Stress An aluminium circular rod of length L = 420 mm has one third of its lengt
ID: 1862197 • Letter: S
Question
Stress
An aluminium circular rod of length L = 420 mm has one third of its length hollow, as seen in Figure 1. The external diameter of the rod is 60 mm and the inner diameter of the inner part is 35 mm. A strain gauge is fixed to the hollow part (axial direction) and the whole rod is compressed by a force P. a) if the strain measured on the hollow section is epsilon h = 470 times 10-6 what is the strain in the solid section of the rod? b) What is the total shortening delta of the bar? If the compressive stress in the rod cannot exceed 48 MPa what is the maximum allowable value of force P?Explanation / Answer
a)force P = areaof hollow part*strain of hollow part =areaof solid part *strain of solid part
=>strain of solid part = (area of hollwo part/area of solid part)*(strain of hollow part)
=((3600-1225)/3600)*(470*10^-6)
= 310*10^-6
b)total shortening =strain of hollow bar*L/3 + strain of solid bar*2l/3 =140*( 470+620)*10^-6 =1090*140*10^-6 =152.6*10^-3mm
c)maxmimum allowable force =maximumallowable force*hollow area =48*10^6 *3.14*10^-4 *(3600-1225) =35796000N