Flexural Stress varies directly linearly with distance from the neutral axis. Thus for a symmetrical section such as wide flange, the compressive and tensile stresses will be the same. This will be desirable if the material is both equally strong in tension and compression. However, there are materials, such as cast iron, which are strong in compression than in tension. It is therefore desirable to use a beam with unsymmetrical cross section giving more area in the compression part making the stronger fiber located at a greater distance from the neutral axis than the weaker fiber. Some of these sections are shown below.
 

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.
 

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 f_b = 1000 psi. (b) Will this spacing be safe if the maximum f_b, (f_b)_max = 1600 psi, and the water reaches its maximum depth of 15 ft?
 

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.
 

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?
 

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

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

$\dfrac{dr}{r} = -4t\,dt$