Solution: The strain in the bar is: ANS: E =6.37 GPa Poison s ration for the material is:

Transcription

1 Problem 10.4 A prismatic bar with length L 6m and a circular cross section with diameter D 0.0 m is subjected to 0-kN compressive forces at its ends. The length and diameter of the deformed bar are measured and determined to be L m and D m. What are the modulus of elasticity and Poisson s ration of the material? The strain in the bar is: ε L L L 5.94 m 6.0 m 6.0 m 0.01 The compressive stress in the bar is: σ P 0, 000 N 63.7 MPa A π(0.01 m) The modulus of elasticity for the material is: E σ 63.7 MPa ε 0.01 E 6.37 GPa Poison s ration for the material is: υ ε LAT ε υ 0.3 (D D)/D ε ( m 0.0 m)/0.0 m 0.01 Problem 10.5 The bar has modulus of elasticity E psi and Poisson s ration ν 0.3. It has a circular cross section with diameter D 0.75 in. What compressive force would have to be eerted on the right end of the bar to increase its diameter to 0.75 in? From the definition of Poisson s ratio: (0.75 in 0.75 in)/0.75 in 0.3 ε The strain which will be produced by the applied load is: ε σ ( )( E P π(0.375 in) lb/in Substituting the above epression for the strain into the epression for Poisson s ratio: (0.75 in 0.75 in)/0.75 in 0.3 ( (P/(π(0.375 in) )) P kip ) lb/in )

2 Problem 11.1 A cube of material is subjected to a pure shear stress τ 9 MPa. The angle β is measured ad determined to be What is the shear modulus G of the material? Diagram: Converting the shear strain angle into radians: γ ( ) 180 π radians Using the definition of the shear modulus: G τ γ N/m G 5.8 GPa Problem 11. If the cube in Problem 11.1 consists of material with shear modulus G psi and the shear stress τ 8000 psi, what is the angle β in degrees? The shear strain will be: γ τ G The angle β is: 8, 000 lb/in lb/in radians β 89.9 β 90 γ Problem 11.3 If the cube in Problem 11.1 consists of aluminum alloy that will safely support a pure stress of 70 MPa and G 6.3 GPa, what is the largest shear strain to which the cube can safely be subjected? The shear strain will be: γ γ τ G N/m N/m

3 Problem The bar has a circular cross section with 15-mm diameter and the shear modulus of the material is G 6GPa. If the torque T 10N m, determine (a) the magnitude of the maimum shear stress in the bar; (b) the angle of twist of the end of the bar in degrees. J π c4 π (0.0075m)4 J m 4 Maimum shear stress in the bar is: τ MAX Tρ J (10 N m) ( m) (a) τ MAX N/m The angle of twist for the bar is: φ LT JG (0.8m)(10 N m) ( m 4 )( N/m ) φ rad Problem If the bar in Problem is subjected to a torque T that causes the end of the bar to rotate 4, what is the magnitude of the maimum shear stress in the bar? The polar moment of inertia for the cross section is: J π c4 π (0.0075) Using the angle of rotation at the end of the bar to determine the applied torque: ( 4 ) 180 (π) rad TL JG (0.8 m)(t ) ( m 4 )( N/m ) T 11.7 N m Maimum shear stress in the cross section is: τ MAX Tc J τ MAX MPa (11.7 N m)( m) m 4

4 Problem Consider the solid circular shaft in Problem The shear modulus of the material is G 80GPa. What angle of twist per unit meter of length is caused by the -MN-m torque? J π c4 π (0.4 m) m 4 The angle of twist per meter of length is: φ L T JG 10 6 N m (0.040 m 4 )( N/m ) φ rad/m degrees/m L Problem If the shaft in Problem has a hollow circular cross section with 0.5-m outer radius and 0.3-m inner radius, what is the maimum shear stress? The polar moment of inertia for the hollow shaft is: J π (ro r i) π [(0.5 m) 4 (0.3 m) 4] J m 4 Maimum shear stress in the shaft is: τ MAX Tρ ( 10 6 J N m ) (0.5 m) m 4 τ MAX 11.7 MPa

