Question

provide a detailed solution showing your work step by step

Do not provide a solution using axial deformation, this class is machine elements and not mechanics of materials

IS NOT THE SOLUTION TO THIS PROBLEM

solution is provided as -28.38mm

PL δ= AE Problem 2 Ans: -28.38mm A solid plastic rod [E = 45MPa; v = 0.33] with a diameter d = 100mm is placed in a hole with diameter D = 101mm with rigid walls. The rod has a length L = 400mm. Determine the change in length of the rod after a load P 32kN is applied. Rigid walls Tube

Answer 1

Piobleu Giuen th Aatic od EP5MP. A slid VE.33 . diaueex d)= 100 ew A hole wiHR diauelkr D- lo1 w. L Do0 w, has a The Yod have to dexuine R chosge Novo we load P= 52 a OA H& 1od apfen qiveu ld)= l00 w qiveu P 3e kN. boye epleuow 3.bst:kaleba 32 10 G Co. 2 32 X1O 32 X10 0.031 32X 10 4.85 X10

IP.0T643 3 MPa. sHvion -e oe tnous IP.036 3S MPa P5 MPa. above cpen. Ing sdudanap 34vian -p.076433 ai-dundenal P-O.090 5871P. straik lalum) diawestor alerap shain Max chonge eigina hileraf) diawestior, lDexa) stroi= latkxa oir - -0.01. w hase to dkwhe H deuge Nao angiluamp a<leraf siraiu rain- L slhaiu-o0905814, Pready dlogide we Jaberal strain = -o.0. bove espreion bA: kude e bitt wfies 4vain=-o.090587 P - 2Eo.01)

O05874P + 02 -O.07058p . Eve) s'gu tdicade shot qiveu L 0O dear XPOD O OTO 5&TL 17 8--28.23 냐96. -28 23 nu louge