Chapter 8

2017- 5- 21

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Chapter 8

2017- 5- 21

Chapter 8

2017- 5- 21

Chapter 8

2017- 5- 21

Chapter 8

2017- 5- 21

Chapter 8

2017- 5- 21

Chapter 8

2017-5-21

Chapter 8

2017-5-21

Chapter 8

2017-5-21

Chapter 8

2017-5-21

Chapter 8

2017-5-21

Chapter 8

2017-5-21

Chapter 8

2017-5-21

Chapter 8

2017-5-21

Chapter 8

SECTION 8.3 Cylindrical Pressure Vessels 661 HORIZONTAL PIPES d1 ϭ 2 ft ϭ 24 in. r1 ϭ 12 in. t1 ϭ 1.0 in. (b) FIND HOOP ST...

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Chapter 8

662 CHAPTER 8 Applications of Plane Stress Solution 8.3-11 d ϭ 18 in. p ϭ 450 psi sallow ϭ 8200 psi (tension) tallow ϭ 300...

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Chapter 8

SECTION 8.3 Cylindrical Pressure Vessels 663 (c) CIRCUMFERENTIAL STRAIN 1 ϭ 3.97 ϫ 10Ϫ4 LONGITUDINAL STRAIN 2 ϭ 9.33 ϫ 1...

Maximum Stresses in Beams When solving the problems for Section 8.4, consider only the in-plane stresses and disregard the...

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Chapter 8

SECTION 8.4 Maximum Stresses in Beams 665 Problem 8.4-2 Solve the preceding problem for the following data: P ϭ 130 kN, b ...

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Chapter 8

666 CHAPTER 8 Applications of Plane Stress Problem 8.4-4 An overhanging beam ABC with a guided support at A is of rectangu...

SECTION 8.4 Maximum Stresses in Beams 667 Problem 8.4-5 Solve the preceding problem if the stress and dimensions are as fo...

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Chapter 8

668 CHAPTER 8 Applications of Plane Stress Problem 8.4-6 A beam of wide-flange cross section (see figure) has the followin...

SECTION 8.4 Maximum Stresses in Beams 669 Problem 8.4-7 A beam of wide-flange cross section (see figure) has the following...

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Chapter 8

670 CHAPTER 8 Applications of Plane Stress Problem 8.4-8 A W 200 ϫ 41.7 wide-flange beam (see Table E-1(b). Appendix E) is...

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Chapter 8

SECTION 8.4 Maximum Stresses in Beams 671 Problem 8.4-9 A W 12 ϫ 14 wide-flange beam (see Table E-1(a), Appendix E) is sim...

672 CHAPTER 8 Applications of Plane Stress Problem *8.4-10 A cantilever beam of T-section is loaded by an inclined force o...

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Chapter 8

SECTION 8.4 Maximum Stresses in Beams 673 EQUIVALENT LOADS AT FREE END OF BEAM N0 ϭ PH N0 ϭ Ϫ3.25 kN V ϭ Pv V ϭ 5.629 kN M...

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Chapter 8

674 CHAPTER 8 Applications of Plane Stress Problem *8.4-12 Solve the preceding problem for a cross section mn located 0.18...

SECTION 8.5 Combined Loadings 675 Solution 8.4-12 M ϭ 65 kN m L ϭ 0.75 m c ϭ 0.18 m b ϭ 20 mm h ϭ 120 mm At section m-n M ...

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Chapter 8

676 CHAPTER 8 Applications of Plane Stress Solution 8.5-1 b1 ϭ 3.6 ft b2 ϭ 2.2 ft P ϭ 1400 lb d2 ϭ 7.5 in. d1 ϭ 6.8 in. VE...

SECTION 8.5 Combined Loadings 677 Solution 8.5-2 Gondola on a ski lift b ϭ 180 mm sallow ϭ 100 MPa (tension) tallow ϭ 50 M...

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Chapter 8

678 CHAPTER 8 Applications of Plane Stress Solution 8.5-3 P ϭ compressive force T ϭ Torque d2 ϭ outer diameter d1 ϭ inner ...

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Chapter 8

SECTION 8.5 Combined Loadings 679 Solution 8.5-4 P ϭ Compressive_force P0 ϭ Power f ϭ frequency d2 ϭ outer diameter d1 ϭ i...

680 CHAPTER 8 Applications of Plane Stress Solution 8.5-5 P ϭ compressive force T ϭ Torque d2 ϭ outer diameter d1 ϭ inner ...

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Chapter 8

SECTION 8.5 Combined Loadings 681 OUT-OF-PLANE SHEAR STRESSES Solving, p2 ϭ 9.60 MPa 60 MPa ϭ 3.125 p Ϫ 28.648 MPa Solving...

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Chapter 8

682 CHAPTER 8 Applications of Plane Stress Problem 8.5-8 The torsional pendulum shown in the figure consists of a horizont...

SECTION 8.5 Combined Loadings 683 Solution 8.5-8 Torsional pendulum L ϭ 2.0 m d ϭ 4.0 mm M ϭ 60 kg G ϭ 80 GPa sallow ϭ 100...

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Chapter 8

684 CHAPTER 8 Applications of Plane Stress Problem 8.5-9 Determine the maximum tensile, compressive, and shear stresses at...

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Chapter 8

SECTION 8.5 Combined Loadings 685 MAXIMUM TENSILE STRESS st ϭ s1 MAXIMUM COMPRESSIVE STRESS sc ϭ s2 MAXIMUM IN-PLANE SHEAR...

686 CHAPTER 8 Applications of Plane Stress Solution 8.5-10 Cylindrical pressure vessel T ϭ 120 kN m r ϭ 300 mm t ϭ 15 mm P...

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Chapter 8

SECTION 8.5 Combined Loadings 687 Problem 8.5-11 An L-shaped bracket lying in a horizontal plane supports a load P ϭ 150 l...

688 CHAPTER 8 Applications of Plane Stress PRINCIPAL STRESSES AND MAXIMUM SHEAR STRESS s1 ϭ 4324 psi s2 ϭ Ϫ1870 psi MAXIMU...

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Chapter 8

SECTION 8.5 Combined Loadings 689 STRESS ELEMENT AT THE TOP OF THE BAR AT B sx ϭ 0 sy ϭ sB txy ϭ tB PRINCIPAL STRESSES: ϭ ...

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Chapter 8

690 CHAPTER 8 Applications of Plane Stress RESULTANT FORCES ACTING ON AB P ϭ weight of AB and BC P ϭ qL ϭ pgLd2 /4 (2) T ϭ...

SECTION 8.5 Combined Loadings 691 Solution 8.5-14 Cylindrical tank r ϭ 50 mm t ϭ 3.0 mm p ϭ 3.5 MPa T ϭ 450 N m sallow ϭ 7...

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Chapter 8

692 CHAPTER 8 Applications of Plane Stress Solution 8.5-15 P ϭ 240 lb b ϭ 5 ft length of arm h ϭ 27 ft height of post r2 ϭ...

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Chapter 8

SECTION 8.5 Combined Loadings 693 Problem 8.5-16 A sign is supported by a pipe (see figure) having outer diameter 110 mm a...

694 CHAPTER 8 Applications of Plane Stress MAXIMUM SHEAR STRESS AT POINT B MAXIMUM SHEAR STRESS AT POINT C sx ϭ 0 sy ϭ 0 t...

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Chapter 8

SECTION 8.5 Combined Loadings 695 Solution 8.5-17 PIPE: d2 ϭ 10.5 in. d1 ϭ 8.5 in. A ϭ 29.845 in.2 I ϭ 340.421 in.4 Ip ϭ 2...

696 CHAPTER 8 Applications of Plane Stress Problem 8.5-18 A horizontal bracket ABC consists of two perpendicular arms AB o...

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Chapter 8

SECTION 8.5 Combined Loadings 697 Problem 8.5-19 A cylindrical pressure vessel with flat ends is subjected to a torque T a...

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Chapter 8

698 CHAPTER 8 Applications of Plane Stress Problem 8.5-20 For purposes of analysis, a segment of the crankshaft in a vehic...

SECTION 8.5 Combined Loadings 699 Solution 8.5-20 P ϭ 1.0 kN b1 ϭ 80 mm b2 ϭ 120 mm b3 ϭ 40 mm PROPERTIES OF THE CROSS SEC...

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Chapter 8

700 CHAPTER 8 Applications of Plane Stress Problem 8.5-21 A moveable steel stand supports an automobile engine weighing W ...

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Chapter 8

SECTION 8.5 Combined Loadings 701 Solution 8.5-21 W ϭ 750 1b Fy ϭ 200 1b b ϭ 2.5 in. d ϭ 30 in. x1 ϭ 24 in. A ϭ b2 Ϫ (b Ϫ ...

702 CHAPTER 8 Applications of Plane Stress Problem 8.5-22 A mountain bike rider going uphill applies force P ϭ 65 N to eac...

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Chapter 8

sx ϭ 17.863 MPa (tensile stress) sy ϭ 0 txy ϭ 2.977 MPa s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx...

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Chapter 8

SECTION 8.5 Combined Loadings 705 Solution 8.5-23 Cross section t ϭ 0.125 in. Thickness d2 ϭ 2 in. Outer width d1 ϭ d2 Ϫ 2...

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Chapter 8

706 CHAPTER 8 Applications of Plane Stress SIDE OF THE CROSS SECTION (AT POINT A) BOTTOM OF THE CROSS SECTION (AT POINT A)...

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Chapter 8

2017-5-21

Chapter 8

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Chapter 8 1. 1. Spherical Pressure Vessels When solving the problems for Section 8.2, assume that the given radius or diameter is an inside dimension and that all internal pressures are gage pressures. Problem 8.2-1 A large spherical tank (see figure) contains gas at a pressure of 450 psi. The tank is 42 ft in diameter and is constructed of high-strength steel having a yield stress in tension of 80 ksi. Determine the required thickness (to the nearest 1/4 inch) of the wall of the tank if a factor of safety of 3.5 with respect to yielding is required. Probs. 8.2-1 and 8.2-2 8 Applications of Plane Stress (Pressure Vessels, Beams, and Combined Loadings) 649 Solution 8.2-1 Radius: r ϭ 252 in. Internal Pressure: p ϭ 450 psi Yield stress: sY ϭ 80 ksi (steel) Factor of safety: n ϭ 3.5 MINIMUM WALL THICKNESS tmin From Eq. (8-1): or sY n ϭ pr 2t smax ϭ pr 2t r ϭ 1 2 42 * 12 t ϭ 2.481 in. to nearest 1/4 inch, tmin ϭ 2.5 in. ; t ϭ prn 2sY 08Ch08.qxd 9/18/08 11:03 AM Page 649 2. 2. 650 CHAPTER 8 Applications of Plane Stress Problem 8.2-2 Solve the preceding problem if the internal pressure is 3.75 MPa, the diameter is 19 m, the yield stress is 570 MPa, and the factor of safety is 3.0. Determine the required thickness to the nearest millimeter. Solution 8.2-2 Radius: r ϭ 9.5 ϫ 103 mm Internal Pressure: p ϭ 3.75 MPa Yield stress: sY ϭ 570 MPa Factor of safety: n ϭ 3 MINIMUM WALL THICKNESS tmin From Eq. (8-1): or sY n ϭ pr 2t smax ϭ pr 2t r ϭ 1 2 (19 m) t ϭ 93.8 mm Use the next higher millimeter tmin ϭ 94 mm ; t ϭ prn 2sY Problem 8.2-3 A hemispherical window (or viewport) in a decompression chamber (see figure) is subjected to an internal air pressure of 80 psi. The port is attached to the wall of the chamber by 18 bolts. Find the tensile force F in each bolt and the tensile stress s in the viewport if the radius of the hemisphere is 7.0 in. and its thickness is 1.0 in. Solution 8.2-3 Hemispherical viewport FREE-BODY DIAGRAM T ϭ total tensile force in 18 bolts FHORIZ ϭ T Ϫ pA ϭ 0 T ϭ pA ϭ p(pr2 ) F ϭ force in one bolt TENSILE STRESS IN VIEWPORT (EQ. 8-1) s ϭ pr 2t ϭ 280 psi ; F ϭ T 18 ϭ 1 18 (ppr 2 ) ϭ 684 lb ; a Radius: r ϭ 7.0 in. Internal pressure: p ϭ 80 psi Wall thickness: t ϭ 1.0 in. 18 bolts 08Ch08.qxd 9/18/08 11:03 AM Page 650 3. 3. SECTION 8.2 Spherical Pressure Vessels 651 Problem 8.2-4 A rubber ball (see figure) is inflated to a pressure of 60 kPa. At that pressure the diameter of the ball is 230 mm and the wall thickness is 1.2 mm. The rubber has modulus of elasticity E ϭ 3.5 MPa and Poisson’s ratio v ϭ 0.45. Determine the maximum stress and strain in the ball. Prob. 8.2-4, 8.2-5 Solution 8.2-4 Rubber ball CROSS-SECTION MAXIMUM STRESS (EQ. 8-1) MAXIMUM STRAIN (EQ. 8-4) ϭ 0.452 ; âmax ϭ pr 2tE (1 Ϫ v) ϭ (60 kPa)(115 mm) 2(1.2 mm)(3.5 MPa) (0.55) ϭ 2.88 MPa ; smax ϭ pr 2t ϭ (60 kPa)(115 mm) 2(1.2 mm) Radius: r ϭ (230 mm)/2 ϭ 115 mm Internal pressure: p ϭ 60 kPa Wall thickness: t ϭ 1.2 mm Modulus of elasticity: E ϭ 3.5 MPa (rubber) Poisson’s ratio: v ϭ 0.45 (rubber) Problem 8.2-5 Solve the preceding problem if the pressure is 9.0 psi, the diameter is 9.0 in., the wall thickness is 0.05 in., the modulus of elasticity is 500 psi, and Poisson’s ratio is 0.45. Solution 8.2-5 Rubber ball CROSS-SECTION Modulus of elasticity: E ϭ 500 psi (rubber) Poisson’s ratio: v ϭ 0.45 (rubber) MAXIMUM STRESS (EQ. 8-1) MAXIMUM STRAIN (EQ. 8-4) ϭ 0.446 ; âmax ϭ pr 2tE (1 Ϫ v) ϭ (9.0 psi)(4.5 in.) 2(0.05 in.)(500 psi) (0.55) ϭ 405 psi ;smax ϭ pr 2t ϭ (9.0 psi) (4.5 in.) 2(0.05 in.) Radius: Internal pressure: p ϭ 9.0 psi Wall thickness: t ϭ 0.05 in. r ϭ 1 2 (9.0 in.) ϭ 4.5 in. 08Ch08.qxd 9/18/08 11:03 AM Page 651 4. 4. 652 CHAPTER 8 Applications of Plane Stress Problem 8.2-6 A spherical steel pressure vessel (diameter 480 mm, thickness 8.0 mm) is coated with brittle lacquer that cracks when the strain reaches 150 ϫ 10Ϫ6 (see figure). What internal pressure p will cause the lacquer to develop cracks? (Assume E ϭ 205 GPa and v ϭ 0.30.) Cracks in coating Solution 8.2-6 Spherical vessel with brittle coating CROSS-SECTION Cracks occur when max ϭ 150 ϫ 10Ϫ6 From Eq. (84): ϭ 2.93 MPa ; P ϭ 2(8.0 mm)(205 GPa)(150 * 10Ϫ6 ) (240 mm)(0.70) ‹ P ϭ 2tE âmax r(1Ϫv) âmax ϭ pr 2tE (1 Ϫ v) r ϭ 240 mm E ϭ 205 GPa (steel) t ϭ 8.0 mm v ϭ 0.30 Problem 8.2-7 A spherical tank of diameter 48 in. and wall thickness 1.75 in. contains compressed air at a pressure of 2200 psi. The tank is constructed of two hemispheres joined by a welded seam (see figure). (a) What is the tensile load f (lb per in. of length of weld) carried by the weld? (b) What is the maximum shear stress tmax in the wall of the tank? (c) What is the maximum normal strain  in the wall? (For steel, assume E ϭ 30 ϫ 106 psi and v ϭ 0.29.) Probs. 8.2-7 and 8.2-8 Weld Solution 8.2-7 r ϭ 24 in. E ϭ 30 ϫ 106 psi t ϭ 1.75 in. v ϭ 0.29 (steel) (a) TENSILE LOAD CARRIED BY WELD T ϭ Total load f ϭ load per inch T ϭ pA ϭ ppr2 c ϭ Circumference of tank ϭ 2pr ϭ26.4 k/in. ; ϭ (2200 psi)(24 in.) 2 f ϭ T c ϭ p1pr 2 2 2pr ϭ pr 2 (b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3) (c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4) ϭ 3.57 * 10Ϫ4 ; âmax ϭ pr (1 Ϫ v) 2tE ϭ (2200 psi)(24 in.)(0.71) 2(1.75 in.)130106 psi2 ϭ 7543 psitmax ϭ pr 4t ϭ (2200 psi)(24 in.) 4(1.75 in.) 08Ch08.qxd 9/18/08 11:03 AM Page 652 5. 5. SECTION 8.2 Spherical Pressure Vessels 653 Problem 8.2-8 Solve the preceding problem for the following data: diameter 1.0 m, thickness 48 mm, pressure 22 MPa, modulus 210 GPa, and Poisson’s ratio 0.29. Solution 8.2-8 r ϭ 0.5 m E ϭ 210 GPa t ϭ 48 mm v ϭ 0.29 (steel) (a) TENSILE LOAD CARRIED BY WELD T ϭ Total load f ϭ load per inch T ϭ pA ϭ ppr2 c ϭ Circumference of tank ϭ 2pr ϭ 5.5 MN/m ; f ϭ T c ϭ p1pr 2 2 2pr ϭ pr 2 ϭ (22 MPa)(0.5 m) 2 (b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3) (c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4) ϭ 3.87 * 10Ϫ4 ; âmax ϭ pr (1 Ϫ v) 2 tE ϭ

