1. Stress in any direction
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Stress in any direction

2. Two dimensional state of stress, and the stress elementX Y F1 F2 F3 F4 Fn M
Stress Element
txy sx sy P Y sy P sx X
txy
ccw
txy cw
When a set of co-planer external forces and moments act on a body,
the stress developed at any point ‘P’ inside the body
can be completely defined by the two dimensional state of stress:
sx = normal stress in X direction,
sy = normal stress in Y direction, and
txy = shear stress which would be equal but opposite in
X (cw) and Y (ccw) directions, respectively.
The 2D stress at point P is described by a box drawn with its faces perpendicular to X & Y directions, and showing all normal and shear stress vectors (both magnitude and direction) on each face of the box. This is called the stress element of point P.

3. The stress formulae that we have learnt thus far, can determine the 2D stresses developed inside a part, ONLY ALONG A RECTANGULAR AXIS SYSTEM X -Y, that is defined by the shape of the part.
For example, X axis for a cantilever beam is parallel to its length,
and Y axis is perpendicular to X. Y
For a combined bending and axial loading (F1, F2 etc.) of this cantilever beam:
the normal and shear stress at a point P, can be determined using the formulae, such as,
sx= Mv/I+P/A,
txy=VQ/(Ib). V
txy sx sy P X f Y F1 U P F2 X
Note that, these formulae can only determine stresses parallel to X and Y axis, and the stress element is aligned with X-Y axis.
The question is, what would be the values of normal and shear stresses at the same point P, if the stresses are measured along another rectangular axis system U-V, rotated at an angle f with the X-Y axis system ?

4. The Problem is: given sx, sy, txy and f,
Knowing the 2D stresses at point P along XY coordinate system,
we want to determine the 2D stresses for the same point P, when measured along a new coordinate system UV,
which is rotated by an angle f with respect to the XY system.
txy X Y sx sy su v
tuv u sv X f F v u X F1 Y F2 P f X Y u f F2 P
The Problem is: given sx, sy, txy and f,
can we determine su, sv, tuv ?

5. THIS IS HOW WE CAN ACHIEVE THAT
2. To maintain static equilibrium, let the internal normal and shear stresses su & tuv, respectively are developed on the cut plane
1. We cut the stress element by an arbitrary plane at an angle f. This plane is normal to u-axis
txy sx sy f sx
txy X Y sy su
tuv f f L
Lsinf
Lcosf U
txy(LBsinf)
sx(LBcosf)
txy(LBcosf)
sy(LBsinf)
su(LB)
tuv(LB) f
4. If the thickness of the element is B, then the force acting on each face of the element will be equal to the stress multiplied by the area of the face.
3. Let, L be the length of the cut side. Then the other two sides are Lsinf & Lcosf

6. CONTINUING 5. Forces acting on the faces = force x area
txyLBsinfcosf f
txyLBsin2f
txyLBcos2f
su(LB)
tuv(LB)
syLBsinfcosf
syLBsin2f
sxLBcos2f
sxLBcosfsinf
CONTINUING
5. Forces acting on the faces = force x area
txy(LBsinf)
sx(LBcosf)
txy(LBcosf)
sy(LBsinf)
su(LB)
tuv(LB) f
6. Resolving each force in u & v directions
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)

7. Replacing the square terms of trigonometric functions by double angle terms and rearranging :
Equations 3, 4 & 5 gives us the 2D stress values, if measured along U-V axis which is at an angle f from X-Y axis.
Since both sets of stresses refer to the stress of the same point, the two sets of stresses are also equivalent.
txy X Y sx sy f su
tuv v sv X f u

8. Mohr Circle

9. implements these three equations
Mohr’s circle
implements these three equations
by a graphical aid, which simplifies computation and visualization of the changes in stress values (su, sv & tuv) with the rotation angle f of the measurement axis. s -s t -t
Mohr circle is plotted on a rectangular coordinate system in which the positive horizontal axis represents positive (tensile) normal stress (s), and the positive vertical axis represents the positive (clockwise) shear stress (t).
Thus the plane of the Mohr circle is denoted as s-t plane.
In this s-t plane, the stresses acting on two faces of the stress element are plotted.
txy sx X cw
txy Y sx sy X
x faces have stress:
(sx & txy)
txy Y sy
ccw
For a stress element
Y faces have stress: (sy,-txy)

