Click here to load reader

Mechanics of structures module4

  • View
    228

  • Download
    2

Embed Size (px)

Text of Mechanics of structures module4

  • 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

Search related