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Chapter 8

diameter of 22 in. is used to store propane gas at a pressure of 2450 psi. The properties of the steel are as follows: yield stress in tension, 140,000 psi; yield stress in shear, 65,000 psi; modulus of elasticity, 30 ϫ 106 psi; and Poisson’s ratio, 0.28. The desired factor of safety with respect to yielding is 2.8. Also, the normal strain must not exceed 1100 ϫ 10Ϫ6 . Determine the minimum permissible thickness tmin of the tank. Solution 8.2-9 r ϭ 11 in. E ϭ 30 ϫ 106 psi p ϭ 2450 psi v ϭ 0.28 (steel) sY ϭ 140000 psi n ϭ 2.8 tY ϭ 65000 psi max ϭ 1100 ϫ 10Ϫ6 MIMIMUM WALL THICKNESS t (1) TENSION (EQ. 8-1) ϭ (2450 psi)(11 in.) 2 140000 psi 2.8 ϭ 0.269 in. t1 ϭ pr 2smax ϭ pr 2a sY n b smax ϭ pr 2t1 (2) SHEAR (EQ. 8-3) (3) STRAIN (EQ. 8-4) t3 Ͼ t2 Ͼ t1 Thus, tmin ϭ 0.294 in. ; ϭ0.294 in. ϭ (2450 psi)(11 in.) 211100 * 10Ϫ6 2130 * 106 psi2 0.72 t3 ϭ pr 2âmax E (1 Ϫ v) âmax ϭ pr 2t3E (1 Ϫ v) t2 ϭ pr 4 tY n ϭ (2450 psi)(11 in.) 4 65000 psi 2.8 ϭ 0.29 in. tmax ϭ pr 4t2 08Ch08.qxd 9/18/08 11:03 AM Page 653 6. 6. 654 CHAPTER 8 Applications of Plane Stress Problem 8.2-10 Solve the preceding problem if the diameter is 500 mm, the pressure is 18 MPa, the yield stress in tension is 975 MPa, the yield stress in shear is 460 MPa, the factor of safety is 2.5, the modulus of elasticity is 200 GPa, Poisson’s ratio is 0.28, and the normal strain must not exceed 1210 ϫ 10Ϫ6 . Solution 8.2-10 r ϭ 250 mm E ϭ 200 GPa p ϭ 18 MPa v ϭ 0.28 (steel) sY ϭ 975 MPa n ϭ 2.5 tY ϭ 460 MPa max ϭ 1210 ϫ 10Ϫ6 MINIMUM WALL THICKNESS t (1) TENSION (EQ. 8-1) ϭ (18 MPa)(250 mm) 2 975 MPa 2.5 ϭ 5.769 mm t1 ϭ pr 2smax ϭ pr 2a sY n b smax ϭ pr 2t1 (2) SHEAR (EQ. 8-3) (3) STRAIN (EQ. 8-4) t3 Ͼ t2 Ͼ t1 Thus, tmin ϭ 6.69 mm ; ϭ 6.694 mm ϭ (18 MPa)(250 mm) 211210 * 10Ϫ6 2(200 GPa) 0.72 t3 ϭ pr 2âmax E (1 Ϫ v) âmax ϭ pr 2t3E (1 Ϫ v) t2 ϭ pr 4 tY n ϭ (18 MPa)(250 mm) 4 460 MPa 2.5 ϭ 6.114 mm tmax ϭ pr 4t2 Problem 8.211 A hollow pressurized sphere having radius r ϭ 4.8 in. and wall thickness t ϭ 0.4 in. is lowered into a lake (see figure). The compressed air in the tank is at a pressure of 24 psi (gage pressure when the tank is out of the water). At what depth D0 will the wall of the tank be subjected to a compressive stress of 90 psi? D0 08Ch08.qxd 9/18/08 11:03 AM Page 654 7. 7. SECTION 8.3 Cylindrical Pressure Vessels 655 Solution 8.2-11 Pressurized sphere under water CROSS-SECTION r ϭ 4.8 in. p1 ϭ 24 psi t ϭ 0.4 in. g ϭ density of water ϭ 62.4 lb/ft3 (1) IN AIR: p1 ϭ 24 psi D0 ϭ depth of water (in.) Compressive stress in tank wall equals 90 psi. (Note: s is positive in tension.) ϭ 1080 in. ϭ 90 ft ; Solve for D0: D0 ϭ 234 0.21667 ϭ 144 Ϫ 0.21667 D0 Ϫ90 psi ϭ (24 psi Ϫ 0.03611 D0)(4.8 in.) 2(0.4 in.) s ϭ pr 2t ϭ ( p1 Ϫ p2)r 2t s ϭ Ϫ90 psi p2 ϭ gD0 ϭ a 62.4 lb/ft3 1728 in.3 / ft3 bD0 ϭ 0.036111 D0 (psi) (2) UNDER WATER: p1 ϭ 24 psi Cylindrical Pressure Vessels When solving the problems for Section 8.3, assume that the given radius or diameter is an inside dimension and that all internal pressures are gage pressures. Problem 8.3-1 A scuba tank (see figure) is being designed for an internal pressure of 1600 psi with a factor of safety of 2.0 with respect to yielding. The yield stress of the steel is 35,000 psi in tension and 16,000 psi in shear. If the diameter of the tank is 7.0 in., what is the minimum required wall thickness? (1) IN AIR (2) UNDER WATER 08Ch08.qxd 9/18/08 11:03 AM Page 655 8. 8. 656 CHAPTER 8 Applications of Plane Stress Solution 8.3-1 Scuba tank Cylindrical pressure vessel p ϭ 1600 psi n ϭ 2.0 d ϭ 7.0 in. r ϭ 3.5 in. sY ϭ 35,000 psi Y ϭ 16,000 psi Find required wall thickness t. (1) BASED ON TENSION (EQ. 8-5) (2) BASED ON SHEAR (EQ. 8-10) Shear governs since t2 Ͼ t1 І tmin ϭ 0.350 in. ; t2 ϭ pr 2tallow ϭ (1600 psi)(3.5 in.) 2(8,000 psi) ϭ 0.350 in. tmax ϭ pr 2t t1 ϭ pr sallow ϭ (1600 psi)(3.5 in.) 17,500 psi ϭ 0.320 in. smax ϭ pr t sallow ϭ sY n ϭ 17,500 psi tallow ϭ tY n ϭ 8,000 psi Problem 8.3-2 A tall standpipe with an open top (see figure) has diameter d ϭ 2.2 m and wall thickness t ϭ 20 mm. (a) What height h of water will produce a circumferential stress of 12 MPa in the wall of the standpipe? (b) What is the axial stress in the wall of the tank due to the water pressure? d h Solution 8.3-2 d ϭ 2.2 m r ϭ 1.1 m t ϭ 20 mm weight density of water height of water h water pressure p ϭ h (a) HEIGHT OF WATER s1 ϭ pr t ϭ 12 MPa ϭ 0.00981h(1.1 m) 20 mm g ϭ 9.81 kN/m3 (b) AXIAL STRESS IN THE WALL DUE TO WATER PRESSURE ALONE Because the top of the tank is open, the internal pressure of the water produces no axial (longitudinal) stresses in the wall of the tank. Axial stress equals zero. ; h ϭ 12(20) 0.00981(1.1) ϭ 22.2 m 08Ch08.qxd 9/18/08 11:03 AM Page 656 9. 9. SECTION 8.3 Cylindrical Pressure Vessels 657 Problem 8.3-3 An inflatable structure used by a traveling circus has the shape of a half-circular cylinder with closed ends (see figure). The fabric and plastic structure is inflated by a small blower and has a radius of 40 ft when fully inflated. A longitudinal seam runs the entire length of the “ridge” of the structure. If the longitudinal seam along the ridge tears open when it is subjected to a tensile load of 540 pounds per inch of seam, what is the factor of safety n against tearing when the internal pressure is 0.5 psi and the structure is fully inflated? Longitudinal seam Solution 8.3-3 Inflatable structure Half-circular cylinder r ϭ 40 ft ϭ 480 in. Internal pressure p ϭ 0.5 psi T ϭ tensile force per unit length of longitudinal seam Seam tears when T ϭ Tmax ϭ 540 lb/in. Find factor of safety against tearing. CIRCUMFERENTIAL STRESS (EQ. 8-5) where t ϭ thickness of fabric Actual value of T due to internal pressure ϭ s1t І T ϭ s1t ϭ pr ϭ (0.5 psi)(480 in.) ϭ 240 lb/in. FACTOR OF SAFETY n ϭ Tmax T ϭ 540 lb/in. 240 lb/in. ϭ 2.25 ; s1 ϭ pr t Problem 8.3-4 A thin-walled cylindrical pressure vessel of radius r is subjected simultaneously to internal gas pressure p and a compressive force F acting at the ends (see figure). What should be the magnitude of the force F in order to produce pure shear in the wall of the cylinder? FF Solution 8.3-4 C ylindrical pressure vessel r ϭ Radius p ϭ Internal pressure STRESSES (SEE EQ. 8-5 AND 8-6): s2 ϭ pr 2t Ϫ F A ϭ pr 2t Ϫ F 2prt s1 ϭ pr t 08Ch08.qxd 9/18/08 11:03 AM Page 657 10. 10. 658 CHAPTER 8 Applications of Plane Stress FOR PURE SHEAR, the stresses s1 and s2 must be equal in magnitude and opposite in sign (see, e.g., Fig. 711 in Section 7.3). І s1 ϭ Ϫs2 OR Solve for F: F ϭ 3ppr2 ; pr t ϭ Ϫ a pr 2t Ϫ F 2prt b Problem 8.3-5 A strain gage is installed in the longitudinal direction on the surface of an aluminum beverage can (see figure). The radius-to-thickness ratio of the can is 200. When the lid of the can is popped open, the strain changes by ʦ0 ϭ 170 ϫ 10Ϫ6 . What was the internal pressure p in the can? (Assume E ϭ 10 ϫ 106 psi and v ϭ 0.33.) 12 FL OZ (355 mL) Solution 8.3-5 Aluminum can E ϭ 10 ϫ 106 psi v ϭ 0.33 0 ϭ change in strain when pressure is released ϭ 170 ϫ 10Ϫ6 Find internal pressure p. r t ϭ 200 STRAIN IN LONGITUDINAL DIRECTION (EQ. 8-11a) Substitute numerical values: p ϭ 2(10 * 106 psi)(170 * 10Ϫ6 ) (200)(1 Ϫ 0.66) ϭ 50 psi ; â2 ϭ â0 ‹ p ϭ 2tEâ0 (r)(1 Ϫ 2v) ϭ 2Eâ0 (r/t) (1 Ϫ 2v) â2 ϭ pr 2tE (1 Ϫ 2v) or p ϭ 2tEâ2 r(1 Ϫ 2v) 08Ch08.qxd 9/18/08 11:03 AM Page 658 11. 11. SECTION 8.3 Cylindrical Pressure Vessels 659 Problem 8.3-6 A circular cylindrical steel tank (see figure) contains a volatile fuel under pressure. A strain gage at point A records the longitudinal strain in the tank and transmits this information to a control room. The ultimate shear stress in the wall of the tank is 84 MPa, and a factor of safety of 2.5 is required. At what value of the strain should the operators take action to reduce the pressure in the tank? (Data for the steel are as follows: modulus of elasticity E ϭ 205 GPa and Poisson’s ratio v ϭ 0.30.) A Cylindrical tank Pressure relief valve Solution 8.3-6 tULT ϭ 84 MPa E ϭ 205 GPa v ϭ 0.3 n ϭ 2.5 tmax ϭ 33.6 MPa Find maximum allowable strain reading at the gage From Eq. (8-10) Pmax ϭ 2ttmax r tmax ϭ s1 2 ϭ pr 2t s2 ϭ pr 2t s1 ϭ pr t tmax ϭ tULT n From Eq. (8-11a) max ϭ 6.56 ϫ 10Ϫ5 âmax ϭ tmax E (1 Ϫ 2v) â2max ϭ pmaxr 2tE (1 Ϫ 2v) ϭ tmax E (1 Ϫ 2v) â2 ϭ pr 2tE (1 Ϫ 2v) Problem 8.3-7 A cylinder filled with oil is under pressure from a piston, as shown in the figure. The diameter d of the piston is 1.80 in. and the compressive force F is 3500 lb. The maximum allowable shear stress tallow in the wall of the cylinder is 5500 psi. What is the minimum permissible thickness tmin of the cylinder wall? (See the figure on the next page.) Probs. 8.3-7 and 8.3-8 Cylinder F Piston p Solution 8.3-7 Cylinder with internal pressure d ϭ 1.80 in. r ϭ 0.90 in. F ϭ 3500 lb tallow ϭ 5500 psi Find minimum thickness tmin. Pressure in cylinder: p ϭ F A ϭ F pr2 Maximum shear stress (Eq. 8-10): Minimum thickness: Substitute numerical values: tmin ϭ 3500 lb 2p(0.90 in.)(5500 psi) ϭ 0.113 in. ; tmin ϭ F 2prtallow tmax ϭ pr 2t ϭ F 2prt 08Ch08.qxd 9/18/08 11:03 AM Page 659 12. 12. 660 CHAPTER 8 Applications of Plane Stress Problem 8.3-8 Solve the preceding problem if d ϭ 90 mm, F ϭ 42 kN, and tallow ϭ 40 MPa. Solution 8.3-8 Cylinder with internal pressure d ϭ 90 mm r ϭ 45 mm F ϭ 42.0 kN tallow ϭ 40 MPa Find minimum thickness tmin. Pressure in cylinder: p ϭ F A ϭ F pr2 Maximum shear stress (Eq. 8-10): Minimum thickness: Substitute numerical values: tmin ϭ 42.0 kN 2p(45 mm)(40 MPa) ϭ 3.71 mm ; tmin ϭ F 2prtallow tmax ϭ pr 2t ϭ F 2prt Problem 8.3-9 A standpipe in a water-supply system (see figure) is 12 ft in diameter and 6 inches thick. Two horizontal pipes carry water out of the standpipe; each is 2 ft in diameter and 1 inch thick. When the system is shut down and water fills the pipes but is not moving, the hoop stress at the bottom of the standpipe is 130 psi. (a) What is the height h of the water in the standpipe? (b) If the bottoms of the pipes are at the same elevation as the bottom of the standpipe,

