Eulers Physics Essay - 275 Words

Euler's Physics (Essay Sample) Content: Question 1 Assignment 1Question 1aPin-ended Strut15798806350P00PUsing Eulerà ¢Ã¢â€š ¬s Formula to calculate load likely to cause buckling, PCr=EIà Ã¢â€š ¬2L (Freund, 1990)PCr=200ÃÆ'109 Ià Ã¢â€š ¬22=9.872IÃÆ'1011I=AL212I=à Ã¢â€š ¬ÃƒÆ'0.03242212=0.00023565M4PCr=232.64MNFor a load with eccentricity,e=0.007m the maximum load inducing a compressive stress of 200mpa can be calculated as follows;742950101600WMPeP00WMPePBalancing the moments on the free-body diagram it gives (Hill, 1950)M=P(W+e)The governing equation for the strutà ¢Ã¢â€š ¬s transverse displacement w can be illustrated as;d2ydx2+PEIW=PeEIM was eliminated using Euler-Bernoulli Theory.The above equation contains a non-homogeneous term -Fe/EI and its general solution is (Lubliner, 1990),Wx=Asinkx+Bcoskx-eWherek2=PEI. This is denoted as à ¢Ã‚  in the formula sheet. This analysis will use k instead.The coefficients A and B depend on the boundary conditions. For a simply supported column the boundary c onditions are,W0=wL=oThe solution for the column's displacement is therefore,W=etankL2sinkx+coskx-1From this the secant formula is derived to give the maximum value of the displacement as in the equation belowWmax=eseckl2-1The maximum stress is therefore given by;à Ã†â€™max =PA1+ecr2secPLEA2rNote c is denoted as y in the formula sheet and is the distance from centroidal axis to the extreme fiber on the concave side of the column.For this question we are given;à Ã†â€™max =20mpaE=200GPaA=à Ã¢â€š ¬d24L=2md=30mme=7mmWe are required to determine P in this equationr=IA, r is the radius of gyrationFrom which;c=7+15=22mm=0.022mA=à Ã¢â€š ¬ÃƒÆ'0.0324=0.00070695m2r=0.00023565M40.00070695m2 =0.5773503m20ÃÆ'106=P0.000706951+0.007ÃÆ'0.0220.5773503ÃÆ'0.5773503secPÃÆ'2200ÃÆ'109 ÃÆ'0.00070695ÃÆ'2ÃÆ'0.577350320ÃÆ'106=P0.000706951+0.000462sec1.22502PÃÆ'10811068.0621620ÃÆ'106=P0.000706951+0.000462sec11068.06216ÃÆ'P20ÃÆ'106=P0.000706951+0.000462cos11068.06216ÃÆ'P20ÃÆ'106=P0.000706 951-0.00047463cosP20ÃÆ'106=P0.00070695-0.67138PcosP14139=P-0.000474634PcosP14139cosP=PcosP-0.000474634PPcosP-14139cosP-0.000474634P=0Using a simple spread sheet program, the value of P can be estimated. The following are some instances of the program results:Iteration for the first three approximationsP P cosP Pcos(P) 14139cosP 0.000474634P Sum 100 10 -0.839071 -83.90715291 -11863.63235 0.0474634 11779.67773 110 10.48808848 -0.485981 -53.45796029 -6871.291823 0.05220974 6817.781653 120 10.95445115 -0.041112 -4.933385736 -581.2761743 0.05695608 576.2858325 Further approximations400 20 0.408082062 163.2328247 5769.872272 0.1898536 -5606.829301 410 20.24845673 0.171050237 70.13059705 2418.479297 0.19459994 -2348.5433 420 20.49390153 -0.07348299 Euler's Physics Essay - 275 Words Euler's Physics (Essay Sample) Content: Question 1 Assignment 1Question 1aPin-ended Strut15798806350P00PUsing Eulerà ¢Ã¢â€š ¬s Formula to calculate load likely to cause buckling, PCr=EIà Ã¢â€š ¬2L (Freund, 1990)PCr=200ÃÆ'109 Ià Ã¢â€š ¬22=9.872IÃÆ'1011I=AL212I=à Ã¢â€š ¬ÃƒÆ'0.03242212=0.00023565M4PCr=232.64MNFor a load with eccentricity,e=0.007m the maximum load inducing a compressive stress of 200mpa can be calculated as follows;742950101600WMPeP00WMPePBalancing the moments on the free-body diagram it gives (Hill, 1950)M=P(W+e)The governing equation for the strutà ¢Ã¢â€š ¬s transverse displacement w can be illustrated as;d2ydx2+PEIW=PeEIM was eliminated using Euler-Bernoulli Theory.The above equation contains a non-homogeneous term -Fe/EI and its general solution is (Lubliner, 1990),Wx=Asinkx+Bcoskx-eWherek2=PEI. This is denoted as à ¢Ã‚  in the formula sheet. This analysis will use k instead.The coefficients A and B depend on the boundary conditions. For a simply supported column the boundary c onditions are,W0=wL=oThe solution for the column's displacement is therefore,W=etankL2sinkx+coskx-1From this the secant formula is derived to give the maximum value of the displacement as in the equation belowWmax=eseckl2-1The maximum stress is therefore given by;à Ã†â€™max =PA1+ecr2secPLEA2rNote c is denoted as y in the formula sheet and is the distance from centroidal axis to the extreme fiber on the concave side of the column.For this question we are given;à Ã†â€™max =20mpaE=200GPaA=à Ã¢â€š ¬d24L=2md=30mme=7mmWe are required to determine P in this equationr=IA, r is the radius of gyrationFrom which;c=7+15=22mm=0.022mA=à Ã¢â€š ¬ÃƒÆ'0.0324=0.00070695m2r=0.00023565M40.00070695m2 =0.5773503m20ÃÆ'106=P0.000706951+0.007ÃÆ'0.0220.5773503ÃÆ'0.5773503secPÃÆ'2200ÃÆ'109 ÃÆ'0.00070695ÃÆ'2ÃÆ'0.577350320ÃÆ'106=P0.000706951+0.000462sec1.22502PÃÆ'10811068.0621620ÃÆ'106=P0.000706951+0.000462sec11068.06216ÃÆ'P20ÃÆ'106=P0.000706951+0.000462cos11068.06216ÃÆ'P20ÃÆ'106=P0.000706 951-0.00047463cosP20ÃÆ'106=P0.00070695-0.67138PcosP14139=P-0.000474634PcosP14139cosP=PcosP-0.000474634PPcosP-14139cosP-0.000474634P=0Using a simple spread sheet program, the value of P can be estimated. The following are some instances of the program results:Iteration for the first three approximationsP P cosP Pcos(P) 14139cosP 0.000474634P Sum 100 10 -0.839071 -83.90715291 -11863.63235 0.0474634 11779.67773 110 10.48808848 -0.485981 -53.45796029 -6871.291823 0.05220974 6817.781653 120 10.95445115 -0.041112 -4.933385736 -581.2761743 0.05695608 576.2858325 Further approximations400 20 0.408082062 163.2328247 5769.872272 0.1898536 -5606.829301 410 20.24845673 0.171050237 70.13059705 2418.479297 0.19459994 -2348.5433 420 20.49390153 -0.07348299