(Solved): Problem II. A simply supported beam is made of W14x53. The beam is subject to a uniformly distribut ...

Problem II. A simply supported beam is made of W14x53. The beam is subject to a uniformly distributed load that is parallel to the minor axis of the beam section. Using the 2004 ASEP Steel Handbook and assuming no local buckling, answer the following questions. II-1) What is the value of the radius of gyration about the minor axis (in mm)? (a) \( 48.01 \) (b) \( 48.77 \) (c) \( 62.23 \) (d) \( 62.48 \) (e) \( 62.99 \) II-3) What is the value of the elastic section modulus about the major axis \( \left(\right. \) in \( \times 10^{3} \mathrm{~mm}^{3} \) )? (a) 1027 (b) 1275 (c) 1511 (d) 1688 (e) 1835 (f) 2016 II-4) What is the value of the plastic moment, Mp (in kN-m)? (a) \( 283.0 \) (b) \( 353.9 \) (c) \( 414.4 \) (d) \( 467.5 \) (e) \( 512.1 \) (f) \( 564.9 \) II-5) Using the short formula, what is the value of Lr in meters? (Round off to 2 decimal places) II-6) What is the value of the lateral-torsional buckling modification factor, Cb? (Round off to 3 decimal places) II-7) If the length of the beam is \( 2 \mathrm{~m} \), what is the maximum factored moment that can be applied (in kN-m)? (a) \( 508.4 \) (b) \( 254.7 \) (c) \( 318.5 \) (d) \( 373.0 \) (e) \( 420.7 \) (f) \( 460.9 \) II-8) If the length of the beam is \( 5 \mathrm{~m} \), what is the maximum factored moment that can be applied (in kN-m)? (a) \( 206.1 \) (b) \( 259.3 \) (c) \( 353.9 \) (d) \( 398.7 \) (e) \( 436.9 \) (f) \( 482.5 \) II-9) If the length of the beam is \( 9 \mathrm{~m} \), what is the maximum factored moment that can be applied (in \( \mathrm{kN} \)-m, rounded off to a whole number)?