August 2011

Problem 03 | Division by t

Problem 03
Find the Laplace transform of   $f(t) = \dfrac{\sin^2 t}{t}$.
 

Solution 03
$f(t) = \dfrac{\sin^2 t}{t}$

$f(t) = \dfrac{\frac{1}{2}(1 - \cos 2t)}{t}$

$f(t) = \dfrac{1}{2} \left[ \dfrac{1}{t} - \dfrac{\cos 2t}{t} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2} \mathcal{L} \left[ \dfrac{1}{t} - \dfrac{\cos 2t}{t} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2} \mathcal{L} \left( \dfrac{1}{t} \right) - \dfrac{1}{2} \left( \dfrac{\cos 2t}{t} \right)$
 

Since

Problem 04 | Division by t

Problem 04
Find the Laplace transform of   $f(t) = \dfrac{\cos 4t - \cos 5t}{t}$.
 

Solution 04
$f(t) = \dfrac{\cos 4t - \cos 5t}{t}$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left[ \dfrac{\cos 4t - \cos 5t}{t} \right]$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left[ \dfrac{\cos 4t}{t} \right] - \mathcal{L} \left[ \dfrac{\cos 5t}{t} \right]$
 

Since
$\mathcal{L} (\cos bt) = \dfrac{s}{s^2 + b^2}$
 

Then,

Analysis of Structures

There are many kinds of structure. This section will limit to those that are pin-connected. Two types of pin-connected structures will be covered here; pin-connected trusses and pin-connected frames. In the actual structure, the joints may be welded, riveted, or bolted to a gusset plate at the joint. However as long as the center-line of the member coincide at the joint, the assumption of a pinned joint maybe used.
 

Method of Joints | Analysis of Simple Trusses

Method of Joints
The free-body diagram of any joint is a concurrent force system in which the summation of moment will be of no help. Recall that only two equilibrium equations can be written
 

$\Sigma F_x = 0$   and   $\Sigma F_y = 0$

 

Problem 001-mj | Method of Joints

Problem
Find the force acting in all members of the truss shown in Figure T-01.
 

001-3m-truss-given.gif

 

Problem 403 | Method of Joints

Problem 403
Determine the force in each bar of the truss shown in Fig. P-403.
 

Tower-like truss

 

Method of Sections | Analysis of Simple Trusses

Method of Sections
In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces we wish to determine. This method permits us to solve directly any member by analyzing the left or the right section of the cutting plane.
 

To remain each section in equilibrium, the cut members will be replaced by forces equivalent to the internal load transmitted to the members. Each section may constitute of non-concurrent force system from which three equilibrium equations can be written.
 

Problem 001-ms | Method of Sections

Problem 001-ms
From the truss in Fig. T-01, determine the force in mebers BC, CE, and EF.
 

001-3m-truss-given.gif

 

Problem 002-mj | Method of Joints

Problem 002-mj
The structure in Fig. T-02 is a truss which is pinned to the floor at point A, and supported by a roller at point D. Determine the force to all members of the truss.
 

Warren truss by method of joints

 

Problem 002-ms | Method of Sections

Problem 002-ms
The roof truss shown in Fig. T-03 is pinned at point A, and supported by a roller at point H. Determine the force in member DG.
 

Roof truss by method of sections

 

Problem 003-ms | Method of Sections

Problem 003-ms
The truss in Fig. T-04 is pinned to the wall at point F, and supported by a roller at point C. Calculate the force (tension or compression) in members BC, BE, and DE.
 

Cantilever truss by method of sections
Figure T-04

 

Problem 003-mj | Method of Joints

Problem 003-mj
Find the force in each member of the truss shown in Fig. T-04.
 

005-cantilever-truss.gif

 

Problem 004-ms | Method of Sections

Problem 004-ms
For the truss shown in Fig. T-05, find the internal fore in member BE.
 

Right Triangular Truss

 

Problem 004-mj | Method of Joints

Problem 004-mj
The truss pinned to the floor at D, and supported by a roller at point A is loaded as shown in Fig. T-06. Determine the force in member CG.
 

Truss with rectangular arrangement of members at the center

 

Problem 005-ms | Method of Sections

Problem 005-ms
The structure shown in Figure T-07 is pinned to the floor at A and H. Determine the magnitude of all the support forces acting on the structure and find the force in member BF.
 

Frame with members under axial force only

 

Problem 005-mj | Method of Joints

Problem 005-mj
Compute the force in all members of the truss shown in Fig. T-08.
 

