For the problem below, what is the difference in load from elastic to plastic? (
ID: 1843045 • Letter: F
Question
For the problem below, what is the difference in load from elastic to plastic? (It should increase, but calculate specific value)
The bar in Fig. 4-28a is made of steel that is assumed to be elastic perfectly plastic, with or 250 MPa. Determine (a) the maximum value of the applied load P that can be applied without causing the steel to yield and (b) the maximum value of P that the bar can support. Sketch the stress distribution at the critical section for each case. SOLUTION Part (a) When the material behaves elastically, we must use a stress-concentration factor determined from Fig. 4-23 that is unique for the bar's geometry. Here 4 mm 4 mm 0.125 h 40 mm 8 mm 40 mm 1.25 (40 mm 8 mm 4 mm From the figure K 1.75. The maximum load, without causing (a) yielding, occurs when or max o r. The average normal stress is oavg P/A. Using Eq. 4-6, have we Kor avg 250(106) Pa 1.75 (0.002 m) (0.032 m Pr 9.14 kN This load has been calculated using the smallest cross section. The resulting stress distribution is shown in Fig. 4-28b. For equilibrium, PP the volume" contained within this distribution must equal 9.14 kN. Part (b). The maximum load sustained by the bar will cause all the material at the smallest cross section t yield. Therefore, as P is increased to the plastic load Pr, it gradually changes the stress distribution from the elastic state shown in Fig. 4-28b to the plastic state shown in Fig. 4-28c. We require 250(106) Pa (0.002 m)(0.032 m) PP 16.0 kN Here P, equals the "volume" contained within the stress distribution which in this case is P 40 mm (b) (c) Fig. 4 8 2 mm GYExplanation / Answer
IN this problem load is a point load which is applied axially to the bar.
In the part a the material assumed to before yielding condition i.e here stress distribution is shown in figure 4-28 b
and the maximum load is 9.14 KN
In part b the material is plastic so to calculate maximum value of P here the stress distibution is uniform throgh out the cross section Area.
and load P is 16 KN
Load difference is 16-9.14=8.86 KN