July 2011

Problem 347 | Equilibrium of Non-Concurrent Force System

Problem 347
Repeat Problem 346 if the cable pulls the boom AB into a position at which it is inclined at 30° above the horizontal. The loads remain vertical.
 

Cable and boom structure

 

Problem 01 | Equations with Homogeneous Coefficients

Problem 01
$3(3x^2 + y^2) \, dx - 2xy \, dy = 0$
 

Solution 01
$3(3x^2 + y^2) \, dx - 2xy \, dy = 0$
 

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

Substitute,
$3(3x^2 + v^2x^2) \, dx - 2vx^2 (v \, dx + x \, dv) = 0$

$3(3 + v^2)x^2 \, dx - 2vx^2 (v \, dx + x \, dv) = 0$
 

Divide by x2,
$3(3 + v^2) \, dx - 2v (v \, dx + x \, dv) = 0$

$9 \, dx + 3v^2 \, dx - 2v^2 \, dx - 2vx \, dv = 0$

$9 \, dx + v^2 \, dx - 2vx \, dv = 0$

$(9 + v^2) \, dx - 2vx \, dv = 0$

Problem 02 | Equations with Homogeneous Coefficients

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

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

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

Substitute,
$(x - 2vx) \, dx + (2x + vx)(v \, dx + x \, dv) = 0$

$x \, dx - 2vx \, dx + 2vx \, dx + 2x^2 \, dv + v^2x \, dx + vx^2 \, dv = 0$

$x \, dx + 2x^2 \, dv + v^2x \, dx + vx^2 \, dv = 0$

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

$x(1 + v^2) \, dx + x^2(2 + v) \, dv = 0$

Problem 03 | Equations with Homogeneous Coefficients

Problem 03
$2(2x^2 + y^2) \, dx - xy \, dy = 0$
 

Solution 03
$2(2x^2 + y^2) \, dx - xy \, dy = 0$
 

Let
$y = vx$

$dy = v \, dx + x \, dv$
 

$2(2x^2 + v^2x^2) \, dx - vx^2(v \, dx + x \, dv) = 0$

$4x^2 \, dx + 2v^2x^2 \, dx - v^2x^2 \, dx - vx^3 \, dv = 0$

$4x^2 \, dx + v^2x^2 \, dx - vx^3 \, dv = 0$

$x^2(4 + v^2) \, dx - vx^3 \, dv = 0$

$\dfrac{x^2(4 + v^2) \, dx}{x^3(4 + v^2)} - \dfrac{vx^3 \, dv}{x^3(4 + v^2)} = 0$

$\dfrac{dx}{x} - \dfrac{v \, dv}{4 + v^2} = 0$

Problem 04 | Equations with Homogeneous Coefficients

Problem 04
$xy \, dx - (x^2 + 3y^2) \, dy = 0$
 

Solution 04
$xy \, dx - (x^2 + 3y^2) \, dy = 0$
 

Let
$x = vy$

$dx = v \, dy + y \, dv$
 

$vy^2(v \, dy + y \, dv) - (v^2y^2 + 3y^2) \, dy = 0$

$vy^2(v \, dy + y \, dv) - y^2(v^2 + 3) \, dy = 0$

$v(v \, dy + y \, dv) - (v^2 + 3) \, dy = 0$

$v^2 \, dy + vy \, dv - v^2 \, dy - 3 \, dy = 0$

$vy \, dv - 3 \, dy = 0$

$v \, dv - \dfrac{3 \, dy}{y} = 0$

$\displaystyle \int v \, dv - 3 \int \dfrac{dy}{y} = 0$

Problem 348 | Equilibrium of Non-Concurrent Force System

Problem 348
The frame shown in Fig. P-348 is supported in pivots at A and B. Each member weighs 5 kN/m. Compute the horizontal reaction at A and the horizontal and vertical components of the reaction at B.
 

Simple Frame Supported in Pivots

 

Problem 349 | Equilibrium of Non-Concurrent Force System

Problem 349
The truss shown in Fig. P-349 is supported on roller at A and hinge at B. Solve for the components of the reactions.
 

Truss supported by a roller and a hinge

 

Problem 01 | Exact Equations

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

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

Test for exactness
$M = x + y$   ;   $\dfrac{\partial M}{\partial y} = 1$

$N = x - y$   ;   $\dfrac{\partial N}{\partial x} = 1$

$\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x}$   ;   thus, exact!
 

Step 1: Let
$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = x + y$
 

Problem 02 | Exact Equations

Problem 02
$(6x + y^2) \, dx + y(2x - 3y) \, dy = 0$
 

Solution 02
$(6x + y^2) \, dx + y(2x - 3y) \, dy = 0$
 

$M = 6x + y^2$

$N = y(2x - 3y) = 2xy - 3y^2$
 

Test for exactness
$\dfrac{\partial M}{\partial y} = 2y$

$\dfrac{\partial N}{\partial x} = 2y$

Exact!
 

Let
$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = 6x + y^2$

$\partial F = (6x + y^2) \, \partial x$
 

Integrate partially in x, holding y as constant

Problem 350 | Equilibrium of Non-Concurrent Force System

Problem 350
Compute the total reactions at A and B for the truss shown in Fig. P-350.
 

Overhang truss at both ends

 

Problem 351 | Equilibrium of Non-Concurrent Force System

Problem 351
The beam shown in Fig. P-351 is supported by a hinge at A and a roller on a 1 to 2 slope at B. Determine the resultant reactions at A and B.
 