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62.4 lb/ft3 ϭ 62.4 1728 lb/in.3 (a) FIND HEIGHT h OF WATER IN THE STANDPIPE p ϭ pressure at bottom of standpipe ϭ gh From Eq. (8-5): Substitute numerical values: ϭ 25 ft ; h ϭ (130 psi)(6 in.) a 62.4 1728 lb/in.3 b(72 in.) ϭ 300 in. s1 ϭ pr t ϭ ghr t or h ϭ s1t gr 08Ch08.qxd 9/18/08 11:03 AM Page 660 13. 13. SECTION 8.3 Cylindrical Pressure Vessels 661 HORIZONTAL PIPES d1 ϭ 2 ft ϭ 24 in. r1 ϭ 12 in. t1 ϭ 1.0 in. (b) FIND HOOP STRESS s1 IN THE PIPES Since the pipes are 2 ft in diameter, the depth of water to the center of the pipes is about 24 ft. h1 Ϸ 24 ft ϭ 288 in. p1 ϭ gh1 Based on the average pressure in the pipes: s1 Ϸ 125 psi ; ϭ 125 psi ϭ a 62.4 1728 lb/in.3 b(288 in.)(12 in.) 1.0 in. s1 ϭ p1r1 t1 ϭ gh1r1 t1 Problem 8.3-10 A cylindrical tank with hemispherical heads is constructed of steel sections that are welded circumferentially (see figure). The tank diameter is 1.25 m, the wall thickness is 22 mm, and the internal pressure is 1750 kPa. (a) Determine the maximum tensile stress sh in the heads of the tank. (b) Determine the maximum tensile stress sc in the cylindrical part of the tank. (c) Determine the tensile stress sw acting perpendicular to the welded joints. (d) Determine the maximum shear stress th in the heads of the tank. Probs. 8.3-10 and 8.3-11 (e) Determine the maximum shear stress tc in the cylindrical part of the tank. Welded seams Solution 8.3-10 d ϭ 1.25 m t ϭ 22 mm p ϭ 1750 kPa (a) MAXIMUM TENSILE STRESS IN HEMISPHERES (EQ. 8-1) sh ϭ 24.9 MPa (b) MAXIMUM STRESS IN CYLINDER (EQ. 8-5) sc ϭ 49.7 MPa ;sc ϭ pr t ;sh ϭ pr 2t r ϭ d 2 (c) TENSILE STRESS IN WELDS (EQ. 8-6) sw ϭ 24.9 MPa (d) MAXIMUM SHEAR STRESS IN HEMISPHERES (EQ. 8-3) th ϭ 12.43 MPa (e) MAXIMUM SHEAR STRESS IN CYLINDER (EQ. 8-10) tc ϭ 24.9 MPa ;tc ϭ pr 2t ;th ϭ pr 4t ;sw ϭ pr 2t Problem 8.3-11 A cylindrical tank with diameter d ϭ 18 in. is subjected to internal gas pressure p ϭ 450 psi. The tank is constructed of steel sections that are welded circumferentially (see figure). The heads of the tank are hemispherical. The allowable tensile and shear stresses are 8200 psi and 3000 psi, respectively. Also, the allowable tensile stress perpendicular to a weld is 6250 psi. Determine the minimum required thickness tmin of (a) the cylindrical part of the tank and (b) the hemispherical heads. 08Ch08.qxd 9/18/08 11:03 AM Page 661 14. 14. 662 CHAPTER 8 Applications of Plane Stress Solution 8.3-11 d ϭ 18 in. p ϭ 450 psi sallow ϭ 8200 psi (tension) tallow ϭ 3000 psi (shear) WELD sa ϭ 6250 psi (tension) (a) FIND MINIMUM THICKNESS O F C YLINDER TENSION tmin ϭ 0.494 in. SHEAR tmin ϭ 0.675 in. WELD s ϭ pr 2t tmin ϭ pr 2tallow tmax ϭ pr 2t tmin ϭ pr sallow smax ϭ pr t r ϭ d 2 tmin ϭ 0.324 in. (b) FIND MINIMUM THICKNESS OF HEMISPHERES TENSION tmin ϭ 0.247 in. SHEAR tmin ϭ 0.338 in. tmin ϭ 0.338 in. ; tmin ϭ pr 4tallow tmax ϭ pr 4t tmin ϭ pr 2sallow smax ϭ pr 2t tmin ϭ 0.675 in. ; tmin ϭ pr 2sa Problem *8.3-12 A pressurized steel tank is constructed with a helical weld that makes an angle a ϭ 55° with the longitudinal axis (see figure). The tank has radius r ϭ 0.6 m, wall thickness t ϭ 18 mm, and internal pressure p ϭ 2.8 MPa. Also, the steel has modulus of elasticity E ϭ 200 GPa and Poisson’s ratio v ϭ 0.30. Determine the following quantities for the cylindrical part of the tank. (a) The circumferential and longitudinal stresses. (b) The maximum in-plane and out-of-plane shear stresses. (c) The circumferential and longitudinal strains. (d) The normal and shear stresses acting on planes parallel and perpendicular to Probs. 8.3-12 and 8.3-13 the weld (show these stresses on a properly oriented stress element). Solution 8.3-12 a ϭ 55 deg r ϭ 0.6 m t ϭ 18 mm p ϭ 2.8 MPa E ϭ 200 GPa v ϭ 0.3 (a) CIRCUMFERENTIAL STRESS s1 ϭ 93.3 MPa LONGITUDIAL STRESS s2 ϭ 46.7 MPa ;s2 ϭ pr 2t ;s1 ϭ pr t (b) IN-PLANE SHEAR STRESS t1 ϭ 23.3 MPa OUT-OF-PLANE SHEAR STRESS t2 ϭ 46.7 MPa ;t2 ϭ s1 2 ;t1 ϭ s1 Ϫ s2 2 Helical weld a 08Ch08.qxd 9/18/08 11:03 AM Page 662 15. 15. SECTION 8.3 Cylindrical Pressure Vessels 663 (c) CIRCUMFERENTIAL STRAIN 1 ϭ 3.97 ϫ 10Ϫ4 LONGITUDINAL STRAIN 2 ϭ 9.33 ϫ 10Ϫ5 (d) STRESS ON WELD u ϭ 90 deg Ϫ a u ϭ 35 deg sx ϭ s2 sx ϭ 46.667 MPa sy ϭ s1 sy ϭ 93.333 MPa txy ϭ 0 ;â2 ϭ s2 E (1 Ϫ 2v) ;â1 ϭ s1 2E (2 Ϫ v) For u ϭ 35 deg sx1 ϭ 62.0 MPa tx1y1 ϭ 21.9 MPa sy1 ϭ sx ϩ sy Ϫ sx1 sy1 ϭ 78.0 MPa ; ; tx1y1 ϭ Ϫ sx Ϫ sy 2 sin (2u) + txy cos (2u) ; + txy sin(2u) sx1 ϭ sx + sy 2 + sx Ϫ sy 2 cos (2u) Problem *8.3-13 Solve the preceding problem for a welded tank with a ϭ 62°, r ϭ 19 in., t ϭ 0.65 in., p ϭ 240 psi, E ϭ 30 ϫ 106 psi, and v ϭ 0.30. Solution 8.3-13 a ϭ 62 deg r ϭ 19 in. t ϭ 0.65 in. p ϭ 240 psi E ϭ 30 ϫ 106 psi v ϭ 0.3 (a) CIRCUMFERENTIAL STRESS LONGITUDIAL STRESS (b) IN-PLANE SHEAR STRESS OUT-OF-PLANE SHEAR STRESS (c) CIRCUMFERENTIAL STRAIN 1 ϭ 1.988 ϫ 10Ϫ4 ;â1 ϭ s1 2E (2 Ϫ v) ;t2 ϭ 3508 psit2 ϭ s1 2 ;t1 ϭ 1754 psit1 ϭ s1 Ϫ s2 2 ;s2 ϭ 3508 psis2 ϭ pr 2t ;s1 ϭ 7015 psis1 ϭ pr t LONGITUDINAL STRAIN 2 ϭ 4.68 ϫ 10Ϫ5 (d) STRESS ON WELD u ϭ 90 deg Ϫ a u ϭ 28 deg sx ϭ s2 sx ϭ 3.508 ϫ 103 psi sy ϭ s1 sy ϭ 7.015 ϫ 103 psi txy ϭ 0 For u ϭ 28 deg sy1 ϭ sx ϩ sy Ϫ sx1 ;sy1 ϭ 6242 psi ;tx1y1 ϭ 1454 psi tx1y1 ϭ Ϫ sx Ϫ sy 2 sin (2u) + txy cos (2u) ;sx1 ϭ 4281 psi + txy sin (2u) sx1 ϭ sx + sy 2 + sx Ϫ sy 2 cos (2u) ;â2 ϭ s2 E (1 Ϫ 2v) 08Ch08.qxd 9/18/08 11:03 AM Page 663 16. 16. Maximum Stresses in Beams When solving the problems for Section 8.4, consider only the in-plane stresses and disregard the weights of the beams Problem 8.4-1 A cantilever beam of rectangular cross section is subjected to a concentrated load P ϭ 17 k acting at the free end (see figure). The beam has width b ϭ 3 in. and height h ϭ 12 in. Point A is located at distance c ϭ 2.5 ft from the free end and distance d ϭ 9 in. from the bottom of the beam. Calculate the principal stresses s1 and s2 and the maximum shear stress tmax at point A. Show these stresses on sketches of properly oriented elements. Probs. 8.4-1 and 8.4-2 P c A b d h Solution 8.4-1 P ϭ 17 k c ϭ 2.5 ft b ϭ 3 in. d ϭ 9 in. h ϭ 12 in. v ϭ 0.3 STRESS AT POINT A I ϭ 432 in.4 M ϭ ϪPc M ϭ Ϫ5.1 ϫ 105 lb-in. V ϭ P V ϭ 1.7 ϫ 104 lb yA ϭ 3 in. sx ϭ 3.542 ϫ 103 psi Q ϭ 40.5 in.3 t ϭ 531.25 psi txy ϭ t sx ϭ 3.542 ϫ 103 psi sy ϭ 0 txy ϭ 531.25 psi PRINCIPAL STRESSES up ϭ 8.35 degup ϭ 1 2 atana 2txy sx Ϫ sy b tan12up2 ϭ 2txy sx Ϫ sy ϭ 2txy sx Ϫ sy ϭ 0.3 t ϭ VQ Ib Q ϭ bd a h 2 Ϫ d 2 b sx ϭ Ϫ MyA I yA ϭ Ϫ h 2 + d I ϭ bh3 12 For u1 ϭ up u1 ϭ 8.35 deg sx1 ϭ 60.306 Mpa For u2 ϭ 90 deg ϩ up u2 ϭ 98.35 deg sx2 ϭ Ϫ77.971 psi Therefore s1 ϭ sx2 up1 ϭ u2 s2 ϭ sx1 up2 ϭ u1 MAXIMUM SHEAR STRESSES us1 ϭ up1 Ϫ 45 deg us2 ϭ us1 ϩ 90 deg savg ϭ 1771 psi ;savg ϭ sx + sy 2 us2 ϭ 143.4 deg ; us1 ϭ 53.4 deg ; tmax ϭ 1849 psi ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 up2 ϭ 8.35 deg ;s2 ϭ 3620 psi ; up1 ϭ 98.4 deg ;s1 ϭ Ϫ78.0 psi ; + txy sin (2u2) sx2 ϭ sx + sy 2 + sx Ϫ sy 2 cos (2u2) + txy sin (2u1) sx1 ϭ sx + sy 2 + sx Ϫ sy 2 cos (2u1) 664 CHAPTER 8 Applications of Plane Stress 08Ch08.qxd 9/18/08 11:04 AM Page 664 17. 17. SECTION 8.4 Maximum Stresses in Beams 665 Problem 8.4-2 Solve the preceding problem for the following data: P ϭ 130 kN, b ϭ 80 mm, h ϭ 260 mm, c ϭ 0.6 m, and d ϭ 220 mm. Solution 8.4-2 P ϭ 130 kN c ϭ 0.6 m b ϭ 80 mm d ϭ 220 mm h ϭ 260 mm v ϭ 0.3 STRESS AT POINT A I ϭ 1.172 ϫ 108 mm4 M ϭ ϪPc M ϭ Ϫ7.8 ϫ 104 N m V ϭ P V ϭ 1.3 ϫ 105 N yA ϭ 90 mm sx ϭ 59.911 MPa Q ϭ 3.52 ϫ 105 mm3 t ϭ 4.882 MPa txy ϭ t sx ϭ 59.911 MPa sy ϭ 0 txy ϭ 4.882 MPa PRINCIPAL STRESSES up ϭ 4.628 degup ϭ 1 2 atana 2txy sx Ϫ sy b tan(2up) ϭ 2txy sx Ϫ sy ϭ 2txy sx Ϫ sy ϭ 0.163 t ϭ VQ Ib Q ϭ bd a h 2 Ϫ d 2 b sx ϭ Ϫ MyA I yA ϭ Ϫ h 2 + d # I ϭ bh3 12 For u1 ϭ up u1 ϭ 4.628 deg sx1 ϭ 60.306 MPa For u2 ϭ 90 deg ϩ up u2 ϭ 94.628 deg sx2 ϭ Ϫ0.395 MPa Therefore s1 ϭ sx1 up1 ϭ u1 s2 ϭ sx2 up2 ϭ u2 MAXIMUM SHEAR STRESSES us1 ϭ up1 Ϫ 45 deg us2 ϭ us1 ϩ 90 deg savg ϭ 30.0 deg ;savg ϭ sx + sy 2 us2 ϭ 49.6 deg ; us1 ϭ Ϫ40.4 deg ; tmax ϭ 30.4 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + t xy 2 up2 ϭ 94.6 deg ;s2 ϭ Ϫ0.395 MPa ; up1 ϭ 4.63 deg ;s1 ϭ 60.3 MPa ; + txy sin (2u2) sx2 ϭ sx + sy 2 + sx Ϫ sy 2 cos (2u2) + txy sin (2u1) sx1 ϭ sx + sy 2 + sx Ϫ sy 2 cos (2u1) Problem 8.4-3 A simple beam of rectangular cross section (width 4 in., height 10 in.) carries a uniform load of 1200 lb/ft on a span of 12 ft (see figure). Find the principal stresses s1 and s2 and the maximum shear stress tmax at a cross section 2 ft from the left-hand support at each of the following locations: (a) the neutral axis, (b) 2 in. above the neutral axis and (c) the top of the beam. (Disregard the direct compressive stresses produced by the uniform load bearing against the top of the beam.) 10 in. 4 in. 1200 lb/ft 2 ft 12 ft 08Ch08.qxd 9/18/08 11:04 AM Page 665 18. 18. 666 CHAPTER 8 Applications of Plane Stress Problem 8.4-4 An overhanging beam ABC with a guided support at A is of rectangular cross section and supports concentrated loads P both at A and at the free end C (see figure). The span length from A to B is L, and the length of the overhang is L/2. The cross section has width b and height h. Point D is located midway between the supports at a distance d from the top face of the beam. Knowing that the maximum tensile stress (principal stress) at point D is s1 ϭ 35 MPa. determine the magnitude of the load P. Data for the beam are as follows: L ϭ 1.75 m, b ϭ 50 mm, Probs. 8.4-4 and 8.4-5 h ϭ 200 mm, and d ϭ 40 mm. Solution 8.4-3 b ϭ 4 in. h ϭ 10 in. A ϭ bh I ϭ 333.333 in.4 c ϭ 2 ft L ϭ 12 ft RA ϭ 7.2 ϫ 103 lb M ϭ 1.44 ϫ 105 lb in. V ϭ RA Ϫ qc V ϭ 4.8 ϫ 103 lb (a) NEUTRAL AXIS sx ϭ 0 sy ϭ 0 txy ϭ Ϫ180 psi Pure shear: s1 ϭ Ϫtxy s2 ϭ Ϫs1 tmax ϭ s1 s1 ϭ 180 psi s2 ϭ Ϫ180 psi (b) 2 IN. ABOVE THE NEUTRAL AXIS y ϭ 2 in. d ϭ 3 in. sx ϭ Ϫ864 psi sy ϭ 0 Q ϭ 42 in.3 Q ϭ bd a h 2 Ϫ d 2 b sx ϭ Ϫ My I tmax ϭ 180 psi ; txy ϭ Ϫ 3V 2A #M ϭ

