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 ? = 50??, ? = 25??, ? = 200??, ℎ = 5??. A concentrated force ? 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 ?.

Answer 1

SAE 1018HR: y Su = 400 MPa,S, = 220 MPa Sut = 482 MPa,S, = 393 MPa E7010 Electrode : Comparing the minimum strength between plate and weld, SAE 1018HR control the design. Strength of SAE 1018HR will be used to find maximum allowable force. S, 220 y - 88 MPa allow 2.5 ny min (0.30S...,0.40S, )= min (0.30(400),0.40(220))= min (120,88) = 88 MPa I mit T min 88 T = = 35.2 MPa allow 2.5

Free body diagram of plate. F b b d RA 1,=0 RA + Rg - F = 0 R+R, = ? .(1) Free body diagram of left weld.

F, F. E. S There are 2 transverse fillet and 1 parallel fillet. Calculate strength of transverse fillet weld. fr Pallow Amroat (2 is due to two transverse fillet weld) = 0 F, = 88(2x throat xb) F, = 88(2*0.707hxb) F, = 88(2*0.707x5x50) F, = 31108 N

Calculate strength of parallel fillet weld. S F. = T allon Ahroat F. = 35.2(throat x d) F. = 35.2(0.707xhxd) F = 35.2(0.707x5x25) F, = 3110.8 N S Total force acting on the left fillet weld RA = F, +F,. R2 =3110843110.8 R, = 34218.8 N Calculate maximum allowable force. Due to symmetry R = R , substitute in equation 1.

Fmax = 2R FM = 2 x 34218.8 F = 68437.6 N = MAX = MAX