Question

3. Show that the length of a bar subjected to an axial load P is AL = PL/AE. 4. A hollow copper cylinder (OD = 10 in and ID = 5 in) contains a solid steel core. Determine the compressive stresses in the steel and the copper when a force of 200,000 lb is applied with a press.

Answer 1

Anx 3 - le A bar Force where of uniform cross- xectional area (A) is applied axially as shown in figure - Axial force = length of the bar A = Uniform Croft - fectional are a directly According to Hookiq how the street is proportional to strain. then Stress (o) a strain (E) TLE > E= Modulus of clasticity T = EE change in length Di hi , e original length then 1 - Ex el

Ane u niven copper Outer diameter (D)= 10 in Inner diameter (d) = 5 In Compressive force (P)= 200000 Ib two componenta, then Sectional areas of Area of the steel core (As) = A (5)2 = 4+25 = 19.63 (In)? Area of the copper (Ac) = ( 102-32) a58.go (In) 2 steel and copper the to be steel a the component and let Js streeree streexer in in strain & in Since the combo henta E- equal = EE sa e a Total load(ele on copper a + load on steel and load o f = P- JA Tex Acap Ts x As + -nin + Tox S&g= 200000 Ist 19.63

from equation Tsx 19.63 + and @ Ec) 589 = 200000 we know that modulux of clasticity for &teel = 29000 000 pxi - 29x to 6 pai copper = 16 7 x 100 pri for TS X 16.7x106 = 200000 200 Ts x 19.63 + R9X16 " Tex 20.21= 20oooo des 200000 0. 21 Ts = gf g 2.12 P8i thin s & Ec to= Es 9292. 12 x 16 7 x ot 29 x 106 =5699.95 Pei