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242_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 242_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: Sec_2.8-2.12.qxd 236 9/25/08 11:44 AM Page 236 CHAPTER 2 Axially Loaded Members Problem 2.11-6 A rigid bar AB, pinned at end A, is supported by a wire CD and loaded by a force P at end B (see figure). The wire is made of high-strength steel having modulus of elasticity E 210 GPa and yield stress Y 820 MPa. The length of the wire is L 1.0 m and its diameter is d 3 mm. The stress-strain diagram for the steel is defined by the modified power law, as follows: C L A s s D B EP 0 … s … sY sY a EP n b s Ú sY sY P 2b b (a) Assuming n 0.2, calculate the displacement B at the end of the bar due to the load P. Take values of P from 2.4 kN to 5.6 kN in increments of 0.8 kN. (b) Plot a load-displacement diagram showing P versus B. Solution 2.11-6 Rigid bar supported by a wire sY s 1/n ab E sY 3P Axial force in wire: F 2 3P F Stress in wire: s A 2A PROCEDURE: Assume a value of P Calculate from Eq. (6) Calculate from Eq. (4) or (5) Calculate B from Eq. (3) From Eq. (2): Wire: E 210 GPa L 3 mm pd2 4 7.0686 mm2 (MPa) Eq. (6) 2.4 3.2 4.0 4.8 5.6 1.0 m d A P (kN) 820 MPa Y 509.3 679.1 848.8 1018.6 1188.4 Eq. (4) or (5) (5) (6) B (mm) Eq. (3) 0.002425 0.003234 0.004640 0.01155 0.02497 3.64 4.85 6.96 17.3 37.5 STRESS-STRAIN DIAGRAM E s (0 En sY a b sY (a) DISPLACEMENT Y) ( (1) Y) (n 3 3 d L 2 2 from stress-strain equations: From Eq. (1): (2) B AT END OF BAR elongation of wire dB Obtain 0.2) sE (0 … s … sY) (3) (4) For 820 MPa: 0.0039048 P 3.864 kN Y (b) LOAD-DISPLACEMENT DIAGRAM B 5.86 mm ...
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