Question

Two brass rods AB and BC, each of uniform diameter, will be
brazed together at B to form a nonuniform rod of total length
100 m which will be suspended from a support at A as shown.
Knowing that the density of brass is 8470 kg/m3, determine (a) the
length of rod AB for which the maximum normal stress in ABC is
minimum, (b) the corresponding value of the maximum normal
stress.

Answer 1

A AB= IxQs) 2 176.71mm² = 176.71810-6 m2 ABC= x (100² 78. Symm² 78 suxirt m² b=loo-a WAB - PgAARLAB-8470x 9,81126.71410bxq = 14,6830 Weca Bog Archoce 8400X9.8.46780 S4X1062x[100ng) WBC 652.59-6.526a PA- WABTWBC 14.6830 + 652.59-6-526a GA = PA 3.6930x106 + 46.160 x 100 a ААВ PB = WBC= 652.59-6.526a obate 65%.59-6.52 a 38030 904106-83.2X103 al ABC - 78.5uxlot Length of rod AB GA-69 lett: 3.69371064 (46.760x10aa (8.30x66 *83.1x1039) Ta=35.71m Ano 6A2366930x106+6460/60-X1032x (35.7D = B 534m palet B-863090 X/06+(-83.904103)x(35.7122 5.34 Mpala

Answer 2

Answer 3

For rod AB

At B we get normal stress due to weight of rod BC

At A we get normal stress due to weight of full rod (AB+BC)

so we can say max stress is at A and min stress at B

then inorder to have min stress we have to consider stress at B should be max normal stress in the rod AB

thats why stress at A=B

Note- Here we are considering min stress at B because it is the weakest point in the rod ABC