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# Test 1 Study Guide coe 3001 I

Georgia Tech

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This 14 page Study Guide was uploaded by Eleanor Notetaker on Tuesday September 22, 2015. The Study Guide belongs to coe 3001 I at Georgia Institute of Technology taught by Dr. Kennedy in Summer 2015. Since its upload, it has received 246 views. For similar materials see Deformable Bodies in Engineering and Tech at Georgia Institute of Technology.

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Date Created: 09/22/15

COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide Questions 1 A hollow tube with d1 of 3 in and d2 of 5 in has a force P100 ib placed on it If the poissons ratio of the material is 3 and the Young39s modulus is 30000 ksi find the shortening of the tube lateral strain and Ad1 and Adz 61 P 62 2 A joint between two concrete slabs A and B is filled with a exible epoxy that bonds securely to the concrete see figure The height of the joint is h100 mm its length is Ll0 m and its thickness is t12 mm Under the action of shear forces V the slabs displace vertically through the distance d0048 mm relative to each other a What is the average shear strain Vaver in the epoxy b What is the magnitude of the forces V if the shear modulus of elasticity G for the epoxy is 960 MPa 3 Given that in the figure below P100 kN EC2068 GPa Es205 GPa L15 m AC5 cm2 and AS10 cm2 find forces Nc and NS stresses 0 Us and elongation 5 W 39 quot fa inningsWm u quotw gt P1Ljrrjl is t 1 taquot 39f39 u t 393 4 quot39 i 39 w i and 5 r u 7 K quot I I n rn 7 I i COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 4 An aluminum bar AD see figure has a crosssectional area of A250 mm2 and is loaded by forces P17560 N P25340 N and P35780 N The lengths of the segments of the bar are a1525 mm b610 mm and c910 mm a Assuming that the modulus of elasticity E72 GPa calculate the change in length of the bar Does the bar elongate or shorten b By what amount P should the load P3 be increased so that the bar does not change in length when the three loads are applied c If P3 remains at 5780 N what reVised crosssectional area for segment AB will result 5 Find the stress of the bar in the figure below AT0 at X0 while AT30 C at XLLooking at in no change of length when all three loads are applied 3 x the diagram we can see that Tx A TB E Assume that 112 X 106 C and E205 GPa for this type of material 6 A rigid bar of weight W3560 N hangs from three equally spaced wires two of steel and one of aluminum see figure The diameter of the wires is 32 mm Before they were loaded all three wires had the same length What temperature increase AT in all three wires will result in the entire load being carried by the steel wires Assume Es205 GPa xs12 X 10396 C and xa24 x 10396 C COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 7 Based off the figure below find the max shear stress arid angle of twist T50 Nm dA10 mm and dB18 mm 50 mm G36 GPa EI 7 A n a r 7 l V r 8 A stepped shaft ACE is held against rotation at ends A and B and subjected to a torque To acting at section C see figure The two segments of the shaft AC and CB have diameters dA and dB respectively and polar moments of inertia IPA and IpB respectively The shaft has length L and segment AC has length a a For what ratio aL will the maximum shear stresses be the same in both segments of the shaft b For what ratio aL will the internal torques be the same in both segments of the shaft i 9 If the catilever support beam shown below has a force P125 N placed on it with a length L25 m draw the internal moment and shear force diagrams of the beam COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 10 The composite shaft shown in the figure is manufactured by shrinkfitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion The outer diameters of the two parts are d140 mm for the brass core and d250 mm for the steel sleeve The shear moduli of elasticity are Gb36 GPa for the brass and GS80 GPa for the steel a