Mechanics of structuresMechanics of structures
Stresses in 2D plane, ColumnsColumns
Dr. Rajesh K. N.Assistant Professor in Civil EngineeringAssistant Professor in Civil EngineeringGovt. College of Engineering, Kannur
Dept. of CE, GCE Kannur Dr.RajeshKN
1
Module IV
Transformation of stresses and strains (two dimensional case only)
Module IV
Transformation of stresses and strains (two-dimensional case only) -equations of transformation - principal stresses - mohr's circles of stress and strain - strain rosettes - compound stresses - superposition
d it li it ti and its limitations
Eccentrically loaded members - columns - theory of columns -buckling theory - Euler's formula - effect of end conditions -eccentric loads and secant formula
Dept. of CE, GCE Kannur Dr.RajeshKN
2
Analysis of plane stress and plane strainy p p
cosnA A =P
LAA
PA
= cosn
nAA =
AAn
A
22cos cos cosn
P PA A = = =cosP
nA A
sin sin cos sin2P P
PsinP An
Dept. of CE, GCE Kannur Dr.RajeshKN
32n nA A
= = =
2 2 +( ) ( )2 22cos sin cos
R n n
= += +
Resultant stress on inclined plane
2 2cos cos sincos
= +=
0 2 cos sin 0ndd = =For maximum normal stress,
0sin2 0 0 = =0When 0 P = = =
0 2 0nd For maximum shear stress
,maxWhen 0 , n A = = =
0 0
0 cos2 0
2 90 45
n
d
= = = =
For maximum shear stress,
Dept. of CE, GCE Kannur Dr.RajeshKN
40
,maxWhen 45 , 2 2nPA
= = =
n
n
R 2 2n n +
,maxn
Dept. of CE, GCE Kannur Dr.RajeshKN
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Element under pure shearB
n
n
cos sinBC AC AB = A C
. .cos . .sinsin cos cos sinsin 2
n
n
BC AC AB
= =
. . .sin . .cosn BC AC ABAC AB
= +sin 2n =
2 2
. .sin . .cos
cos sin
n
n
AC ABBC BC
= +=
Dept. of CE, GCE Kannur Dr.RajeshKN
6cos2
n
n =
Element under biaxial normal stress
Bn
y
nn
xx
A C
x
ysin cosBC AB AC = cos sinBC AB AC = +
y. . .sin . .cos
. .sin . .cos
n x y
n x y
BC AB ACAB ACBC BC
==
. .cos . .sin
. .cos . .sin
n x y
n x y
BC AB ACAB ACBC BC
= += +
( )cos sin sin cos
sin 2n x y
BC BC
= 2 2cos sin
y
n x y
BC BC = +
Dept. of CE, GCE Kannur Dr.RajeshKN
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( ) 2n x y =
Element under a general two-dimensional stress
xyxy B n
nx
xy
x
xyxy
A C
x
xy
. .cos . .sin . .cos . .sinn x y xy xyBC AB AC AC AB = + yxy y
. .cos . .sin . .cos . .sinn x y xy xyAB AC AC ABBC BC BC BC
= + 2 2cos sin sin 2n x y xy = + +
Dept. of CE, GCE Kannur Dr.RajeshKN
8cos2 sin 2
2 2x y x y
n xy
+ = +
i iBC AB AC AB AC . . .sin . .cos . .cos . .sin
sin cos cos sin
n x y xy xyBC AB AC AB AC
AB AC AB AC
= +
= + 2 2
. .sin . .cos . .cos . .sin
cos sin sin cos cos sin
n x y xy xy
n x y xy xy
BC BC BC BC
= +
= + n x y xy xy
sin 2 cos22
x yn xy
= +2n xy
Note: In the above derivations,
Sign convention for : Anticlockwise angle is +ve.g g
Sign convention for shear stress: With respect to a point inside the element, clockwise shear stress is +ve.
Dept. of CE, GCE Kannur Dr.RajeshKN
9
the element, clockwise shear stress is +ve.
+ cos2 sin 2
2 2x y x y
n xy
+= +
Th b ti i th l d t ti l ( h ) t
sin 2 cos22
x yn xy
= +The above equations give the normal and tangential (shear) stresses on any plane inclined at with the vertical.
