- Compute the section modulus
Determine the section modulus of the rectangular cross section with respect to each axis.
Thus Sx = (Vejbd^2 - (%)(18)(24)^2 = 1728 in^3 (28,321.9 cm^3 ); Sy = (%)(24)(18)^2 = 1296 in^3
(21,241 cm^3 ). - Compute the stresses produced
Compute the uniform stress caused by the concentric load and the stresses at the edges
caused by the bending moments. Thus/! = PIA = 100,000/[ 18(24)] = 231 lb/in
2
(1592.7
kPa);/ = MxISx = 200,000/1728 = 116 lb/in
2
(799.8 kPa);}j, = MyISy = 100,000/1296 =
77 lb/in
2
(530.9 kPa). - Determine the stress at each corner
Combine the results obtained in step 3 to obtain the stress at each corner. Thus fA = 231 +
116 + 77 = 424 lb/in^2 (2923.4 kPa);/ 5 = 231 + 116 - 77 = 270 lb/in^2 (1861.5 kPa);/c =
231 - 116 + 77 = 192 lb/in^2 (1323.8 kPa);/^ = 231 - 116 - 77 = 38 lb/in^2 (262.0 kPa).
These stresses are all compressive because a positive stress is considered compressive,
whereas a tensile stress is negative. - Check the computed corner stresses
Use the following equation that applies to the special case of a rectangular cross section:/
= (PIA)(I ± 6ex/dx + 6ey/dy), where ex and ey = eccentricity of load with respect to the x
and y axes, respectively; dx and dy = side of rectangle, in (mm), normal to x m\dy axes, re-
spectively. Solving for the quantities within the brackets gives 6CxIdx = 6(2)/24 = 0.5;
6ey/dy = 6(1)718 = 0.33. Then/4 = 231(1 + 0.5 + 0.33) = 424 lb/in
2
(2923.4 kPa);)i =
231(1 + 0.5 - 0.33) = 270 lb/in
2
(1861.5 kPa);/c = 231(1 - 0.5 + 0.33) = 192 lb/in
2
(1323.8 kPa);/D = 231(1 - 0.5 - 0.33) = 38 lb/in
2
(262.0 kPa). These results veriiy those
computed in step 4.
FIGURE 32. Transverse section of
a post
Section X-XFIGURE 33. Curved member in bending.