Five-meter high truss

 

Problem 01 | Linear Equations

Problem 01
$(x^5 + 3y) \, dx - x \, dy = 0$
 

Solution 01
$(x^5 + 3y) \, dx - x \, dy = 0$

$\dfrac{(x^5 + 3y) \, dx}{x \, dx} - \dfrac{x \, dy}{x \, dx} = 0$

$\dfrac{x^5 + 3y}{x} - \dfrac{dy}{dx} = 0$

$x^4 + \dfrac{3y}{x} - \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} - \dfrac{3}{x}y = x^4$       → linear in y
 

Hence,
$P = -\dfrac{3}{x}$

$Q = x^4$
 

Integrating factor:
$e^{\int P \, dx} = e^{\int \left( -\frac{3}{x} \right)dx}$

$e^{\int P \, dx} = e^{-3\int \frac{dx}{x}}$

Problem 02 | Linear Equations

Problem 02
$2(2xy + 4y - 3) \, dx + (x + 2)^2 \, dy = 0$
 

Solution 02
$2(2xy + 4y - 3) \, dx + (x + 2)^2 \, dy = 0$

$2[ \, 2y(x + 2) - 3 \, ] \, dx + (x + 2)^2 \, dy = 0$

$\dfrac{2[ \, 2y(x + 2) - 3 \, ] \, dx}{(x + 2)^2 \, dx} + \dfrac{(x + 2)^2 \, dy}{(x + 2)^2 \, dx} = 0$

$\dfrac{4y(x + 2) - 6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$

$\dfrac{4y(x + 2)}{(x + 2)^2} - \dfrac{6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$

$\left( \dfrac{4}{x + 2} \right) y - \dfrac{6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$

Problem 03 | Linear Equations

Problem 03
$y' = x - 2y$
 

Solution 03
$y' = x - 2y$

$\dfrac{dy}{dx} + 2y = x$       → linear in y
 

Hence,
$P = 2$

$Q = x$
 

Integrating factor:
$\displaystyle e^{\int P \, dx} = e^{\int 2 \, dx}$

$\displaystyle e^{\int P \, dx} = e^{2x}$
 

Thus,
$\displaystyle ye^{\int P\,dx} = \int Qe^{\int P\,dx} \, dx + C$

$\displaystyle ye^{2x} = \int xe^{2x} \, dx + c$
 

Using integration by parts
$u = x$,   $du = dx$

Problem 04 | Linear Equations

Problem 04
$(y + 1) \, dx + (4x - y) \, dy = 0$
 

Solution 04
$(y + 1) \, dx + (4x - y) \, dy = 0$

$\dfrac{(y + 1) \, dx}{(y + 1) \, dy} + \dfrac{(4x - y) \, dy}{(y + 1) \, dy} = 0$

$\dfrac{dx}{dy} + \dfrac{4x - y}{y + 1} = 0$

$\dfrac{dx}{dy} + \dfrac{4x}{y + 1} - \dfrac{y}{y + 1} = 0$

$\dfrac{dx}{dy} + \left( \dfrac{4}{y + 1} \right)x = \dfrac{y}{y + 1}$       → linear in x
 

Hence,
$P = \dfrac{4}{y + 1}$

$Q = \dfrac{y}{y + 1}$
 

Integrating factor:

Additional Topics on the Equations of Order One

Integrating factors by inspection
Determination of integrating factors
Substitution as suggested by the equation
Bernoulli's equation
Coefficient linear in the two variables
Solutions involving elementary integrals
 

Topics available so far...

Integrating Factors Found by Inspection

This section will use the following four exact differentials that occurs frequently.

  1. $d(xy) = x \, dy + y \, dx$
     
  2. $d\left( \dfrac{x}{y} \right) = \dfrac{y \, dx - x \, dy}{y^2}$
     
  3. $d\left( \dfrac{y}{x} \right) = \dfrac{x \, dy - y \, dx}{x^2}$
     
  4. $d\left( \arctan \dfrac{y}{x} \right) = \dfrac{x \, dy - y \, dx}{x^2 + y^2}$
  5.  

  6. $d\left( \arctan \dfrac{x}{y} \right) = \dfrac{y \, dx - x \, dy}{x^2 + y^2}$

 

Problem 02 | Integrating Factors Found by Inspection

Problem 02
$y(y^3 - x) \, dx + x(y^3 + x) \, dy = 0$
 

Solution 02
$y(y^3 - x) \, dx + x(y^3 + x) \, dy = 0$

$y^4 \, dx - xy \, dx + xy^3 \, dy + x^2 \, dy = 0$

$(y^4 \, dx + xy^3 \, dy) - (xy \, dx - x^2 \, dy) = 0$

$y^3(y \, dx + x \, dy) - x(y \, dx - x \, dy) = 0$

$y^3 \, d(xy) - x(y \, dx - x \, dy) = 0$
 

Divide by y3
$d(xy) - \dfrac{x}{y} \left( \dfrac{y \, dx - x \, dy}{y^2} \right) = 0$

$d(xy) - \dfrac{x}{y} \, d\left( \dfrac{x}{y} \right) = 0$

Pages