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0 psi s2 ϭ sx s2 ϭ Ϫ2.16 ϫ 103 psi tmax ϭ 1080 psi ; tmax ϭ A a sx 2 b 2 + txy 2 sx ϭ Ϫ M a h 2 b I tmax ϭ 458 psi ; tmax ϭ A a sx 2 b 2 + txy 2 s2 ϭ Ϫ890 psi ; s2 ϭ sx 2 Ϫ A a sx 2 b 2 + txy 2 s1 ϭ 25.7 psi ; s1 ϭ sx 2 + A a sx 2 b 2 + txy 2 txy ϭ Ϫ VQ Ib B P P h b D d C A L 2 L 2 — L 2 L 2 — L 2 L 2 — 08Ch08.qxd 9/18/08 11:04 AM Page 666 19. 19. SECTION 8.4 Maximum Stresses in Beams 667 Problem 8.4-5 Solve the preceding problem if the stress and dimensions are as follows: s1 ϭ 2450 psi, L ϭ 80 in., b ϭ 2.5 in., h ϭ 10 in., and d ϭ 2.5 in. Solution 8.4-4 L ϭ 1.75 m b ϭ 50 mm h ϭ 200 mm d ϭ 40 mm Maximum principal stress at point D: RB ϭ 0 VD ϭ P STRESS AT POINT D I ϭ 3.333 ϫ 107 mm4 y ϭ 60 mm sy ϭ 0sx ϭ Ϫ My I ϭ Ϫ(ϪPL)y I ϭ (3150P) N>m2 y ϭ h 2 Ϫ d I ϭ bh3 12 MD ϭ Ϫ 3 2 PL + P L 2 ϭ ϪPL MA ϭ 3 2 PLMA ϭ PL + P L 2 Q ϭ 1.6 ϫ 105 mm3 PRINCIPAL STRESSES With s1 ϭ 35 MPa P ϭ 11.10 kN ; P ϭ s1 3.153 * 103 ϭ sx 2 + A a sx 2 b 2 + txy 2 ϭ (3.153 * 103 )P s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ VQ Ib ϭ (96P) N>m2 Q ϭ bd a h 2 Ϫ d 2 b Solution 8.4-5 L ϭ 80 in. b ϭ 2.5 in. h ϭ 10 in. d ϭ 2.5 in. Maximum principal stress at point D: RB ϭ 0 VD ϭ P STRESS AT POINT D I ϭ 208.333 in.4 y ϭ 2.5 in. sy ϭ 0 sx ϭ Ϫ My I ϭ Ϫ(Ϫ PL)y I ϭ (0.96P) lb in.2 y ϭ h 2 Ϫ d I ϭ bh3 12 MD ϭ Ϫ 3 2 PL + P L 2 ϭ ϪPL MA ϭ 3 2 PLMA ϭ PL + P L 2 Q ϭ 23.438 in.3 PRINCIPAL STRESSES With s1 ϭ 2450 psi P ϭ 2.55 k ; P ϭ s1 0.962 ϭ sx 2 + A a sx 2 b 2 + txy 2 ϭ (0.962P) 1b in.2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ VQ Ib ϭ (0.045P) lb/in.2 Q ϭ bd a h 2 Ϫ d 2 b 08Ch08.qxd 9/18/08 11:04 AM Page 667 20. 20. 668 CHAPTER 8 Applications of Plane Stress Problem 8.4-6 A beam of wide-flange cross section (see figure) has the following dimensions: b ϭ 120 mm, t ϭ 10 mm, h ϭ 300 mm, and h1 ϭ 260 mm. The beam is simply supported with span length L ϭ 3.0 m. A concentrated load P ϭ 120 kN acts at the midpoint of the span. At a cross section located 1.0 m from the left-hand support, determine the principal stresses s1 and s2 and the maximum shear stress tmax at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis. Probs. 8.4-6 and 8.4-7 h1 h t b Solution 8.4-6 Simply supported beam P ϭ 120 kN L ϭ 3.0 m c ϭ 1.0 m b ϭ 120 mm t ϭ 10 mm h ϭ 300 mm h1 ϭ 260 mm (a) TOP OF THE BEAM (POINT A) I ϭ bh3 12 Ϫ (b Ϫ t)h1 3 12 ϭ 108.89 * 106 mm4 V ϭ P 2 ϭ 60 kNM ϭ Pc 2 ϭ 60 kN # m (b) TOP OF THE WEB (POINT B) (c) NEUTRAL AXIS (POINT C) sx ϭ 0 sy ϭ 0 ϭ 420.5 * 103 mm3 Q ϭ ba h 2 b a h 4 b Ϫ (b Ϫ t)a h1 2 b a h1 4 b tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 ϭ 40.3 MPa ; s1 ϭ 4.5 MPa, s2 ϭ Ϫ76.1 MPa ; ϭ Ϫ35.82 ; 40.32 MPa s1,2 ϭ sx + sy 2 ; A a sx Ϫ sy 2 b 2 + txy 2 ϭ Ϫ18.51 MPa txy ϭ Ϫ VQ It ϭ Ϫ (60 kN)(336 * 103 mm3 ) (108.89 * 106 mm4 )(10 mm) Q ϭ (b)a h Ϫ h1 2 b a h + h1 4 b ϭ 336 * 103 mm3 sy ϭ 0 ϭ Ϫ71.63 MPa sx ϭ Ϫ My I ϭ Ϫ (60 kN # m)(130 mm) 108.89 * 106 mm4 Uniaxial stress: s1 ϭ 0 s2 ϭ Ϫ82.7 MPa tmax ϭ 41.3 MPa K ; sy ϭ 0 txy ϭ 0 ϭ Ϫ82.652 MPa sx ϭ Mc I ϭ Ϫ (60 kN # m)(150 mm) 108.89 * 106 mm4 08Ch08.qxd 9/18/08 11:04 AM Page 668 21. 21. SECTION 8.4 Maximum Stresses in Beams 669 Problem 8.4-7 A beam of wide-flange cross section (see figure) has the following dimensions: b ϭ 5 in., t ϭ 0.5 in., h ϭ 12 in., and h1 ϭ 10.5 in. The beam is simply supported with span length L ϭ 10 ft and supports a uniform load q ϭ 6 k/ft. Calculate the principal stresses s1 and s2 and the maximum shear stress tmax at a cross section located 3 ft from the left-hand support at each of the following locations: (a) the bottom of the beam, (b) the bottom of the web, and (c) the neutral axis. ϭ Ϫ23.17 MPa txy ϭ Ϫ VQ It ϭ Ϫ (60 kN)(420.5 * 103 mm3 ) (108.89 * 106 mm4 )(10 mm) Pure shear: s1 ϭ 23.2 MPa, s2 ϭ Ϫ23.2 MPa tmax ϭ 23.2 MPa K ; Solution 8.4-7 Simply supported beam q ϭ 6.0 k/ft L ϭ 10 ft ϭ 120 in. cϭ3ftϭ36in. b ϭ 5.0 in. t ϭ 0.5 in. h ϭ 12 in. h1 ϭ 10.5 in. (a) BOTTOM OF THE BEAM (POINT A) Uniaxial stress: s1 ϭ 15,870 psi, s2 ϭ 0 tmax ϭ 7930 psi d ; sy ϭ 0 txy ϭ 0 ϭ 15,866 psi sx ϭ Ϫ Mc I ϭ (756,000 lb-in.)(Ϫ6.0 in.) 285.89 in.4 I ϭ bh3 12 Ϫ (b Ϫ t)h1 3 12 ϭ 285.89 in.4 V ϭ qL 2 Ϫ qc ϭ 12,000 lb M ϭ qLc 2 Ϫ qc2 2 ϭ 756,000 lb-in. (b) BOTTOM OF THE WEB (POINT B) ϭ 6941.5 ; 7163.9 psi s1,2 ϭ sx + sy 2 ; A a sx Ϫ sy 2 b 2 + txy 2 ϭ Ϫ1771 psi txy ϭ Ϫ VQ It ϭ Ϫ (12,000 lb)(21.094 in.3 ) (285.89 in.4 )(0.5 in.) sy ϭ 0 Q ϭ ba h Ϫ h1 2 b a h + h1 4 b ϭ 21.094 in.3 ϭ 13,883 psi sx ϭ Ϫ My I ϭ Ϫ (756,000 lb-in.)(Ϫ5.25 in.) 285.89 in.4 08Ch08.qxd 9/18/08 11:04 AM Page 669 22. 22. 670 CHAPTER 8 Applications of Plane Stress Problem 8.4-8 A W 200 ϫ 41.7 wide-flange beam (see Table E-1(b). Appendix E) is simply supported with a span length of 2.5 m (see figure). The beam supports a concentrated load of 100 kN at 0.9 m from support B. At a cross section located 0.7 m from the left-hand support, determine the principal stresses s1 and s2 and the maximum shear stress tmax at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis. (c) NEUTRAL AXIS (POINT C) sx ϭ 0 sy ϭ 0 ϭ 27.984 in.3 Q ϭ ba h 2 b a h 4 b Ϫ (b Ϫ t)a h1 2 b a h1 4 b tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 ϭ 7160 psi ; s1 ϭ 14,100 psi, s2 ϭ Ϫ220 psi ; Pure shear: s1 ϭ 2350 psi, s2 ϭ Ϫ2350 psi, tmax ϭ 2350 psi K ; ϭ Ϫ2349 psi txy ϭ Ϫ VQ It ϭ Ϫ (12,000 lb)(27.984 in.3 ) (285.89 in.4 )(0.5 in.) D 100 kN BA W 200 × 41.7 0.9 m 2.5 m 0.9 m0.7 m Solution 8.4-8 RB ϭ 64 kN (upward) RA ϭ 100 kN Ϫ RB RA ϭ 36 kN (upward) At the point D M ϭ RA(0.7 m) M ϭ 25.2 kN m V ϭ RA V ϭ 36 kN # RB ϭ 100 kN a 0.7 + 0.9 2.5 b (a) TOP OF THE BEAM (POINT A) sx ϭ Ϫ63.309 MPa sy ϭ 0 txy ϭ 0 Uniaxial stress: s1 ϭ sy s2 ϭ sx (b) TOP OF THE WEB (POINT B) sx ϭ Ϫ56.021 MPa sy ϭ 0 Q ϭ 1.892 ϫ 105 mm3 Q ϭ ba h Ϫ h1 2 b a h + h1 4 b sx ϭ Ϫ M a h1 2 b I tmax ϭ 31.7 MPa ; s2 ϭ Ϫ63.3 MPa ;s1 ϭ 0 ; tmax ϭ ` sx 2 ` sx ϭ Ϫ M a h 2 b I W200 ϫ 41.7 I ϭ 40.8 ϫ 106 mm4 b ϭ 166 mm tw ϭ 7.24 mm tf ϭ 11.8 mm h ϭ 205 mm h1 ϭ h Ϫ 2tf h1 ϭ 181.4 mm 08Ch08.qxd 9/18/08 11:04 AM Page 670 23. 23. SECTION 8.4 Maximum Stresses in Beams 671 Problem 8.4-9 A W 12 ϫ 14 wide-flange beam (see Table E-1(a), Appendix E) is simply supported with a span length of 120 in. (see figure). The beam supports two anti-symmetrically placed concentrated loads of 7.5 k each. At a cross section located 20 in. from the right-hand support, determine the principal stresses s1 and s2 and the maximum shear stress tmax at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis. txy ϭ Ϫ23.061 MPa tmax ϭ 36.3 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ Ϫ64.3 MPa ; s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ 8.27 MPa ; s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Ϫ VQ Itw (c) NEUTRAL AXIS (POINT C) sx ϭ 0 sy ϭ 0 Q ϭ 2.19 ϫ 105 mm3 txy ϭ Ϫ26.69 MPa Pure shear: s2 ϭ Ϫs1 tmax ϭ txy tmax ϭ Ϫ26.7 Mpa ; s2 ϭ Ϫ26.7 Mpa ; s1 ϭ 26.7 Mpa ;s1 ϭ ƒtxyƒ txy ϭ Ϫ VQ Itw Q ϭ b a h 2 b a h 4 b Ϫ 1 b Ϫ tw2 a h1 2 b a h1 4 b D 7.5 k 7.5 k W 12 × 14 20 in. 20 in. 120 in. 40 in.40 in. Solution 8.4-9 RB ϭ 2.5 k (upward) RA ϭ ϪRB RA ϭ Ϫ 2.5 k (downward) At Section C-C M ϭ RB 20 in. M ϭ 50 k in. V ϭ ϪRB V ϭ Ϫ2.5 k ## RB ϭ 7.5 # 80 Ϫ 7.5 # 40 120 k (a) TOP OF THE BEAM (POINT A) sx ϭ Ϫ3.361 ϫ 103 psi sy ϭ 0 txy ϭ 0 Uniaxial stress: s1 ϭ sy s2 ϭ sx (b) TOP OF THE WEB (POINT B) sx ϭ Ϫ3.234 ϫ 103 psi sy ϭ 0 sx ϭ Ϫ M a h1 2 b I tmax ϭ 1680 psi ; s2 ϭ Ϫ3361 psi ;s1 ϭ 0 ; tmax ϭ ` sx 2 ` sx ϭ Ϫ M a h 2 b I W12 ϫ 14 I ϭ 88.6 in.4 b ϭ 3.970 in. tw ϭ 0.200 in. tf ϭ 0.225 in. h ϭ 11.91 in. h1 ϭ h Ϫ 2tf h1 ϭ 11.46 in. 08Ch08.qxd 9/18/08 11:04 AM Page 671 24. 24. 672 CHAPTER 8 Applications of Plane Stress Problem *8.4-10 A cantilever beam of T-section is loaded by an inclined force of magnitude 6.5 kN (see figure). The line of action of the force is inclined at an angle of 60° to the horizontal and intersects the top of the beam at the end cross section. The beam is 2.5 m long and the cross section has the dimensions shown. Determine the principal stresses s1 and s2 and the maximum shear stress tmax at points A and B in the web of the beam near the support. Q ϭ 5.219 in.3 txy ϭ 736.289 psi tmax ϭ 1777 psi ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ Ϫ3393 psi ; s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ 159.8 psi ; s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Ϫ VQ Itw Q ϭ ba h Ϫ h1 2 b a h + h1 4 b (C) NEUTRAL AXIS (POINT C) sx ϭ 0 sy ϭ 0 Q ϭ 8.502 in3 txy ϭ 1.2 ϫ 103 psi Pure shear: s2 ϭ Ϫs1 tmax ϭ txy tmax ϭ 1200 psi ; s2 ϭ Ϫ1200 psi ; s1 ϭ 1200 psi ;s1 ϭ ƒtxyƒ txy ϭ Ϫ VQ I tw Q ϭ ba h 2 b a h 4 b Ϫ 1 b Ϫ tw2a h1 2 b a h1 4 b 2.5 m 80 mm 80 mm 160 mm 25 mm C 60° 6.5 kN A B 25 mm z y Solution 8.4-10 P ϭ 6.5 kN L ϭ 2.5 m A ϭ 2(160 mm)(25 mm) A ϭ 8 ϫ 103 mm2 b ϭ 160 mm t ϭ 25 mm Location of centroid C From Eq. (12-7b) in Chapter 12: c2 ϭ 126.25 mm c1 ϭ 185 mm Ϫ c2 c1 ϭ 58.75 mm MOMENT OF INTERTIA IZ ϭ 2.585 ϫ 107 mm4 IZ ϭ 1 3 t c2 3 + 1 3 bc1 3 Ϫ 1 3 ( b Ϫ t) ( c1 Ϫ t)3 c2 ϭ (160 mm) (25 mm) a160ϩ 25 2 b mm + (160 mm) (25 mm) (80 mm) A c2 ϭ ©( yi Ai) A 08Ch08.qxd 9/18/08 11:04 AM Page 672 25. 25. SECTION 8.4 Maximum Stresses in Beams 673 EQUIVALENT LOADS AT FREE END OF BEAM N0 ϭ PH N0 ϭ Ϫ3.25 kN V ϭ Pv V ϭ 5.629 kN M ϭ ϪMc Ϫ Pv L M ϭ Ϫ1.388 ϫ 104 N m# Stress at point A (bottom of web) sx ϭ Ϫ31.951 MPa sy ϭ 0 txy ϭ 0 Uniaxial stress: s1 ϭ sy s2 ϭ sx Stress at point B (top of web) sx ϭ 17.715 MPa sy ϭ 0 Q ϭ 1.85 ϫ 105 mm3 txy ϭ Ϫ1.611 MPa tmax ϭ 9.00 MPa ;tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ Ϫ0.145 MPa ;s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ 17.86 MPa ;s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Ϫ VQ Izt Q ϭ bt ac1 Ϫ t 2 b sx ϭ N0 A Ϫ M1c1 Ϫ t2 Iz

