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# Undeterminate problems

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### Undeterminate problems

1. 1. Problem Number (1) A 250-mm bar of 15 * 30-mm rectangular cross section consists of two aluminum layers, 5-mm think, brazed to a center brass layer of the same thickness. If it is subjected to centric forces of magnitude P = 30KN, and knowing that Ea = 70GPa and Eb = 105GPa, determine the normal stress (a) in the aluminum layers, (b) in the brass layers. Solution: Fa : is the force in the aluminum layers. Fs : is the force in the steel layer. Fa + Fs = 30KN (1) β π΄ = πΉπ πΏ π πΈ π π΄ π = πΉπ (250)(10)β3 (70)(10)9(10)(30)(10)β6 = 1.19(10)β8 πΉπ ππ Ξ π  = πΉπ  πΏ π  πΈπ  π΄ π  = πΉπ (250)(10)β3 (105)(10)9(5)(30)(10)β6 = 1.587(10)β8 πΉπ  ππ Compatibility condition: βa = βs β΄ πΉπ = 1.334 πΉπ  From (1) :
2. 2. β΄ πΉπ  = 12.85 πΎπ β΄ ππ  = 85.67 πππ β΄ πΉπ = 17.15 πΎπ β΄ ππ = 57.17 MPa Problem Number (2) Compressive centric forces of 160KN are applied at both ends of the assembly shown by means of rigid plates. Knowing that Es = 200 GPa and Ea = 70GPa, determine (a) the normal stresses in the steel core and the aluminum shell, (b) the deformation of the assembly. Solution: Let the force Ps be the force that affect on the steel core and the force Pa be the force that affect on the Aluminum shell. Then, Ps + Pa = P β΅ ΞL = P L A E , ΞLs = ΞLa β΄ Ps L As Es = Pa L Aa Ea As = 3.14 * 12.5 * 12.5 * 10-6 = 4.9 * 10-4 m Aa = 3.14 * ((31)2 β (12.5)2) * 10-6 = 2.527 * 10-3 m
3. 3. Ps = P - Pa β΄ (P β Pa) L As Es = Pa L Aa Ea β΄ Pa = AaEa P AaEa + AsEs = 102848.9 N β΄ Οa = 40.7 MPa β΄Ps = 160000 β 102848.9 = 57151.1 N β΄ Οs = 116.3 MPa The deformation of the assembly = 102848.9 Γ 0.25 2.527 Γ 10β3 Γ 70 Γ 109 = 0.145 mm Problem Number (3) Three steel rods E = 200 GPa support a 36-KN load P. Each of the rods AB and CD has a 200-mm2 cross sectional area and rod EF has a 625-mm2 cross sectional area. Neglecting the deformation of rod BED, determine (a) the change in length of rod EF, (b) the stress in each rod.
4. 4. Let the force P1 affect on the rod EF of L1 = 0.4m and of cross section area 625 * 10-6 m2 and the force P2 affect on the rod CD of L2 = 0.5 m and of cross section area 200 * 10-6 m2 and the force P3 affect on the rod AB of L3 = 0.5 m and of cross section area 200 * 10-6 m2 . Then, P = P1 + P2 + P3 Since, P2 = P3 , then P = P1 + 2P2 , then P2 = Pβ P1 2 ΞL1 = ΞL2 π1 πΏ1 π΄1 πΈ = π2 πΏ2 π΄2 πΈ P1 L1 A1 = (P β P1) L2 2A2 P1 = P L2 A1 L2 A1 + 2A2 L1 = 36000 β 0.5 β 625 β 10β6 (0.5 β 625 β 10β6) + (2 β 200 β 0.4 β 10β6) = 23.81 KN ΞπΏ1 = π1 πΏ1 π΄1 πΈ = 23810 β 0.4 625 β 200 β 103 = 0.0762 mm π1 = π1 π΄1 = 23810 625 β 10β6 = 38.1 MPa π2 = π3 = π2 π΄2 = (36 β 23.81) β 103 2 β 200 β 10β6 = 30.475 MPa
5. 5. Problem Number (4) Two cylindrical rods, one of steel and the other of brass, are joined to C and restrained by rigid supports at A and E. for the loading shown and knowing that Es = 200 GPa and Eb = 105 GPa, determine (a) the reactions at A and E, (b) the deflection of point C. By dividing the rod to four parts 1) DE = 0.1m , 2) CD = 0.1m , 3) BC = 0.12m , 4) AB = 0.18m, and by taking the force at point B to direction of the force at D. Ξ1 = 0 Ξ2 = 40 β 1000 β 0.1 15 β 15 β 3.14 β 105 β 103 = 5.4 β 10β5 π Ξ3 = 40 β 1000 β 0.12 20 β 20 β 3.14 β 200 β 103 = 1.91 β 10β5 π Ξ4 = 100 β 1000 β 0.18 20 β 20 β 3.14 β 200 β 103 = 7.166 β 10β5 π β = Ξ1 + Ξ2 + Ξ3 + Ξ4 = 0.14476 mm RA + RE = 100KN By dividing the rod to two parts 1) CE and 2) AC Ξ1 + Ξ2 = 0.14476 ππ
6. 6. Ξ1 = π πΈ β 0.2 15 β 15 β 3.14 β 105 β 103 β (1) Ξ2 = π πΈ β 0.3 20 β 20 β 3.14 β 200 β 103 β (2) π‘βππ, π πΈ ( 0.2 15 β 15 β 3.14 β 105 β 103 + 0.3 20 β 20 β 3.14 β 200 β 103 ) = 0.14476 Then, RE = 37.21 KN , RA = 62.79 KN The deflection of the point C is : ( 1.91 * 10-5 ) + ( 7.166 * 10-5 ) β ( 4.443 * 10-5 ) = 46.3 ππ