The composite bar BCD in the figure is fixed at the wall B (x=0), and has constant cross sectional area A0. The bar is composed by joining, at section C, two segments (BC and CD) of equal length L. Segment CD is homogeneous, made of copper. Segment BC is obtained by joining two identical wedges, a copper wedge and a steel wedge, as indicated in the figure. Along segment BC the cross sectional areas of copper and steel are given, respectively, by:
Ac=A0xL,
As=A0(1−xL),
which sum to give the constant cross-sectional area A0 at each x within bar segment BC.
The bar is subjected to a uniform distributed load per unit length of magnitude f0 in the direction indicated in the figure.
The Young’s modulus of copper is EC=E0, and the modulus of steel is ES=2E0.
The given (known) quantities are: L[m], A0[m2], E0[Pa], and f0[Nm].
NOTE: careful! The total length of the bar is 2L.
1) Obtain a symbolic expression for the axial force resultant N(x) in terms of L, x and f0 (enter this as f_0):
N(x)=
2) Obtain an expression for the axial strain in the bar, ϵa(x) in terms of x, L, E0, A0 and f0 (enter E0, A0 and f0 as, respectively, E_0, A_0 and f_0):
for 0≤x≤L, ϵa(x)=
for L≤x≤2L, ϵa(x)=
3) Obtain a symbolic expression for the normal stress in the steel at the midspan of segment BC, σnsteel(x=L2), in terms of L, A0, and f0 (enter A0 and f0 as, respectively, A_0 and f_0).
σnsteel(x=L2)=
4) Obtain an expression for the displacement field in the bar, ux(x) in terms of x, L, E0, A0 and f0 (enter E0, A0 and f0 as, respectively, E_0, A_0 and f_0):
for 0≤x≤L, ux(x)=
for L≤x≤2L, ux(x)=
5) Obtain a symbolic expression for the elongation of entire bar BD, δBD, in terms of L, E0, A0 and f0 (enter E0, A0 and f0 as, respectively, E_0, A_0 and f_0):.
δBD=
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