MEE 320 Problems Chapter 2, Spring 2018

PROBLEM 2.3 An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam. It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that E  200 GPa, determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire. SOLUTION (a) , or PL AE P AE L     with 2 2 6 2 1 1 (0.005) 19.6350 10 m 4 4        A d 6 2 9 2 (0.045 m)(19.6350 10 m )(200 10 N/m ) 9817.5 N 18 m      P P  9.82 kN  (b) 6 6 2 9817.5 N 500 10 Pa 19.6350 10 m        P A   500 MPa  PROBLEM 2.6 A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN. Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the smallest diameter rod that should be used, (b) the corresponding maximum length of the rod. SOLUTION (a) 2 ; 4   P d A A   Substituting, we have 2 3 6 3 4 4 4(4 10 N) (180 10 Pa) 5.3192 10 m                      P P d d d d d  5.32 mm  (b) ;       E L Substituting, we have        E E L L 9 3 6 (105 10 Pa)(3 10 m) (180 10 Pa)      L L 1.750m  0.4 m 0.5 m P Q 20-mm diameter 60-mm diameter A B C PROBLEM 2.19 Both portions of the rod ABC are made of an aluminum for which E  70 GPa. Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that the deflection at A is zero, (b) the corresponding deflection of B. SOLUTION (a) 2 2 6 2 (0.020) 314.16 10 m 4 4 AB AB A  d       2 2 3 2 (0.060) 2.8274 10 m 4 4 BC BC A  d       Force in member AB is P tension. Elongation: 3 6 9 6 (4 10 )(0.4) 72.756 10 m (70 10 )(314.16 10 )           AB AB AB PL EA Force in member BC is Q  P compression. Shortening: 9 9 3 ( ) ( )(0.5) 2.5263 10 ( ) (70 10 )(2.8274 10 )             BC BC BC Q P L Q P Q P EA For zero deflection at A, BC AB    9 6 3 2.5263 10 ( ) 72.756 10 28.8 10 N           Q P Q P 3 3 3 28.3 10 4 10 32.8 10 N Q       Q  32.8 kN  (b) 72.756 10 6 m AB BC B         0.0728 mm    AB  Problem 2.2 Each of the links AB and CD is made of aluminum ( E = 75 GPa) and has a cross-sectional area of 258 mm2. Knowing that they support the rigid member BC, determine the deflection of point E. Solution Use member BC as a free body 3 3 0: (0.64) (0.44)(5 10 ) 0 3.4375 10 N C AB AB M F F Σ = − + × = = × 3 3 0: (0.64) (0.20)(5 10 ) 0 1.5625 10 N B CD CD M F F Σ = − × = = × For links AB and CD 2 6 2 3 6 9 6 3 6 9 6 258 mm 258 10 m (3.4375 10 )(0.36) 63.953 10 m (75 10 )(258 10 ) (1.5625 10 )(0.36) 29.07 10 m (75 10 )(258 10 ) AB AB AB B CD CD CD C A F L EA F L EA δ δ δ δ − − − − − = = × × = = = × = × × × = = = × = × × 6 6 6 6 6 34.883 10 Slope 54.505 10 rad 0.64 29.07 10 (0.44)(54.505 10 ) 53.05 10 m 0.0531 mm B C BC E C EC l l δ δ θ δ δ θ − − − − − − × = = = × = + = × + × = × =  7 300 mm 60 mm Aluminium shell E � 70 GPa Brass core E � 105 GPa 25 mm PROBLEM 2.37 An axial force of 200 kN is applied to the assembly shown by means of rigid end plates. Determine (a) the normal stress in the aluminum shell, (b) the corresponding deformation of the assembly. SOLUTION Let Pa = Portion of axial force carried by shell. Pb = Portion of axial force carried by core. , or a a a a a a P L E A P E A L     , or b b b b b b P L E A P E A L     Thus, ( ) a b a a b b P P P E A E A L      with 2 2 3 2 2 3 2 [(0.060) (0.025) ] 2.3366 10 m 4 (0.025) 0.49087 10 m 4 a b A A            9 3 9 3 6 [(70 10 )(2.3366 10 ) (105 10 )(0.49087 10 )] 215.10 10             P L P L Strain: 3 3 6 6 200 10 0.92980 10 215.10 10 215.10 10            P L (a) 9 3 6 (70 10 )(0.92980 10 ) 65.1 10 Pa          a Ea 65.1 MPa a   (b) (0.92980 10 3 )(300 