Problem_Chapter2_Solution_materials

Give Me Liberty!: An American History (First Edition, Seagull Edition) (Vol. 2)

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PROBLEMS CHAPTER 2 Changes in Lengths of Axially Loaded Members 2.2-12 A hollow, circular, steel column ( E = 210 GPa) is subjected to a compressive load P , as shown in the figure. The column has length L = 2.5 m and outside diameter d = 200 mm. The load P = 500 kN. If the allowable compressive stress is 55 MPa and the allowable shortening of the column is 0.60 mm, what is the required wall thickness t min ? PROB.2.2-12 Solution: (a) The compressive strength condition () 1 22 1 1 2 2 1 1 4 4 500 4 1 1 0.843 200 55 in allow allow d P d d kN P d mm MPa σα πα α πσ π == =− = × (b) The compressive deformation condition ( ) ( ) 2 2 2 2 2 1 1 4 4 500 2.5 4 1 1 0.827 210 200 0.60 in allow allow d PL d Ed kN m PL GPa mm mm δα πδ ×− = × (c) The required wall thickness is ( ) min min 11 1 200 1 0.827 17.3 td m m m = = m Nonuniform Bars 2.3-4 A reinforced concrete pedestal ( E = 25 GPa) having dimensions L 1 = 2 m and L 2 = 1.5 m is shown in the figure. The loads applied to the pedestal are P 1 = 400 kN and P 2 = 650 kN. Under the action of these loads, the maximum permissible shortening of the pedestal is 1.0 mm. Let A 1 and A 2 represent the cross-sectional areas of the upper
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and lower parts, respectively, of the pedestal. (a) If the area A 2 is three times the area A 1 , what is the minimum permissible area A 1 ? (b) If the areas A 1 and A 2 are such that the compressive stresses in both parts of the pedestal are the same, what is the minimum permissible area A 1 ? PROB.2.3-4 Solution: (a) The internal force in the upper and lower parts are 11 21 2 400 1050 NP k N P k =− =− =− − =− N According to the deformation condition we get () ( ) ( ) ( ) 22 121 1min 1 1 max 32 1 3 1 3 1050 1.5 1 400 2 25 1.0 3 53 10 NL NL NL EA EA EA AN L E kN m kN m GPa mm mm δ ⎛⎞ =+ = + ⎜⎟ ⎝⎠ + (b) The relationship of the area A 1 and A 2 is 12 2 1 1 1050 2.625 400 NN AA kN N 1 A Nk N = == = A Similarly, ( ) ( ) ( ) 1min 1 1 max 1 2.625 1050 1.5 1 400 2 25 1.0 2.625 56 10 L E kN m kN m GPa mm mm + 2.3-6 A steel bar 2.4 m long has a circular cross section of diameter d l = 20 mm over
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one-half of its length and diameter d 2 = 12 mm over the other half (see figure). The modulus of elasticity E = 205 GPa. (a) How much will the bar elongate under a tensile load P = 22 kN? (b) If the same volume of material is made into a bar of constant diameter d and length 2.4 m, what will be the elongation under the same load P ? PROB.2.3-6 Solution: (a) The internal forces in the left and right parts are same. 11 22 NNP k N = == The elongation of the bar is () 2 2 1 2 12 1 2 22 22 1.2 1.2 205 20 12 44 1.55 NL NL P L L EA EA E A A kN mm MPa mm mm mm δ ππ ⎛⎞ =+ = + ⎜⎟ ⎝⎠ = (b) For the same volume of material, we can get the diameter. 2 1 2 2 4 2 20 12 16.49 dd d LL L L d d d mm mm dm π ×+ ×= × + = + + = m Then, the elongation of the bar is ( )( ) 2 2 22 2.4 1.21 1 205 16.49 4 NL k N m mm EA MPa mm = 2.3-8 A wood pile, driven into the earth, supports a load P entirely by friction along its sides (see figure). The friction force f per unit length of pile is assumed to be uniformly distributed over the surface of the pile. The pile has length L , cross-sectional area A , and modulus of elasticity E .
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Problem_Chapter2_Solution_materials - PROBLEMS CHAPTER 2...

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