Question

Please don't copy other solutions, and show steps clearly.

For all problems assume LRFD and A992 steel unless otherwise stated.

Answer 1

Select the suitable W- shape for the beam after applying checks for moment and shear. Follow the procedure as follows: Using Load Resistance and Factored Design Method Check for Bending Compute the factored loads acting on the beam as follows: Calculate the factored concentrated load, P, as follows: Pi=1.2P, +1.6P, = (1.2x 40+1.6x65) kips = 152 kips 4 For the given loading, the maximum bending moment acts at the centre and is given by: MQ=2,2 _152 kips 30 ft 4 =1140 ft kips Check the value of nominal flexural strength of W- Sections from the table 3-2 of AISC Z, table. Try a W14x176 Check for flexural strength excluding the beam weight. For a W14x176: 0;M, = 1200 ft-kips

Here the nominal flexural strength is Ø, M,. Compare the value of M, and @M,. 1140 ftkips < 1200 ft kips M, <0,M, Therefore, the section is safe in flexure if the self-weight is not taken into consideration. Check for Beam Weight Calculate the factored uniformly distributed load, w, as follows: From tables self-weight of W14 x176 section is 0.176 k/ft W, =((1.2x(W. + wy))+(1.6xw,)) kips. ft-1 w; =((1.2x(W. + w))+(1.6xwz )) =((1.2x(0+0.176))+(1.6x0)) kips-ft- = 0.2112 kips-ft-? Compute the maximum bending moment as follows: 8 4 - + mw, , P.,L 8 4 0.2112 kips . ftx(30ft) 8 =1163.76 ft kips 152 kips x 30 ft - +-

Check for flexural strength: For a W14x176: 0;M, = 1200 ft-kips Here the nominal flexural strength is 0M, Compare the value of M, and 6 M,. 1163 ft. kips < 1200 ft kips M, <0,M, Therefore, the section is safe in flexure if the self-weight is taken into consideration. Hence, the beam is safe is flexure. Hence Use W14x176