mm) L         0.279 mm  Problem 2.44 The rigid bar AD is supported by two steel wires of 1.5 mm diameter ( E = 200 GPa) and a pin and bracket at D. Knowing that the wires were initially taught, determine (a) the additional tension in each wire when a 1.0 kN load P is applied at D, (b) the corresponding deflection of point D. Solution Let θ be the rotation of bar ABCD Then 9 6 300 600 10 200 4 (1.5)(300 ) 10 250 B C BE BE B BE BE BE P L AE EA P L δ θ δ θ δ π θ δ = = =   ×     = = 3 424.1 10 CF CF C P L EA θ δ = × = 9 6 3 10 200 4 (1.5)(600 ) 10 450 471.25 10 CF CF CF EA P L π θ δ θ   ×     = = = × Using free body ABCD 0 300 600 900 0 A BE CF M P P P Σ = + − = 3 3 6 5 (300)(421.1 10 ) (600)(471.25 10 ) (900)(1000) 0 409.08 10 (9 10 ) 0.00220 rad θ θ θ θ × + × − = × = × ∴ = (a) 3 (424.1 10 )(0.00220) 933 N PBE = × = 3 (471.25 10 )(0.00220) 1036.7 N C P = × =  (b) 900 (900)(0.00220) 1.98 mm δD θ = = =  Problem 2.5 A rod consisting of two cylindrical portions AB and BC is restrained at both ends. Portion AB is made of brass ( 105 GPa, b E = 20.9 10 6 / C) αb − = × ° and portion BC is made of aluminum ( 72 GPa, s E = 23.9 10 6 /°C). αs − = × Knowing that the rod is initially unstressed, determine (a) the normal stresses induced in portions AB and BC by a temperature rise of 42°C, (b) the corresponding deflection of point B. Solution 2 2 3 2 3 2 2 2 3 2 3 2 (60) 2.8274 10 mm 2.8274 10 m 4 4 (40) 1.2566 10 mm 1.2566 10 m 4 4 AB AB BC BC A d A d π π π π − − = = = × = × = = = × = × Free thermal expansion. 6 6 3 ( ) ( ) (1.1)(20.9 10 )(42) (1.3)(23.9 10 )(42) 2.2705 10 m T AB b BC a L T L T δ α α − − − = Δ + Δ = × + × = × Shortening due to induced compressive force 9 3 9 3 9 1.1 1.3 (105 10 )(2.8274 10 ) (72 10 )(1.2566 10 ) 18.074 10 BC AB P b AB a BC PL PL E A E A P P P δ − − − = = = + × × × × = × For zero net deflection, P T δ = δ 9 3 3 18.074 10 2.2705 10 125.62 10 N P P − − × = × = × (a) 3 6 3 125.62 10 44.4 10 Pa 44.4 MPa 2.8274 10 AB AB P A σ − × = − = − = − × = − ×  3 6 3 125.62 10 100.0 10 Pa 100.0 MPa 1.2566 10 BC BC P A σ − × = − = − = − × = − ×  (b) 3 6 9 3 6 ( ) (125.62 10 )(1.1) (1.1)(20.9 10 )(42) (105 10 )(2.3274 10 ) 500 10 m 0.500 mm AB B AB b b AB PL L T E A δ α − − − = + − Δ × = − × × × = − × = − i.e. 0.500 mm  2 Problem 2.58 Knowing that a 0.5-mm gap exists when the temperature is 24 C, ° determine (a) the temperature at which the normal stress in the aluminum bar will be equal to −75 MPa, (b) the corresponding exact length of the aluminum bar. Solution 6 6 75 MPa (75 10 )(1800 10 ) 135 kN a a a P A σ σ − = − = − = − × × = − Shortening due to P: 9 6 9 6 6 (135000)(0.35) (135000 0.45) (105 10 )(1500 10 ) (73 10 )(1800 10 ) 762.3 10 m b a P b b a a PL PL E A E A δ − − − = + × = + × × × × = × Available elongation for thermal expansion: 6 3 (0.0005 762.3 10 ) 1.2623 10 m δT − − = + × = × But ( ) ( ) T b b a a L T L T δ α α = Δ + Δ 6 6 6 (0.35)(21.6 10 )( ) (0.45)(23.2 10 )( ) (18 10 ) T T T − − − = × Δ + × Δ = × Δ Equating, 6 3 (18 10 ) 1.2623 10 70.1 C T T − − × Δ = × Δ = ° (a) hot cold 24 70.1 94.1 C T T T = + Δ = + = °  (b) ( ) a a a a a a PL L T E A δ α = Δ − 6 3 9 6 (135000)(0.45) (0.45)(23.2 10 )(70.1) 0.27 10 m (73 10 )(1800 10 ) − − − = × − = × × × 3 exact 0.45 0.27 10 0.45027m L − = + × =  Problem 2.6 A 20-mm-diameter rod made of an experimental plastic is subjected to a tensile force of magnitude P = 6 kN. Knowing that an elongation of 14 mm and a decrease in diameter of 0.85 mm are observed in a 150-mm length, determine the modulus of elasticity, the modulus of rigidity, and Poisson’s ratio for the material. Solution Let the y-axis be along the length of the rod and the x-axis be transverse. 