Question

Problem 3 A simple beam AB is loaded as shown in the figure. (a) Calculate the required section modulus Sif Gallow - 18,000 psi, L-32 ft. P-2900 lb, and q- 450 lb/ft. Then select a suitable I-beam (S shape) from Table F-2(a). Appendix F. and recalculate S taking into account the weight of the beam. Select a new beam size if necessary (b) What is the maximum load P that can be applied to your final beam selec tion in part (a)? LLL

Answer 1

--usy.in So, Ilma 4223.75 lb Ausi

. 4 КА1 4 4 1 4 , А ее Wow, taking taking moments about A we get. * Селско ) ax 14 x 4 sehat at Px 4 / + q xL x 7/8 L = LXRB) 2) в. = 1 + 1 = 4солі 22- an c TOP e2 29 до Uо 2 4 2 x4 x 4ppx 3- 3. 3 TL- 3- ft 9x10 $ 850 +31 - Es 2 4 + (4 до 5 15х1. x 103 | xto 2 So, Bending moment at the mid setion is goixers RB x L _ Q x L x 3 2 2 Male) in. Me antiche in = nex - 4 х 4 уз- 9 XSL 4 SOS М. : gos - 430 М. 2 76 cm b-fe

Now, from pure bending equation we get M= s.o M=Bending mona o- stress 2) here, s=section modulus. - inch. ME 37600 lift on 37600x12 6 = 180oo psi 6. Po Sen oner und conve Chera andidate es inch = 451200 M 451200 225.07, in Aptitude co Design, Draw of Energy and 7. Basics of based tools and Engineering ote Tae pape enginering may ra all. so required section modulus is = 25. So, from table F-2 (a) the suitable I bean is -510x35 invina and weight of the beam perfrst is = 35 16 Now, if we consider the weight of the beans then the total weight of the beam is wa 35x32211 20'ls als which acts on the middle of the beam. (assuming So, the loading on the beam becomes, A 4A | LA LA LAB

So, By taking woment about A we get Roxla qx x 3 + P +W) x 2 + 9 +2 x 2 RB = 450*32 + (29 600 + 1120) x2 + 450 x 32 x 2 2 450.+2010+ 3150 25610 lb. So, Bending moment at c B Me = 56100 32 – 450 * 3 2 x 3 x 22 Me = 46560 lb-ft an, Me= 46560x12=558720 us-in. So, the 6 while taleing the account the beam weight is - S = = 558 S = 31004 in3 Aus So from the table F-2 (a) we can choose the new Beam size as I (512x35) in na 68 espending to Casout t- boxin) as 38.1 p. in 3

6 so, for the So, for the final Beans section the maximum moment should be linnited to, s dall. Mmax = 5 bal. 2 38.1 x 180 2 685800 lbrin' So, let maximum load to applied should be W RB XL llo = 40 so from the fig. 2 we get RB XL = qx 2 x 1 4 + (PmtW) x 2 +qxkaXL RB2 450 x + (Pm+1120) xh +450x3 7 2 450+ Pmit 520+ 3150 RB 2 4160 + Pus b = 1.67 x be : 3. na pato 3.63x10 34 + 68.27 91.95 99.73% Muaxe = ( 41607 Pum) x 2 - 450 x 1 x L O Manax a 4160816+ 8pm - 450x3432 66560 + 8pm - 432000 66 685800 2 12 57150 2 8 lmt 22360 7 8pms 33700