PROBLEM #1.1 (4 + 4 points, no partial credit A thermal switch consists of a copper bar which under elevation of temperature closes a gap and closes an electrical circuit. The copper bar possesses a length L, modulus E, and coefficient of thermal epansion a. At room temperature T 0 the gap is d. (1 The gap is closed at a temperature T C calculated from equation (circle one: At T 0 L d (a d = α(t C T 0 (b d = Lα(T C (c d = Lα(T C T 0...correct (d d = Eα(T C T 0 (2 As the temperature is increased beond T C, the there eist stress in the bar equal to (circle one: (a σ = Eα(T T 0 (b σ = Eα(T T C...correct (c σ = Eα(T T 0 (d σ = ELα(T T 0 1 ME 323 Final Eam, Sample & Solutions
PROBLEM #1.2 (4 + 4 points, no partial credit (1 A clindrical pressure vessel with diameter D=2.5 m and wall thickness t=10 mm is filled with gas at an internal pressure of p=2400 kpa. The absolute maimum shear stress in the vessel is (circle one (a 300 MPa (b 150 MPa. correct (c 75 MPa (d 37.5 MPa (2 A propane tank with shape of a clindrical pressure vessel has a diameter D=12 in. and a wall thickness of t=1/8 in. The tank is pressuried to p=200 psi. Choose the Mohr s circle diagram that corresponds to the state of stress!!!!! (circle one. (a is correct! 2 ME 323 Final Eam, Sample & Solutions
!! PROBLEM #1.3 (4 + 4 points, no partial credit A circular torsion bar consists of a core material (diameter D core, shear modulus G core and a sleeve material (outer diameter D sleeve, G sleeve bonded firml together at the interface. An eternal torque T is applied to the composite shaft. (1 In the analsis of this problem one needs to consider that (circle one: (a The stress distribution is possesses a jump (i.e. is discontinuous at the interface. Correct (b The strain distribution is possesses a jump (i.e. is discontinuous at the interface. (c Both stress and strain possess a jump (i.e. are discontinuous at the interface. 3 ME 323 Final Eam, Sample & Solutions
(d Neither stress and strain possess a jump (i.e. are discontinuous at the interface. (2 Which ones of the following statements are true regarding the maimum shear stress τ ma and its location in the bar? (circle one (a The maimum shear stress τ ma alwas occurs in the sleeve at the outer surface of the sleeve. (b If the cross sectional area of the sleeve is greater than that of the core, then the maimum stress τ ma occurs on the outer surface of the sleeve; and if the cross sectional area of the core is greater than that of the sleeve then τ ma occurs on the outer surface of the core. (c If GsleeveDsleeve > GcoreD then τ core ma occurs on the outer surface of the sleeve; and if GsleeveD sleeve <GcoreD then τ core ma occurs on the outer surface of the core. Correct (d If GsleeveJsleeve > GcoreJ then τ core ma occurs on the outer surface of the sleeve; and if GsleeveJ sleeve <GcoreJ then τ core ma occurs on the outer surface of the core. (e The maimum shear stress τ ma alwas occurs at the outer surface of the core. (f None of the above. PROBLEM #2 (25 points A truss structure consists of two members AB and BC which are connected b pin joints. The rigid member AB is subjected to a distributed load of magnitude 10kN/m. The deformable member BC possesses a circular cross section of diameter D. Determine the diameter of the truss BC in order to prevent buckling of the member BC with a factor of safet FS=2.9. Consider that the modulus of the material used for the truss BC is E=210GPa. 4 ME 323 Final Eam, Sample & Solutions
10 kn/m 1 m 2 m 5 ME 323 Final Eam, Sample & Solutions
6 ME 323 Final Eam, Sample & Solutions
