# Mechanics of structures module4

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• 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

5

• 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

7

( ) 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

10

• 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

16

• 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

17

• 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

21

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

22

• 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

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