10. t s -s -t Y V X X sx txy su U tuv sy sv X (sx,txy) su U sx txy sy tuv
DRAWING MOHR CIRCLE Y sx
txy sy X su U V
tuv sv X f
Start by drawing the original stress element with its sides parallel to XY axis, and show the normal and the shear stress vectors on the element.
Draw the s-t rectangular axis and label them.
On the s-t plane, plot X with normal and shear stress values of sx and txy, and Y with values sy and –txy.
Join X and Y points by a straight line, which intersects the horizontal s axis at C. C denotes the average normal stress savg=(sx+sy)/2 .
The line CX denotes X axis, and line CY denotes Y axis in Mohr circle. Name them.
Draw the Mohr circle using C as the center, and XY line as the diameter.
To find stress along the new UV axis system, draw a line UV rotated at an angle 2f from the XY line. CU line denotes U axis, and CV denotes V axis.
The normal and shear stress values of the points U and V on the s-t plane denote the stresses in U and V directions, respectively.
This way we can find stresses for an element rotated at any desired angle f. t su
X (sx,txy)
Shear stress axis (t) U sx
txy sy
X axis 2f
tuv s -s C
tuv
Normal stress axis (s)
txy
Y axis V sv
Y(sy,-txy)
savg=(sx+sy)/2 -t

11. PROOF -s X (sx,txy) Shear stress axis (t) U (su,tuv) sx txy X sy
X axis 2f
Shear stress axis (t)
Y(sy,txy) sx sy 2q
Normal Stress axis (s)
Y axis
savg=(sx+sy)/2
U (su,tuv)
V (sv,txy) a
X (sx,txy) X Y sx
txy sy X su U V
tuv sv X f

12. Principal Normal Stresses s1 & s2, and Max Shear Stress tmaxY X
In the Mohr circle, for a rotation of 2q angle, the XY axis line becomes horizontal. In the rotated axis s1-s2, the shear stress vanishes.
The element will have only normal stresses s1 & s2, and s1 being the maximum normal stress. s1 & s2 are called the Principal normal stresses. s2 q s1 x Y
tmin
tmax s2
savg s1
tmax q’
Y(sy,-txy) sx sy
savg -s -t o
txy
X axis
Y axis t s
X (sx,txy)
tmax
2q’
(savg,tmax)
(savg,-tmax)
Similarly, if the XY axis line is rotated by an angle 2q ‘ to make it vertical, then the shear stress maximizes and the element will have normal stress = savg and Maximum shear stress = tmax 2q s2 s1

13. t -s s o -t Formulea for Principal Normal Stresses & Max Shear StressY sx
txy sy X
tmax
(savg,tmax)
X (sx,txy)
X axis
txy sy
2q’ -s 2q sx s o s2 s1
savg
txy
Y axis
Y(sy,-txy) -t
(savg,-tmax) s1 s2 Y q X
tmin
tmax
savg x Y f
Principal normal stress element
Maximum shear stress element

14. t s -s -t Determining su, sv & tuv Given sx, sy, txy & f Y sx txy sy X
X (sx,txy)
U (su,tuv) sx
X axis
txy sy 2f 2q
tuv s -s C
tuv
txy
Y axis
V (sv,txy) sv
Y(sy,-txy)
savg=(sx+sy)/2 -t

15. Mohr Circle
Example

16. Y X
For a stress element with
sx=20,000 psi,
sy= psi, and
txy= 5000 psi.
4,000 psi
5,000 psi
20,000 psi
Draw the Mohr Circle and, draw two stress elements properly oriented for (i) the principal normal stresses, and (ii) max shear stresses element. t -t
Draw the stress element along XY axis.
Draw the s-t axes for mohr circle
Plot point X for sx=20K, txy=5k
Plot point Y for sy= -4K, txy=-5k
Draw line XY and show X & Y axes.
Draw the circle with XY as the diameter
(8k,13k)
tmax
X (20k,5k)
X axis 5k s -s o
67.4
20k
22.6
s2= -5k
-4k 8k
R=13K
s1=21k 5k
Y axis
Y(-4k,-5k)
(8k,-13k)
This completes the Mohr circle. Next, the stress elements

17. t -s o s -t Y X Y s2 X tmax s1 X (20k,5k) tmin tmax x
5,000 psi
20,000 psi
4,000 psi
The principal normal stress axis will be rotated CW
Draw the principal stress axis 11.3o CW from XY axis.
Show the principal stresses.
PRINCIPAL NORMAL STRESS ELEMENT Y X s2 t -t 5k
(8k,13k)
21k
tmax
q=11.3 s1 5k
X (20k,5k)
STRESS ELEMENT FOR tMAX
X axis 5k s -s o
67.4
The tmax axis will be rotated CCW
Draw the tmax stress axis 33.7o CCW from XY axis.
Show the the stresses.
20k
22.6
s2= -5k
-4k 8k
R=13K
s1=21k 5k Y
Y axis
tmin
Y(-4k,-5k) 8k
tmax
13k 8k
33.7 x
(8k,-13k) 8k
13k 8k
That completes the drawing of the two stress elements

18. This ends the presentation
and thanks for watching it