Question

Can you please show me a detailed solution to this problem.?

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 Joon d d kb.

Answer 1

Solution: For E7010 r = 70 kpsi Z .. T= 241.32 Nimme: =482.65 tall = I 2.5 9653 Nimm 2 We are analysing one plate_by considering lood p = E P Direct shear = (50+5025) t=h 5 -3.53 mm 12 ra Direct sbeava P Nimm2 12543-53 Now we find Izzwhich is Ixx + Igy! I = 5oXtX25+50x+x25+25x+xo 50xt + 50xt+25+ Td 22.61° 30 = 20 mm 12.5 3257 22.61 X Ixx = 250x+x(12.5)2] +253 12 = 5.98 x 104 mm Tyy = 2 tX503 +50x2x52) +257x202 12 =11:76x104 mm 4

Page :: Iz2 - Inzt Iyy - = 17748104 mm 4 MEDX23.01 Imax will be at a= mr J DX230X32-5 17.748104 KV302412.52 T= Id?+72 +2 Id I Coso (96.53) + 441-2 4192 (0.042P) + 2x()(0.042) 44125 X LOS 2001 ={5:14710417631057 1871 75X10-4P 1.986X10-3 p? 2 P = 2167 N .: PEF 2 :. F= 4335N