HW5 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 04 Question 6.3: A specimen of aluminum having a rectangular cross section 10 mm 12.7 mm (0.4 in. 0.5 in.) is pulled in tension with 35,500 N (8000 lb f ) force, producing only elastic deformation. Calculate the resulting strain. Solution: This problem calls for us to calculate the elastic strain that results for an aluminum specimen stressed in tension. The cross-sectional area is just (10 mm) (12.7 mm) = 127 mm 2 (= 1.27 10 -4 m 2 = 0.20 in. 2 ); also, the elastic modulus for Al is given in Table 6.1 as 69 GPa (or 69 10 9 N/m 2 ). Combining Equations 6.1 and 6.5 and solving for the strain yields = E = F A 0 E = 35,500 N ( 1.27 10 4 m 2 )( 69 10 9 N/m 2 ) = 4.1 10 -3 -------------------------------------------------------------------------------------------------------------------------------------------- Question 6.5: A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled in tension with a load of 89,000 N (20,000 lb f ), and experiences an elongation of 0.10 mm (4.0 10 -3 in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel. Solution: This problem asks us to compute the elastic modulus of steel. For a square cross-section, A 0 = b 0 2 , where b 0 is the edge length. Combining Equations 6.1, 6.2, and 6.5 and solving for E , leads to E ±±=± F A 0 l l 0 Fl 0 b 0 2 l (89,000 N) ( 100 10 3 m ) (20 10 3 m ) 2 (0.10 10 3 m ) = 223 10 9 N/m 2 = 223 GPa (31.3 10 6 psi)
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Question 6.7: For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 10 6 psi). (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm 2 (0.5 in. 2 ) without plastic deformation? (b) If the original specimen length is 115 mm (4.5 in.), what is the maximum length to which it may be stretched without causing plastic deformation? Solution: (a) This portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation ( F y ). Taking the yield strength to be 275 MPa, and employment of Equation 6.1 leads to F y = y A 0 ( 275 10 6 N/m 2 )( 325 10 -6 m 2 ) = 89,375 N (20,000 lb f ) (b) The maximum length to which the sample may be deformed without plastic deformation is determined from Equations 6.2 and 6.5 as l i l 0 1 E = (115 mm) 1 275 MPa 115 10 3 MPa = 115.28 mm (4.51 in.) -------------------------------------------------------------------------------------------------------------------------------------------- Question 6.13: In Section 2.6 it was noted that the net bonding energy E N between two isolated positive and negative ions is a function of interionic distance r as follows: E N  A r B r n (6.25) where A, B, and n are constants for the particular ion pair. Equation 6.25 is also valid for the bonding energy between adjacent ions in solid materials. The modulus of elasticity E is proportional to the slope of the interionic force–separation curve at the equilibrium interionic separation; that is,
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E dF dr r o Derive an expression for the dependence of the modulus of elasticity on these A, B, and n parameters (for the two-ion system) using the following procedure: 1. Establish a relationship for the force F as a function of r, realizing that F dE N dr 2. Now take the derivative dF/dr.
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This note was uploaded on 09/11/2011 for the course ENG 407 taught by Professor Tsakal during the Spring '11 term at Rutgers.

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HW5 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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