Beam with roller support on a slope

 

Problem 03 | Exact Equations

Problem 03
$(2xy - 3x^2) \, dx + (x^2 + y) \, dy = 0$
 

Solution 03
$(2xy - 3x^2) \, dx + (x^2 + y) \, dy = 0$
 

$M = 2xy - 3x^2$

$N = x^2 + y$
 

Test for exactness
$\dfrac{\partial M}{\partial y} = 2x$

$\dfrac{\partial N}{\partial x} = 2x$

Exact!
 

Let
$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = 2xy - 3x^2$

$\partial F = (2xy - 3x^2) \, \partial x$
 

Integrate partially in x, holding y as constant

Problem 04 | Exact Equations

Problem 04
$(y^2 - 2xy + 6x) \, dx - (x^2 - 2xy + 2) \, dy = 0$
 

Solution 04
$(y^2 - 2xy + 6x) \, dx - (x^2 - 2xy + 2) \, dy = 0$

$M = y^2 - 2xy + 6x$

$N = -x^2 + 2xy - 2$
 

Test for exactness
$\dfrac{\partial M}{\partial y} = 2y - 2x$

$\dfrac{\partial N}{\partial x} = -2x + 2y$

Exact!
 

Let
$\dfrac{\partial F}{\partial x} = M$

$\dfrac{\partial F}{\partial x} = y^2 - 2xy + 6x$

$\partial F = (y^2 - 2xy + 6x) \, \partial x$
 

Integrate partially in x, holding y as constant

Problem 352 | Equilibrium of Non-Concurrent Force System

Problem 352
A pulley 4 ft in diameter and supporting a load 200 lb is mounted at B on a horizontal beam as shown in Fig. P-352. The beam is supported by a hinge at A and rollers at C. Neglecting the weight of the beam, determine the reactions at A and C.
 

Pulley mounted at the midspan of simple beam

 

Problem 353 | Equilibrium of Non-Concurrent Force System

Problem 353
The forces acting on a 1-m length of a dam are shown in Fig. P-353. The upward ground reaction varies uniformly from an intensity of p1 kN/m to p2 kN/m at B. Determine p1 and p2 and also the horizontal resistance to sliding.
 

Gravity dam

 

Problem 354 | Equilibrium of Non-Concurrent Force System

Problem 354
Compute the total reactions at A and B on the truss shown in Fig. P-354.
 

354-roof-truss.gif

 

Problem 355 | Equilibrium of Non-Concurrent Force System

Problem 355
Determine the reactions at A and B on the Fink truss shown in Fig. P-355. Members CD and FG are respectively perpendicular to AE and BE at their midpoints.
 

Fink truss with support at -30 degree slope

 

Advance Engineering Mathematics

  • Laplace transform, inverse Laplace transform, and application of Laplace transform to differential equations.
  • Simultaneous ordinary differential equations.
  • Infinite series, Maclaurin's series, power series, Taylor's series.
  • Matrices and determinants.
  • Complex numbers.

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Engineering Economy

Simple Interest, Compounded Interest, Annuity, Capitalized Cost, Annual Cost, Depreciation, Depletion, Capital Recovery, Property Valuation or Appraisal, Principles of Accounting, Cost Accounting, Break-even Analysis, Minimum Cost Analysis, Public Economy, Inflation and Deflation, Risk and Uncertainty.

Topics so far...
 

Laplace Transform

Definition of Laplace Transform

Let   $f(t)$   be a given function which is defined for   $t \ge 0$. If there exists a function   $F(s)$  so that
 

$\displaystyle F(s) = \int_0^\infty e^{-st} \, f(t) \, dt$,

 

then   $F(s)$   is called the Laplace Transform of   $f(t)$, and will be denoted by   $\mathcal{L} \left\{f(t)\right\}$.   Notice the integrator   $e^{-st} \, dt$   where   $s$   is a parameter which may be real or complex.
 

Laplace Transform by Direct Integration

To get the Laplace transform of the given function   $f(t)$,   multiply   $f(t)$   by   $e^{-st}$   and integrate with respect to   $t$   from zero to infinity. In symbol,
 

$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$.

 

Problem 01 | Laplace Transform by Integration

Problem 01
Find the Laplace transform of   $f(t) = 1$   when   $t > 0$.
 

Solution 01
$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$

$\displaystyle \mathcal{L} (1) = \int_0^\infty e^{-st} (1) \, dt$

$\displaystyle \mathcal{L} (1) = \int_0^\infty e^{-st} \, dt$

$\displaystyle \mathcal{L} (1) = -\dfrac{1}{s} \int_0^\infty e^{-st} \, (-s \, dt)$

$\displaystyle \mathcal{L} (1) = -\dfrac{1}{s} \left[ e^{-st} \right]_0^\infty$

Problem 02 | Laplace Transform by Integration

Problem 02
Find the Laplace transform of   $f(t) = e^{at}$.
 

Solution 02
$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$

$\displaystyle \mathcal{L} (e^{at}) = \int_0^\infty e^{-st} e^{at} \, dt$

$\displaystyle \mathcal{L} (e^{at}) = \int_0^\infty e^{-st + at} \, dt$

$\displaystyle \mathcal{L} (e^{at}) = \int_0^\infty e^{-(s - a)t} \, dt$

$\displaystyle \mathcal{L} (e^{at}) = -\dfrac{1}{s - a} \int_0^\infty e^{-(s - a)t} \, [ \, -(s - a) \, dt \, ]$

Problem 03 | Laplace Transform by Integration

Problem 03
Find the Laplace transform of   $f(t) = \sin bt$.
 

Problem 03
$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$

$\displaystyle \mathcal{L} (\sin bt) = \int_0^\infty e^{-st} \sin bt \, dt$
 

For   $\displaystyle \int_0^\infty e^{-st} \sin bt \, dt$.

Using integration by parts:   $\displaystyle \int u\,dv = uv - \int v\, du$.   Let

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