Question

bjected to the forces shown. Two of the rods are made aterial with Poisson's ratio , and two of the rods have on's ratio , as shown in the figure. Part B-Calculate the lateral strain Anew structure witha simlar setup but different materials is buit as shown in the image (Figure 2). This time, rods /and l are made from glass with E60 GPa and 02. Rods W and IV are made from bronze with E100 GPa and 034. The applied force is P 15 kN, each rod has a cross-sectional area of A 5 cm, and the vertical connecting bars are rigid and cannot rotate. What are the lateral strains in rods and in Express your comma separated answers to three significant figures View Available Hint(s) vec <2 of 2 ,3 III v Submit IV v P Next X Provide Feedback

Answer 1

I = 0.34 L E₂= 100 GPa P= 15 KN I E = 60 GPa Ug = 0.2 Cross-sectional area A=5 cm² > In this problem, rod-I and rod-IV is of no use. So, I have represented by Force directly. Given, vertical bars are rigid and connot rotate that means, longitudinal elongation im rod II and I will be equal. i.e, longitudinal strain in sod II and II will be same. Eg = Eo Eg Eb = Fg = 0.6Fb But, Fg + F = 15 kN - - 0

we get Solving and Fy=5.625 KN Fi = 9.375 KN dow Eye Care Ep-. 19.375*103 5810-4 Eg=/5.625x103 5x10-4 60x109 1oox 109 > Eb = 1.875 X10-4 - Eg = 1.875 810-4 < But, v= - Elateral Elongitudinal - Esateral = -V. E longitudinal =-0.2 X1-975 x164 = -3.75 810-5 But, y=- Elateral Elongitudinal Elateral =-)). Elongitud and = -0.34 x1.875x10-4 --6.375X10-5