Question

A long steel (SAE 1018HR) plate is welded on both ends as shown in the following figure. The plate thickness is 10mm. The dimensions shown blow are b = 50mm, d 25mm, c = 200mm, h = 5mm. A concentrated force F is applied at the center. The electrode material is E7010. A factor of safety is assumed to be 2.5 for both plate and welding. Find the maximum allowable force F. F wall *** b

Answer 1

Solution-

F de! A BIa C C RA RB Hence force is acting at the centre, su due ta sym mometric RA= RB E F/2

To determine the maximum allowable force,we have to find out maximum allowable support reaction.

Free bady diagram of support А - D т A 1 (10 100t 1 T RA Su ір with підріd tо suppant reaction RA, there w to Hпад усло and come parallel jaint

The permissible temsile streda in weld material is 440 MPa & Permissible shear stress of wella material is 22ompa after 220m Pa after compidening the factor of safety. Syt = 440 fos yt: 440 2.5 I = syt 176 mpa Sys 220 fos sya y - 220 2.5 sys = 88 MPa The strength of trams veroe fillet wella is denoted by RA

RA = 0.707 xhxd x 20 ubot RA, (0.707 xhxo x5 y 1) x2 RAI (0.707 45x50x176) x 2 RA, = 62216N The strength of parallel fillet weld is denated by RAL 0.707 X 5 X 25 X 8 8 RA, ㅋㅋㅋㅋN Tatal support reacticum =) RA =RA,TR, (.62216+ 7777)N 6 9 9 9 3N maximum allowable force & RA= RB = F/2 Fmax = 2X RA=2869993 Fmar 139.986 KN Amower

So,the maximum allowable force F_max is finally found which is 2 x 69.993 = 139.986 KN