07-02 Chap Gere

07-02 Chap Gere - SECTION 7.3 Principal Stresses and...

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Unformatted text preview: SECTION 7.3 Principal Stresses and Maximum Shear Stresses 443 Data for 7.3-15 x 3000 psi, y 12,000 psi, xy 6000 psi Solution 7.3-15 Plane stress 3000 psi 12,000 psi x y 6000 psi xy (a) PRINCIPAL STRESSES tan 2up 2 2 p p (b) MAXIMUM SHEAR STRESSES t max B p1 p1 ¢ x y 2 2 45 45 t 2 xy 7500 psi 7500 psi 7500 psi 2t xy sx y us1 1.3333 us2 s aver 18.43 and 71.57 and y } x 53.13 and p 26.57 233.13 and p 116.57 x y x y 2 7500 psi s x1 2 2 cos 2u t xy sin 2u y 7500 psi 7500 psi s2 For 2 p 53.13 : s x1 0 15,000 psi For 2 p 233.13 : x1 Therefore, 0 and p1 26.57 1 15,000 psi and up2 116.57 2 71.57 x } O y 15,000 psi 7500 psi p1 26.57 O x Data for 7.3-16 x 100 MPa, y 50 MPa, xy 50 MPa Solution 7.3-16 Plane stress 100 MPa 50 MPa x y (a) PRINCIPAL STRESSES tan 2up 2 2 p p xy 50 MPa 2t xy sx y 0.66667 65.1 MPa y 33.69 and p 16.85 213.69 and p 106.85 x y x y 115.1 MPa s x1 2 2 cos 2u t xy sin 2u O p2 16.85 For 2 p 33.69 : s x1 115.1 MPa For 2 p 213.69 : x1 65.1 MPa Therefore, 65.1 MPa and up1 106.85 1 115.1 MPa and up2 16.85 2 x } 444 CHAPTER 7 Analysis of Stress and Strain (b) MAXIMUM SHEAR STRESSES t ¢ x y 2 y 25.0 MPa 25.0 MPa max B p1 p1 2 45 45 t 2 xy 90.1 MPa 90.1 MPa 90.1 MPa s1 us2 aver 61.85 and 151.85 and y } O 90.1 MPa s1 61.85 x x 2 25.0 MPa Problem 7.3-17 At a point on the surface of a machine component the stresses acting on the x face of a stress element are x 6500 psi and xy 2100 psi (see figure). What is the allowable range of values for the stress y if the maximum shear stress is limited to 0 2900 psi? O y y xy = 2100 psi x = 6500 psi x Solution 7.3-17 Allowable range of values 6500 psi xy 2100 psi ? x y Find the allowable range of values for y if the maximum allowable shear stresses is 0 2900 psi. t max B ¢ ¢ x y 2 Substitute numerical values: sy 6500 psi 2 ( 2900 psi) 2 ( 2100 psi) 2 6500 psi 4000 psi Therefore, 2500 psi y GRAPH OF max 2 x y 2 t 2 xy Eq. (1) 10,500 psi or t2 max 2 y t 2 xy Eq. (2) From Eq. (1): t max B ¢ 6500 2 y 2 ( 2100) 2 Eq. (3) SOLVE FOR sy x 2 t 2 max t 2 xy 6 max (ksi) 4 Eq. (3) 2.9 ksi ( 2 2.5 6.5 5 10 2.1 ksi o) 10.5 15 y (ksi) 5 0 SECTION 7.3 Principal Stresses and Maximum Shear Stresses 445 Problem 7.3-18 At a point on the surface of a machine component the stresses acting on the x face of a stress element are x 45 MPa and xy 30 MPa (see figure). What is the allowable range of values for the stress y if the maximum shear stress is limited to 0 34 MPa? O y y xy= x 30 MPa = 45 MPa x Solution 7.3-18 Allowable range of values 45 MPa xy 30 MPa ? x y Find the allowable range of values for y if the maximum allowable shear stresses is 0 34 MPa. t max B ¢ ¢ x y 2 SOLVE FOR sy x y 2 t 2 max t 2 xy 2 x y 2 t 2 xy Eq. (1) Substitute numerical values: sy 45 MPa 2 ( 34 MPa) 2 ( 30 MPa) 2 45 MPa 32 MPa Therefore, 13 MPa y GRAPH OF max or t2 max 2 t 2 xy Eq. (2) 77 MPa From Eq. (1): t max B ¢ 45 2 y 2 ( 30) 2 Eq. (3) Eq. (3) 40 max (MPa) 30 20 10 13 20 0 20 40 45 60 77 80 34 MPa ( 30 MPa o) 100 y (MPa) Problem 7.3-19 An element in plane stress is subjected to stresses 6500 psi and xy 2800 psi (see figure). It is known that one x of the principal stresses equals 7300 psi in tension. (a) Determine the stress y. (b) Determine the other principal stress and the orientation of the principal planes; then show the principal stresses on a sketch of a properly oriented element. y y 6500 psi O 2800 psi x 446 CHAPTER 7 Analysis of Stress and Strain Solution 7.3-19 Plane stress 6500 psi xy 2800 psi ? x y One principal stress 7300 psi (tension) (a) STRESS y (b) PRINCIPAL STRESSES tan 2up 2t xy sx y 0.62222 p p y Because x is smaller than the given principal stress, we know that the given stress is the larger principal stress. 