!1!

On =

} (10+10+1) = 7

I ( 77 .) i or = ( toB.°g )

b) 1

of -

XII =o

4=-6 ; xlz=-3 ; X3 =

9

air ) = ( § ) ; new =±d f)I (1/3) = ¥ ( to )

!2!

a) Balance Equations ✓

b) • Boundary z= he

or ( x ) = 0;

Exz ( x ) =o

→ Bottom boundary stress free .

• Z=- {

of( × ) = -

B is Txz = O

⇒ Top boundaryhas constant

stress in z -direction

.

( say ) Pz= -

E=B

• x = lz

• × = - e- )similar to above .

2

→Both of

these Left S right )

boundaries have shear S normal

stresses.

The resultant forces S moments on the lefts right boundaries are

↳

h,

Normal

±N ( f ) = Ntl ) = t {not dz

- 0-2

Shear : VC f) = ✓ ( £ ) =

= [÷z2×zdz

= ItBe

moment :

my ) =m( if ) =tf¥±z Edt

= At.

!3!

Jx = O,

'

Jy =- 250 j Jxy = lxy

= 3 of

E- =

IF( E -

Fitr ( E) E)

1 6 0=¥l::;)Unit rectors in IBs AT directions

e. fins}:) =÷f}l ; ⇐ need

Strain In FB direction

4,

= EI E- E,

=

2¥53Also 4 ,

= eye ⇒l El Gtzr E)

Deformed angle between AT SAI

x= Is - Hz ;

where Vez = 2 EIEe-

=> Nz = 1,69 If

!4!

§ is Bioharmonic

on

=j5×I= 2C

; on =- IIm = -

B

ozz=2¥ = 2A

oxi

at xz=o ;

¥= = a = (But ) ; with I = (9)

at xz=c ; tz=⇐B )

at x. =o ; Is - (F) ; atx. =b ; tu=(2§)

b) 033 = a ( on + on ) =2 ✓ ( CEA )

6,3 = 523 = °

Gz = 93 = 53=0 ( a plain strain )

Ex = 1¥ ( G- Non - ✓ on ]= 2k¥

) ( Cle . D- ✓A)

similarly Ezz,

E 72.

AXZ 2An n n n

^

<

§<c c < c >

c✓ >

2C K. > 2C

- >

<

~ )

c> >

> > > > B"

×

,

ZA v

u -✓

v