2017-5-21

Chapter 8

tmax ϭ 15.98 MPa ;s2 ϭ Ϫ32.0 MPa ;s1 ϭ 0 ; tmax ϭ ` sx 2 ` sx ϭ N0 A + Mc1 IZ PH ϭ ϪP cos (60 deg) PH ϭ Ϫ3.25 kN Pv ϭ P sin (60 deg) Pv ϭ 5.629 kN Mc ϭ PH c1 Mc ϭ Ϫ190.938 N m STRESS RESULTANTS AT FIXED END OF BEAM # 08Ch08.qxd 9/18/08 11:04 AM Page 673 26. 26. 674 CHAPTER 8 Applications of Plane Stress Problem *8.4-12 Solve the preceding problem for a cross section mn located 0.18 m from the support if L ϭ 0.75 m, M ϭ 65 kN m, h ϭ 120 mm, and b ϭ 20 mm.# 2×103 1×103 –1×103 σ.1(y) σ.2(y) τmax (y) Prin.stresses(psi) –2×103 –3 –3 –2.5 –2 –1.5 –1 –0.5 0 y dist. y (in.) 0.5 1 1.5 2 2.5 3 3 0 Problem *8.4-11 A simple beam of rectangular cross section has span length L ϭ 62 in. and supports a concentrated moment M ϭ 560 k-in at midspan (see figure). The height of the beam is h ϭ 6 in. and the width is b ϭ 2.5 in. Plot graphs of the principal stresses s1 and s2 and the maximum shear stress tmax, showing how they vary over the height of the beam at cross section mn, which is located 24 in. from the left-hand support. Probs. 8.4-11 and 8.4-12 L 2 —M at h L b m n Solution 8.4-11 M ϭ 560 k in. L ϭ 62 in. c ϭ 24 in. b ϭ 2.5 in. h ϭ 6 in. I ϭ 45 in4 RA ϭ 9.032 k (upward) RB ϭ ϪRA RB ϭ Ϫ9.032 k (downward) At section m-n M ϭ RAc M ϭ 216.774 k in. V ϭ RA V ϭ 9.032 k # RA ϭ M L I ϭ bh3 12 # sy ϭ 0 PRINCIPAL STRESSES s1(3 in.) ϭ 0 psi s2(3 in.) ϭ Ϫ14,452 psi tmax(3 in.) ϭ 7226 psi s1(0) ϭ 903 psi s2(0) ϭ Ϫ903 psi tmax(0) ϭ 903 psi s1(Ϫ3 in.) ϭ 14,452 psi s2(Ϫ3 in.) ϭ 0 psi tmax(Ϫ3 in.) ϭ 7226 psi tmax(y) ϭ A a sx(y) 2 b 2 + txy(y)2 s2(y) ϭ sx(y) 2 Ϫ A a sx(y) 2 b 2 + txy(y)2 s1(y) ϭ sx(y) 2 + A a sx(y) 2 b 2 + txy(y)2 txy(y) ϭ VQ(y) Ib Q(y) ϭ ba h 2 Ϫ yb a 1 2 b a h 2 + yb sx(y) ϭ Ϫ My I 08Ch08.qxd 9/18/08 11:04 AM Page 674 27. 27. SECTION 8.5 Combined Loadings 675 Solution 8.4-12 M ϭ 65 kN m L ϭ 0.75 m c ϭ 0.18 m b ϭ 20 mm h ϭ 120 mm At section m-n M ϭ RA c M ϭ 15.6 kN m V ϭ RA V ϭ 86.667 kN # RB ϭ ϪRA RB ϭ Ϫ86.667 kN (downward) RA ϭ M L RA ϭ 86.667 kN (upward) I ϭ bh3 12 I ϭ 2.88 * 106 mm4 # PRINCIPAL STRESSES s1(60 mm) ϭ 0 MPa s2(60 mm) ϭ Ϫ325 MPa tmax(60 mm) ϭ 162.5 MPa s1(0) ϭ 54.2 MPa s2(0) ϭ Ϫ54.2 MPa tmax(0) ϭ 54.2 MPa s1(Ϫ60 mm) ϭ 325 MPa s2(Ϫ60 mm) ϭ 0 MPa tmax(Ϫ60 mm) ϭ 162.5 MPa tmax( y) ϭ A a sx( y) 2 b 2 + txy( y)2 s2( y) ϭ sx( y) 2 Ϫ A a sx( y) 2 b 2 + txy( y)2 s1( y) ϭ sx( y) 2 + A a sx( y) 2 b 2 + txy( y)2 txy( y) ϭ VQ( y) Ib Q( y) ϭ b a h 2 Ϫ yb a 1 2 b a h 2 + yb sx( y) ϭ My I sy ϭ 0Prin.stresses(MPa) dist.y(mm) Combined Loadings The problems for Section 8.5 are to be solved assuming that the structures behave linearly elastically and that the stresses caused by two or more loads may be superimposed to obtain the resultant stresses acting at a point. Consider both in-plane and out-of-plane shear stresses unless otherwise specified. Problem 8.5-1 A bracket ABCD having a hollow circular cross section consists of a vertical arm AB, a horizontal arm BC parallel to the x0 axis, and a horizontal arm CD parallel to the z0 axis (see figure). The arms BC and CD have lengths b1 ϭ 3.6 ft and b2 ϭ 2.2 ft, respectively. The outer and inner diameters of the bracket are d2 ϭ 7.5 in. and d1 ϭ 6.8 in. A vertical load P ϭ 1400 lb acts at point D. Determine the maximum tensile, compressive, and shear stresses in the vertical arm. z0 y0 A x0 B C D P b1 b2 08Ch08.qxd 9/18/08 11:04 AM Page 675 28. 28. 676 CHAPTER 8 Applications of Plane Stress Solution 8.5-1 b1 ϭ 3.6 ft b2 ϭ 2.2 ft P ϭ 1400 lb d2 ϭ 7.5 in. d1 ϭ 6.8 in. VERTICAL ARM AB MAXIMUM TENSILE STRESS st ϭ 5100 psi ;st ϭ Ϫ P A + M a d2 2 b I M ϭ P 2b1 2 + b2 2 M ϭ 7.088 * 104 lb # in. I ϭ p 64 ad2 4 Ϫ d1 4 b I ϭ 50.36 in.4 A ϭ p 4 ad2 2 Ϫ d1 2 b A ϭ 7.862 in.2 MAXIMUM COMPRESSIVE STRESS MAXIMUM SHEAR STRESS Uniaxial stress tmax ϭ |sc| tmax ϭ 5456 psi ; sc ϭ Ϫ5456 psi ; sc ϭ Ϫ P A Ϫ M a d2 2 b I Problem 8.5-2 A gondola on a ski lift is supported by two bent arms, as shown in the figure. Each arm is offset by the distance b ϭ 180 mm from the line of action of the weight force W. The allowable stresses in the arms are 100 MPa in tension and 50 MPa in shear. If the loaded gondola weighs 12 kN, what is the minimum diameter d of the arms? (a) (b) b d W W 08Ch08.qxd 9/18/08 11:06 AM Page 676 29. 29. SECTION 8.5 Combined Loadings 677 Solution 8.5-2 Gondola on a ski lift b ϭ 180 mm sallow ϭ 100 MPa (tension) tallow ϭ 50 MPa Find dmin MAXIMUM TENSILE STRESS or a pst 4W bd3 Ϫ d Ϫ 8b ϭ 0 st ϭ W A + M S ϭ 4W pd2 + 32 Wb pd3 A ϭ pd2 4 S ϭ pd3 32 W ϭ 12 kN 2 ϭ 6 kN SUBSTITUTE NUMERICAL VALUES: 8b ϭ 1.44 m 13,090 d3 Ϫ d Ϫ 1.44 ϭ 0 (d ϭ meters) Solve numerically: d ϭ 0.04845 m І dmin ϭ 48.4 mm MAXIMUM SHEAR STRESS Since tallow is one-half of sallow, the minimum diameter for shear is the same as for tension. tmax ϭ st 2 (uniaxial stress) ; pst 4W ϭ psallow 4W ϭ p(100 Mpa) 4(6 kN) ϭ 13,089.97 1 m2 Problem 8.5-3 The hollow drill pipe for an oil well (see figure) is 6.2 in. in outer diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the pipe (due to the weight of the pipe) is 62 k and the torque (due to drilling) is 185 k-in. Determine the maximum tensile, compressive, and shear stresses in the drill pipe. 08Ch08.qxd 9/18/08 11:06 AM Page 677 30. 30. 678 CHAPTER 8 Applications of Plane Stress Solution 8.5-3 P ϭ compressive force T ϭ Torque d2 ϭ outer diameter d1 ϭ inner diameter P ϭ 62 k T ϭ 185 k in. d2 ϭ 6.2 in. t ϭ 0.75 in. d1 ϭ d2 Ϫ 2t d1 ϭ 4.7 in. A ϭ 12.841 in.2 Ip ϭ 97.16 in.4 STRESSES AT THE OUTER SURFACE Ip ϭ p 32 ad2 4 Ϫ d1 4 b A ϭ p 4 ad2 2 Ϫ d1 2b # txy ϭ 5903 psi PRINCIPAL STRESSES s1 ϭ 3963 psi s2 ϭ Ϫ8791 psi MAXIMUM TENSILE STRESS st ϭ s1 MAXIMUM COMPRESSIVE STRESS sc ϭ s2 MAXIMUM IN-PLANESHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. tmax ϭ 6377 psi ; tmax ϭ A a sx Ϫ sy 2 b + txy2 sc ϭ Ϫ8791 psi ; st ϭ 3963 psi ; s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy2 txy ϭ T a d2 2 b Ip Problem 8.5-4 A segment of a generator shaft is subjected to a torque T and an axial force P, as shown in the figure. The shaft is hollow (outer diameter d2 ϭ 300 mm and inner diameter d1 ϭ 250 mm) and delivers 1800 kW at 4.0 Hz. If the compressive force P ϭ 540 kN, what are the maximum tensile, compressive, and shear stresses in the shaft? P T T PProbs. 8.5-4 and 8.5-5 sy ϭ Ϫ4828 psi sx ϭ 0 sy ϭ Ϫ P A 08Ch08.qxd 9/18/08 11:06 AM Page 678 31. 31. SECTION 8.5 Combined Loadings 679 Solution 8.5-4 P ϭ Compressive_force P0 ϭ Power f ϭ frequency d2 ϭ outer diameter d1 ϭ inner diameter f ϭ 4.0 Hz P ϭ 540 kN P0 ϭ 1800 kW T ϭ 7.162 ϫ 104 N m d2 ϭ 300 mm d1 ϭ 250 mm A ϭ 2.16 ϫ 104 mm2 Ip ϭ 4.117 ϫ 108 mm4 STRESSES AT THE OUTER SURFACE Ip ϭ p 32 ad 2 4 Ϫ d1 4 bA ϭ p 4 ad2 2 Ϫ d1 2 b #T ϭ P0 2pf T ϭ torgue ϭ P0 2pf sy ϭ Ϫ25.002 MPa sx ϭ 0 txy ϭ 26.093 MPa PRINCIPAL STRESSES s1 ϭ 16.432 MPa s2 ϭ Ϫ41.434 MPa MAXIMUM TENSILE STRESS st ϭ s1 MAXIMUM COMPRESSIVE STRESS sc ϭ s2 MAXIMUM INPLANESHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. tmax ϭ 28.9 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 sc ϭ Ϫ41.4 MPa ; st ϭ 16.43 MPa ; s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ T a d2 2 b Ip sy ϭ Ϫ P A Problem 8.5-5 A segment of a generator shaft of hollow circular cross section is subjected to a torque T ϭ 240 k-in. (see figure). The outer and inner diameters of the shaft are 8.0 in. and 6.25 in., respectively. What is the maximum permissible compressive load P that can be applied to the shaft if the allowable in-plane shear stress is tallow ϭ 6250 psi? 08Ch08.qxd 9/18/08 11:06 AM Page 679 32. 32. 680 CHAPTER 8 Applications of Plane Stress Solution 8.5-5 P ϭ compressive force T ϭ Torque d2 ϭ outer diameter d1 ϭ inner diameter T ϭ 240 k in. d2 ϭ 8 in. d1 ϭ 6.25 in. A ϭ 19.586 in.2 tallow ϭ 6250 psi Ip ϭ 252.321 in.4 STRESSES AT THE OUTER SURFACE Ip ϭ p 32 ad 2 4 Ϫ d1 4 b A ϭ p 4 ad2 2 Ϫ d1 2 b # sx ϭ 0 txy ϭ 3805 psi MAXIMUM IN-PLANESHEAR STRESS tmax ϭ tallow sy ϭ 9917 psi P ϭ sy A NOTE: The maximum in-plane shear is larger than the maximum out-of-plane shear stress. P ϭ 194.2 k ; sy ϭ 11tmax 2 Ϫ txy 2 2 4 tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ T a d2 2 b Ip sy ϭ Ϫ P A Problem 8.5-6 A cylindrical tank subjected to internal pressure p is simultaneously compressed by an axial force F ϭ 72 kN (see figure). The cylinder has diameter d ϭ 100 mm and wall thickness t ϭ 4 mm. Calculate the maximum allowable internal pressure pmax based upon an allowable shear stress in the wall of the tank of 60 MPa. F F Solution 8.5-6 Cylindrical tank with compressive force F ϭ 72 kN p ϭ internal pressure d ϭ 100 mm t ϭ 4 mm tallow ϭ 60 MPa CIRCUMFERENTIAL STRESS (TENSION) Units: s1 ϭ MPa p ϭ MPa s1 ϭ pr t ϭ p(50 mm) 4 mm ϭ 12.5 p 08Ch08.qxd 9/18/08 11:06 AM Page 680 33. 33. SECTION 8.5 Combined Loadings 681 OUT-OF-PLANE SHEAR STRESSES Solving, p2 ϭ 9.60 MPa 60 MPa ϭ 3.125 p Ϫ 28.648 MPa Solving, p3 ϭ 28.37 MPa CASE 2, OUT-OF-PLANE SHEAR STRESS GOVERNS pmax ϭ 9.60 MPa ; Case 3: tmax ϭ s2 2 ϭ 3.125 p Ϫ 28.648 MPa Case 2: tmax ϭ s1 2 ϭ 6.25 p; 60 MPa ϭ 6.25 p Problem 8.5-7 A cylindrical tank having diameter d ϭ 2.5 in. is subjected to internal gas pressure p ϭ 600 psi and an external tensile load T ϭ 1000 lb (see figure). Determine the minimum thickness t of the wall of the tank based upon an allowable shear stress of 3000 psi. TT LONGITUDINAL STRESS (TENSION) Units: s2 ϭ MPa p ϭ MPa BIAXIAL STRESS IN-PLANE SHEAR STRESS (CASE 1) 60 MPa ϭ 3.125 p ϩ 28.648 MPa Solving, p1 ϭ