Assuming that the allowable shear stresses in the brass and steel are Tb48 MPa and quot880 MPa respectively determine the maximum permissible torque Tlmx that may be applied to the shaft j COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide Answers 1 2u 5D12x10 6E5D23x10665 a P d 2 d 2 5 2 3 2 A 2 1 217 392 H 2 2 2 2 5080 P 100 39 2 63662 39 O A 6 15708 pquot Note The stress is negative due to the fact that the force is acting down on the object causing it to be a compressive force So it is just as accurate to say the stress is quot 63663 psiquot or quot66663 psi in compressionquot 039 63662 6 oEa gta 2 10 2 2 a E 30000000 X Lateral straining 8 I 6 DzT ns eu 3 2x10 63663u I 6D 6D1 I 39 6 6 gt 60 2 d163663x10 32x10 a D0 d1 1 6D2 39 6 2 gt6D2a d23x10 er 2 Shortening a gt6La L O P Eg ngz 0 E O A 52L g 2 A P E EA EA COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide yavgz 004 V 2384 kN 2 assuming ytany law of small angles d 0000 48 2 004 V Wg t 012 Magnitude of internal shear force Vy GhL004960x10611384kN avg COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 3 NC 33528 kN Ns 66472 kN 062 67056 kPa ms 66472 kPa 62 5 um N L N L N A E 66 6s 66 C as S NS C S S A E ASES ACEC C C Internal forces AS S NCNC ACEC P NC PECAC 1000002068x10905 E33528 EcAcEsAsi 2068x10905205x1091 NS PESAS 100000205x1091 E66472 ECACESAS 2068x10905205x1091 PEC 1000002068x109 67056kPa c ECACESAS 2068x10905205x1091 39 PES 100000205x109 66472kpa OSECACESAS 2068x10905205X1091 2 5pm 5255 PL 10000015 C s EcAcEsAs 2068x10905205x1091 COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 4 622961umAP317503NAAB49103 m2 Total elongation 5HP1P2 P3aP2 P3b P3cl 1 75605340 5780 1525 5340 578061 578091 72x10925 Z Z 2961um 02 P1P2P3aP2P3bP3Cl0 P3abcP1P2 61sz P1P2aP2b 7560 5340 1525 534061 abc 15256191 275303 N P3 A P3 75303 5780 17503 N P3 must increase by 17503 N in order for the bar to feel no elongation 0 A P P P AAB a 1 2 325152575605340 578049103m2 cP3 P2 P3 b 915780 5340 578061 The new crosssectional area should be 49103 mm3 COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 5 021845x 107 Pa RAL EA 6R L 5TxATLf OLA Txdx o 3 X ATxATBF L 3 4 L L 6 2 AT x dxz xAT x xAT T J 3L3 B4L3 0 B4 R L AT L 66R6T0 A H B EA 4 1 RAzzchTBEA gtRA RB from equilibrium NX RA Nix RA 2QATBEi12x 10 630205x1091845x 107 Pa COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 6 A T 89969 C SST elongationE steel bars due 3 heat 55 elongation E steel bars due 3 weight SAT 2 elongation E aluminum die 3 heat 6ST6SP26AT RL x ATchx ATL 2ESA S A A T W 3560 89969 C 2ESAocA ocs 2205x 109H016224x106 12x 10 6 COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 7 Tm 25465x 108 Pa pAB02442 rad Max shear stress 16T16T1650 max 25465x108Pa Hd3 HdA3 11013 max shear stress occurs at dA we know by looking at the figure Twist angle Ipixi3 2dixi4 P4 1 1 d d L T 32T 32TL I dx 4 x 0 GIPX 0 G iabxi 3HGdBdA 1 1 d 7 L 1 L f 4dx 0 abx 3dBdA 325005 3n36x109018 01 11 i01f i018f pABz 202442 rad COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 2 dA 2 IPA 8 L dAdB LIPAIPB Shear stresses T TACdA TCBdB 2110A 21103 L aIP T 2T A AC 0 L aIPAaIPB all T 2T B CB 0 L aIPAaIPB L aI a T0 PA dA T0 PB dB L aIPAaIPB L aIPAaIPB IPA PB L adAadBLdA adA Internal torques L aIP T 2T 2T A 2T AC BC 0 LaIPAaIPB 6111 B L aIPAaIPB 0 L aIPAaIPB COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 9 P125 N laced on it with a len th L25 m RA P 0 RAP125N MA LPO MAz LPz 25125 3125 Nm VxRAP125N MxMA RAxO MxRAx MAPx L125x 25Nm Shear and bending moment diagrams COE 3001 Mechanics of Deformable Bodies Test 1 Study Guide 10 Tmax15212Nm P1p2 Gprb T T b 0 GbIPbGSIP GsIP T T 0 GbIPbGSIP S GbIPb9O478N mZGSIPS28981Nm2 solid16T Gb Pb maxsolid d3 quotWk GbIPbGSIPS 16 T Tm dnd3 GprbGsts 48x106H043 9047828981 max 16 Gb Pb 16 90478 325353 Nm hollow quotmquot Hd24 dl4 GSIPS Tmaxhollow d24d14 quotW GbIPbGSIP 16d2 TmathZlowHd24 d14 GbIPGSIP T b s quotW5 16d2 GSIP Z 80x106 054 044 9047828981 15212Nm 1605 28981 15212 lt 25353 Tmax is 15212

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