02-06ChapGere.0012

02-06ChapGere.0012 - d k v d k v k 5 8.0MN/m W 5 545kN d 5...

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

View Full Document Right Arrow Icon
SECTION 2.8 Impact Loading 155 W 5 100lb A 5 0.080 in. 2 E 5 21 3 10 6 psi h 5 45in s allow 5 s max 5 70ksi Find minimum length L min S TATIC STRESS s st 5 W A 5 100 lb 0.080 in. 2 5 1250 psi M INIMUM LENGTH L min or Square both sides and solve for L : S UBSTITUTE NUMERICAL VALUES : 5 500 in. L min 5 2(21 3 10 6 psi)(45 in.)(1250 psi) (70,000 psi)[70,000 psi 2 2(1250 psi)] L 5 L min 5 2 Eh s st s max ( s max 2 2 s st ) s max s st 2 1 5 ¢ 1 1 2 hE L s st 1 / 2 Eq. (2-59): Ê s max 5 s st B 1 1 ¢ 1 1 2 hE L s st 1 / 2 R Problem 2.8-10 A bumping post at the end of a track in a railway yard has a spring constant k 5 8.0 MN/m (see figure). The maximum possible displacement d of the end of the striking plate is 450 mm. What is the maximum velocity v max that a railway car of weight W 5 545 kN can have without damaging the bumping post when it strikes it? Solution 2.8-10 Bumping post for a railway car
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: d k v d k v k 5 8.0MN/m W 5 545kN d 5 maximum displacement of spring d 5 d max 5 450mm Find v max K INETIC ENERGY BEFORE IMPACT KE 5 Mv 2 2 5 Wv 2 2 g S TRAIN ENERGY WHEN SPRING IS COMPRESSED TO THE MAXIMUM ALLOWABLE AMOUNT C ONSERVATION OF ENERGY S UBSTITUTE NUMERICAL VALUES : 5 5400 mm / s 5 5.4 m / s v max 5 (450 mm) B 8.0 MN / m (545 kN) / (9.81 m / s 2 ) v 5 v max 5 d B k W / g KE 5 U Wv 2 2 g 5 kd 2 2 v 2 5 kd 2 W / g U 5 k d 2 max 2 5 kd 2 2 Solution 2.8-9 Slider on a cable h L W A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
View Full Document

Ask a homework question - tutors are online