Question

3. No image for this problem. Assume we have a composite cylinder under torsion. The inner diameter is 2 inches and the outer is 4 inches. There are two materials with two different shear moduli: G1-2000 ksi and G2-8000 ksi. Consider two cases: a. The inner (core) material is composed of G1 and the shell of G2. The stress at a radius of 2 inches is 1 ksi. Plot the shear stress as a function of the radius. Note: This is really easy if we do in these steps: divide by G2 to get the strain at p-2, then since strain is linear get the strain at the interface. Then just multiply the strain by G1 and G2 to get the values of shear stress on either side of the interface. Finally, use τ_Tpl to get the total torque required to generate a shear stress of 1 ksi at ρ Now flip the materials. The inner (core) material is composed of G2 and the shell of G1. The stress at a radius of 2 inches is 1 ksi. Plot the shear stress as a function of the radius and compute the total torque. Notice the difference in torque required as we move the stiffer (higher G) material to the shell b.

Answer 1

stress at r = 2 in. Strain at r= 2 in. = 1 ksi 8000 ksi =0.000125 = 0.000125 0.0000625 n =1 in. r = 2 in. strain diagram Calculate the Strain at r = lin. by using the property of similar triangles: n1 0.0001250 2 1 0 =0.0000625 The strain at the interface of two composite materials is 0.0000625.

Strain of the material having modulus of rigidity G: TA=0xG TA=0.0000625Ⓡ 2000 TA=0.125ksi Strain of the material having modulus of rigidity G: T3 = 4G Tg = 0.0000625x 8000 Tg = 0.5ksi Shera stress as a function of radius: 5 = 1 ksi Tg = 0.5 ksi T = 0.125 ksi ri = 1 in. r = 2 in. stress diagram

Now, for series combination: T =T+T2 TE GJEJ R R 1x- 0.125x - T= 2 1 t = 0.1963 +11.78 T=11.97 lbfin. very BEST

(b) stress at r = 2 in. Strain at r = 2 in. =- G 1 ksi 2000 ksi = 0.0005 A = 0.0005 02 = 0.00025 r2 = 1 in. r = 2 in. strain diagram Calculate the Strain at r = lin. by using the property of similar triangles: 11 12 0.0005_02 2 1 6 = 0.00025 The strain at the interface of two composite materials is 0.00025.

Strain of the material having modulus of rigidity G: TA = 0, G, 74 = 0.00025x8000 TA = 2 ksi Strain of the material having modulus of rigidity G: 73 = 02xG Iz = 0.00025Ⓡ 2000 Tg = 0.5ksi Shera stress as a function of radius: 25 = 1 ksi to = 0.5 ksi 71 = 2 ksi r2 = 1 in. r; = 2 in. stress diagram

Now, for series combination: T=T; +T I-TJEJ CRR Ax(24 – 14)Ax14 1x 2 - T=_ 2 t=3.141+11.78 T=14.92 lbf.in.