The composite bar BCD in the figure is fixed at the wall B (x=0), and has constant cross sectional area A_0. 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:
A_c=A_0*x/L,
A_s=A_0(1−x/L),
which sum to give the constant cross-sectional area A_0 at each x within bar segment BC.
The bar is subjected to a uniform distributed load per unit length of magnitude f_0 in the direction indicated in the figure.
The Young’s modulus of copper is E_C=E_0, and the modulus of steel is E_S=2E_0.
NOTE: careful! The total length of the bar is 2L.
Find:
1. expression for the axial force resultant N(x) in terms of L, x and f_0.
2. expression for the axial strain in the bar, ϵ_a(x) in terms of x, L, E_0, A_0
3. expression for the normal stress in the steel at the midspan of segment BC, σ_n steel(x=L2), in terms of L, A0, and f0
4. expression for the displacement field in the bar, u_x(x) in terms of x, L, E_0, A_0 and f_0
for 0≤x≤L, ux(x)=
for L≤x≤2L, ux(x)=
5. expression for the elongation of entire bar BD, δ_BD, in terms of L, E_0, A_0 and f_0
0 answers