Undeterminate problems
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1) The document contains 4 engineering problems involving the calculation of stresses, forces, and deformations in mechanical structures under applied loads. 2) In problem 1, the normal stresses in the aluminum and brass layers of a composite bar are calculated. 3) Problem 2 determines the stresses in the steel core and aluminum shell of an assembly under compression, as well as the total deformation. 4) Problem 3 finds the change in length, stress, and distributed forces in three supporting rods with different cross-sectional areas. 5) The final problem calculates the reactions, stresses, and deflection at point C in a structure composed of steel and brass rods.
Undeterminate problems
- 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) ∆ 𝐴 = 𝐹𝑎 𝐿 𝑎 𝐸 𝑎 𝐴 𝑎 = 𝐹𝑎 (250)(10)−3 (70)(10)9(10)(30)(10)−6 = 1.19(10)−8 𝐹𝑎 𝑚𝑚 Δ 𝑠 = 𝐹𝑠 𝐿 𝑠 𝐸𝑠 𝐴 𝑠 = 𝐹𝑠(250)(10)−3 (105)(10)9(5)(30)(10)−6 = 1.587(10)−8 𝐹𝑠 𝑚𝑚 Compatibility condition: ∆a = ∆s ∴ 𝐹𝑎 = 1.334 𝐹𝑠 From (1) :
- 2. ∴ 𝐹𝑠 = 12.85 𝐾𝑁 ∴ 𝜎𝑠 = 85.67 𝑀𝑃𝑎 ∴ 𝐹𝑎 = 17.15 𝐾𝑁 ∴ 𝜎𝑎 = 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 ∵ ΔL = P L A E , ΔLs = ΔLa ∴ 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 – (12.5)2) * 10-6 = 2.527 * 10-3 m
- 3. Ps = P - Pa ∴ (P − Pa) L As Es = Pa L Aa Ea ∴ Pa = AaEa P AaEa + AsEs = 102848.9 N ∴ σa = 40.7 MPa ∴Ps = 160000 – 102848.9 = 57151.1 N ∴ σs = 116.3 MPa The deformation of the assembly = 102848.9 × 0.25 2.527 × 10−3 × 70 × 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. 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 = P− P1 2 ΔL1 = ΔL2 𝑃1 𝐿1 𝐴1 𝐸 = 𝑃2 𝐿2 𝐴2 𝐸 P1 L1 A1 = (P − P1) L2 2A2 P1 = P L2 A1 L2 A1 + 2A2 L1 = 36000 ∗ 0.5 ∗ 625 ∗ 10−6 (0.5 ∗ 625 ∗ 10−6) + (2 ∗ 200 ∗ 0.4 ∗ 10−6) = 23.81 KN Δ𝐿1 = 𝑃1 𝐿1 𝐴1 𝐸 = 23810 ∗ 0.4 625 ∗ 200 ∗ 103 = 0.0762 mm 𝜎1 = 𝑃1 𝐴1 = 23810 625 ∗ 10−6 = 38.1 MPa 𝜎2 = 𝜎3 = 𝑃2 𝐴2 = (36 − 23.81) ∗ 103 2 ∗ 200 ∗ 10−6 = 30.475 MPa
- 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. Δ1 = 0 Δ2 = 40 ∗ 1000 ∗ 0.1 15 ∗ 15 ∗ 3.14 ∗ 105 ∗ 103 = 5.4 ∗ 10−5 𝑚 Δ3 = 40 ∗ 1000 ∗ 0.12 20 ∗ 20 ∗ 3.14 ∗ 200 ∗ 103 = 1.91 ∗ 10−5 𝑚 Δ4 = 100 ∗ 1000 ∗ 0.18 20 ∗ 20 ∗ 3.14 ∗ 200 ∗ 103 = 7.166 ∗ 10−5 𝑚 ∆ = Δ1 + Δ2 + Δ3 + Δ4 = 0.14476 mm RA + RE = 100KN By dividing the rod to two parts 1) CE and 2) AC Δ1 + Δ2 = 0.14476 𝑚𝑚
- 6. Δ1 = 𝑅 𝐸 ∗ 0.2 15 ∗ 15 ∗ 3.14 ∗ 105 ∗ 103 → (1) Δ2 = 𝑅 𝐸 ∗ 0.3 20 ∗ 20 ∗ 3.14 ∗ 200 ∗ 103 → (2) 𝑡ℎ𝑒𝑛, 𝑅 𝐸 ( 0.2 15 ∗ 15 ∗ 3.14 ∗ 105 ∗ 103 + 0.3 20 ∗ 20 ∗ 3.14 ∗ 200 ∗ 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 ) – ( 4.443 * 10-5 ) = 46.3 𝜇𝑚