Lecture 2 principal stress and strain

principal planes and principal stresses

2. Unit 1- Stress and Strain
 Lecture -1 - Introduction, state of plane stress
 Lecture -2 - Principle Stresses and Strains
 Lecture -3 - Mohr's Stress Circle and Theory of
Failure
 Lecture -4- 3-D stress and strain, Equilibrium
equations and impact loading
 Lecture -5 - Generalized Hook's law and Castigliono's
Topics Covered

3. Stresses and strains
 In last lecture we looked at stresses were acting in a
plane that was at right angles/parallel to the action of
force.
Tensile Stress Shear Stress

4. Stresses and strains
Compressive load Failure in shear
Stresses are acting normal to the surface yet the material failed in a different plane

5. Principal stresses and
strains
 What are principal stresses.
 Planes that have no shear stress are called as principal
planes.
 Principal planes carry only normal stresses

6. Stresses in oblique plane
 In real life stresses does not act in normal direction but
rather in inclined planes.
Normal Plane Oblique Plane

7. Stresses in oblique plane
€
σ
n
€
σ
t
Unit depth
€
σ
n
=σcos
2
θ
€
σ
t
=
σ
2
sin2θ
€
σ=
P
A
P =Axial Force
A=Cross-sectional area
perpendicular to force
€
θ

8. Stresses in oblique plane
 Member subjected to direct stress in one plane
 Member subjected to direct stress in two mutually
perpendicular plane
 Member subjected to simple shear stress.
 Member subjected to direct stress in two
mutually perpendicular directions + simple shear
stress
€
σ
1
€
σ
1
€
σ
1
€
σ
1
€
σ
2
€
σ
2
€
τ
€
τ
€
τ
€
τ
€
σ
1
€
σ
2
€
σ
2
€
σ
1
€
τ
€
τ
€
τ
€
τ

9. Stresses in oblique plane
 Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
€
σ
n
=
σ
1
+σ
2
2
+
σ
1
−σ
2
2
cos2θ+τsin2θ
€
σ
t
=
σ
1
−σ
2
2
sin2θ−τcos2θ

10. Stresses in oblique plane
 Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
 POSITION OF PRINCIPAL PLANES
 Shear stress should be zero
€
tan2θ=
2τ
σ
1
−σ
2
€
σ
t
=
σ
1
−σ
2
2
sin2θ−τcos2θ=0

11. Stresses in oblique plane
 Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
 POSITION OF PRINCIPAL PLANES
€
tan2θ=
2τ
σ
1
−σ
2
€
2τ
€
σ
1
−σ
2
€
θ
€
sin2θ=
2τ
σ
1
−σ
2
( )2
+4τ
2
€
cos2θ=
σ
1
−σ
2
( )
σ
1
−σ
2
( )2
+4τ
2

12. Stresses in oblique plane
 Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
€
=
σ
1
+σ
2
2
+
σ
1
−σ
2
2
$
%
&
'
(
)
2
+τ
2
Major principal Stress
€
=
σ
1
+σ
2
2
−
σ
1
−σ
2
2
$
%
&
'
(
)
2
+τ
2
Minor principal Stress

13. Stresses in oblique plane
 Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
 MAX SHEAR STRESS
€
d
dθ
σ
1
−σ
2
2
sin2θ−τcos2θ
&
'
(
)
*
+
=0
€
d
dθ
σ
t
()
=0
€
tan2θ=
σ
1
−σ
2
2τ

14. Stresses in oblique plane
 Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress
 MAX SHEAR STRESS
€
tan2θ=
σ
1
−σ
2
2τ
€
σ
t
=
σ
1
−σ
2
2
sin2θ−τcos2θ
€
σ
t
()
max
=
1
2
σ
1
−σ
2
( )
2
+4τ
2
Evaluate the following equation at

15. Stresses in oblique plane
 Member subjected to direct stress in one plane
 Member subjected to direct stress in two mutually
perpendicular plane
 Member subjected to simple shear stress.
 Member subjected to direct stress in two mutually
perpendicular directions + simple shear stress

16. Stresses in oblique plane
 Member subjected to direct stress in one plane
€
σ
n
=
σ
1
+σ
2
2
+
σ
1
−σ
2
2
cos2θ+τsin2θ
€
σ
t
=
σ
1
−σ
2
2
sin2θ−τcos2θ
Stress in one direction and no shear stress
€
σ
2
=0
€
τ=0
€
σ
n
=
σ
1
2
+
σ
1
2
cos2θ=σ
1
cos
2
θ
€
σ
t
=
σ
1
2
sin2θ

17. Stresses in oblique plane
 Member subjected to direct stress in two mutually
perpendicular plane
€
σ
n
=
σ
1
+σ
2
2
+
σ
1
−σ
2
2
cos2θ+τsin2θ
€
σ
t
=
σ
1
−σ
2
2
sin2θ−τcos2θ
Stress in two direction and no shear stress
€
τ=0
€
σ
n
=
σ
1
+σ
2
2
+
σ
1
−σ
2
2
cos2θ
€
σ
t
=
σ
1
−σ
2
2
sin2θ

18. Stresses in oblique plane
 Member subjected to simple shear stress.
€
σ
n
=
σ
1
+σ
2
2
+
σ
1
−σ
2
2
cos2θ+τsin2θ
€
σ
t
=
σ
1
−σ
2
2
sin2θ−τcos2θ
No stress in axial direction but only shear stress
€
σ
1
=σ
2
=0
€
σ
n
=τsin2θ
€
σ
t
=−τcos2θ

19. Principal stresses and
strains
 PROBLEM- The tensile stresses at a point across
two mutually perpendicular planes are 120N/mm
2

and 60 N/mm
2
. Determine the normal, tangential
and resultant stresses on a plane inclined at 30deg to
the minor stress.

20. Principal stresses and
strains
 PROBLEM- A rectangular block of material is
subjected to a tensile stress of 110 N/mm
2
on one
plane and a tensile stress of 47 N/mm
2
on the plane
at right angles to the former. Each of the above
stresses is accompanied by a shear stress of 63 N/
mm
2
and that associated with the former tensile
stress tends to rotate the block anticlockwise. Find
1)The direction and magnitude of each of the principal
stress.
2) Magnitude of the greatest shear stress.