Problem

Rework Example 6.7a. using Eq. 6.60 to calculate Jb. using Eq. 6.62 to calculate Jc. using...

Rework Example 6.7

a. using Eq. 6.60 to calculate J
b. using Eq. 6.62 to calculate J
c. using the value J = 6.077 × 106 mm4 as published in A1SC (1997)
d. Compare the results obtained in parts (a), (b), and (c) to the results obtained in Example 6.7. Comment on the differences.

K1 is the value of ϕ on the inner elliptic surface, A1 is the area bounded by the inner ellipse, and R is the solid region bounded by the inner and outer ellipses.

Example 6.7

The nominal dimensions of a steel wide-flange section (W760×220) are shown in Figure E6.7. The beam is subjected to a twisting moment T = 5000 N · m.

(a) Determine the maximum shear stress τmax and its location. Ignore the fillets and stress concentrations.
(b) Determine the angle of twist per unit length for the applied twisting moment.

FIGURE E6.7

For the flanges b/h = 8.867 < 10. So, for a flange, k1 = 0.308 by interpolation from Table 6.1. Therefore, for two flanges

Jf = 2[k1(bf)(tf)3] = 4,424,000 mm4

For the web, b/h = 43.58 > 10. Therefore, for the web k1 = 0.333 and

Jw = k1(d − 2tf)(t3w) = 1,076,600 mm4

Hence, the torsional constant for the section is

J = Jf + Jw = 5,500,700 mm4 5.501 × 10−6 m4

(a) By Eq. 6.63, the maximum shear stress is
and it is located along the vertical line of symmetry on the outer edge of the top and bottom flanges.
(b) By the second of Eqs. 6.63 or the first of Eqs. 6.64. the angle of twist per unit length is