PROBLEM #3 (25 points A Diameter d=1 cm Diameter D=3 cm C 100 N A force is applied at point C to a (ver thin circular rigid plate with diameter D=3cm. The plate is attached to an elastic shaft (AB of diameter d=1cm and length L=10cm. All the strain energ is stored in the shaft (none in the plate since it is rigid. Also assume that strain energ stored due to transverse shear is negligible. The material is Aluminum (E= 73 GPa, G=27 GPa Using the work-energ principle determine the deflection of point C in the direction of the applied force (i.e. along +. Both torsion and bending are to be considered. B 1 2 Pδ = U el R = D / 2 U el = T 2 L 2GJ + M 2 2EI d = L U el = (P R2 L 2GJ 0 + P2 2EI L 0 (P R 2 L 2GJ 2 d = + L 0 (P R 2 L 2GJ (P 2 2EI 1 2 Pδ = P2 R 2 L 2GJ + P2 L 3 6EI δ = PR2 L GJ + PL3 3EI d = + P2 2EI L 3 3 = P2 R 2 L 2GJ + P2 L 3 6EI 7 ME 323 Final Eam, Sample & Solutions
σ M = 2 [(σ 2 σ 2 + (σ σ 2 + (σ σ 2 + 6(τ 2 + τ 2 + τ 2 ] For A and C σ M = 2 [(σ 2 2 + (σ 2 + 6τ 2 ] For B and D σ M = 2 2 [(σ 2 + (σ 2 + 6τ 2 ] 8 ME 323 Final Eam, Sample & Solutions
9 ME 323 Final Eam, Sample & Solutions
PROBLEM #4 (26 points At a particular cross section of a shaft the internal resultants were determined to consist of a torque (T = 2500 lb.in, a bending moment (M = 2000 lb.in and a shear force (V = 300 lb. The shaft possesses a circular cross-section with radius r = 0.5 in. Determine the states of stress at locations A,B,C,D on the cross-section. Document our answer b drawing the respective material elements at each location with the stress components clearl indicated b vectors and stress magnitudes. Determine the factor of safet using the vom Mises criterion, considering that the ield strength is 30 ksi. A V = 300 lb B D T = 2500 lb.in C 0.5 in M = 2000 lb.in 10ME 323 Final Eam, Sample & Solutions
The stresses at points A, B, C, and D will be due to torsion bending moment shear force A Torsion D T = T = 2500 lb.in π 4 3 4 J = c = 98.17 10 in 2 Tc ( 2500 lb.in( 0.5 in τ = = = 12.73 ksi (CCW 4 J 98.17 in ( B C T ρ τ = J @ A, @ B, @ C, @ D, τ τ = τ = 12.73 ksi (towards +Z = τ = 12.73 ksi (towards -Y τ = τ = 12.73 ksi (towards -Z τ = τ = 12.73 ksi (towards +Y compressive A Bending moment M = 2000 lb.in π 4 3 4 I = c = 49.09 10 in 4 M 2500 lb.in 0.5 in + @A, σ = = I 49.09 10 3 in 4 @B, @C, @B, ( ( ( = 20.37 ksi (compressive M ( 2500 lb.in( 0 in σ = = = 0 ksi I 49.09 10 3 in 4 σ M = = I ( ( 2500 lb.in( 0.5 in ( 49.09 10 3 in 4 =+ 20.37 ksi (tensile M ( 2500 lb.in( 0 in σ = = = 0 ksi I 49.09 10 3 in 4 ( B C D M σ = I tensile 11ME 323 Final Eam, Sample & Solutions
Shear force V = 300 lb π 4 3 4 I = c = 49.09 10 in 4 3 @A & C, Q = 0 in τ 0 ksi =± 0.5 in = 4r π @B & D, Q = 8.33 10 in 0 in i Ai = r = = 3π 2 3 3 VQ ( 300 lb( 8.33 10 in τ = = 3 4 I t 49.09 10 in 2 0.5 in ' ' 2 3 3 ( ( = 0.51 ksi (downwards B A C VQ τ = I t D The combined state of stress at points A, B, C, and D will be the resultant linear superposition of the individual stress values due to torsion, bending, and shear force. Point element A Point element B τ = 12.73 ksi σ = 20.37 ksi (C τ = (12.73 + 0.51 ksi =13.24 ksi Point element C Point element D τ = 12.73 ksi τ = (12.73 0.51 ksi =12.22 ksi σ = 20.37 ksi (T 12ME 323 Final Eam, Sample & Solutions
σ M = 2 2 [(σ σ 2 + (σ σ 2 + (σ σ 2 + 6(τ 2 + τ 2 + τ 2 ] For A and C σ M = 2 [2σ 2 2 + 6τ 2 ] σ M ( A = σ M (C = 42.5ksi FS = σ ield / σ M For B and D σ M = 2 2 [6τ 2 ] σ M (B = 32ksi σ M (C = 30ksi FS = σ ield / σ M 13ME 323 Final Eam, Sample & Solutions
14ME 323 Final Eam, Sample & Solutions