# Question: The mechanical properties of a 4ﬂ30-carbon steel have been measured in the lab and are as follows: E = 205 GPa; cry = 860 MPa; UTS = 1230 MPa; v = 0.29. A 22 mm diameter bar of this alloy is used as a structural member in an engineering design. The length of the bar is 2 m -Free Course Hero Question Answer.

Question Description: he mechanical properties of a 4ﬂ30-carbon steel have been measured in the lab and are as follows: E = 205 GPa; cry = 860 MPa; UTS = 1230 MPa; v = 0.29. A 22 mm diameter bar of this alloy is used as a structural member in an engineering design. The length of the bar is 2 m. 3.1) The structural load on the bar is the last ﬁve digits of your SID in newtons. For example, if your SID is 123456239, the load applied is 56,2891″. If the 5’11 digit of your SID is 0 (eg. 123406789) then replace this 0 with a I so the load applied becomes 16,789N. 3.1.1) What will be the length of the bar under this structural load? Justify the use of any equaﬁnn. 3.1.2) What will be the diameter of the bar under this structural load? Justify the use of any equation. 3.2) The design engineer is considering a structural change that will increase the tensile load on this member. What is the maximum tensile load that can be permitted without inducing plastic deformation of the bar?

3.1.1) Length of the bar under the structural load of 56789 N = 2.001458 m

3.1.2) Diameter of the bar under the structural load of 56789 N = 0.0220046 m

3.2) Maximum tensile load that can be permitted without inducing plastic deformation of bar = 326748.4 N

Step-by-step explanation

Image transcription text

`E = 205 GPa = 205 x 10 N / MZ Jy: 860 M Pa - 860 x 106 N/mz UTS = 1280 MPa = 1280x106 N/mz V = 0 . 29 Diameter of bag = 2 2 mm = 0. 022 m Length of bay = 2 m Load = 56789 N Stress o : 56789 TI * (0 1 022 ) 2 - 149, 46 x 10 N / M2 Linear Strain. E = O 149. 46x10 E 205 x 109 = 0'00 0 729 m (ie. Strain corresponds to the load of 5678 9 N) Change in length, SQ = E &amp; = 0.00 0 72 9 x 2 = 0:00 / 45 8 m Length of bay under? = 2 + 0 00 1 458 structural load of 5678 9N = 2100 1458 m Reduction in diameter: Lateral Strain x diameter = Linear Strain x Y x diameter = 0.00 0729 x 0 . 29 x 0. 022 = 0.00 00 0 46 m Diameter of bas under? 10102 +0. 00 00 0 46 structural load of 56789NJ = 0. 0220046 m`