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UCI ENGRMAE 150 - SolutionProject2

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Project 2 SolutionFAMGiven:σij=8 2 −52 −4 3−5 3 6ksi• Direction of the element of design (weld):lxn=1√2, lyn=12, lzn= negative• Design is based on the normal stress acting on the weld• Max normal and shear stresses in the whole structure and safe• Tables includes several materialsRequired:The selection of an appropriate material to meet design requirementsAssumptions:uniform stresses, isotropic, elastic materialsApproach:•l2xn+ l2yn+ l2zn= 1• Calculate the normal stress acting on the weld• Calculate principal normal and shear stresses• Using the value of the normal stress on the weld along with those of the principalnormal and shear stresses to select an appropriate material that is ductile enough and thatcan keep the structure safeAnalysis:Let’s divide the analysis into the following items11 (a) lznl2xn+ l2ynl2zn= 112+14+ l2zn= 1l2zn=14(1)but since lznis negativethen lzn= −122 (b) Normal stress on the weldUse equation (1.14) in FAM’s noteσnr= σxxlxnlxr+ σyxlynlxr+ σzxlznlxr+ σxylxnlxr+ σyylynlyr+ σzylznlyr+ σxzlxnlzr+ σyzlynlzr+ σzzlznlzr(2)replace r in the above equation by nσnn= σxxlxnlxn+ σyxlynlxn+ σzxlznlxn+ σxylxnlxn+ σyylynlyn+ σzylznlyn+ σxzlxnlzn+ σyzlynlzn+ σzzlznlzn(3)using the values of stresses in the above tensor and the values of direction cosinesσnn=82+−44+64+2√2+ 2−34+ 252√2(4)σnn= 7.95 ksi (5)3 (c) principal stressesσ3p− 3σIσ2p−13σIIσp− σIII(6)Calculate σI, σII, σIIIσI=13σx+ σy+ σz=138 + (−4) + 6=103ksi (7)σII= 3− σxσy− σyσz− σzσx+ τ2xy+ τ2yz+ τ2zx= 3− 8(−4) − (−4)6 − (8)(6) + 22+ 32+ (−5)2σII= 138 ksi2(8)2σII= σxσyσz+ 2τxyτyzτzx− σzτ2xy− σxτ2yz− σyτ2zx= −248 ksi3(9)f(σp) = σ3p− 10σ2p− 46σp+ 248 (10)plot f(σp) against σp-10 -5 0 5 10 15σp (ksi)-1500-1000-50005001000f(σp)Figure 1:From figure 1, the three roots areσp1= 12.107 ksi, σp2= 3.59 ksi, σp1= −5.7 ksi (11)The largest principal normal stress is σ = 12.1 ksiThe largest principal shear stressτ31=−5.7 − 12.12=17.82= 8.9 ksi (12)now using σp1= 12.1 ksi and τ31= 8.9 ksi along with σnn= 7.95 ksi and the data inthe table, only materials B,D,E pass the inspection. Material E is too expensive. MaterialB is too brittle. Therefore we choose:MAT ERIAL D34 AppendixFrom the cosine we knowlxn=1√2, lyn=12, lzn= −12(13)Replace r by n for each component we getσxxlxnlxn=82= 4σyxlynlxn= 2121√2=1√2σzxlznlxn= −5−121√2=52√2σxylxnlyn= 2121√2=1√2σyylynlyn=−44= −1σzylznlyn= 3−121√2=−34σxzlxnlzn= −5−121√2=52√2σyzlynlzn= 3−121√2=−34σzzlznlzn= 61212=64(14)so the normal component σnnisσnn= 3 +7√2= 7.95


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UCI ENGRMAE 150 - SolutionProject2

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