5 Problem The propeller of the wind generator is supported by a hollow circular shaft with 0.4-m outer radius and 0.3-m inner radius. The shear modulus of the material is G 80GPa. If the propeller eerts an 840- kn-m torque on the shaft, what is the resulting maimum shear stress? J [(0.4 π m) 4 (0.3 m) 4] J m 4 Maimum shear stress in the shaft is: τ MAX Tc J τ MAX 1. MPa (840, 000 N m) (0.4 m) m 4 Problem 11.0 In Problem 11.19, what is the angle of twist of the propeller shaft per meter of length? J [(0.4 π m) 4 (0.3 m) 4] J m 4 Angle of twist for the shaft is: φ LT JG (1 m) (840, 000 N m) (0.075 m 4 ) ( N/m ) φ rad 0.019

6 Problem 11.1 In designing a new shaft for the wind generator in Problem 11.19, the engineer wants to limit the maimum shear stress in the shaft to 10 MPa, but design constraints require retaining the 0.4-m outer radius. What new inner radius should she use? J [(0.4 π m) 4 (r i ) 4] J ri 4 Maimum shear stress in the shaft is: τ MAX Tc J (840, 000 N m) (0.4 m) ( r 4 i ) N/m r i m Problem 11. The bar has a circular cross section with 1-in. diameter and the shear modulus of the material is G psi. If the torque T 1000 in lb, determine (a) the magnitude of the maimum shear stress in the bar; (b) the magnitude of the angle of twist of the right end of the bar relative to the wall in degrees. Maimum torque in the shaft is 1,000 in-lb. J π (0.5 in)4 J in 4 (a) Maimum shear stress in the shaft is: τ MAX TC J τ MAX lb/in (1, 000 in lb) (0.5 in) in 4 (b) The angle of twist in the 8-inch section of the bar is: φ 8in (8 in)(500 in lb)) (0.098 in 4 )( in 4 ) φ 8in rad 0.40 The angle of twist in the 6-inch section of the bar is: φ 6in (6 in)(1,000 in lb) (0.098 in 4 )( lb/in ) φ 6in rad Total angle of twist for the bar is: φ φ 8in + φ 6in φ 1.006

7 Problem 11.3 For the bar in Problem 11., what value of the torque T would cause the angle of twist of the end of the bar to be zero? The torque in the 8-inch section of the bar is (T 500 in lb). The torque in the 6-inch section of the bar is T. The equation for total angle of twist for the bar is: 0 L 8inT 8in JG + L 6inT 6in JG Solving the equation for T : T 86 in lb (8 in) (T 500 in lb) (6 in) T + JG JG Problem 11.4 Part A of the bar has a solid circular cross section and Part B has a hollow circular cross section. The shear modulus of the material is G psi. Determine the magnitudes of the maimum shear stresses in parts A and B of the bar. The torque in the solid section of the bar is 50,000 in-lb. The torque in the hollow section of the bar is 100,000 in-lb. Polar moment of inertia for the solid section of the bar is: J S π ( in) in 4 Polar moment of inertia for the hollow section of the shaft is: J H π [ ( in) 4 (1 in) 4] 3.56 in 4 Maimum shear stress in the solid section of the bar is: (τ MAX ) S T Sc S J S (50, 000 in lb) ( in) 5.13 in lb (τ MAX ) S 19, lb/in ksi Maimum shear stress in the hollow section of the bar is: (τ MAX ) H T Hc H J H (100, 000 in lb) ( in) 3.56 in 4 (τ MAX ) H 8, lb/in 8.49 ksi

