Question

Question 1 (6.7 points) AW12 x 50 steel beam is used as a simply supported beam on span of 24 feet. The beam supports a uniformly distributed load of 1000 lb/ft. Calculate the maximum bending stress. a) 9,560 psi Ob) 11,280 psi Oc) 13,460 psi d) none of the above

Answer 1

Question 1.  

solution:-

First find the dimensions of W12*50 beam from the databook.

then draw FBD of beam then find the maximum bending moment.

hence corresonding to that we can find maximum bending streess.

see below complete solution.

solution - Griven Beam W12x 50 Question 1) -= zu ff (UDL) w= 1000 lbl ft FBD i = 1000 lbft A moramo B L RP RA 1000 X24 eb. RAF = 12000 2 RB.= 1000 X 24 12000 lb 2 W 12x50 fro rom Data Book We find that has following diamensions: Beam tf ff=0.64 in xd=12.19 in Ixx 3gy in bf=8.08 in Note :- Moment of Inertia about x-x-axis for the IN 12x5o Beam is. Ixx = 3gy in

for simply supported Beam with uniformly distributed load maximum bending moment occurs at mid of the beam, We can poove it as followsin FBD 1000 lb/tt no RB Х RA taking xxx-section Bending moment is 1000Xu? BiM RAXU- 2 Differentiate once with respect to a. 1000 X u 1000 X24 2 d (B,M.) du for manimum Bending moment d (B.M.) du =0 1000Xn 1000 X 24 2 n = 12 ft Maximum Bonding momonti point. 1000X2 4812 1000X123 at u=12 ff. ) Man 2 2 (B.mo), lemi) E72000 lb.ft man

Now for maximum bending stress. (B.M.) max |d/2) d/a (86) man 3 Ixx * los man = 72.050 X 128/1349) ab/in? 394 = 13365.6852 13365.6852 lb in? (06)man psi this is nearest to option (c), or or if being precise about answer then may chose (d).