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 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 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 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 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 534 | Economic Sections

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

Problem 01 | Separation of Variables

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

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

$\dfrac{dr}{r} = -4t\,dt$
 

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
 

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

 

Exact Equations | Equations of Order One

The differential equation
 

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

 

is an exact equation if
 

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

 

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.
 

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