8 Problem 11.5 For the bar in Problem 11.4, determine the magnitude of the angle of twist of the end of the bar in degrees. The torque in the solid section of the bar is 50,000 in-lb. The torque in the hollow section of the bar is 100,000 in-lb. Polar moment of inertia for the solid section of the bar is: J S π ( in) in 4 Polar moment of inertia for the hollow section of the shaft is: J H π [ ( in) 4 (1 in) 4] 3.56 in 4 The angle of twist for the solid section of the shaft is: φ L ST S (7 in) (50, 000 in lb) ( J S G S 5.13 in 4 )( lb/in ) φ S rad 1.05 The angle of twist for the hollow section of the shaft is: φ L HT H (14 in) (100, 000 in lb) ( J H G H 3.56 in 4 )( lb/in ) φ H rad Total angle of twist for the shaft is: φ 1.95 φ t φ S + φ H Problem 11.6 For the bar in Problem 11.4, determine the magnitude of the maimum shear stresses in parts A and B of the bar and the magnitude of the angle of twist of the end of the bar in degrees if the 150 in-kip couple acts in the opposite direction. Polar moments of inertia for the two sections of the bar are: J A π (in) in 4 J B π [ (in) 4 (1in) 4] 3.56 in 4 From the FBD we see that the torque in the sections of the bar is: T A 50, 000 in lb T B 100, 000 in lb Maimum shear stresses in the sections of the bar are: (50, 000 in lb)( in) (100, 000 in lb)( in) (τ A ) MAX 5.13 in 4 (τ B ) MAX 3.56 in 4 (τ A ) MAX 3, 980 lb/in (τ B ) MAX 8, lb/in 8.49 ksi The angles of twist in each of the sections of the bar are: (50, 000 in lb)(7 in) (100, 000 in lb)(14 in) φ A (5.13 in 4 )( lb/in φ B ) (3.56 in 4 )( lb/in ) φ A radians Total angle of twist is: φ B radians φ φ A + φ B rad rad φ rad 0.686

9 Problem 7.6 Determine the y coordinate of the centroid of the area in Problem 7.5. Let y in Equation (7.7) be the height of the midpoint of a vertical strip: R 1 [ ] yda A y h (R ) 1/ (R ) 1/ d. da da A The upper integral is R 1 h (R ) d 1 ] R [R 3 3 h 1 ( R 3 A 3 R h + h3 3 ). y h y ½(R ) ½ d From the solution of Problem 7.5, [ A da R ) 1/ ( ) πr (1 ] A h h h R R arcsin. R ( ) The centroid is y 1 R 3 R A 3 h + h3 3 Determine the coordinates of the cen- Problem 7.7 troids. y Let us solve this by parts. 40 mm A 1 A h b 60 mm l 40 mm h 40 mm 60 mm 40 mm b l b 60mm l 40mm h 40mm y A 1 1 bh 1 (60)(40) 100 mm 40 mm A lh (40)(40) 1600 mm A 1 + A 800 mm From the tables and inspection 60 mm 40 mm 1 3 b b + l/z y h y 1 h 1 40mm 80mm y mm y 0mm For the composite, substituting, 1A 1 + A A 1 + A y y 1A 1 + y A A 1 + A 6.9 mm 17.1 mm

10 Determine the coordinates of the cen- Problem 7.8 troids. y 0 mm 60 mm 30 mm 70 mm Let us solve this problem by using symmetry and by breaking the composite shape into parts. l 1 0 mm y A 1 h 1 0 mm 60 mm A h l 1 70 mm h 1 70 mm l 70 mm h 70 mm 60 mm l 30 mm l 1 70mm 70 mm h 1 0mm l 30mm h 60mm A 1 l 1 h mm A l h 1800 mm By symmetry, y 1 70mm y 30mm For the composite, 1A 1 + A 0+0 A 1 + A 30 mm 0 y y 1A 1 + y A A 1 + A (70)(1400) + (30)(1800) y 300 y 47.5mm