7300 psi 1 t2 xy 2 B 2 Substitute numerical values and solve for 2500 psi y 1 x y 2 2 p p 31.891 and 148.109 and x y x 15.945 74.053 cos 2u t xy s x1 For 2 For 2 p p 2 2 sin 2u ¢ x y 2 31.891 : s x1 148.109 : x1 7300 psi 3300 psi 15.95 74.05 y: Therefore, 7300 psi and p1 1 3300 psi and up2 2 y 3300 psi } 7300 psi p2 74.05 O x Problem 7.3-20 An element in plane stress is subjected to stresses 68.5 MPa and xy 39.2 MPa (see figure). It is known that x one of the principal stresses equals 26.3 MPa in tension. (a) Determine the stress y. (b) Determine the other principal stress and the orientation of the principal planes; then show the principal stresses on a sketch of a properly oriented element. y y 39.2 MPa O 68.5 MPa x Solution 7.3-20 Plane stress 68.5 MPa xy 39.2 MPa ? x y One principal stress 26.3 MPa (tension) (a) STRESS y 1 26.3 MPa x y s1 Because x is smaller than the given principal stress, we know that the given stress is the larger principal stress. t2 xy 2 B 2 Substitute numerical values and solve for 10.1 MPa y ¢ x y 2 y: SECTION 7.4 Mohr’s Circles 447 (b) PRINCIPAL STRESSES tan 2up 2 2 p p 2t xy sx y 0.99746 p p y 44.93 and 135.07 and x y x 22.46 67.54 cos 2u t xy y 26.3 MPa s x1 For 2 For 2 p p 2 2 sin 2u 84.7 MPa p1 44.93 : s x1 84.7 MPa 135.07 : 26.3 MPa x1 67.54 O x Therefore, 1 2 26.3 MPa and p1 67.54 84.7 MPa and up2 22.46 } y Mohr’s Circles The problems for Section 7.4 are to be solved using Mohr’s circle. Consider only the in-plane stresses (the stresses in the xy plane). Problem 7.4-1 An element in uniaxial stress is subjected to tensile stresses x 14,500 psi, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a counterclockwise angle 24° from the x axis and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. 14,500 psi O x Solution 7.4-1 Uniaxial stress 14,500 psi 0 xy 0 x y (a) ELEMENT AT 2 48 Point C: s x1 24 (All stresses in psi) 7250 psi Point D: s x1 R R cos 2u 12,100 psi t x1y1 R sin 2u 5390 psi Point D : s x1 t x1y1 R R cos 2u 5390 psi y 2400 psi 24 R 7250 psi S2 D( 2 R C O B( 90 ) R D' S1 14,500 x1y1 s2 24 ) 2400 psi D' D 12,100 psi 24 2 2 s1 = A 90 ( 0) x1 O x 5390 psi 448 CHAPTER 7 Analysis of Stress and Strain y (b) MAXIMUM SHEAR STRESSES Point S1: 2us1 90 45 s1 R 7250 psi max Point S2: 2us2 90 us2 45 R 7250 psi min R 7250 psi aver 7250 psi S2 7250 psi 45 s2 O x 7250 psi S1 Problem 7.4-2 An element in uniaxial stress is subjected to tensile stresses x 55 MPa, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at an angle 30° from the x axis (minus means clockwise) and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. y 55 MPa O x Solution 7.4-2 55 MPa x (a) ELEMENT AT 2 60 Point C: s1 x Uniaxial stress 0 xy 0 y 30 (All stresses in MPa) R 27.5 MPa D' y 13.8 MPa 30 27.5 MPa S2 D' 23.8 MPa O x –30 41.2 MPa R B( O 90 ) C 2 = 60 R S1 55 MPa x1y1 D (b) MAXIMUM SHEAR STRESSES A( 0) x1 Point S1: 2us1 ( D 30 ) 90 45 s1 R 27.5 MPa max Point S2: 2us2 90 us2 45 R 27.5 MPa min R 27.5 MPa aver y Point D: s x1 t x1y1 Point D : s x1 t x1y1 R R cos ƒ 2u ƒ R (1 cos 60 ) 41.2 MPa R sin ƒ 2u ƒ R sin 60 23.8 MPa R R cos ƒ 2u ƒ R sin ƒ 2u ƒ 13.8 MPa 23.8 MPa O S2 27.5 MPa s2 45 x 27.5 MPa S1 27.5 MPa SECTION 7.4 Mohr’s Circles 449 Problem 7.4-3 An element in uniaxial stress is subjected to compressive stresses of magnitude 5600 psi, as shown in the figure. Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a slope of 1 on 2 (see figure) and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. y 1 2 O 5600 psi x Solution 7.4-3 Uniaxial stress 5600 psi 0 xy 0 x y (a) ELEMENT AT A SLOPE OF 1 ON 2 (All stresses in psi) u 1 2 1120 psi y D 4480 psi 26.57 O x 2240 psi arctan 2 R 1 2 26.565 D' 53.130 26.57 2800 psi Point C: s x1 2800 psi (b) MAXIMUM SHEAR STRESSES S2 D' Point S1: 2us1 max 90 R 45 s1 2800 psi 2 ( A 0) 2= 53.13 90 s2 = C R B( O 90 ) x1 Point S2: 2us2 min aver R 90 us2 45 R 2800 psi 2800 psi y 2 D s1 S1 x1y1 2800 psi S2 S1 2800 psi 45 5600 Point D: s x1 t x1y1 Point D : s x1 t x1y1 R R cos 2u 4480 psi R sin 2u 2240 psi R R cos 2u 1120 psi R sin 2u 2240 psi s1 O x 2800 psi Problem 7.4-4 An element in biaxial stress is subjected to stresses 60 MPa and y 20 MPa, as shown in the figure. x Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a counterclockwise angle 22.5° from the x axis and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. O y 20 MPa 60 MPa x 450 CHAPTER 7 Analysis of Stress and Strain Solution 7.4-4 Biaxial stress 60 MPa 20 MPa x y (a) ELEMENT AT 22.5 y xy 0 8.28 MPa D' D 48.28 MPa 22.5 (All stresses in MPa) 2 45 22.5 2R 60 20 80 MPa Point C: s x1 20 MPa S2 R A 0) C 2 2 ( 22.5 ) D 40 60 O R 40 MPa O x 28.28 MPa D' (b) MAXIMUM SHEAR STRESSES Point S1: 2us1 B( 90 ) x1 ( 90 45 s1 R 40 MPa max Point S2: 2us2 90 us2 45 R 40 MPa min 20 MPa aver y s1 S1 20 20 x1y1 20 MPa 40 MPa 20 MPa s1 45 Point D: s x1 20 R cos 2u 48.28 MPa t x1y1 R sin 2u 28.28 MPa Point D : s x1 R cos 2u 20 8.28 MPa t x1y1 R sin 2u 28.28 MPa O S1 x S2 20 MPa Problem 7.4-5 An element in biaxial stress is subjected to stresses 6000 psi and y 1500 psi, as shown in the figure. x Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a counterclockwise angle 60° from the x axis and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. O y 1500 psi 6000 psi x Solution 7.4-5 Biaxial stress ( D R 60 ) S 2 x 6000 psi y 1500 psi 60 xy 0 B( 90 ) O 60 C 2 = 120 A( R 30 S1 2250 1500 6000 D' 0) x1 (a) ELEMENT AT (All stresses in psi) 2 120 60 2R 7500 psi R 3750 psi Point C: s x1 2250 psi SECTION 7.4 Mohr’s Circles 451 Point D: s x1 2250 R cos 60 375 psi t x1y1 R sin 60 3248 psi Point D : s x1 t x1y1 2250 R cos 60 4125 psi R sin 60 3248 psi y 375 psi 4125 psi D 60 (b) MAXIMUM SHEAR STRESSES Point S1: 2us1 90 45 s1 R 3750 psi max Point S2: 2us2 90 us2 45 R 3750 psi min y 2250 psi S2 aver 2250 psi 2250 psi s2 45 O x 3250 psi O x 3750 psi S1 D' y Problem 7.4-6 An element in biaxial stress is subjected to stresses 24 MPa and y 63 MPa, as shown in the figure. x Using Mohr’s circle, determine (a) the stresses acting on an element oriented at a slope of 1 on 2.5 (see figure) and (b) the maximum shear stresses and associated normal stresses. Show all results on sketches of properly oriented elements. 63 MPa 1 2.5 O 24 MPa x Solution 7.4-6 Biaxial stress 24 MPa 63 MPa x y xy 0 (a) ELEMENT AT A SLOPE OF 1 ON 2.5 1 21.801 (All stresses in MPa) u arctan 2.5 2 43.603 1 21.801 2R 87 MPa 2.5 R 43.5 MPa Point C: s x1 19.5 MPa R cos 2u 19.5 12 MPa Point D: s x1 t x1y1 R sin 2u 30 MPa S2 D' R A 0) C O 2 R D 19.5 24 x1y1 Point D : s x1 t x1y1 19.5 R cos 2u 51 MPa R sin 2u 30 MPa y D 12 MPa 21.80 O x 30 MPa 51 MPa D' (b) MAXIMUM SHEAR STRESSES Point S1: 2us1 90 45 s1 R 43.5 MPa max Point S2: 2us2 90 us2 45 R 43.5 MPa min 19.5 MPa aver y S1 19.5 MPa S2 19.5 MPa s1 43.603 B( 90 ) x1 ( 45 S1 63 O x 43.5 MPa ...
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