Transcript Stress in any direction

Two dimensional state of stress and stress element
Y
Fn
sx
txy
sx
sx
txy
txy
sx
Z
txy
txy
X
F2
sy
X
txy
F3
sy
F4
txy
Y
Y
txy
F1
sy
sy
Z
If a body is subjected by external forces F1, F2, ….Fn, all acting on a plane parallel to XY
plane, then stress will be developed inside the elastic body. The stress at an arbitrary
point within the body can be described by drawing an microscopic cubic stress
element (dx, dy, dz) aligned to an XYZ coordinate system around the point and
determining the normal and shear stresses on each face of the cube. If the external
forces are co-planer and parallel to XY plane, the state of stress can be completely
defined by:
sx = normal stress in X direction,
sy = normal stress in Y direction, and
txy = shear stress which would be equal in magnitude X and Y direction
X
Previously, we have learnt how to calculate normal (sx & sy) and shear (txy)
stresses in different types of loading conditions (axial, bending and torsion).
Y sy
sx= Mv/I+P/A,
txy
txy
For example, for a bending load
on a simply-supported beam, we
know the stress condition at the
point shown, can be calculated
from the formulas such as,
txy
sx
X
txy
Y
sy
F1
F2
sy = 0, and
txy=VQ/(Ib).
Note that these formulas calculate stresses parallel to X and Y
axis, which is predefined by the geometry of the part.
sx
X
Now, knowing the stress condition at a point along XY coordinate
system, we want to determine the stress condition for the same point,
along a new coordinate system, which is rotated by an arbitrary angle f
Y sy
txy
txy
txy
sx f
X
txy
Y
sy
F
X
sx
Y
F
f
X
f
X
ie., given sx, sy, txy and f, find su, sv, tuv
2. To maintain the equilibrium of
the cut piece, let the normal and
shear stresses in the cut plane
su & tuv, respectively are
developed.
1. Our original stress
element is cut by an arbitrary
plane at an angle f
sy
Y sy
txy
txy
sx
f
txy(LBcosf)
sy
f
Lsinf
X
txy
sx
Lcosf
sx
f
txy
txy
txy
sy(LBsinf)
txy(LBsinf)
3. Let, L be the
length of the cut
sx(LBcosf) f
side. Then the other
two sides are Lsinf
& Lcosf
4. Assuming the thickness of
the element is B, then forces
on each side of the cut
element are stress multiplied
by the area of the face.
txy(LBcosf)
5. Forces
acting on the
cut element
f
sy(LBsinf)
txy(LBsinf)
f
f
sx(LBcosf) f
f
f
Equating forces in u-direction:
suLB = sxLBcos2f + syLBsin2f + 2txyLBsinfcosf
Or,
su = sxcos2f + sysin2f + 2txysinfcosf ………..(1)
Equating forces in v-direction:
tuvLB = txyLBcos2f - txyLBsin2f - sxLBsinfcosf+ syLBsinfcosf
Or,
tuv = txy(cos2f - sin2f) – (sx-sy) sinfcosf ……. (2)
6. Resolving
the forces in u
& v directions
Replacing the square terms of trigonometric functions by double
angle terms and rearranging the equations:
s u  s x cos 2 f + s y sin 2 f + 2t xy sin f cos f
su 
2
(1 + cos 2f ) +
s x +s y
sy
s x s y
2
(1  cos 2f ) + t xy sin 2f
cos 2f + t xy sin 2f ..........(3)
2
2
 t xy (cos 2 f  sin 2 f )  s x sin fcosf + s y sin fcosf

t uv
sx
+
 t xy cos 2f 
s x s y
sin 2f ....................(4)
2
Also , if we cut the stress element by a plane perpendicu lar
to the v  axis, then it can be shown that
s x +s y s x s y
sv 

cos 2f  t xy sin 2f .......(5)
2
2
Thus knowing a stress condition (sx, sy & txy) at a point in a given
orthogonal axis system (xy), we can use equations 3, 4 & 5 to
determine stress condition (su, sv & tuv) at the same point along a
new axis system (uv) which is rotated by an angle f. Since both set
of stresses refer to the same point, the two sets of stresses are also
equivalent.
su 
sx +sy
sx s y
cos 2f + t xy sin 2f
2
sx sy
t uv  t xy cos 2f 
sin 2f
2
s +s y sx s y
sv  x

