 0
x = P/A
2x
max
1 = 02= x
Pure TorsionT
T
J
cTxy
xymax
1 = xy2 = -xy
CHALK1
Brittle materials tend to fail in TENSION.
Uniaxial Tension & Torsional Shear Stresses
• Rotating shaft with axial load.
• Basis for design of shafts.
x = P/A
xy = Tc/J
Rmax
1 = x/2+Rx/2
x, xy)
2 = x/2-R
0, yx)
maxxyxR
2
2
2