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: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:heT
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
ᵟ
X=0 X=L
So, w(x) =
δ
+
δ
M(x) = EIw''(x) = . +2 = EI(
δ
+
δ
)
S(x) = = M'(x) =
δ
>>> constant
:=0reaction forces and moments at XheT
S(0) =
δ𝐸𝐼
𝐿
and M(0) =
δ𝐸𝐼
𝐿
:=Lreaction forces and moments at XheT
S(L) =
δ𝐸𝐼
𝐿
and M(L) = EI(
δ
𝐿
𝐿+
δ
𝐿
) = 0
e force and moment diagrams:hT
ᵟ
S(0) =
δ𝐸𝐼
𝐿
S(L) =
δ𝐸𝐼
𝐿
M(0) =
δ𝐸𝐼
𝐿
M(L) =

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

  • 1. :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:heT 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 ᵟ X=0 X=L
  • 2. So, w(x) = δ + δ M(x) = EIw''(x) = . +2 = EI( δ + δ ) S(x) = = M'(x) = δ >>> constant :=0reaction forces and moments at XheT S(0) = δ𝐸𝐼 𝐿 and M(0) = δ𝐸𝐼 𝐿 :=Lreaction forces and moments at XheT S(L) = δ𝐸𝐼 𝐿 and M(L) = EI( δ 𝐿 𝐿+ δ 𝐿 ) = 0 e force and moment diagrams:hT ᵟ S(0) = δ𝐸𝐼 𝐿 S(L) = δ𝐸𝐼 𝐿 M(0) = δ𝐸𝐼 𝐿 M(L) =