32 3.0 26 24 A 22 20 13 1.3 12 1.6 0 0. 2 3 0405 06 0 rD (a) Flat bars with hole

Question

32 3.0 26 24 A 22 20 13 1.3 12 1.6 0 0. 2 3 0405 06 0 rD (a) Flat bars with holes nat bars with fi ts Fig. 2.60 Stress concentration factors for flat bars under axial loading shown in Fig, 2.60. To determine the maximum stress occurring near a discontinuity in a given member subjected to a given axial load P, the designer needs orly to compnite the avenge stress-= P/A in the critical section, and multiply the result obtained by the appropriate value fh valid Nole " at the average stress mus, be computed ocross the norrowes, secion: er"-P/k where t is the thickness of the bor factor K You should note, owever, that this is valid only as long as --dos not exeed the proportional of the material, since the values of K plotted in Fig 260 were obtained by assuming a linear relation between stress and strain. EXAMPLE 2.12 Deternine the largest axial load P that can be safely supported by a flat steel har consisting of two portiottboth 10 mm thick and, respectively 40 and 60 mm wide, connected bllets of radiusr8 mm. Assume an allowable mormal stress of 165 We first compute the rati D 60 d 40 5 m 40 mm 0.20 Using the cuve in Fig 2.60% cor Aponding to Dld -1.50, we find that the value of the tor/d=0.20 is K 182 Carrying this value into Pepe!as) and salvinr for … we have 152 But ". cannot exceed the alknvalle stress -165 MPa Substituting this value for owe find that the average stress in the narrower portion (d = 40 mm) of the bar should not exeed the value 63 MP 163 MP 182 90.7 MPa Recalling that a," =P/A, we have p " A (401mmx10mm)(90.7 MPa) = 36.3 × 103 N P-363kN 116 York 197

Explanation / Answer

The graph is just for giving an insight that the behavior is of decreasing nature. In exams , you are usually provided with specific value or tables

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