March 2009

Solution to Problem 529 | Economic Sections

Problem 529
A 10-m beam simply supported at the ends carries a uniformly distributed load of 16 kN/m over its entire length. What is the lightest W shape beam that will not exceed a flexural stress of 120 MPa? What is the actual maximum stress in the beam selected?

Solution to Problem 530 | Economic Sections

Problem 530
Repeat Prob. 529 if the distributed load is 12 kN/m and the length of the beam is 8 m.

Solution to Problem 531 | Economic Sections

Problem 531
A 15-ft beam simply supported at the ends carries a concentrated load of 9000 lb at midspan. Select the lightest S section that can be employed using an allowable stress of 18 ksi. What is the actual maximum stress in the beam selected?

Solution to Problem 532 | Economic Sections

Problem 532
A beam simply supported at the ends of a 25-ft span carries a uniformly distributed load of 1000 lb/ft over its entire length. Select the lightest S section that can be used if the allowable stress is 20 ksi. What is the actual maximum stress in the beam selected?

Solution to Problem 533 | Economic Sections

Problem 533
A beam simply supported on a 36-ft span carries a uniformly distributed load of 2000 lb/ft over the middle 18 ft. Using an allowable stress of 20 ksi, determine the lightest suitable W shape beam. What is the actual maximum stress in the selected beam?
 

Elementary Differential Equations

Families of Curves
Equations of Order One
Elementary Applications
Additional Topics on Equations of Order One
Linear Differential Equations
Linear Equations with Constant Coefficients
Nonhomogeneous Equations: Undetermined Coefficients
Variation of Parameters
Inverse Differential Operators
Applications

Topics so far...

Separation of Variables | Equations of Order One

Given the differential equation
 

$M(x, y)\,dx + N(x, y)\,dy = 0 \,\, \to \,\,$ Equation (1)

 
where $\,M\,$ and $\,N\,$ may be functions of both $\,x\,$ and $\,y\,$. If the above equation can be transformed into the form
 

$f(x)\,dx + f(y)\,dy = 0\,\, \to \,\,$ Equation (2)

 
where $\,f(x)\,$ is a function of $\,x\,$ alone and $\,f(y)\,$ is a function of $\,y\,$ alone, equation (1) is called variables separable.
 

Derivation of Formula for Sum of Years Digit Method (SYD)

The depreciation charge and the total depreciation at any time m using the sum-of-the-years-digit method is given by the following formulas:
 

Depreciation Charge:

$d_m = (FC - SV) \dfrac{n - m + 1}{SYD}$

 

Total depreciation at any time m

$D_m = (FC - SV) \dfrac{m(2n - m + 1)}{2 \times SYD}$

 

Where:
FC = first cost
SV = salvage value
n = economic life (in years)
m = any time before n (in years)
SYD = sum of years digit = 1 + 2 + ... + n = n(1 + n)/2
 

Homogeneous Functions | Equations of Order One

If the function f(x, y) remains unchanged after replacing x by kx and y by ky, where k is a constant term, then f(x, y) is called a homogeneous function. A differential equation
 

$M \, dx + N \, dy = 0 \,\, \to \,\,$ Equation (1)

 
is homogeneous in x and y if M and N are homogeneous functions of the same degree in x and y.
 

Exact Equations | Equations of Order One

The differential equation
 

$M(x, y) \, dx + N(x, y) \, dy = 0$

 

is an exact equation if
 

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

 

Linear Equations | Equations of Order One

Linear Equations of Order One
Linear equation of order one is in the form
 

$\dfrac{dy}{dx} + P(x) \, y = Q(x).$

 

The general solution of equation in this form is
 

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

 

Problem 01 | Separation of Variables

Problem 01
$\dfrac{dr}{dt} = -4rt$,   when   $t = 0$,   $r = r_o$
 

Solution 01
$\dfrac{dr}{dt} = -4rt$

$\dfrac{dr}{r} = -4t\,dt$
 

Solution to Problem 534 | Economic Sections

Problem 534
Repeat Prob. 533 if the uniformly distributed load is changed to 5000 lb/ft.
 

Solution to Problem 535 | Economic Sections

Problem 535
A simply supported beam 24 ft long carries a uniformly distributed load of 2000 lb/ft over its entire length and a concentrated load of 12 kips at 8 ft from left end. If the allowable stress is 18 ksi, select the lightest suitable W shape. What is the actual maximum stress in the selected beam?
 

Solution to Problem 536 | Economic Sections

Problem 536
A simply supported beam 10 m long carries a uniformly distributed load of 20 kN/m over its entire length and a concentrated load of 40 kN at midspan. If the allowable stress is 120 MPa, determine the lightest W shape beam that can be used.
 

Solution to Problem 538 | Floor Framing

Problem 538
Floor joists 50 mm wide by 200 mm high, simply supported on a 4-m span, carry a floor loaded at 5 kN/m2. Compute the center-line spacing between joists to develop a bending stress of 8 MPa. What safe floor load could be carried on a center-line spacing of 0.40 m?
 

Solution to Problem 539 | Floor Framing

Problem 539
Timbers 12 inches by 12 inches, spaced 3 feet apart on centers, are driven into the ground and act as cantilever beams to back-up the sheet piling of a coffer dam. What is the maximum safe height of water behind the dam if water weighs = 62.5 lb/ft3 and ( fb )max = 1200 psi?
 

Solution to Problem 540 | Floor Framing

Problem 540
Timbers 8 inches wide by 12 inches deep and 15 feet long, supported at top and bottom, back up a dam restraining water 9 feet deep. Water weighs 62.5 lb/ft3. (a) Compute the center-line spacing of the timbers to cause fb = 1000 psi. (b) Will this spacing be safe if the maximum fb, (fb)max = 1600 psi, and the water reaches its maximum depth of 15 ft?
 

Solution to Problem 541 | Floor Framing

Problem 541
The 18-ft long floor beams in a building are simply supported at their ends and carry a floor load of 0.6 lb/in2. If the beams have W10 × 30 sections, determine the center-line spacing using an allowable flexural stress of 18 ksi.
 

Solution to Problem 542 | Floor Framing

Problem 542
Select the lightest W shape sections that can be used for the beams and girders in Illustrative Problem 537 of text book if the allowable flexural stress is 120 MPa. Neglect the weights of the members.
 

For Girder (G - 1) | Solution to Problem 542

For Girder (G - 1)
542-girder-g1.jpg$S_{live-load} = \dfrac{M}{f_b} = \dfrac{40(1000^2)}{120}$
$S_{live-load} = 333.33 \times 10^3 \, \text{mm}^3$
 

For Beams (B - 2) | Solution to Problem 542

For Beams (B - 2)

542-beam-b2.jpg$\Sigma M_{R2} = 0$
$6R_1 = 20(4) + 10(2)(5) + 15(4)(2)$
$R_1 = 50 \, \text{kN}$
 
$\Sigma M_{R1} = 0$
$6R_2 = 20(2) + 10(2)(1) + 15(4)(4)$
$R_2 = 50 \, \text{kN}$
 

For Beams (B - 3) | Solution to Problem 542

For Beams (B - 3)

542-beam-b3.jpg$M_{max} = \frac{1}{8}(20)(62)$
$M_{max} = 90 \, \text{kN}\cdot\text{m}$
 
$S_{required} = \dfrac{M_{max}}{f_b} = \dfrac{90(1000^2)}{120}$
$S_{required} = 750 \times 10^3 \, \text{mm}^3$
 

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