Question

In the assembly shown below, the three-piece rod is made of steel, E = 200 GPa. If the length of section M and N is LM=LN = 400 mm, and their cross-section areas are, AM = 40 mm?, An = 60 mm², determine: 1. how much force P should be to just close the gap S= 4 mm, and 2. if we double the force P, find the reactions at points A and B, and determine the displacement of point C (in the middle of rod 2) in this case. Note: You must construct element matrices and assemble them to solve the problem. Divide each rod into two elements and solve the problem. You can use the node numbers arbitrarily. Apply B.C. as instructed. ܠܠܠ N . : M

Answer 1

Given, E = 200 Gea m=Ly=400mm = 4mm AM = 40mm² AN= 60 mm² - + it it - VA X, For us finding forces in various elements let cut each setions one by one. FER=0 we know For=P , FAD = P. that. Rotal = DAD + doet (FFE=0) 0 0s, 4516: t .. (1987 4X AD or, 4x 163 - Px0.4 40x10 200x109 Px0,4 Göxlox200x109 P- 48000 N= 48 KN.

(2.) If p is doubled P'= 96kN. then If this Force will act then the point will touch the E be point B and there wil no gap. O B det, RA and RB be the syns as shown in. the sig. & RAT RB = P - : d=0 - dag + DE + DED=0 ors or * (184**P)*) AD AE JEB or (RAX 0.4 ( 40x 200 10 RAX 0.4 +/(RA- 96000) 0,4)=0 40*200* 10 bor 200X163 RAF 36000N = 36 KN RB= 60000N = 60 KN.

Displacement of point ce consider AO ,DE and EC parts of section de = SAD + DE + geo ("F&c=0) de = (PE) A+ ( 2 ) DE 3 / 96000x0.4 1 40 x 10 x 200) 196000x 0-4 (60 x 103 x 200 de = 8x153 m² 8mm