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Solution 8.5-7 Cylindrical tank with tensile load T ϭ 1000 lb t ϭ thickness p ϭ 600 psi d ϭ 2.5 in. tallow ϭ 3000 psi CIRCUMFERENTIAL STRESS (TENSION) Units: s1 ϭ psi t ϭ inches s2 ϭ psi s1 ϭ pr t ϭ (600 psi)(1.25 in.) t ϭ 750 t 08Ch08.qxd 9/18/08 11:06 AM Page 681 34. 34. 682 CHAPTER 8 Applications of Plane Stress Problem 8.5-8 The torsional pendulum shown in the figure consists of a horizontal circular disk of mass M ϭ 60 kg suspended by a vertical steel wire (G ϭ 80 GPa) of length L ϭ 2 m and diameter d ϭ 4 mm. Calculate the maximum permissible angle of rotation fmax of the disk (that is, the maximum amplitude of torsional vibrations) so that the stresses in the wire do not exceed 100 MPa in tension or 50 MPa in shear. M = 60 kg d = 4 mm L = 2 m fmax LONGITUDINAL STRESS (TENSION) BIAXIAL STRESS ϭ 375 t + 1000 lb 2p(1.25 in.)t ϭ 375 t + 127.32 t + 502.32 t s2 ϭ pr 2t + T A ϭ pr 2t + T 2prt OUT-OF-PLANE SHEAR STRESSES Solving, t2 ϭ 0.125 in. Solving, t3 ϭ 0.0837 in. CASE 2, OUT-OF-PLANE SHEAR STRESS GOVERNS tmin ϭ 0.125 in. ; Case 3: tmax ϭ s2 2 ϭ 251.16 t ; 3000 ϭ 251.16 t Case 2: tmax ϭ s1 2 ϭ 375 t ; 3000 ϭ 375 t Solving, t1 ϭ 0.0413 in. 3000 psi ϭ 123.84 t IN-PLANE SHEAR STRESS (CASE 1) tmax ϭ s1 Ϫ s2 2 ϭ 247.68 2t ϭ 123.84 t 08Ch08.qxd 9/18/08 11:06 AM Page 682 35. 35. SECTION 8.5 Combined Loadings 683 Solution 8.5-8 Torsional pendulum L ϭ 2.0 m d ϭ 4.0 mm M ϭ 60 kg G ϭ 80 GPa sallow ϭ 100 MPa tallow ϭ 50 MPa W ϭ Mg ϭ (60 kg)(9.81 m/s2 ) ϭ 588.6 N A ϭ pd2 4 ϭ 12.5664 mm2 PRINCIPAL STRESSES Note that s1 is positive and s2 is negative. Therefore, the maximum in-plane shear stress is greater than the maximum out-of-plane shear stress. MAXIMUM ANGLE OF ROTATION BASED ON TENSILE STRESS s1 ϭ maximum tensile stress sallow ϭ 100 MPa (100 Ϫ 23.420)2 ϭ 5316 ϭ fmax ϭ 0.9114 rad ϭ 52.2° MAXIMUM ANGLE OF ROTATION BASED ON INPLANE SHEAR STRESS (50)2 ϭ Solving, fmax ϭ 0.5522 rad ϭ 31.6° SHEAR STRESS GOVERNS fmax ϭ 0.552 rad ϭ 31.6° ; (23.420)2 + 6400fmax 2 1(23.420)2 + 6400fmax 2tallow ϭ 50 MPa 50 ϭ tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 ϭ 1(23.420)2 + 6400fmax 2 6400fmax 2 123.42022 + 6400fmax 2 ‹ 100 MPa ϭ 23.420 ; 1123.42022 + 6400fmax 2 s1,2 ϭ 23.420 ; 1123.42022 + 6400fmax 2 ( MPa) s1,2 ϭ sx + sy 2 ; A a sx Ϫ sy 2 b 2 + txy 2 TORQUE: (EQ. 315) SHEAR STRESS: (EQ. 3-11) t ϭ 80 fmax Units: t ϭ MPa fmax ϭ radians t ϭ a GIpfmax L b a r IP b ϭ Grfmax L ϭ (80 * 106 Pa)fmax t ϭ Tr Ip T ϭ GIpfmax L TENSILE STRESS sx ϭ 0 sy ϭ st ϭ 46.839 MPa txy ϭ Ϫ80 fmax (MPa) sx ϭ W A ϭ 46.839 MPa 08Ch08.qxd 9/18/08 11:06 AM Page 683 36. 36. 684 CHAPTER 8 Applications of Plane Stress Problem 8.5-9 Determine the maximum tensile, compressive, and shear stresses at points A and B on the bicycle pedal crank shown in the figure. The pedal and crank are in a horizontal plane and points A and B are located on the top of the crank. The load P ϭ 160 lb acts in the vertical direction and the distances (in the horizontal plane) between the line of action of the load and points A and B are b1 ϭ 5.0 in., b2 ϭ 2.5 in. and b3 ϭ 1.0 in. Assume that the crank has a solid circular cross section with diameter d ϭ 0.6 in. b1 b2 b3 b3 ϫ P = 160 lb P A B d = 0.6 in. Crank Top view b2 = 2.5 in. b3 = 1.0 in. b3 A b1 = 5.0 in. B Solution 8.5-9 P ϭ 160 lb d ϭ 0.6 in. b1 ϭ 5.0 in. b2 ϭ 2.5 in. b3 ϭ 1.0 in. STRESS RESULTANTS on cross section at point A: Torque: TA ϭ Pb2 Moment: MA ϭ Pb1 S hear Force: VA ϭ P STRESS RESULTANTS at point B: TB ϭ P(b1 ϩ b3) MB ϭ P(b2 ϩ b3) VB ϭ P S ϭ p d3 32 STRESS AT POINT A: t ϭ 9.431 ϫ 103 psi s ϭ 3.773 ϫ 104 psi (The shear force V produces no shear stresses at point A) PRINCIPAL STRESSES AND MAXIMUM SHEAR STRESS sx ϭ 0 sy ϭ s txy ϭ Ϫt s1 ϭ 3.995 ϫ 104 psi s2 ϭ Ϫ2.226 ϫ 103 psi A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + s ϭ MA S t ϭ 16TA pd3 08Ch08.qxd 9/18/08 11:06 AM Page 684 37. 37. SECTION 8.5 Combined Loadings 685 MAXIMUM TENSILE STRESS st ϭ s1 MAXIMUM COMPRESSIVE STRESS sc ϭ s2 MAXIMUM INPLANE SHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. STRESS AT POINT B t ϭ 2.264 ϫ 104 psi s ϭ 2.641 ϫ 104 psi (The shear force V produces no shear stresses at point A) PRINCIPAL STRESSES AND MIXIMUM SHEAR STRESS sx ϭ 0 sy ϭ s txy ϭ Ϫt A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + s ϭ MB S t ϭ 16TB pd3 tmax ϭ 21,090 psi ; A a sx Ϫ sy 2 b 2 + txy 2 tmax ϭ sc ϭ Ϫ2226 psi ; st ϭ 39,950 psi ; s1 ϭ 3.941 ϫ 104 psi s2 ϭ Ϫ1.3 ϫ 104 MAXIMUM TENSILE STRESS st ϭ s1 MAXIMUM COMPRESSIVE STRESS sc ϭ s2 MAXIMUM IN-PLANE SHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. tmax ϭ 26,210 psi ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 sc ϭ Ϫ13,000 psi ; st ϭ 39,410 psi ; s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 Problem 8.5-10 A cylindrical pressure vessel having radius r ϭ 300 mm and wall thickness t ϭ 15 mm is subjected to internal pressure p ϭ 2.5 MPa. In addition, a torque T ϭ 120 kN m acts at each end of the cylinder (see figure). (a) Determine the maximum tensile stress smax and the maximum in- plane shear stress tmax in the wall of the cylinder. (b) If the allowable in-plane shear stress is 30 MPa, what is the maximum allowable torque T? # T T 08Ch08.qxd 9/18/08 11:06 AM Page 685 38. 38. 686 CHAPTER 8 Applications of Plane Stress Solution 8.5-10 Cylindrical pressure vessel T ϭ 120 kN m r ϭ 300 mm t ϭ 15 mm P ϭ 2.5 MPa STRESSES IN THE WALL OF THE VESSEL (EQ. 3-11) (EQ. 3-18) txy ϭ Ϫ T 2pr2 t ϭ Ϫ14.147 MPa Ip ϭ 2pr3 t txy ϭ Ϫ Tr Ip sx ϭ pr 2t ϭ 25 MPa sy ϭ pr t ϭ 50 MPa # s1 ϭ 56.4 MPa s2 ϭ 18.6 MPa ‹ smax ϭ 56.4 MPa MAXIMUM IN-PLANE SHEAR STRESS (b) MAXIMUM ALLOWABLE TORQUE T tallow ϭ 30 MPa (in-plane shear stress) (1) Units: txy ϭ MPa T ϭ N m Substitute into Eq. (1): Square both sides, rearrange, and solve for T: (30)2 ϭ (12.5)2 ϩ (117.893 ϫ 10Ϫ6 )2 T2 Tmax ϭ 231 kN m ;# T ϭ 231.3 * 103 N # m T2 ϭ 743.750 13,899 * 10Ϫ12 ϭ 53,512 * 106 (N # m)2 ϭ 2(Ϫ12.5 MPa)2 + (Ϫ117.893 * 10Ϫ6 T)2 tmax ϭ tallow ϭ 30 MPa # txy ϭ Ϫ T 2pr2 t ϭ Ϫ117.893 * 10Ϫ6 T sx ϭ pr 2t ϭ 25 MPa sy ϭ pr t ϭ 50 MPa tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 ϭ 18.9 MPa ; ; (a) PRINCIPAL STRESSES ϭ 37.5 ; 18.878 MPa s1,2 ϭ sx + sy 2 ; A a sx Ϫ sy 2 b 2 + txy 2 08Ch08.qxd 9/18/08 11:06 AM Page 686 39. 39. SECTION 8.5 Combined Loadings 687 Problem 8.5-11 An L-shaped bracket lying in a horizontal plane supports a load P ϭ 150 lb (see figure). The bracket has a hollow rectangular cross section with thickness t ϭ 0.125 in. and outer dimensions b ϭ 2.0 in. and h ϭ 3.5 in. The centerline lengths of the arms are b1 ϭ 20 in. and b2 ϭ 30 in. Considering only the load P, calculate the maximum tensile stress st, maximum compressive stress sc, and maximum shear stress tmax at point A, which is located on the top of the bracket at the support. P = 150 lb A b1 = 20 in. b2 = 30 in. h = 3.5 in. b = 2.0 in. t = 0.125 in. Solution 8.5-11 L-shaped bracket P ϭ 150 lb b1 ϭ 20 in. b2 ϭ 30 in. t ϭ 0.125 in. h ϭ 3.5 in. b ϭ 2.0 in. FREE-BODY DIAGRAM OF BRACKET STRESSES AT POINT A ON THE TOP OF THE BRACKET (The shear force V produces no stresses at point A.) STRESS ELEMENT AT POINT A (This view is looking downward at the top of the bracket.) s ϭ Mc I ϭ (3000 lb-in.)(1.75 in.) 2.1396 in.4 ϭ 2454 psi t ϭ T 2tAm ϭ 4500 lb-in. 2(0.125 in.)(6.3281 in.2 ) ϭ 2844 psi STRESS RESULTANTS AT THE SUPPORT Torque: T ϭ Pb2 ϭ (150 lb)(30 in.) ϭ 4500 lb-in. Moment: M ϭ Pb1 ϭ (150 lb)(20 in.) ϭ 3000 lb-in. Shear force: V ϭ P ϭ 150 lb PROPERTIES OF THE CROSS SECTION For torsion: Am ϭ (b Ϫ t)(h Ϫ t) ϭ (1.875 in.)(3.375 in.) ϭ 6.3281 in.2 For bending: ϭ 2.1396 in.4 ϭ 1 12 (2.0 in.)(3.5 in.)3 Ϫ 1 12 (1.75 in.)(3.25 in.)3 I ϭ 1 12 (bh3 ) Ϫ 1 12 (b Ϫ 2t)(h Ϫ 2t)3 c ϭ h 2 ϭ 1.75 in. sx ϭ 0 sy ϭ s ϭ 2454 psi txy ϭ Ϫt ϭ Ϫ2844 psi 08Ch08.qxd 9/18/08 11:06 AM Page 687 40. 40. 688 CHAPTER 8 Applications of Plane Stress PRINCIPAL STRESSES AND MAXIMUM SHEAR STRESS s1 ϭ 4324 psi s2 ϭ Ϫ1870 psi MAXIMUM TENSILE STRESS: st ϭ 4320 psi ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 ϭ 3097 psi ϭ 1227 psi ; 3097 psi s1,2 ϭ sx + sy 2 ; A a sx Ϫ sy 2 b 2 + txy 2 MAXIMUM COMPRESSIVE STRESS: sc ϭ Ϫ1870 psi MAXIMUM SHEAR STRESS: tmax ϭ 3100 psi NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear stress is larger than the maximum out-of-plane shear stress. ; ; Problem 8.5-12 A semicircular bar AB lying in a horizontal plane is supported at B (see figure). The bar has centerline radius R and weight q per unit of length (total weight of the bar equals pqR). The cross section of the bar is circular with diameter d. Obtain formulas for the maximum tensile stress st, maximum compressive stress sc, and maximum in-plane shear stress tmax at the top of the bar at the support due to the weight of the bar. d B O R A Solution 8.5-12 Semicircular bar d ϭ diameter of bar R ϭ radius of bar q ϭ weight of bar per unit length W ϭ weight of bar ϭ pqR Weight of bar acts at the center of gravity From Case 23, Appendix D, with b ϭ p/2, we get Bending moment at B: MB ϭ Wc ϭ 2qR2 Torque at B: TB ϭ WR ϭ pqR2 (Shear force at B produces no shear stress at the top of the bar.) y ϭ 2R p ‹ c ϭ 2R p STRESSES AT THE TOP OF THE BAR AT B tB ϭ TB(d/2) IP ϭ (pqR2 )(d/2) pd4 /32 ϭ 16qR2 d3 sB ϭ MB(d/2) I ϭ (2qR2 )(d/2) pd4 /64 ϭ 64qR2 pd3 08Ch08.qxd 9/18/08 11:06 AM Page

