h.w (2) ... advanced material course ... aero494

2
: problem ) 2 ( .W H For the given beam: find out a relation for M(x), S(x) and w(x) and find the reaction forces and moments at each support. : ssumptions A 1. Geometric material 2. Linearity Let w(x) = . + . + . + boundary conditions are: he T at X=0 >> it is fixed support >> so, w(0) = 0 and w'(0) = 0 at X=L >> it is roller support >> so, w(L) = δ and EIw''(L)=0 Form w(x) equation, we have that w'(x) = . +2 . + w''(x) = . +2 and w'''(x) = B.C's: Apply w(0) =0 >> .+ .+ .+ =0 >> =0 w'(0) =0 >> .+2 .+ =0 >> =0 w(L) = δ >> . + . = δ .. (1) EIw''(L) =0 >> w''(L)=0 >> . +2 =0 >> = - . .. (2) from 1,2 >> . +- . . = δ >> = δ and = δ X=0 X=L

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Page 1: H.W (2) ... advanced material course ... aero494

:problem )2( .WH

For the given beam: find out a relation for M(x), S(x) and w(x)

and find the reaction forces and moments at each support.

:ssumptionsA

1. Geometric material

2. Linearity

Let w(x) = . + . + . +

boundary conditions are:he T

at X=0 >> it is fixed support >> so, w(0) = 0 and w'(0) = 0

at X=L >> it is roller support >> so, w(L) = δ and EIw''(L)=0

Form w(x) equation, we have that w'(x) = . +2 . +

w''(x) = . +2 and w'''(x) =

B.C's: Apply

w(0) =0 >> . + . + . + =0 >> =0

w'(0) =0 >> . +2 . + =0 >> =0

w(L) = δ >> . + . = δ ….. (1)

EIw''(L) =0 >> w''(L)=0 >> . +2 =0 >> = - . .. (2)

from 1,2 >> . +- . . = δ >> = δ

and = δ

X=0 X=L

Page 2: H.W (2) ... advanced material course ... aero494

So, w(x) = δ

+

δ

M(x) = EIw''(x) = . +2 = EI( δ

+

δ

)

S(x) =

= M'(x) =

δ

>>> constant

:=0reaction forces and moments at Xhe T

S(0) = δ𝐸𝐼

𝐿 and M(0) =

δ𝐸𝐼

𝐿

:=Lreaction forces and moments at Xhe T

S(L) = δ𝐸𝐼

𝐿 and M(L) = EI(

δ

𝐿 𝐿+

δ

𝐿 ) = 0

e force and moment diagrams:hT

S(0) = δ𝐸𝐼

𝐿 S(L) =

δ𝐸𝐼

𝐿

M(0) = δ𝐸𝐼

𝐿 M(L) =