A brass rod with a length of 1.40 m and a cross-sectional

Solution 93P Step 1 of 7: In the given problem, two rods of brass and nickel are fastened end to end with force subjected to compound rod as F=4× 10 N equal and opposite. Where length and cross 2 4 2 sectional area for the brass rod is L1= 1.4 mand A = 21cm = 2 × 10 m with young’s 10 modulus Y 1 9 × 10 Pa, for nickel rod with length and area L 2 Land A = 1 cm = 1 × 10 m with young’s modulus Y = 21 × 10 Pa. 10 2 2 In part (a) we need to calculate the length of the nickel rod L, when the elongations in two rods are equal (also forces are equal) using definition of stress. Where in Part (b) and (c) we need to estimate the stress and strain in each rod using basic equations.