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УДК 539.3
MODELING STABLE EQUILIBRIUM SHAPE RECTILINEAR ROD AT ITS COMPRESSION AND BENDING
A. A. Chistyakov, M. S. Strelchenko Scientific Supervisor - R. A. Sabirov
Reshetnev Siberian State Aerospace University 31, Krasnoyarsky Rabochy Av., Krasnoyarsk, 660037, Russian Federation E-mail: [email protected]
Completed physical experiment of compression and bending of the rod, with the deflection measurements at individual points. Found deflection function using the Lagrange interpolation formula, normalized form is attached for solving boundary value problem of longitudinal-transverse bending in the differential setting. We calculate the longitudinal force at a given point of the rod.
Keywords: longitudinal force, Hooke's law, Resistance of materials.
МОДЕЛИРОВАНИЕ УСТОЙЧИВОЙ РАВНОВЕСНОЙ ФОРМЫ ПРЯМОЛИНЕЙНОГО СТЕРЖНЯ ПРИ ЕГО СЖАТИИ И ИЗГИБАНИИ
А. А. Чистяков, М. С. Стрельченко Научный руководитель - Р. А. Сабиров
Сибирский государственный аэрокосмический университет имени академика М. Ф. Решетнева Российская Федерация, 660037, г. Красноярск, просп. им. газ. «Красноярский рабочий», 31
Е-mail: [email protected]
Выполнен физический эксперимент сжатия и изгиба стержня, с замерами прогиба в отдельных точках. Найдена функция прогиба с использованием интерполяционной формулы Лагранжа, нормированный вид которой приложен для решения краевой задачи продольно-поперечного изгиба в дифференциальной постановке. Вычислена продольная сила в заданной точке стержня.
Ключевые слова: продольная сила, закон Гука, сопротивление материалов.
We compress elastic ruler, the length of which is equal to L = 0,3 m. Taking the origin at the left edge of the ruler, measure the deflection v = v(x) on the interval 0 < x < L in the points
v1 = v(L/5) = 0,8• 10-2m; v2 = v(2L/5) = 2,4-10-2m;
v3 = v(3L/5) = 3,3-10-2m.; v4 = v(4L /5) = 2,3-10-2m.
At the ends v0 = v(0) = 0 and vL = v(L) = 0 . Note that these deflections were measured manually and have a certain error. Required to find the longitudinal force that occurs in this range.
We find the deflection function. To do this, we define a deflection range in the form of the interpolation polynomial.
v( x) = a0 + ax x + a2 x2 +a3 x3 +a4 x4 +a5 x5 +a6 x6 +a7 x7 (1)
and write for (1) boundary conditions: v0 = 0; dv(x) / dx = 0, which give, respectively, a0 = 0 and ax = 0 ; On the right side x = L deflection is v(L) = 0 and the bending moment M(L) = 0 is absent, from which it
follows that d 2v / dx2 = M (L) / EJ = 0 [1].
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We calculate the coefficients (1), satisfying the boundary conditions and deflections in the points with the specified v1 - v4. Set up a system of equations
L2 L3 L4 L5 L6 L7
25 125 625 3125 15625 78125
4L2 8L3 16L4 32L5 792L6 128L7
25 125 625 3125 15625 78125
L2 27 L3 81L4 243L5 729L6 2187L7
25 125 625 3125 15625 78125
L2 64L3 256L4 1024L5 4096L6 16384L7
25 125 625 3125 15625 78125
L2 L3 L4 L5 L6 L7
2 6 L 12 L2 2° L 30L4 42L5
a2 v1
a3 v2
a4 v3
a5 v4
a6 v(L)
a7 d v / dx
x=L
(2)
Normalize the right side of (2), dividing the deflections of ruler by v3 = 3,3 • 10 m; we obtain a dimensionless vector of right side. The solution of system of equations (2) gives form to the elastic line
v(x) = 9,81 • 10-3x2 -5,3• 10-4x3 + 4,3•lO-5x4 -3,2• 10-6x5 + 9,74• 10-8x6 -M0-9x7 , (3)
(picture 2)
We write the equation of equilibrium for a deformed line element (picture 3), where the sum of the moments about the point, gives
M (x) + N (0)v( x) = 0. Presenting the ruler as a girder, adding (3) Hooke's law and geometric equation, we get
(4)
EJ ~~v = N (°)v( x).
dx
(5)
Differentiating (5) twice, we have the differential equation of equilibrium for any contour conditions.
d 4v
EJ— = N (0)
dx
d 2v
dx
2 '
(6)
Fig. 3 Deformed line
Substituting the second and fourth derivatives of (3) in equation (6). At a given point is obtained from (6) the value of the longitudinal force.
N(0) = 582,82EJ . (7)
Cross-sectional dimensions of ruler 0,03м x 1,5 -10-3 m , Young's modulus E = 650MPa, then J = 8,43 -10-12 m4; EJ = 0,00548Nm2 and
N(0) = 0,018N. (8)
This force value can be compared with the value of critical Euler's force for the reduced length of the rod, equal Lpr = 0,7 L = 0,21m
n2 FJ
Pr =nLFL = 223,8EJ .
Lpr
Comparing (7) and (9) to note that the strength of the same order, and are distinguished by a factor of 2.6, which is satisfactory. If you create a special mechanical support to hold the line, and do not hold it out as shown in the picture (Pic. 1), and more precisely measure the deflection, then the value of the forces will be found more accurately.[1-3]
Библиографические ссылки
1. Сабиров Р. А. Расчет устойчивости пластины вариационно-разностным методом от действия сил инерции // Актуальные проблемы авиации и космонавтики : сб. науч. ст. Красноярск, 2014. С.31-34.
2. Тимошенко С. П. Устойчивость стержней, пластин и оболочек. М. : Наука, 1971. 807 с.
3. Пановко Я. Г., Губанова И. И. Устойчивость и колебания упругих систем: Современные концепции, ошибки и парадоксы. 3-е изд., перераб. М. : Наука. Гл. ред. физ.-мат. лит., 1979, 384 с.
© Чистяков А. А., Стрельченко М. С., 2016
