02-07 Chap Gere

# 02-07 Chap Gere - 160 CHAPTER 2 Axially Loaded Numbers...

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Stress Concentrations The problems for Section 2.10 are to be solved by considering the stress-concentration factors and assuming linearly elastic behavior. Problem 2.10-1 The flat bars shown in parts (a) and (b) of the figure are subjected to tensile forces P 5 3.0 k. Each bar has thickness t 5 0.25 in. (a) For the bar with a circular hole, determine the maximum stresses for hole diameters d 5 1 in. and d 5 2 in. if the width b 5 6.0 in. (b) For the stepped bar with shoulder fillets, determine the maximum stresses for fillet radii R 5 0.25 in. and R 5 0.5 in. if the bar widths are b 5 4.0 in. and c 5 2.5 in. Solution 2.10-1 Flat bars in tension 160 CHAPTER 2 Axially Loaded Numbers P P P P b d b (a) c R P 5 3.0 k t 5 0.25 in. (a) B AR WITH CIRCULAR HOLE ( b 5 6 in.) Obtain K from Fig. 2-63 F OR d 5 1 in.: c 5 b 2 d 5 5 in. d / b 5 K < 2.60 s max 5 k s nom < For d 5 2 in.: c 5 b 2 d 5 4 in. d / b 5 K < 2.31 s max 5 K s nom < 6.9 ksi 1 3 s nom 5 P ct 5 3.0 k (4 in.)(0.25 in.) 5 3.00 ksi 6.2 ksi 1 6 s nom 5 P ct 5 3.0 k (5 in.)(0.25 in.) 5 2.40 ksi (b) S TEPPED BAR WITH SHOULDER FILLETS b 5 4.0 in. c 5 2.5 in.; Obtain k from Fig. 2-64 F OR R 5 0.25 in.: R / c 5 0.1 b / c 5 1.60 k < 2.30 s max 5 K s nom < F OR R 5 0.5 in.: R / c 5 0.2 b / c 5 1.60 K < 1.87 s max 5 K s nom < 9.0 ksi 11.0 ksi s nom 5 P ct 5 3.0 k (2.5 in.)(0.25 in.) 5 4.80 ksi P P P P b d b (a) (b) c R = radius Probs. 2.10-1 and 2.10-2

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SECTION 2.10 Stress Concentrations 161 Problem 2.10-2 The flat bars shown in parts (a) and (b) of the figure are subjected to tensile forces P 5 2.5 kN. Each bar has thickness t 5 5.0 mm. (a) For the bar with a circular hole, determine the maximum stresses for hole diameters d 5 12 mm and d 5 20 mm if the width b 5 60 mm. (b) For the stepped bar with shoulder fillets, determine the maximum stresses for fillet radii R 5 6 mm and 5 10 mm if the bar widths are b 5 60 mm and c 5 40 mm. Solution 2.10-2 Flat bars in tension P P P P b d b (a) (b) c R = radius P 5 2.5 kN t 5 5.0 mm (a) B AR WITH CIRCULAR HOLE ( b 5 60 mm) Obtain k from Fig. 2-63 F OR d 5 12 mm: c 5 b 2 d 5 48 mm d / b 5 K < 2.51 s max 5 K s nom < F OR d 5 20 mm: c 5 b 2 d 5 40 mm d/b 5 K < 2.31 s max 5 K s nom < 29 MPa 1 3 s nom 5 P ct 5 2.5 kN (40 mm)(5 mm) 5 12.50 MPa 26 MPa 1 5 s nom 5 P ct 5 2.5 kN (48 mm) 5 10.42 MPa (b) S TEPPED BAR WITH SHOULDER FILLETS b 5 60 mm c 5 40 mm; Obtain K from Fig. 2-64 F OR R 5 6 mm: R/c 5 0.15 b/c 5 1.5 K < 2.00 s max 5 K s nom < F OR R 5 10 mm: 5 0.25 b/c 5 1.5 K < 1.75 s max 5 K s nom < 22 MPa 25 MPa s nom 5 P ct 5 2.5 kN mm) 5 12.50 MPa
162 CHAPTER 2 Axially Loaded Numbers Problem 2.10-4 A round brass bar of diameter d 1 5 20 mm has upset ends of diameter d 2 5 26 mm (see figure). The lengths of the segments of the bar are L 1 5 0.3 m and L 2 5 0.1 m. Quarter-circular fillets are used at the shoulders of the bar, and the modulus of elasticity of the brass is E 5 100 GPa. If the bar lengthens by 0.12 mm under a tensile load P , what is the maximum stress s max in the bar? L 1 d 1 d 2 d 2 L 2 L 2 P P P P b d t 5 thickness s t 5 allowable tensile stress Find P max Find K from Fig. 2-64 Because s t , b , and t are constants, we write: P * 5 P max s t bt 5 1 K ¢ 1 2 d b 5 s t K bt ¢ 1 2 d b P max 5 s nom ct 5 s max K ct 5 s t K ( b 2 d ) t We observe that P max decreases as increases.

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## This note was uploaded on 03/19/2008 for the course E M 316 taught by Professor Korkolis during the Spring '08 term at University of Texas.

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02-07 Chap Gere - 160 CHAPTER 2 Axially Loaded Numbers...

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