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.
 

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.
 

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,

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

Greek Alphabets

The following are Greek alphabets commonly used in science and mathematics.

Greek Symbol Greek Letter Name English Equivalent
Upper Case Lower Case
Α α Alpha a
Β β Beta b
Γ γ Gamma g
Δ δ Delta d
Ε ε Epsilon e
Ζ ζ Zeta z
Η η Eta h
Θ θ Theta th
Ι ι Iota i
Κ κ Kappa k
Λ λ Lambda l
Μ μ Mu m
Ν ν Nu n
Ξ ξ Xi x
Ο ο Omicron o
Π π Pi p
Ρ ρ Rho r
Σ σ Sigma s
Τ τ Tau t
Υ υ Upsilon u
Φ φ Phi ph
Χ χ Chi ch
Ψ ψ Psi ps
Ω ω Omega o

Problem 02 | Division by t

Problem 02
Find the Laplace transform of   $f(t) = \dfrac{e^{4t} - e^{-3t}}{t}$.
 

Solution 03
$f(t) = \dfrac{e^{4t} - e^{-3t}}{t}$

$f(t) = \dfrac{e^{4t}}{t} - \dfrac{e^{-3t}}{t}$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left\{ \dfrac{e^{4t}}{t} \right\} - \mathcal{L} \left\{ \dfrac{e^{-3t}}{t} \right\}$
 

Since
$\mathcal{L} (e^{4t}) = \dfrac{1}{s - 4}$   and

$\mathcal{L} (e^{-3t}) = \dfrac{1}{s + 3}$
 

Then,

Problem 01 | Division by t

Problem 01
Find the Laplace transform of   $f(t) = \dfrac{\sin t}{t}$.
 

Solution 01
$\mathcal{L} (\sin t) = \dfrac{1}{s^2 + 1}$
 

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \int_s^\infty \dfrac{du}{u^2 + 1}$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \big[ \arctan u \big]_s^\infty$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty}\big[ \arctan u \big]_s^a$