cos 2f  t xy sin 2f
2
2
2
+
These relationships between the
stresses in different axis system can
be conveniently depicted by Mohr’s
circle.
Sign convention for Mohr Circle:
Normal stress: Tensile = positive
Shear stress: Producing clockwise rotation of the element= positive
Angle: Clockwise from positive X axis is positive.
sx
For the stress element shown:
Normal and shear stresses in X-direction are (sx & txy)
Normal and shear stresses in Y-direction are (sy,-txy)
Since the pair of txy in Y-direction producing CCW
rotation so txy is negative for Y-direction.
The angle f of direction U is positive.
txy
Y
sy
sx X
txy
sy
txy
sx
X
X
sy
Shear stress
axis (t)
su
tuv
sy
sx
txy
1. The axis system of Mohr circle is st axis.
Draw the st axis and name them.
sx
2. Plot the point X with a coordinate sx and txy
along the st coordinate system.
3. Mark another point Y with the coordinate sy
and –txy
4. Join the XY line. Let at point C, XY
intersects the horizontal axis. The point C
denotes the average normal stress. The line
CX denotes X axis and line CY denotes Y
axis in mohr circle.
5. Note CX and CY are making 1800 angle
s
with each other, whereas in reality the X
and Y axes are at an angle 900.
6. RULE: ALL ANGLES IN MOHR’S CIRCLE
IS TWICE THE REAL ANGLE.
7. Use C as the center, and draw a circle with
XY as the diameter.
8. To find stress in a direction U-V, which is f
angle CW from X-Y axis, draw a line UV
through C at an angle 2f CW from XY line.
9. The coordinate values of U & V denote the
normal and shear stress in UV direction.
txy
Y
sy
C
txy
tuv
CONSTRUCTION OF MOHR’S CIRCLE
FOR A GIVEN STRESS ELEMENT
sv
Y(sy,-txy)
savg(sx+sy)/2
Normal stress
axis (s)
sx
sy
txy
Shear stress axis (t)
s
 a sin( 2  2f )
PROOF
 a (sin 2 cos 2f  cos 2 sin 2f )
t
s s y
 a ( xy cos 2f  x
sin 2f )
a
2a
s s
 t xy cos 2f  x y sin 2f
2
Normal Stress axis (s)
 a cos( 2  2f )
 a (cos 2 cos 2f + sin 2 sin 2f )
s s
t
 a ( x y cos 2f + xy sin 2f )
2a
a
s s
 x y cos 2f + t xy sin 2f
2
Y(sy,txy)
savg(sx+sy)/2
txy
sx
Y
sy
txy
sy
sx
X
X
su 
sx +sy
2
+
sx s y
t uv  t xy cos 2f 
sv 
sx +sy
2

cos 2f + t xy sin 2f
2
sx sy
2
sx s y
2
sin 2f
cos 2f  t xy sin 2f
Principal Normal Stresses: s1 & s2
&
Maximum Shear Stress: tmax
Similarly if we measure stress at an angle 2f from XY
axes, then we get the maximum shear stress tmax
Y
txy
tmax
sy
sx
txy
(savg,tmax)
If we measure the stress
at an angle 2 from XY
X (sx,txy)
axes, then the normal
txy
stress will maximize, and
2f
sy
2 sx
no shear stress will be
savg
s2
s1
present. Normal stresses
txy
s1 and s2 are called
principal normal stresses.
Y(sy,-txy)
X
sx
sy
s
o
t
s avg 
Y
(savg,-tmax)
s x +s y
s 1  s avg + R
2
 s x s y 
2
R 
 + t xy
 2 
2
f
x
 2t xy
2  tan 
 s x s y

1



s 2  s avg  R
t max  R
In design, often we want to
know the max normal (s1)
and max shear (tmax)
stresses to predict failure of
a part.
Finding Principal
Normal Stresses (s1 &
s2) and Max Shear
Stress (tmax)
Y
4,000 psi
20,000 psi
t
tmax
(8k,13k)
X
X (20k,5k)
20,000 psi
5,000 psi
4,000 psi

o
s
s2-5k
4k
22.6
 11.3 o
2
2f67.4
222.6
5k
20k
s121k
8k
s
5k
Y(-4k,-5k)
11.3
t
s x + s y 20k  4k
s


 8k psi
67.4
avg
2
2
f
 33.7 o
2
2
2
Y
 s x s y 
 20 + 4 
2
2
R 

2
 + t xy  


2
 + 5  13k psi

s 1  s avg + R  8 + 13  21Kpsi
33.7
(8k,-13k)
t max  R  13Kpsi
2f  90  2
 90  22.6  67.4o
x s 2  s avg  R  8  13  5Kpsi
 2t xy
2  tan 1 
s s
y
 x

  tan 1  2  5   tan 1 (0.417)  22.6 o

 20 + 4 

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