Question

9) The aluminum rod AB and bronze rod BC are joined at the collar B and fixed connected at their ends. An axial load, P, is applied at point B. Neglect any effects of the collar at B itself. The cross-sectional area of cach rod is A. Determine the reactions (with respect to the reaction vectors shown). Given: Pg = 20 N (applied at point B) EAB-73 GPa (aluminum); EBc 103 GPa (bronze) Cross-sectional areas: AAB = 250 mm: ABC-100 mm2 10mn 1 2m RA Вс Determine the reactions (with respect to the reaction vectors shown). A) R,--5.2 N, Re-148 N C) RA -20 N, Re-ON B) RA 14.8 N, Rc 14.8 N D) RA-5.2N, Rc 14.8 N

Answer 1

consider after the application of P_B, section AB is under compression, and section BC is under tension. Due to tension in section BC and compression in section AB, there would be change in length of section AB and BC.

$\delta _{AB}$ = change length of section AB.

$\delta _{BC}$ = change in length of section BC.

  

R_A + R_C = P_B,

Net compressive force acting on AB = R_A ,

Net tensile force acting on BC = R_C

we know that $\delta _{BC}=\delta _{AB}$ ,

$\therefore \frac{R_{A}L_{AB}}{A_{AB}E_{AB}}=\frac{R_{C}L_{BC}}{A_{BC}E_{BC}}$

$\frac{(P_{B}-R_{C})L_{AB}}{A_{AB}E_{AB}}=\frac{R_{C}L_{BC}}{A_{BC}E_{BC}}$

$R_{C}=\frac{20\times 100\times103\times10}{250\times73\times2+100\times103\times10}$

R_C=14.767 N, R_A= P_B -R_C = 5.23 N,