Question

A simply supported beam as shown in the figure. The beam section is W18x211. The beam must support its own weight and must carry the following loading: Super-imposed distributed dead load = 0.25 kip/ft Distributed live load = 1 kip/ft Concentrated dead load = 12 kip The beam span L = 26 ft and the distance of the concentrated load from the right support a=6 ft. Consider analy- sis of beam subjected to load combination 1.2 dead + 1.6 live. (i) Including its self weight compute factored distributed load on the beam. Answer: w, = 2.15 kip/ft (ii) Compute factored concentrated load on the beam. Answer: P. = 14.4 kip (iii) Compute support reactions. Answers: Left end reaction - 31.3 kip. Right-end reaction = 39.1 kip (iv) Compute bending moment and shear force at the mid-span. Answers: Shear force = 3.32 kip. Bending moment = 225 kip-ft. (v) Compute and draw shear force and bending moment diagrams. What are the maximum shear and maximum bending moment values and their locations? Answers: Maximum shear force = 39.1 kip at the right end. Maximum bending moment - 227.7 kip-ft at 14.54 ft from the left end. Solution

Answer 1

Solytim:- Given W18 X211 Self wt. of beam 0:21 leip / ft Supen- imposed cdistributeed cteael lond kiplfr live loned :1kip Ift 0.25 Distribyteed Concentrateed loned = 12 kip G) Totall cleacl lonc 0.46 kip lft 0:21 +0.25 live load I kip Ift Factoredd disteibyteed load Wy = 1.2 dead + 1.6 live 1.2 x o.46 + 1.6 x1 wy = 2.15 kiplft (i) Factosred concentoateed loned Py = 1.2 p 1.2 Py= 14.4 kip

Py: 144 kip Wy: 2.15 kipl ft (ii) D (mid) 20 ft ARB RA t RR -14.4-2.15sx26 =0 70:3 kip -0 RA + RB = 2.15X26 X 26-14.4x6 = 0 RA X 26 - 31.3 kip Rn = 31.27 from o RR = 39.0 3 ? 39.1 kip liv) fonce (sit) -sheay forom A 5.F. in Ac (n distance S.F tve Su = +RA - wyn SAIM:0) = tRn SA = 31:3 kip at 31.3-2 15 4=0 Se lx=2044) K=14.5s fal 31.3 - 2.15 x20 Se just kft -11.7 kip %3D Sp = 3 13.-2.15 X13 mid at Jpan S.F Sp=3:35 kip

(* fnom ) SF. in CB Su: - RR + wyn Sain:0) - 39 eip Scln-6) -31 + 2.15 x6 %3D Se= -261 k ip > Bending (Bm) moment Hegging sagging (-ve Bm.) (+ve Bm.) AB In faom A) B.m. in MAin:0) Wy x 13 x13 nolneis) 31.3 x 13 - Mo Imid span) 225.225 kip-ft at mid span 225 kip.t ly. sr2 2. 31-3x 14.55- 2:11X M max (n- 14.55 223.그 eip~f+ 227.83 Mmax = B) ( k from B.m. in - wy -2.1r * 31 kip- f1 mc (u=6)= 195.7

(v) mavimym shean fosie Smay = Sa: 39.1 kip bending maximum is moment at disdance 14.51 ft from a ight end IA) 227.7 kip -ft Pu : 14.4 ki? wy:2:11/4 linean 31.1 SFD 14.55 (-) 26.1 39.1 : 227.7 eip.t Mmar Pasaboli c Mmied 22f'aip-ft 14.55 ft