(a) Explanation Given information: The initial length of each wire ( L ) is 3 m. The cross sectional area of each wire ( A ) is 4 mm 2 . The yield stress of the steel wire ( σ Y ) st is 120 MPa. The yield stress of the aluminum wire ( σ Y ) al is 70 MPa. The modulus of the elasticity of the steel wire ( E st ) is 200 GPa. The modulus of the elasticity of the aluminum wire ( E al ) is 120 GPa. Calculation: Draw the Free Body Diagram of the system: Refer to Figure 1: Assumption: Upward force is positive. Calculate the sum of forces in vertical direction: Apply the vertical equilibrium equation. Σ F y = 0 F al + F st − W = 0 (1) Calculate the load displacement equation as both wires behave elastically: δ al = δ st (2) Here, δ al is the displacement in aluminum and δ st is the displacement in steel. Substitute F al L A E al for δ al and F st L A E st for δ st in Equation (2). F al L A E al = F st L A E st F al E al = F st E st (3) Here, F al is the tensile force in the aluminum wire and F st the tensile force in the steel wire. Substitute 70 GPa for E al and 200 GPa for E st in Equation (3). F al 70 = F st 200 F al = 0 (b) To determine To calculate: The force in each wire if the weight is 720 N.

Mechanics of Materials, Student Value Edition (10th Edition)

10th Edition

ISBN: 9780134321189

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 4.9, Problem 4.96P

(a)

To determine

To calculate: The force in each wire if the weight is 600 N.

(b)

To determine

To calculate: The force in each wire if the weight is 720 N.

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Chapter 4.9, Problem 4.95P

Chapter 4.9, Problem 4.97P

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Mechanics of Materials, Student Value Edition (10th Edition)

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The force in each wire if the weight is 600 N.

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