Question

Got it wrong 3 times unsure if the 4th one is correct

Required information Consider the beam shown in the given figure. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 1800 lbf 300 lbfin 1 in |-10 insta— 30in— Í S Cross section (enlarged) Determine the maximum tensile and compressive bending stresses. Take w= 300 lbf/in, F = 1800 lbf, C1 = 10 in, and C2 = 30 in. The maximum tensile bending stress is 5294 psi. The maximum compressive bending stress is 4473.529 psi. The maximum compressive bending stress is 7456 psi. The maximum compressive bending stress is 3176.470 psi. The maximum compressive bending stress is -7455.88 psi.

Answer 1

0 Given that 1800 lbf 300 lbflin. Roxo Rea o k loin L RBy = 6900 lbf Reg= 3900 lbf 15100 Ibe 11800 lbs 27 in -1800 - 3900 lbf 25,35o in lbf Y o in lbf 1-18.000 in-ibp a from the eauilibrium Conditions EMBO -> (Rcy) (30) +(1800) (10) = (300) (30) (15) -> Rcy = 3900 168 E Fyzo - RBY + Rey = (Boo) (30) + 1800. - RBy = 6900 lbf) Efn=0 - Since there is no hoarizontal load acting on the beam RBn= lonzo lbf

0 Given that 1800 lbf 300 lbflin. Roxo Rea o k loin L RBy = 6900 lbf Reg= 3900 lbf 15100 Ibe 11800 lbs 27 in -1800 - 3900 lbf 25,35o in lbf Y o in lbf 1-18.000 in-ibp a from the eauilibrium Conditions EMBO -> (Rcy) (30) +(1800) (10) = (300) (30) (15) -> Rcy = 3900 168 E Fyzo - RBY + Rey = (Boo) (30) + 1800. - RBy = 6900 lbf) Efn=0 - Since there is no hoarizontal load acting on the beam RBn= lonzo lbf

Section bw A2B ? Considen a section between AGB at a distance - n from SFx 11800 lbf Mr Sfx- - 1800 lbf Ma = - 1800 (2) at noin SFA = -1800 lbf : MA= 0 lof.in. at > = loin —> SFB = -1800 lbf ; MBP - 18,000 in-1bf Section blw Blo: Consider a section between B2C at a distance 1800 lbf x' from c. 1 SFX 300 lbflin. Sn = Gaoo - 1800 - 300 (n-10) k loin Mic= 6900 (2-10) - 1800 (2-0) - 300(n-10)2 to 6900 Ibe Me الب at na loin SFB = 5100 lbf ; MB = -18,000 in-lbe at na noin -> Sfo = -3900 lbf ; Mczo in lof & Shear fonce changing it's sign between B2 5гго 6960 - 1800 - 300 (n-10) zo n=27m - at x=27m from enda' Shear force is zego. when shear force is serio - Bending moment is min/man B Miman = 6900 (27-10) - 1800 (-o) ~150 (27-10) MOM >) Mimar = 25,350 in 1bf

- forom the Bending moment cacation INA INA Area m. I op lin Since the plane is y-anis; centawid of the y-axis plane aboot Neubral anis Symmetrical about the plane lies on 3 in Ay, Azyz from n-aois i A+Az k_ Bin C) = 3 inz zin A = y=% 1+3/2=215 in (Iccha bed= (193) = 2.25 in 4 3 in Az=(3) (s) a 3 in? Ilin " Y2 = 1/2 = 0.5 in . (I cal₂ = (3) (1)3 = 0.25 in" 5 = (3) (25) + (3) Cois) = 115 in from a anis. 3+3 5 = 1.5 in forom y-axis. :) INA: - Anea m'I of plane amina about Neutral anis INA = It Iz som of m'I's of the individual plane lamina's about N.A INA - [Ica,+ m (1-9)"). (fieca) t (4-5)

INA = (225 + (3)(2-5-15) 2 ) + (0-25 + 3) (05-15)) INA = 5.25 +3.25 INA = 8.5 iny Compressive Storess At maximum Bending moments; (At point o] o uppeor da - a bottom layer subjected. to tensile stress Mman = 25,350 in Ibe manimum tensile stress gging moment) tensile Stress [sman - Mmax) (Upa) = (25,350 (35) tensile INA (otensin) = (25,350) (15) = 4493.5 psi 8.5 o o comp) = 18060251350)(25) 8.5 745588 PS tensile AL point Bi | (18000)(215) [otensile] = 45 '5 = 5294.ll psi a - 5894all an (densia Compressin fotome 7 (18000) (15), 3176.47 Psi_ (come) from 0 0 2 - it is obsequed that [Stensiu]> [otensire tensiu] =5294. u psi) mon Cot B to come to <[o Comp Ja - man psi Tuss. 88 at D .