2 2 6 2 3 3 6 6 (20) 314.16 mm 314.16 10 m 6 10 N 4 6 10 19.0985 10 Pa 314.16 10 14 mm 0.093333 150 mm y y y A P P A L π σ δ ε − − = = = × = × × = = = × × = = = Modulus of elasticity: 6 6 19.0985 10 204.63 10 Pa 0.093333 y y E σ ε × = = = × E = 205 MPa  0.85 0.0425 20 x x d ε = δ = − = − Poisson’s ratio: 0.0425 0.093333 x y v ε ε − = − = − v = 0.455  Modulus of rigidity: 6 6 204.63 10 70.31 10 Pa 2(1 ) (2)(1.455) E G v × = = = × + G = 70.3 MPa  1 Problem 2. A fabric used in air-inflated structures is subjected to a biaxial loading that results in normal stresses 120 x σ = MPa and 160 z σ = MPa. Knowing that the properties of the fabric can be approximated as E = 87 GPa and v = 0.34, determine the change in length of (a) side AB, (b) side BC, (c) diagonal AC. Solution 6 6 6 6 6 9 6 6 3 9 120 10 Pa, 0, 160 10 Pa 1 1 ( ) [120 10 (0.34)(160 10 )] 754.02 10 87 10 1 1 ( ) [ (0.34)(120 10 ) 160 10 ] 1.3701 10 87 10 σ σ σ ε σ σ σ ε σ σ σ − − = × = = × = − − = × − × = × × = − − + = − × + × = × × x y z x x y z z x y z v v E v v E (a) 6 ( ) (100 mm)(754.02 10 ) δ ε − = = × AB x AB = 0.0754 mm  (b) 6 ( ) (75 mm)(1.3701 10 ) δ ε − = = × BC z BC = 0.1028 mm  Label sides of right triangle ABC as a, b, and c. 2 2 2 c a b = + Obtain differentials by calculus. 2 2 2 = + = + c dc a da b db a b dc da db c c But 2 2 100 mm, 75 mm, (100 75 ) 125 mm 0.0754 mm 0.1370 mm δ δ = = = + = = = = = AB BC a b c da db (c) 100 75 (0.0754) (0.1028) 125 125 δ = = + AC dc = 0.1220 mm  68 200 mm b a P PROBLEM 2.79 2. An elastomeric bearing (G = 0.9 MPa) is used to support a bridge girder as shown to provide flexibility during earthquakes. The beam must not displace more than 10 mm when a 22 kN lateral load is applied as shown. Determine (a) the smallest allowable dimension b, (b) the smallest required thickness a if the maximum allowable shearing stress is 420 kPa. SOLUTION Shearing force P = 22 ¥ 103 N Shearing stress t = 420 ¥ 103 Pa t = P A \ A = P t = 22 10 420 10 3 3 ¥ ¥ = 52.381 ¥ 10–3 m2 = 52.381 ¥ 103 mm2 A = (200 mm) (b) b = A 200 = 52 381 10 200 3 . ¥ = 262 mm g = t G = 420 10 0 9 10 3 6 ¥ ¥ . = 466.67 ¥ 10–3 But g = d a \ a = d g = 10 mm 466.67 ¥ 10 3 - = 21.4 mm PROBLEM 2.80 2.8 For the elastomeric bearing in Prob. 2.81 with b = 220 mm and a = 30 mm, determine the shearing modulus G and the shear stress t for a maximum lateral load P = 19 kN and a maximum displacement d = 12 mm. SOLUTION Shearing force P = 19 ¥ 103 N Area A = (200 mm)(220 mm) = 44 ¥ 103 mm2 = 44 ¥ 10–3 m2 d P P a 200 mm b a P 79 0 rA 5 20 mm rB 5 15 mm B A 64 mm 88 mm P t PROBLEM 2.98 For P  100 kN, determine the minimum plate thickness t required if the allowable stress is 125 MPa. SOLUTION At the hole: 20 mm 88 40 48 mm A A r d     2 2(20) 0.455 88 A A r D   From Fig. 2.60a, K  2.20 max net max 3 3 6 (2.20)(100 10 N) 36.7 10 m 36.7 mm (0.048 m)(125 10 Pa) A A KP KP KP t A d t d t              At the fillet: 88 88 mm, 64 mm 1.375 64 B B D D d d     15 15 mm 0.2344 64 B B B r r d    From Fig. 2.60b, K  1.70 max min B KP KP A d t    3 3 6 max (1.70)(100 10 N) 21.25 10 m 21.25 mm (0.064 m)(125 10 Pa) B KP t d          The larger value is the required minimum plate thickness.  t  36.7 mm 