Euler's Physics (Essay Sample)

Question 1 Assignment 1
Question 1a
Pin-ended Strut
15798806350P00P
Using Euler’s Formula to calculate load likely to cause buckling, PCr=EIπ2L (Freund, 1990)
PCr=200×109 Iπ22=9.872I×1011
I=AL212
I=π×0.03242212=0.00023565M4
PCr=232.64MN
For a load with eccentricity,e=0.007m the maximum load inducing a compressive stress of 200mpa can be calculated as follows;
742950101600WMPeP00WMPeP
Balancing the moments on the free-body diagram it gives (Hill, 1950)
M=P(W+e)
The governing equation for the strut’s transverse displacement w can be illustrated as;
d2ydx2+PEIW=PeEI
M 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-e
Wherek2=PEI. This is denoted as ” 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 conditions are,
W0=wL=o
The solution for the column's displacement is therefore,
W=etankL2sinkx+coskx-1
From this the secant formula is derived to give the maximum value of the displacement as in the equation below
Wmax=eseckl2-1
The maximum stress is therefore given by;
σmax =PA1+ecr2secPLEA2r
Note 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;
σmax =20mpa
E=200GPa
A=Ï€d24
L=2m
d=30mm
e=7mm
We are required to determine P in this equation
r=IA, r is the radius of gyration
From which;
c=7+15=22mm=0.022m
A=π×0.0324=0.00070695m2
r=0.00023565M40.00070695m2 =0.5773503m
20×106=P0.000706951+0.007×0.0220.5773503×0.5773503secP×2200×109 ×0.00070695×2×0.5773503
20×106=P0.000706951+0.000462sec1.22502P×108
11068.06216
20×106=P0.000706951+0.000462sec11068.06216×P
20×106=P0.000706951+0.000462cos11068.06216×P
20×106=P0.000706951-0.00047463cosP
20×106=P0.00070695-0.67138PcosP
14139=P-0.000474634PcosP
14139cosP=PcosP-0.000474634P
PcosP-14139cosP-0.000474634P=0
Using 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 approximations
P

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 approximations
400

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