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41. 41. SECTION 8.5 Combined Loadings 689 STRESS ELEMENT AT THE TOP OF THE BAR AT B sx ϭ 0 sy ϭ sB txy ϭ tB PRINCIPAL STRESSES: ϭ 16qR2 pd3 (2 ; 24 + p2 ) ϭ 32qR2 pd3 ; A a 32qR2 pd3 b 2 + a 16qR2 d3 b 2 ϭ sB 2 ; A a Ϫ sB 2 b 2 + tB 2 s1,2 ϭ sx + sy 2 ; A a sx Ϫ sy 2 b 2 + txy 2 MAXIMUM TENSILE STRESS MAXIMUM COMPRESSIVE STRESS MAXIMUM IN-PLANE SHEAR STRESS (EQ. 7-26) tmax ϭ 1 2 (s1 Ϫ s2) ϭ 16qR2 pd3 24 + p2 ; ϭ Ϫ8.78 qR2 d3 ; sc ϭ s2 ϭ 16qR2 pd3 (2 Ϫ 24 + p2 ) ϭ 29.15 qR2 d3 ; st ϭ s1 ϭ 16qR2 pd3 (2 + 24 + p2 ) Problem 8.5-13 An arm ABC lying in a horizontal plane and supported at A (see figure) is made of two identical solid steel bars AB and BC welded together at a right angle. Each bar is 20 in. long. Knowing that the maximum tensile stress (principal stress) at the top of the bar at support A due solely to the weights of the bars is 932 psi, determine the diameter d of the bars. A B C z y x Solution 8.5-13 Horizontal arm ABC L ϭ length of AB and BC d ϭ diameter of AB and BC A ϭ cross-sectional area ϭ pd2 /4 g ϭ weight density of steel q ϭ weight per unit length of bars ϭ gA ϭ pgd2 /4 (1) 08Ch08.qxd 9/18/08 11:06 AM Page 689 42. 42. 690 CHAPTER 8 Applications of Plane Stress RESULTANT FORCES ACTING ON AB P ϭ weight of AB and BC P ϭ qL ϭ pgLd2 /4 (2) T ϭ torque due to weight of BC (3) MA ϭ bending moment at A MA ϭ PL ϩ PL/2 ϭ 3PL/2 ϭ 3pgL2 d2 /8 (4) STRESSES AT THE TOP OF THE BAR AT A sA ϭ normal stress due to MA (5) tA ϭ shear stress due to torque T (6) STRESS ELEMENT ON TOP OF THE BAR AT A s1 ϭ principal tensile stress (maximum tensile stress) (7)s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 tA ϭ T(d/2) Ip ϭ T(d/2) pd 4 /32 ϭ 16T pd 3 ϭ 2gL2 d sA ϭ M(d/2) I ϭ M(d/2) pd4 /64 ϭ 32M pd3 ϭ 12gL2 d T ϭ (qL)a L 2 b ϭ qL2 2 ϭ pgL2 d2 8 sx ϭ 0 sy ϭ sA txy ϭ ϪtA (8) Substitute (8) into (7): (9) Substitute from (5) and (6) and simplify: (10) SOLVE FOR d (11) SUBSTITUTE NUMERICAL VALUES INTO EQ. (11): g ϭ 490 lb/ft3 ϭ 0.28356 lb/in.3 L ϭ 20 in. s1 ϭ 932 psi d ϭ 1.50 in. ; d ϭ 2gL2 s1 (3 + 110) ; s1 ϭ gL2 d (6 + 140) ϭ 2gL2 d (3 + 110) s1 ϭ sA 2 + A a sA 2 b 2 + tA 2 Problem 8.5-14 A pressurized cylindrical tank with flat ends is loaded by torques T and tensile forces P (see figure). The tank has radius r ϭ 50 mm and wall thickness t ϭ 3 mm. The internal pressure p ϭ 3.5 MPa and the torque T ϭ 450 N m. What is the maximum permissible value of the forces P if the allowable tensile stress in the wall of the cylinder is 72 MPa? # T T PP 08Ch08.qxd 9/18/08 11:06 AM Page 690 43. 43. SECTION 8.5 Combined Loadings 691 Solution 8.5-14 Cylindrical tank r ϭ 50 mm t ϭ 3.0 mm p ϭ 3.5 MPa T ϭ 450 N m sallow ϭ 72 MPa CROSS SECTION A ϭ 2prt ϭ 2p(50 mm)(3.0 mm) ϭ 942.48 mm2 IP ϭ 2pr3 t ϭ 2p(50 mm)3 (3.0 mm) ϭ 2.3562 ϫ 106 mm4 STRESSES IN THE WALL OF THE TANK # Units: sx ϭ MPa, P ϭ newtons MAXIMUM TENSILE STRESS or Square both sides and simplify: 494.21 ϭ 0.014501 P SOLVE FOR P P ϭ 34,080 N OR Pmax ϭ 34.1 kN ; ϭ 2(Ϫ14.583 + 0.00053052 P)2 + 91.189 28.250 Ϫ 0.00053052P + 2[Ϫ14.583 + (530.52 * 10Ϫ 6 )P]2 + (Ϫ9.5493)2 72 ϭ 43.750 + (530.52 * 10Ϫ6 )P ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 smax ϭ sallow ϭ 72 MPa ϭ Ϫ9.5493 MPa txy ϭ Ϫ Tr IP ϭ Ϫ (450 N # m)(50 mm) 2.3562 * 106 mm4 sy ϭ pr t ϭ 58.333 MPa ϭ 29.167 MPa + 1.0610 * 10Ϫ3 P ϭ (3.5 MPa)(50 mm) 2(3.0 mm) + P 942.48 mm2 sx ϭ pr 2t + P A Problem 8.5-15 A post having a hollow circular cross section supports a horizontal load P ϭ 240 lb acting at the end of an arm that is 5 ft long (see figure). The height of the post is 27 ft, and its section modulus is S ϭ 15 in.3 Assume that outer radius of the post, r2 ϭ 4.5 in., and inner radius r1 ϭ 4.243 in. (a) Calculate the maximum tensile stress smax and maximum in-plane shear stress tmax at point A on the outer surface of the post along the x-axis due to the load P. Load P acts in a horizontal plane at an angle of 30° from a line which is parallel to the (Ϫx) axis. (b) If the maximum tensile stress and maximum in-plane shear stress at point A are lim- ited to 16,000 psi and 6000 psi, respectively, what is the largest permissible value of the load P? A P = 240 lb z y x 5 ft 30° 27 ft 08Ch08.qxd 9/18/08 11:06 AM Page 691 44. 44. 692 CHAPTER 8 Applications of Plane Stress Solution 8.5-15 P ϭ 240 lb b ϭ 5 ft length of arm h ϭ 27 ft height of post r2 ϭ 4.5 in. r1 ϭ 4.243 in. t ϭ r2 Ϫ r1 t ϭ 0.257 in. A ϭ p(r2 2 Ϫ r1 2 ) A ϭ 7.059 in.2 I ϭ 67.507 in.4 Ip ϭ 2I Ip ϭ 135.014 in.4 Q ϭ 9.825 in.3 REACTIONS AT THE SUPPORT M ϭ P cos(30deg)h M ϭ 6.734 ϫ 104 lb in T ϭ ϪP cos(30deg)b T ϭ Ϫ1.247 ϫ 104 lb in Vx ϭ P cos(30deg) Vx ϭ 207.846 lb Vy ϭ ϪP sin(30deg) Vy ϭ Ϫ120 lb STRESSES AT POINT A # # Q ϭ 2 3 1 r2 3 Ϫ r1 3 2 I ϭ p 4 1 r2 4 Ϫ r1 4 2 (a) MAXIMUM TENSILE STRESS AND MAXIMUM SHEAR STRESS sx ϭ 0 sy ϭ 4.489 ϫ 103 psi txy ϭ t s1 ϭ 4.534 ϫ 103 psi s2 ϭ Ϫ44.593 psi MAXIMUM TENSILE STRESS smax ϭ s1 MAXIMUM SHEAR STRESS (b) ALLOWABLE LOAD P sallow ϭ 16000 psi tallow ϭ 6000 psi The stresses at point A are proportional to the load P. Based on tensile stress: Pallow ϭ 847.01lb Based on shear stress: Pallow ϭ 629.07 lb Pallow ϭ 629 lb ; Pallow ϭ tallowP tmax Pallow P ϭ tallow tmax Pallow ϭ sallowP smax Pallow P ϭ sallow smax tmax ϭ 2289 psi ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 smax ϭ 4534 psi ; s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 sx ϭ 0 sy ϭ 4.489 ϫ 103 psi t ϭ Ϫ449.628 psi (The shear force Vx produces no stress at point A) t ϭ Tr2 Ip + VyQ I2t sy ϭ Mr2 I 08Ch08.qxd 9/18/08 11:06 AM Page 692 45. 45. SECTION 8.5 Combined Loadings 693 Problem 8.5-16 A sign is supported by a pipe (see figure) having outer diameter 110 mm and inner diameter 90 mm. The dimensions of the sign are 2.0 m ϫ 1.0 m, and its lower edge is 3.0 m above the base. Note that the center of gravity of the sign is 1.05 m from the axis of the pipe. The wind pressure against the sign is 1.5 kPa. Determine the maximum in-plane shear stresses due to the wind pressure on the sign at points A, B, and C, located on the outer surface at the base of the pipe. B A C B A C XX Rose’s Editing Co. Section X-X Pipe 110 mm 2.0 m 1.05 m to c.g. 1.0 m 3.0 m Solution 8.5-16 PIPE: d2 ϭ 110 mm d1 ϭ 90 mm t ϭ 10 mm I ϭ 3.966 ϫ 106 mm4 Ip ϭ 2I Ip ϭ 7.933 ϫ 106 mm4 SIGN: A ϭ 2m2 Area Height from the base to the center of gravity of the sign b ϭ 1.05 m horizontal distance from the center of gravity of the sign to the axis of the pipe WIND PRESSURE pw ϭ 1.5 kPa P ϭ pw A P ϭ 3 kN STRESS RESULTANTS AT THE BASE M ϭ Ph M ϭ 10.5 kN m T ϭ Pb T ϭ 3.15 kN m V ϭ P V ϭ 3 kN # # h ϭ a3 + 1 2 b m Q ϭ 5.017 * 104 mm3 Q ϭ 1 12 1d2 3 Ϫ d1 3 2 I ϭ p 64 (d 2 4 Ϫ d1 4 ) MAXIMUM SHEAR STRESS AT POINT A sx ϭ 0 sy ϭ 145.603 MPa txy ϭ 21.84 MPa tmax ϭ 76.0 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Td2 2Ip sy ϭ Md2 2I 08Ch08.qxd 9/18/08 11:06 AM Page 693 46. 46. 694 CHAPTER 8 Applications of Plane Stress MAXIMUM SHEAR STRESS AT POINT B MAXIMUM SHEAR STRESS AT POINT C sx ϭ 0 sy ϭ 0 txy ϭ 19.943 MPa Pure shear tmax ϭ 19.94 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Td2 2Ip Ϫ VQ I(2t) sx ϭ 0 sy ϭ 0 txy ϭ 23.738 MPa Pure shear tmax ϭ 23.7 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Td2 2Ip + VQ I(2t) Problem 8.5-17 A sign is supported by a pole of hollow circular cross section, as shown in the figure. The outer and inner diameters of the pole are 10.5 in. and 8.5 in., respectively. The pole is 42 ft high and weighs 4.0 k. The sign has dimensions 8 ft ϫ 3 ft and weighs 500 lb. Note that its center of gravity is 53.25 in. from the axis of the pole. The wind pressure against the sign is 35 lb/ft2 . (a) Determine the stresses acting on a stress element at point A, which is on the outer surface of the pole at the “front” of the pole, that is, the part of the pole nearest to the viewer. (b) Determine the maximum tensile, compressive, and shear stresses at point A. 8 ft 3 ft 42 ft 10.5 in. 8.5 in. Section X-X A A X X Hilda’s Office 08Ch08.qxd 9/18/08 11:06 AM Page 694 47. 47. SECTION 8.5 Combined Loadings 695 Solution 8.5-17 PIPE: d2 ϭ 10.5 in. d1 ϭ 8.5 in. A ϭ 29.845 in.2 I ϭ 340.421 in.4 Ip ϭ 2I Ip ϭ 680.842 in.4 Q ϭ 45.292 in.3 W1 ϭ 4000 lb SIGN: As ϭ (8)(3) ft2 Area As ϭ 24 ft2 Height from the base to the center of gravity of the sign Horizontal distance from the center of gravity of the sign to the axis of the pipe b ϭ 53.25 in. W2 ϭ 500 lb WIND PRESSURE P ϭ pw As P ϭ 840 lb STRESS RESULTANTS AT THE BASE M ϭ Ph M ϭ 4.082 ϫ 105 lb in. T ϭ Pb T ϭ 4.473 ϫ 104 lb in. V ϭ P V ϭ 840 lb Nz ϭ W1 ϩ W2 Nz ϭ 4.5 ϫ 103 lb # # pw ϭ 35 lb/ft2 b ϭ a(4)12 + 10.5 2 b in. h ϭ a42 Ϫ 3 2 b ft Q ϭ 1 12 1d2 3 Ϫ d1 3 2 I ϭ p 64 1d 2 4 Ϫ d 1 4 2 A ϭ p 4 1d2 2 Ϫ d1 2 2 c ϭ d2 2 (a) STRESSES AT POINT A (b) MAXIMUM STRESSES AT POINT A Max. tensile stress Max. compressive stress Max. shear stress NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. tmax ϭ 3092 psi ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ Ϫ19.30 psi ; s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ 6164 psi ; s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ 345 psi ;sxy ϭ Td2 2Ip sy ϭ 6145 psi ;sy ϭ Ϫ Nz A + Md2 2I sx ϭ 0 ; 08Ch08.qxd 9/18/08 11:06 AM Page 695 48. 48. 696 CHAPTER 8 Applications of Plane Stress Problem 8.5-18 A horizontal bracket ABC consists of two perpendicular arms AB of length 0.5 m, and BC of length of 0.75 m. The bracket has a solid circular cross section with diameter equal to 65 mm. The bracket is inserted in a frictionless sleeve at A (which is slightly

