Question

A steel bar of thickness his subjected to a bending force F. The vertical support is stepped such that the horizontal welds are bị and b2 long. Determine F for the following two cases if the maximum allowable shear stress is Tallowable. F -b2 h ... b b2 d h 2 in 30 mm 4 in 50 mm 6 in 150 mm 4 in 50 mm la in 5 mm Tallow 25 kpsi 140 MPa

Answer 1

solution:- ba C d ha Tallow bi zin fini sin 4เก 5/6in 25 kpsi 3omm 50 mm 150 mm 50 mm 5mm 140 MPa 0 yn ด ) ou دم r3 - Y 82 672 TX Throat area for weld -1: 0.707h (62) AI . where, Ai Area of first weld h thickness ba y length of weldaa of bar h = 5m, b 2 = 50mm Here, I think b = 30mm & b2 = 50 mm Refer diagram,

I think there is mis-poisit in the diagram. Actual diagram: Кы, - F, d ba " And all other dimension is same. AT 5 0.707 xhxb, Floorih. Į pielea bisita) | 00707 X 5 X 30 ادرار ، Ara 1 Ala 106.05 mm2 Throat Area of weld weld - 2 A2 Oo 707 x habz 0.707* 5* 50, A2 = 176.75 mm2 al A3 Throat Area for Weld-3 0.707xdxn 111.1.3.11 = 0.707 x 5650 - A3= 176.75 mm

Total Ihreal Area =) A = Ait Azt Az A = 106.75 + 176.75+176.75 459-55 mm2r 459.55X10 55x10m2 A- Jotal length Length of of the weld : li bit batd :ل 30+504 50 130mm of centroid Location along a-corection : - ř - Bibs + 12 b2+363d I bitbatd) XI 15x30 + 25x 50 + 0x50 130 13.0mm Location of centroid aleng y-direction:- Y = Gibit yaba+yad bit batd 50X300x50 + 25x50 » 21.154 om 130 Distance between weld & weld. b. 1 x)²+(d-F)? -/ 13.071)2 + (50- 30 + (50-21-154)2 =) 28.91mm Vi= 28.91mm

Distance between weld & weld-2: V, 5/652 -3)*+ (9) 50 -13.077) ? +(21-154)? 12 - : 125 24.285 mm Distance bir weld & weld - 38 - 335 ()2+(-)? /(15.047)2 + 5.9.-21.154)? 83 13.63imon Polas moment of inertia of weldes about ital Centroid :- 3 Jos 0.70 hbi 12 J .. 0-70 7X 5.x.(30) x.(301.7953-75 mnogo (2.12 Similarly :- I'm = D.Io #xh hab 5x(503 3 =) 0:707 X 5 X 1 2 12 Jai 36822.91 mma 0.707x hx d3 5 0.707* 5x (5033 12 12 J3 36822.919 mmt

polos moment of inertia of the total weld :- J = (J,+ A1972)+( 12 + A2 732+(33+A303) J: (7953.75 + 106.05X(28-912) +(36822.917+176.75*24.283) + | 36822.917 +176-75% 13.631 -637) J: 307298.925 mm Distance is where maximum secondary shear stress Occurs. 15 / ()* + (1 -89 - /(21-159.92 + (50 -13.071)2 ps 42. 553 mm Angle of go with ac-axis: on. tono da tany Ý ri (62-x) tariyat La tani' 21.154 - L:29.809° 50-13.077) Moment about weld conter: FC C+b ) Fl.150.+50 -13.077 ) 186.923 KNmm Primary shear strees in F KN I' =) 2176 FMP a 459.55xjöbo Secondary sheas treed: 186.923 x 42.553 =) 25.884 MPa 307298•25 А m2 : ااح

Maximum where streedi- Tmax = 1 (T"get(T"47")? 2 (25. 25.884 F dior 29.8090J2+(25.884 FCOS 29.809+2017CF) Tmax = 270 793 FM Pa Now, We can calculate Force 'f' by safe condition, tmax & Tallow 276793 F : 140 5.037 KN Hence, The maximum allowable force is 5.037KN.