To find maximum/minimum value of normal stress
( )0n = ( )sin 2 2 cos2 0x y xy =2
tan 2 xy ( )tan 2 yx y =
Dept. of CE, GCE Kannur Dr.RajeshKN
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2 sin 2 xy =xy
( )
2
2x y
xy
+ 2
22
sin 2
2
y
x yxy
= + ( )
2x y
2cos2 x y
=2
222
x yxy
+
cos2 sin 22 2
x y x yn xy
+ = + 2
2max 2 2
x y x yxy
+ = + + 2
2i
x y x y + = + Dept. of CE, GCE Kannur Dr.RajeshKN
11
min 2 2 xy +
2222
max,min 1,3 2 2x y x y
xy
+ = = + ( )2
tan 2 xyx y
=
22
max,min 1,3 2 2x y x y
xy
+ = = + ( )2
tan 2 xyx y
=
Principal stresses Principal planes
Maximum shear stress
We know, sin 2 cos22
x yn xy
= +
F i h t 0nFor maximum shear stress, 0n =( )cos2 2 sin 2 0x y xy =( )x y xy
( )tan 2 x y
=
Dept. of CE, GCE Kannur Dr.RajeshKN
12
tan 22 xy
2( )2
For max normal stress, tan 2 xynx y
=
( )For max shear stress, tan 2
2x y
s
= 1
tan 2=
2 xy tan 2 n
( )0cot 2 tan 2 90n n = = ( )n n02 2 90s n = s n
045s n =
Hence, planes of maximum shear stress are at 450 to the principal planes
Dept. of CE, GCE Kannur Dr.RajeshKN
13
p
To get maximum shear stress
( )2
2
2x y
xy
+ 2
2
cos2 xysx y
= +
g
( )2
x y 2y
2 s2
yxy+
sin 2 x y =xy 2
2
sin 2
22
s
x yxy
= +
,max 2 22 2
22
x y x y xyn xy
x y x y
= + + +
2
22 2xy xy
+ + 2 2
,max 2x y
n xy
= + 2
,max,min 2x y
n xy
= +
Dept. of CE, GCE Kannur Dr.RajeshKN
14
2 21,3 2
We kno2
w, x y x y xy + = +
22
1 3 ,max2 22x y
xy n
= + = 1 3
,max 2n =
2
To get normal stress on planes of maximum shear stress
2 22 2
2 2x y x y xy x y
n xy
x y x y
+ = + 2 222 2
x y x yxy xy
+ + x y + th l f h t
Dept. of CE, GCE Kannur Dr.RajeshKN
152x y
n = , on the planes of max shear stressaverage=
Problem: Find the principal stresses (including principal planes) and maximum shear stress (including its plane)
260 N
280 N mmPrincipal stresses
2120 N mm
260 N mm
2120 N mm2
2max,min 1,3 2 2
x y x yxy
+ = = +
Principal stresses
280 N mm
260 N mm
2
2max min 1 3
120 80 120 80 60 + = = + max,min 1,3 602 2 +
100 63 24 = = max,min 1,3 100 63.24 = = 2
max 1 163.24 N mm = = max 12
min 3and 36.75 N mm = =
Dept. of CE, GCE Kannur Dr.RajeshKN
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2tan 2 xy
=Principal planes 2 60 = 3=( )tan 2 x y = Principal planes ( )120 80= + 3=( )12 tan 3 71 57 = = D 35 78 = D( )2 tan 3 71.57 = = 35.78 =
1 35.78 = D13 35.78 90 125.78 = + = D
xyxy
35 78D
xxy 3135.78D
yxy
Dept. of CE, GCE Kannur Dr.RajeshKN
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Maximum shear stress
1 3,max 2n
=Maximum shear stress
163.25 36.752
+= 263.25 N mm= 2 2
Planes of maximum shear stress
( )tan 2 x y
= 120 80 +=tan 22s xy
= 2 60= xy
xy 1 120 802 t + 18.43= D
xy
9.22D
12 tan2 60
= 18.43
x
xyxy ,maxn
9.22 = D
Dept. of CE, GCE Kannur Dr.RajeshKN
18yxy
cos2 sin 2x y x y + = +
Mohrs circle
cos2 sin 22 2n xy
= +
cos2 sin 2x y x y + = i
sin 2 cos22
x yn xy
= +cos2 sin 2
2 2n xy = i
ii2n xy
Squaring and adding the above equations,
( ) ( )2 2 222 2x y x yn n xy + + = + 2 2 ( ) ( )2 2 20n av n R + =
This is equation of a circle with centre and radius ( )2 22x y xy + ( ),0av
Dept. of CE, GCE Kannur Dr.RajeshKN
19Let us draw this circle!
Mohrs circlexy( ),y xy
xxy
xyxy
22
2y x
xy
+ yxy
y
xy2
y x +
yx 3 12
y x
xy
( ),x xy 1 2tan 2xyy x
= =
Dept. of CE, GCE Kannur Dr.RajeshKN
20: Principal stresses 1, 3measured clockwise
Mohrs circlexy
xxy
y
xy( ),x xy yxy
22
2y x
xy
+
yx
xy
3 x
3 12
y x +2
y x
xy2
1 2tan 2xyy x
= =( ),y xy
Dept. of CE, GCE Kannur Dr.RajeshKN
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y
measured anticlockwise
Mohrs circle is a graphical representation of the state of stress in an Mohr s circle is a graphical representation of the state of stress in an element.
E i h i l h l d h Every point on the circle represents the normal and shear stress on a plane.
While x-coordinate of a point on the circle represents the normal stress on a plane, y-coordinate represents the shear stress on that plane.
Procedure for construction of Mohrs circle
Dept. of CE, GCE Kannur Dr.RajeshKN
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Maximum shear stress from Mohrs circlexy( ),y xy
xxy
xyxy
Max shear 2
2
2y x
xy
+ yxy
Max shear stress
1 3max 2n
=
y
