6b - 6.44 This problem asks that we determine the value of...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6.44 This problem asks that we determine the value of ε T for the onset of necking assuming that necking begins when d σ T d ε T = σ T Let us take the derivative of Equation (6.19), set it equal to σ T , and then solve for ε T from the resulting expression. Thus d K ε T ( 29 n d ε T = Kn ε T ( 29 (n- 1) = σ T However, from Equation (6.19), σ T = K ( ε T ) n , which, when substituted into the above expression, yields Kn ε T ( 29 (n - 1) = K ε T ( 29 n Now solving for ε T from this equation leads to ε T = n as the value of the true strain at the onset of necking. 6.45 This problem calls for us to utilize the appropriate data from Problem 6.29 in order to determine the values of n and K for this material. From Equation (6.38) the slope and intercept of a log σ T versus log ε T plot will yield values for n and log K , respectively. However, Equation (6.19) is only valid in the region of plastic deformation to the point of necking; thus, only the 7th, 8th, 9th, and 10th data points may be utilized. The log-log plot with these data points is given below. 113 The slope yields a value of 0.136 for n , whereas the intercept gives a value of 2.7497 for log K , and thus K = 562 MPa. 6.46 (a) In order to determine the final length of the brass specimen when the load is released, it first becomes necessary to compute the applied stress using Equation (6.1); thus σ = F A o = F π d o 2 2 = 11,750 N π 10 x 10- 3 m 2 2 = 150 MPa (22,000 psi) Upon locating this point on the stress-strain curve (Figure 6.12), we note that it is in the linear, elastic region; therefore, when the load is released the specimen will return to its original length of 120 mm (4.72 in.). (b) In this portion of the problem we are asked to calculate the final length, after load release, when the load is increased to 23,500 N (5280 lb f ). Again, computing the stress σ = 23, 500 N π 10 x 10- 3 m 2 2 = 300 MPa (44, 200 psi) 114 The point on the stress-strain curve corresponding to this stress is in the plastic region. We are able to estimate the amount of permanent strain by drawing a straight line parallel to the linear elastic region; this line intersects the strain axis at a strain of about 0.012 which is the amount of plastic strain. The final specimen length l i may be determined from Equation (6.2) as l i = l o (1 + ε ) = (120 mm)(1 + 0.012) = 121.44 mm (4.78 in.) 6.47 (a) We are asked to determine both the elastic and plastic strains when a tensile force of 110,000 N (25,000 lb f ) is applied to the steel specimen and then released. First it becomes necessary to determine the applied stress using Equation (6.1); thus σ = F A o = F b o d o where b o and d o are cross-sectional width and depth (19 mm and 3.2 mm, respectively). Thus σ = 110,000 N 19 x 10- 3 m ( 29 3.2 x 10- 3 m ( 29 = 1.810 x 10 9 N / m 2 = 1810 MPa (265,000 psi) From Figure 6.24, this point is in the plastic region so there will be both elastic and plastic strains From Figure 6....
View Full Document

This note was uploaded on 01/22/2012 for the course ME 2733 taught by Professor Meng during the Fall '10 term at LSU.

Page1 / 30

6b - 6.44 This problem asks that we determine the value of...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online