Record the pertinent beam data
Thus/; = 2500 lb/in^2 (17,237.5 kPa); /. n = 10; A 3 = 2.20 in^2 (14.194 cm^2 ); nAs = 22.0 in^2
(141.94 cm^2 ). Then M= 62,000(12) = 744,000 in-lb (84,057.1 N-m).
Transform the given section to an equivalent homogeneous
section, as in Fig. 13b
Locate the neutral axis of the member
The neutral axis coincides with the centroidal axis of the transformed section. To locate
the neutral axis, set the static moment of the transformed area with respect to its cen-
troidal axis equal to zero: l2(kdf/2 - 22.0(19.5 - kd) = O; kd = 6.82; d - kd = 12.68 in
(322.072 mm).
Calculate the moment of inertia of the transformed section
Then evaluate the flexural stresses by applying the stress equation: /= (^1 /3)(12)(6.82)^3 +
22.0(12.68)^2 = 4806 in^4 (200,040.6 cm^4 );/c = MMII = 744,000(6.82)74806 = 1060 lb/in^2
(7308.7 kPa);/ 5 , = 10(744,00O)(12.68)74806 = 19,600 lb/in^2
Alternatively, evaluate the stresses by computing the resultant
forces C and T
Thus jd = 19.5 - 6.82/3 = 17.23 in (437.642 mm); C = T=MfJd = 744,000/17.23 = 43,200
Ib (192,153.6 N). But C=^1 / 2 /c(6.82)12; :.fc = 1060 lb/in^2 (7308.7 kPa); and T= 2.20/ 5 ; /.
fs = 19,600 lb/in^2 (135,142 kPa). This concludes part a of the solution. The next step con-
stitutes the solution to part b.
Compute pn and then apply the basic equations
in the proper sequence
Thus p = As/(bd) = 2.20/[12(19.5)] = 0.00940; pn = 0.0940. Then by Eq. 30, k = [0.188 +
(0.094)^2 ]^0 5 - 0.094 = 0.350. By Eq. 22,j = 1 - 0.350/3 = 0.883. By Eq. 23,/c = 2MI(lqbf)
= 2(744,000)/[0.350(0.883)(12)(19.5)^2 ] = 1060 lb/in^2 (7308.7 kPa). By Eq. 25, fs =
MI(AJd) = 744,000/[2.20(0.883)(19.5)] = 19,600 lb/in^2 (135,142 kPa).
