250_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

250_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.8-2.12.qxd 244 9/25/08 11:44 AM Page 244 CHAPTER 2 Axially Loaded Members STRESSES From Eq. (3): FB FC sC sC 2sB (7) A A Wire C has the larger stress. Therefore, it will yield first. 2( sB (a) YIELD LOAD C FC YA sY 2 (From Eq. 7) 1 sA 2Y FB 1 2 a sYA b + 3(sYA) 2 PY YA YA) 5 sA 4Y 4P ; FBL sY L EA E From Eq. (2): dB dD From Eq. (3): P PP 3( From Eq. (4): sC 2 sB Y P YA) dP 2dB 2sYL E ; (c) LOAD-DISPLACEMENT DIAGRAM 4P PP ; From Eq. (4): P 5 P 4Y 2 Y FB L sY L dB EA 2E From Eq. (2): dD dY sY L E 2dB ; (b) PLASTIC LOAD At the plastic load, both wires yield. B Y C FB FC YA Problem 2.12-10 Two cables, each having a length L of approximately 40 m, support a l oaded container of weight W (see figure). The cables, which have effective cross-sectional area A 48.0 mm2 and effective modulus of elasticity E 160 GPa, are identical except that one cable is longer than the other when they are hanging separately and unloaded. The difference in lengths is d 100 mm. The cables are made of steel having an elastoplastic stress-strain diagram with Y 500 MPa. Assume that the weight W is initially zero and is slowly increased by the addition of material to the container. L (a) Determine the weight WY that first produces yielding of the shorter cable. Also, determine the corresponding elongation Y of the shorter cable. (b) Determine the weight WP that produces yielding of both cables. Also, determine the elongation P of the shorter cable when the weight W just reaches the value WP. (c) Construct a load-displacement diagram showing the weight W as ordinate and the elongation of the shorter cable as abscissa. (Hint: The load displacement diagram is not a single straight line in the region 0 W WY.) W ...
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