Question

8:08 Done 8 of 9 Problem nl. 40 pts PyAn overhanging W760 257 rolled-steel beam supports a load- 600kN and B-400 N as shown Knowing that a mb 25 D 350 MPa and 170 MPa. 1. The maimum wale of the normal stress in the bea 2. the maximum value of the principal stress and the maximum value of the sharing stress at the junction of the face and web, 3. State if the specified shape is weather acceptable or not as far as these two stresses and are concerned. 4. Then determine the angles, and and draw the principle planes stress and the marimum in-plane shearing stress. 5. Draw Mohr's circle showing all characteristic points 6. Show . T...in Mohr's circle

Answer 1

Draw FBD of beam 19 RA IP = Goolen P2.400 les K 3 (2.5m) (5m) 9 15m) k Re RA and Rc are reaction forces" refpro :D -RA + Rc -600-40020 1) Re- RA - 1020 low -0 Pɛ Mezo =D - RA (5) – 600 (265) +460 (5) 20 1 5RA = Sco RA = 100 kn fromegro Rc-RA = como 2 Rc = The = rewtrA 1100 en We Need Consider a to draw first sf and Bry diagrame section x-x at a distance of e from A. b/w A and B RA = 100 kN x 7ca -lcow 7ocn22-sm aren) = -lo ne lenom I T sf remain same and by varies lineonly " with (2) at azom Point (A) VCA) - 100 KN BM(A) = -1wxo = olen m

at x = 2.5m Point (B) VG) --Zcoin MIB) = -1 cox 2.5 = - 250 kwam at 2.5m poput FB) consider section b/w B and c TRAFI ODIN - Cole IX V(X) = -100 - 600 - - 700 len 2 MCM) = -160K-600 CH-205) 2 SLK2S í 2.5m (2-25) | v remain sane and my changes with (2) linearey cet x = 2.sm cpoint B) VLG) - - 700 lew.' 6 MG) = -16x9.5600(2.5-2.5) - 250 kn-m at = 5m Point (C) VC): -700 IN MC) = -10015) – 600 (5-2.5) = -200 leim a Now consider TRA= lookN 2.5m section b/w c and P = born (2-2-5) LY 51 5x 210 m ReallokN Very - - 160 - 600 + 11co - 400 kw M(x) = -100<x>- 600 (-25) +1100 (2-5) cet n=5m point (C) ve),- 4001 MCC) --1015)- 66015-2.5) +1100(5-5) = – 2000 lenom at n=lom point (a) UCO - 400 kw T -W110) - 600 (10-2.5 +10o (10-5) so

ARA wel sf and By diagram 160kW 1 6oolen =P 1 400w =12 A Re=11coln 4oor 4 Loan leon -lookN - Ecol - 250 kw.me c =-2020 knom

(1) Manimum value of normal stress om in beam Given section is W7608 2 57 rolled sted. properties of secrion are d = 772 mm to = 16-6 mm br = 381 mm ter - 27.2 mm y =d/ Iq , moment of inertig = 3430 . xidmms Ix = 3430x10 mm 4 > ktw - be maximum normal stress occur a maximum BM. - (BM/max = -2000 KN-m (at a c} one sign shoe Top part is in tension and part is in compression, Sottom (Smex) at y = d/ at outer fibre Onmeyx? Moy - Scr0 x 10®(Namm) 7 772 (mm) 2x 3430 x 10b (many) 24 Condmaa = 225.072 /um2 or MPa Charax = 225.072. MPM ork Amst SUS

Maximum value of Principal strees Smax and maximum value of shearing stress łmax at junctim of flange and web :- te te? d hum 9.44 =(877-P) we are performing at section-c wax : -20 0 0 (4 (Shear force) v = 4001N Ix = 3430x100 mm tf=27*2 mm bp = 381 mm d=772 mm te - 2712 mm we first calculate normal streess que t (shear stress) at juntion of flange and web on web- normal stress ( at juntion of web and Flange) o= Moy { due to M is Eve) this is above? 2 Noto so it will be in tensions IX y a distance from No to = d-atf-8772-2x 2702 2 Y. 358.9 mm SOBE y = 2orox 106 w-mm 61 - 2000 x 106 x 358.6 ? = 20.92 leima 340x106 - 02 M1qTensile 2 Mle 209 212 To = 209 212 Meg I EWS

Sherr stress on juntom of web and flange on web T - VoG Ix b Ge first moment of area about Noto QE AT = (bexte) x-258 Q = (301x27-2) « [ 772-21272 + 27,27 G = 3859255.68 mm be width of section at which shear stress to be calculate = tw = 16.6mm T 2 VG 400x1ő (N) x 38 5925569 (mm) 3430x10° (mm) x 16-6(mm) Ixb T = 27.11 wImma emoura T a 27.11 DOM PG] stress element Be-6 - 209.212 mRq Try =ł --27.11 Ming > ( -ue sig shows act on tve x Plone but go along ove y-aris y

Principal stress op, or op = Getry + / 18x-on jet au PA or Bez = 209.212.70 1 V1209:312-03 427.252 Tame + sign PR = 209,212 + (2.09.212.J27112 Trei = 212 667 MPa 7 Take Eve)sign - 209.212 //209.212 12 2 E-V1209,212.)+270112 -3.45 MPa] 15p2 = Criax - 212.667 - (-3.45) - - 108.OS Mlg 2 So Maximum principal stress = p = 212.667 MRA imax = 108.05 MRq W AYA

0 since max (212-667J MRa Coal = 350mpg Cux (10~ 6s) ч, 2 Сац = 70 ме, Ther fore given shape W 760x287 is acceptable. © Determine (op) Principe stress plane locations . (@s) o Tmax Plane location Le On = 2d 209.212 MPa Eye - 0 27.11 MRq Cave since act on ry ad tver Ral plane but go along -ve yaxis ney tanzeep = 2 Try on-cay tan zeep - 2 (-27-111 209.212-0 20p - - 14.529° Top = -7026° / Eve sign show cw rotasions) Up (-3-451 129 Cops -7.24° 89, = 212 667M89 (Principal stress element) ARRAY SEUNSEOA

Maximum - x shearing stress plane @s - Cap +945º Os = -7.26 + 45° Os = 37.7353° Ifue sign show cow) Imax = logos MPa and quasege in plane stress on Imax plane Cang. Outry. 209.212 = 104.606 MBA ③ Draw mour circle Scale (20419 = 1cm] 2 ya 7 Suavy Os = 37.73 I Maximum SHEAR Stress element л Стал Daug C 3 SYS Tmax (108.05 mph) - 205 - - - L . 20p - omin utk - oman k Sang (104 GOM Ah