Elasticity Solved Example

ELASTICITY SOLVED EXAMPLE Ex.1 A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate (a) stress, (b) elongation, and (c) strain on the rod. Young’s modulus, of structural steel is 2.0 ×1011 N m-2. Ans. We assume that the rod is held by a clamp at one end, and the force F is applied at the other end, parallel to the length of the rod. Then the stress on the rod is given by F F Stress   A r 2 00  103N = 3.18 × 108 N m–2 3.14  102 m2 The elongation, (F / A)L 3.18  108 Nm2  1m  Y 2  1011Nm2 –3 = 1.59 ×10 m = 1.59 mm The strain is given by Strain = L/L = (1.59 ×10–3 m)/(1m) = 1.59 × 10–3 = 0.16 % L  Ex.2 A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied. Ans. The copper and steel wires are under a tensile stress because they have the same tension (equal to the load W) and the same area of cross-section A. we have stress = strain × Young’s modulus. Therefore W/A = Yc × (L c/L c) = Y s × (L s/L s) where the subscripts c and s refer to copper and stainless steel respectively. Or, Lc/Ls = (Ys/Yc) × (Lc/Ls) Given Lc = 2.2 m, Ls = 1.6 m, Yc = 1.1 × 1011 N.m–2, and Ys = 2.0 × 1011 N.m–2. Lc/Ls = (2.0 × 1011/1.1 × 1011) × (2.2/1.6) = 2.5. The total elongation is given to be Lc + Ls = 7.0 × 10-4 m Solving the above equations, Lc = 5.0 × 10-4 m, and Ls = 2.0 × 10-4 m. Therefore W = (A × Yc × Lc)/Lc = (1.5 × 10-3)2 × [(5.0×10-4×1.1×1011)/2.2] = 1.8 × 102 N Ex.3 In a human pyramid in a circus, the entire weight of the balanced group is supported by the legs of a performer who is lying on his back (as shown in Fig. 9.5). The combined mass of all the persons performing the act, and the tables, plaques etc. involved is 280 kg. The mass of the performer lying on his back at the bottom of the pyramid is 60 kg. Each thighbone (femur) of this performer has a length of 50 cm and an effective radius of 2.0 cm. Determine the amount by which each thighbone gets compressed under the extra load. Ans. Total mass of all the performers, tables, plaques etc. = 280 kg Mass of the performer = 60 kg Mass supported by the legs of the performer at the bottom of the pyramid = 280 – 60 = 220 kg Weight of this supported mass = 220 kg wt. = 220 × 9.8 N = 2156 N. Weight supported by each thighbone of the performer = ½ (2156) N = 1078 N. the Young’s modulus for bone is given by Y = 9.4 × 109 N m–2. Length of each thighbone L = 0.5 m the radius of thighbone = 2.0 cm Thus the cross-sectional area of the thighbone A =?× (2 × 10-2)2 m2 = 1.26 × 10-3 m2. Using Eq. (9.8), the compression in each thighbone (?L) can be computed as L = [(F × L)/(Y × A)] = [(1078 × 0.5)/(9.4 × 109 × 1.26 × 10-3)] = 4.55 × 10-5 m or 4.55 × 10-3 cm. This is a very small change! The fractional decrease in the thighbone is L/L = 0.000091 or 0.0091%. Ex.4 A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 ×104 N. The lower edge is riveted to the floor. How much will the upper edge be displaced? Ans. The lead slab is fixed and the force is applied parallel to the narrow face as shown in Fig. 9.7. The area of the face parallel to which this force is applied is A = 50 cm ×10 cm = 0.5 m × 0.1 m = 0.05 m2 Therefore, the stress applied is = (9.4 ×104 N/0.05 m2) = 1.80 ×106 N.m–2 We know that shearing strain = (x/L)= Stress /G. Therefore the displacement x = (Stress XL)/G = (1.8 ×106 N m–2 ×0.5m)/(5.6 ×109 N m–2) = 1.6 × 10–4 m = 0.16 mm Ex.5 The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, V/ V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 ×109 N m–2. (Take g = 10 m s–2) Ans. The pressure exerted by a 3000 m column of water on the bottom layer p = hg = 3000 m ×1000 kg m–3 × 10 m s–2 = 3 ×107 kg m–1 s–2 = 3 ×107 N m–2 Fractional compression V/V, is V/V = stress/B = (3 ×107 N m–2)/(2.2 ×109 N m–2) = 1.36 ×10–2 or 1.36 %