242_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

242_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

Ask a homework question - tutors are online