Method of Members | Frames Containing Three-Force Members

A three-force member is in general a non-axial member that is not simply in tension or compression. A member of this kind has shear forces perpendicular to the member and subjected to bending loads. If forces are applied to more than two positions on the member, it is three-force member. Any beam is a three-force member according to the above definition.
 

Frames are pin-connected structures with some or all members are three-force members. To analyze a frame, we can disconnect the three-force member from the structure and draw the free-body diagram of the member. This approach is called the method of members.
 

Elimination of Arbitrary Constants

Properties

  • The order of differential equation is equal to the number of arbitrary constants in the given relation.
  • The differential equation is consistent with the relation.
  • The differential equation is free from arbitrary constants.

Example
Eliminate the arbitrary constants c1 and c2 from the relation   $y = c_1 e^{-3x} + c_2 e^{2x}$.
 

Solution

Problem 05 | Substitution Suggested by the Equation

Problem 05
$dy/dx = \sin (x + y)$
 

Solution 05
$dy/dx = \sin (x + y)$

$dy = \sin (x + y)~dx$
 

Let
$z = x + y$

$dz = dx + dy$

$dy = dz - dx$
 

$dz - dx = \sin z~dx$

$dz = \sin z~dx + dx$

$dz = (\sin z + 1)~dx$

$\dfrac{dz}{\sin z + 1} = dx$

$\dfrac{dz}{1 + \sin z} \cdot \dfrac{1 - \sin z}{1 - \sin z} = dx$

$\dfrac{(1 - \sin z)~dz}{1 - \sin^2 z} = dx$

$\dfrac{(1 - \sin z)~dz}{\cos^2 z} = dx$

$\dfrac{~dz}{\cos^2 z} - \dfrac{\sin z~dz}{\cos^2 z} = dx$

Problem 04 | Bernoulli's Equation

Problem 04
$y' = y - xy^3e^{-2x}$
 

Solution 04
$y' = y - xy^3e^{-2x}$

$\dfrac{dy}{dx} - y = -xe^{-2x}y^3$

$dy - y~dx = -xe^{-2x}y^3~dx$       → Bernoulli's equation

$dy + Py~dx = Qy^n~dx$
 

From which
$P = -1$

$Q = -xe^{-2x}$

$n = 3$

$(1 - n) = -2$

$z = y^{1 - n} = y^{-2}$
 

Integrating factor,
$u = e^{(1 - n)\int P~dx} = e^{-2\int (-1)~dx}$

$u = e^{2\int dx} = e^{2x}$
 

Thus,
$\displaystyle zu = (1 - n)\int Qu~dx + C$

Problem 03 | Substitution Suggested by the Equation

Problem 03
$dy/dx = (9x + 4y + 1)^2$
 

Solution 03
$dy/dx = (9x + 4y + 1)^2$

$dy = (9x + 4y + 1)^2~dx$
 

Let
$z = 9x + 4y + 1$

$dz = 9~dx + 4~dy$

$dy = \frac{1}{4}(dz - 9~dx)$
 

$\frac{1}{4}(dz - 9~dx) = z^2~dx$

$dz - 9~dx = 4z^2~dx$

$dz = 4z^2~dx + 9~dx$

$dz = (4z^2 + 9)~dx$

$\dfrac{dz}{4z^2 + 9} = dx$

$\dfrac{dz}{(2z)^2 + 3^2} = dx$

$\displaystyle \dfrac{1}{2} \int \dfrac{2dz}{(2z)^2 + 3^2} = \int dx$

Problem 02 | Substitution Suggested by the Equation

Problem 02
$\sin y(x + \sin y)~dx + 2x^2 \cos y~dy = 0$
 

Solution 02
$\sin y(x + \sin y)~dx + 2x^2 \cos y~dy = 0$
 

Let
$z = \sin y$

$dz = \cos y~dy$
 

Hence,
$z(x + z)~dx + 2x^2~dz = 0$       → homogeneous equation
 

Let
$z = vx$

$dz = v~dx + x~dv$
 

$vx(x + vx)~dx + 2x^2(v~dx + x~dv) = 0$

$x^2(v + v^2)~dx + 2vx^2~dx + 2x^3~dv = 0$