02-01ChapGere.0007

02-01ChapGere.0007 - stiffness k 1 5 300 N/m and natural...

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

View Full Document Right Arrow Icon
Problem 2.2-9 An aluminum wire having a diameter d 5 2 mm and length L 5 3.8 m is subjected to a tensile load P (see figure). The aluminum has modulus of elasticity E 5 75 GPa. If the maximum permissible elongation of the wire is 3.0 mm and the allowable stress in tension is 60 MPa, what is the allowable load P max ? SECTION 2.2 Changes in Lengths of Axially Loaded Members 69 P d P L Solution 2.2-9 Aluminum wire in tension P d P L d 5 2mm L 5 3.8 m E 5 75 GPa M AXIMUM LOAD BASED UPON ELONGATION d max 5 3.0 mm Ê d 5 PL EA A 5 p d 2 4 5 3.142 mm 2 5 186 N M AXIMUM LOAD BASED UPON STRESS 5 189 N A LLOWABLE LOAD Elongation governs. Ê P max 5 186 N P max 5 A s allow 5 (3.142 mm 2 )(60 MPa) s allow 5 60 MPa Ê s 5 P A 5 (75 GPa)(3.142 mm 2 ) 3.8 m (3.0 mm) P max 5 EA L d max Problem 2.2-10 A uniform bar AB of weight W 5 25 N is supported by two springs, as shown in the figure. The spring on the left has
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: stiffness k 1 5 300 N/m and natural length L 1 5 250 mm. The corresponding quantities for the spring on the right are k 2 5 400 N/m and L 2 5 200 mm. The distance between the springs is L 5 350 mm, and the spring on the right is suspended from a support that is distance h 5 80 mm below the point of support for the spring on the left. At what distance x from the left-hand spring should a load P 5 18 N be placed in order to bring the bar to a horizontal position? P W x h L A k 1 L 1 k 2 L 2 B A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
View Full Document

Ask a homework question - tutors are online