2017-5-21

Chapter 8

direction and M2 ϭ 1.0 kN m acts in the (Ϫz) direction. Considering only the moments M1 and M2, calculate the maximum tensile stress st, the maximum compressive stress sc, and the maximum in-plane shear stress tmax at point p, which is located at support A on the side of the bracket at midheight. # # z0 B C M2 M1 O y0 x0 p 65 mm Cross section at A 0.5 m 0.75 m Frictionless sleeve embedded in support A x0 y0 p Solution 8.5-18 d ϭ 65 mm b1 ϭ 0.5 m length of arm BC b2 ϭ 0.75 m length of arm AB M1 ϭ 1.5 kN m M2 ϭ 1.0 kN m PROPERTIES OF THE CROSS SECTION d ϭ 65 mm A ϭ 3.318 ϫ 103 mm2 I ϭ 8.762 ϫ 105 mm4 Ip ϭ 2I Ip ϭ 1.752 ϫ 106 mm4 Q ϭ 2.289 ϫ 104 mm3 STRESS RESULTANTS AT SUPPORT A Nz ϭ 0 Axial force My ϭ 0 Mx ϭ 0Mx ϭ ϪM2 b2 b1 + M1 Q ϭ 2 3 r 3 I ϭ p 64 d 4 A ϭ p 4 d 2 r ϭ d 2 # # T ϭ 0 Torsional frictionless sleeve at support A (MZ) sx ϭ 0 sy ϭ 0 txy ϭ Ϫ0.804 MPa Pure Shear STRESSES AT POINT p ON THE SIDE OF THE BRACKET tmax ϭ 0.804 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ VyQ Id Vy ϭ Ϫ M2 b1 08Ch08.qxd 9/18/08 11:06 AM Page 696 49. 49. SECTION 8.5 Combined Loadings 697 Problem 8.5-19 A cylindrical pressure vessel with flat ends is subjected to a torque T and a bending moment M (see figure). The outer radius is 12.0 in. and the wall thickness is 1.0 in. The loads are as follows: T ϭ 800 k-in., M ϭ 1000 k-in., and the internal pressure p ϭ 900 psi. Determine the maximum tensile stress st, maximum compressive stress sc, and maximum shear stress tmax in the wall of the cylinder. z0 x0 y0 T T MM Solution 8.5-19 Cylindrical pressure vessal Internal pressure: p ϭ 900 psi Bending moment: M ϭ 1000 k-in. Torque: T ϭ 800 k-in. Outer radius: r2 ϭ 12 in. Wall thickness: t ϭ 1.0 in. Mean radius: r ϭ r2 Ϫ t/2 ϭ 11.5 in. Outer diameter: d2 ϭ 24 in. Inner diameter: d1 ϭ 22 in. MOMENT OF INERTIA Ip ϭ 2I ϭ 9574.0 in.4 NOTE: Since the stresses due to T and p are the same everywhere in the cylinder, the maximum stresses occur at the top and bottom of the cylinder where the bending stresses are the largest. PART (a). TOP OF THE CYLINDER Stress element on the top of the cylinder as seen from above. I ϭ p 64 (d 2 4 Ϫ d 1 4 ) ϭ 4787.0 in.4 PRINCIPAL STRESSES MAXIMUM SHEAR STRESSES In-plane: t ϭ 3970 psi Out-of-plane: І tmax ϭ 5240 psi MAXIMUM STRESSES FOR THE TOP OF THE CYLINDER st ϭ 10,480 psi sc ϭ 0 (No compressive stresses) tmax ϭ 5240 psi t ϭ s1 2 or s2 2 t ϭ s1 2 ϭ 5240 psi s1 ϭ 10,479 psi s2 ϭ 2540 psi ϭ 6509.1 psi ; 3969.6 psi s1,2 ϭ sx + sy 2 ; A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Tr2 Ip ϭ Ϫ1002.7 psi sy ϭ pr t ϭ 10,350 psi ϭ 2668.2 psi sx ϭ pr 2t Ϫ Mr2 I ϭ 5175.0 psi Ϫ 2506.8 psi 08Ch08.qxd 9/18/08 11:06 AM Page 697 50. 50. 698 CHAPTER 8 Applications of Plane Stress Problem 8.5-20 For purposes of analysis, a segment of the crankshaft in a vehicle is represented as shown in the figure. Two loads P act as shown, one parallel to (Ϫx0) and another parallel to z0; each load P equals 1.0 kN. The crankshaft dimensions are b1 ϭ 80 mm, b2 ϭ 120 mm, and b3 ϭ 40 mm. The diameter of the upper shaft is d ϭ 20 mm. (a) Determine the maximum tensile, compressive, and shear stresses at point A, which is located on the surface of the upper shaft at the z0 axis. (b) Determine the maximum tensile, compressive, and shear stresses at point B, which is located on the surface of the shaft at the y0 axis. PART (b). BOTTOM OF THE CYLINDER Stress element on the bottom of the cylinder as seen from below. MAXIMUM SHEAR STRESSES In-plane: t ϭ 1669 psi Out-of-plane: І tmax ϭ 5340 psi MAXIMUM STRESSES FOR THE BOTTOM OF THE CYLINDER st ϭ 10,680 psi sc ϭ 0 (No compressive stresses) tmax ϭ 5340 psi PART (c). ENTIRE CYLINDER The largest stresses are at the bottom of the cylinder. st ϭ 10,680 psi sc ϭ 0 (No compressive stresses) tmax ϭ 5340 psi ; ; ; t ϭ s1 2 or s2 2 t ϭ s1 2 ϭ 5340 psi PRINCIPAL STRESSES s1 ϭ 10,685 psi s2 ϭ 7347 psi ϭ 9015.9 psi ; 1668.9 psi s1,2 ϭ sx + sy 2 ; A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Ϫ Tr2 Ip ϭ Ϫ1002.7 psi sy ϭ pr t ϭ 10,350 psi ϭ 7681.8 psi sx ϭ pr 2t + Mr2 I ϭ 5175.0 psi + 2506.8 psi P = 1.0 kN P b3 = 40 mm b1 = 80 mm y0 z0 x0 B A b2 = 120 mm d = 20 mm 08Ch08.qxd 9/18/08 11:06 AM Page 698 51. 51. SECTION 8.5 Combined Loadings 699 Solution 8.5-20 P ϭ 1.0 kN b1 ϭ 80 mm b2 ϭ 120 mm b3 ϭ 40 mm PROPERTIES OF THE CROSS SECTION d ϭ 20 mm A ϭ 314.159 mm2 I ϭ 7.854 ϫ 103 mm4 Ip ϭ 2I Ip ϭ 1.571 ϫ 104 mm4 Q ϭ 666.667 mm3 STRESS RESULTANTS AT THE SUPPORT Vx ϭ P (Axial force in X-dir.) Vy ϭ 0 (Shear force in Y-dir.) Vz ϭ P (Shear force in Z-dir.) Mx ϭ Pb2 (Torsional Moment) Mx ϭ 120 kN mm My ϭ P(b1 ϩ b3) (Bending Moment) My ϭ 120 kN mm Mz ϭ Pb2 (Bending Moment) Mz ϭ 120 kN mm (a) STRESSES AT POINT A # # # Q ϭ 2 3 r3 I ϭ p 64 d4 A ϭ p 4 d2 r ϭ d 2 sx ϭ Ϫ155.972 MPa (compressive) sy ϭ 0 txy ϭ 76.394 MPa MAX. TENSILE STRESS MAX. COMPRESSIVE STRESS MAX. SHEAR STRESS (b) STRESSES AT POINT B sx ϭ Ϫ Vx A + Mz I r tmax ϭ 109.2 MPa ; s2 ϭ Ϫ187.2 MPa ; s1 ϭ 31.2 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx Ϫ sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Mxd 2Ip sx ϭ 149.606 MPa (tensile) sy ϭ 0 txy ϭ 80.639 MPa tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Mxd 2Ip + VzQ Id sx ϭ Ϫ Vx A Ϫ My I r 08Ch08.qxd 9/18/08 11:08 AM Page 699 52. 52. 700 CHAPTER 8 Applications of Plane Stress Problem 8.5-21 A moveable steel stand supports an automobile engine weighing W ϭ 750 lb as shown in figure part (a). The stand is constructed of 2.5 in. ϫ 2.5 in. ϫ 1/8 in. thick steel tubing. Once in position the stand is restrained by pin supports at B and C. Of interest are stresses at point A at the base of the vertical post; point A has coordinates (x ϭ 1.25, y ϭ 0, z ϭ 1.25) (inches). Neglect the weight of the stand. (a) Initially, the engine weight acts in the (Ϫz) direction through point Q which has coordinates (24, 0, 1.25); find the maximum tensile, compressive, and shear stresses at point A. (b) Repeat (a) assuming now that, during repair, the engine is rotated about its own longitudinal axis (which is parallel to the x axis) so that W acts through QЈ (with coordinates (24, 6, 1.25)) and force Fy ϭ 200 lb is applied parallel to the y axis at distance d ϭ 30 in. x D B C y A Q D C O B Q' y 24 in. 1.25 in. 12 in. 17 in. 17 in. 6 in. (b) Top view 2.5 in. ϫ 2.5 in. ϫ 1/8 in. Dz Cz Cy Cx 24 in. 17 in. 36 in. (a) Q x d = 30 in. W z Fy A MAX. TENSILE STRESS MAX. COMPRESSIVE STRESS s2 ϭ Ϫ35.2 MPa ; s1 ϭ 184.8 MPa ; MAX. SHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. tmax ϭ 110.0 MPa ; 08Ch08.qxd 9/18/08 11:08 AM Page 700 53. 53. SECTION 8.5 Combined Loadings 701 Solution 8.5-21 W ϭ 750 1b Fy ϭ 200 1b b ϭ 2.5 in. d ϭ 30 in. x1 ϭ 24 in. A ϭ b2 Ϫ (b Ϫ 2t)2 A ϭ 1in.2 Am ϭ (b Ϫ t)2 Am ϭ 6 in.2 y1 ϭ 6 in. I ϭ 1 in.4 (a) ENGINE WEIGHT ACTS THROUGH X-AXIS (POINT Q) Mx ϭ 0 My ϭ Wx1 SHEAR & NORMAL STRESSES AT A s ϭ Ϫ20730 psi t ϭ 0 PRINCIPAL STRESSES & MAX SHEAR STRESS sx ϭ 0 sy ϭ s txy ϭ t s1 ϭ 0 s2 ϭ Ϫ20730 psi tmax ϭ 10365 psi (b) ENGINE WEIGHT ACTS THROUGH POINT Q & FORCE FY ACTS IN Y-DIR My ϭ 18000 in.-1b Mx ϭ Wy1 SHEAR & NORMAL STRESSES AT A s ϭ Ϫ20730 psi same since element A lies on NA for bending about x-axis s ϭ ϪW A Ϫ Myc I ¿ ; ; ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 s ϭ ϪW A Ϫ Myc I I ϭ 1 12 [ b4 Ϫ ( b Ϫ 2t)4 ] c ϭ b 2 . t ϭ 1 8 in. SHEAR STRESS DEPENDS O N TRANSVERSE SHEAR DUE TO FY & TORSION DUE TO MZ (MAX TRANSVERSE SHEAR IN WEB-SEE PROB. #5.10-11) b1 ϭ b Ϫ 2t b1 ϭ 2 in. Q ϭ 1 in.3 tt ϭ 378 psi SHEAR DUE TO TORSIONAL MOMENT MZ USING APPROX. THEORY (EQU. 3-65) Mz ϭ Fy d Mz ϭ 6000 in.-1b tT ϭ 4255 psi t ϭ tt ϩ tT t ϭ 4633 psi PRINCIPAL STRESSES & MAX SHEAR STRESS sx ϭ 0 sy ϭ s txy ϭ t s1 ϭ 988 psi s2 ϭ Ϫ21719 psi tmax ϭ 11354 psi ; ; ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 tT ϭ Mz 2tAm tt ϭ FyQ I(2t) Q ϭ 1 8 1 b3 Ϫ b1 3 2 08Ch08.qxd 9/18/08 11:08 AM Page 701 54. 54. 702 CHAPTER 8 Applications of Plane Stress Problem 8.5-22 A mountain bike rider going uphill applies force P ϭ 65 N to each end of the handlebars ABCD, made of aluminum alloy 7075-T6, by pulling on the handlebar extenders (DF on right handlebar segment). Consider the right half of the handlebar assembly only (assume the bars are fixed at the fork at A). Segments AB and CD are prismatic with lengths L1 and L3 and with outer diameters and thicknesses d01, t01 and d03, t03, respectively, as shown. Segment BC of length L2, however, is tapered and, outer diameter and thickness vary linearly between dimensions at B and C. Consider shear, torsion, and bending effects only for segment AD; assume DF is rigid. Find maximum tensile, compressive, and shear stresses adjacent to support A. Show where each maximum stress value occurs. (b) Section D–F d03 Handlebar extension 45° Handlebar P F y z D d = 100 mm (a) Handlebar extension d03 = 22 mm t03 = 2.95 mm d01 = 32 mm t01 = 3.15 mm L2 = 30 mmL1 = 50 mm A C B F x y D L3 = 220 mm Solution 8.5-22 P ϭ 65 N L1 ϭ 50 mm L2 ϭ 30 mm L3 ϭ 220 mm d01 ϭ 32 mm d03 ϭ 22 mm d ϭ 100 mm PROPERTIES O F THE CROSS S ECTION AT POINT A t01 ϭ 3.15 mm A ϭ 150.543 mm2 I ϭ 1.747 ϫ 104 mm4 Ip ϭ 2I Ip ϭ 3.493 ϫ 104 mm4 Q ϭ 729.625 mm3 Q ϭ 1 12 [ d01 3 Ϫ 1 d01 Ϫ t0123 ] I ϭ p 64 [d01 4 Ϫ 1d01 Ϫ

2017-5-21

Chapter 8

Vmax. ϭ 0.065 kN T ϭ Pd (Torsional Moment) T ϭ 6.5 kN mm M ϭ P(L1 ϩ L2 ϩ L3) (Bending Moment) M ϭ 19.5 kN mm# # Section at point A y z x45° 45° T M A3 Vm ax A2A1 08Ch08.qxd 9/18/08 11:08 AM Page 702 55. 55. sx ϭ 17.863 MPa (tensile stress) sy ϭ 0 txy ϭ 2.977 MPa s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Td01 2 Ip sx ϭ M I r SECTION 8.5 Combined Loadings 703 sx ϭ Ϫ17.863 MPa (compressive stress) sy ϭ 0 txy ϭ 2.977 MPa MAX. TENSILE STRESS MAX. COMPRESSIVE STRESS MAX. SHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. STRESS RESULTANTS AT POINT A2 tmax ϭ 9.42 MPa ; s2 ϭ Ϫ18.35 MPa ; s1 ϭ 0.483 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ T d01 2Ip sx ϭ Ϫ M I r sx ϭ 0 sy ϭ 0 txy ϭ 3.408 MPa MAX. TENSILE STRESS MAX. COMPRESSIVE STRESS MAX. SHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. STRESS RESULTANTS AT POINT A3 tmax ϭ 3.41 MPa ; s2 ϭ Ϫ3.41 MPa ; s1 ϭ 3.41 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ Td01 2Ip + Vmax.Q I2 t01 08Ch08.qxd 9/18/08 11:08 AM Page 703 56. 56. 704 CHAPTER 8 Applications of Plane Stress Problem 8.5-23 Determine the maximum tensile, compressive, and shear stresses acting on the cross section of the tube at point A of the hitch bicycle rack shown in the figure. The rack is made up of 2 in. ϫ 2 in. steel tubing which is 1/8 in. thick. Assume that the weight of each of four bicycles is distributed evenly between the two support arms so that the rack can be represented as a cantilever beam (ABCDEF) in the x-y plane. The overall weight of the rack alone is W ϭ 60 lb. directed through C, and the weight of each bicycle is B ϭ 30 lb. MAX. TENSILE STRESS s1 ϭ 18.35 MPa ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 MAX. COMPRESSIVE STRESS MAX. SHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. tmax ϭ 9.42 MPa ; s2 ϭ Ϫ0.483 MPa ; Bike loads B at each tie down point Fixed support 2 in. ϫ 2 in. ϫ in. steel tube 1 8 — B 2 — B 2 — F E A B C D 2 in. 2 in. in. 1 8 — MAZ Ay Ax y x A B C E F W D 17 in. 6 in. 33 in. 7 in. 3 @ 4 in. 2 in. 4 loads, each B 08Ch08.qxd 9/18/08 11:08 AM Page 704 57. 57. SECTION 8.5 Combined Loadings 705 Solution 8.5-23 Cross section t ϭ 0.125 in. Thickness d2 ϭ 2 in. Outer width d1 ϭ d2 Ϫ 2t Inner width A ϭ d2 2 Ϫ d2 1 A ϭ 0.938 in.2 I ϭ 0.552 in.4 Q ϭ 0.33 in.3 The distance between point A and the center of load W b1 ϭ 17 in. The distance between the point A and the center of a bike load B W ϭ 60 lb B ϭ 30 lb REACTIONS AT SUPPORT A MAz ϭ Wb1 ϩ 4Bb2 MAz ϭ 4.74 ϫ 103 lb in. Ay ϭ W ϩ 4B Ay ϭ 180 lb Ax ϭ 0 # b2 ϭ a17 + 2 + 6 + 3.4 2 b in. Q ϭ (d2 Ϫ 2t) t a d2 2 Ϫ t 2 b + 2 d2 2 t d2 4 I ϭ d4 2 12 Ϫ d4 1 12 TOP OF THE CROSS SECTION (AT POINT A) sx ϭ 8.591 ϫ 103 psi (tensile stress) sy ϭ 0 txy ϭ 0 MAX. TENSILE STRESS MAX. COMPRESSIVE STRESS MAX. SHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. tmax ϭ 4295 psi ; s2 ϭ 0 ; s1 ϭ 8591 psi ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 sx ϭ MAz d2 2I 08Ch08.qxd 9/18/08 11:08 AM Page 705 58. 58. 706 CHAPTER 8 Applications of Plane Stress SIDE OF THE CROSS SECTION (AT POINT A) BOTTOM OF THE CROSS SECTION (AT POINT A) sx ϭ 0 sy ϭ 0 txy ϭ 430.726 psi MAX. TENSILE STRESS MAX. COMPRESSIVE STRESS MAX. SHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. tmax ϭ 431 psi ; s2 ϭ Ϫ431 psi ; s1 ϭ 431 psi ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 txy ϭ AyQ I2t sx ϭ Ϫ8.591 ϫ 103 psi (compressive stress) sy ϭ 0 txy ϭ 0 MAX. TENSILE STRESS MAX. COMPRESSIVE STRESS MAX. SHEAR STRESS NOTE: Since the principal stresses have opposite signs, the maximum in-plane shear is larger than the maximum out-of-plane shear stress. tmax ϭ 4295 psi ; s2 ϭ Ϫ8591 psi ; s1 ϭ 0 ; tmax ϭ A a sx Ϫ sy 2 b 2 + txy 2 s2 ϭ sx + sy 2 Ϫ A a sx Ϫ sy 2 b 2 + txy 2 s1 ϭ sx + sy 2 + A a sx Ϫ sy 2 b 2 + txy 2 sx ϭ Ϫ MAzd2 2I 08Ch08.qxd 9/18/08 11:08 AM Page 706 Recommended

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Chapter 8




Strength of materials 4th ed. by ferdinand l. singer andrew pytel 2 saleem232

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Chapter 8

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