22_ch 03 Mechanical Design budynas_SM_ch03

22_ch 03 Mechanical Design budynas_SM_ch03 -...

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

View Full Document Right Arrow Icon
Chapter 3 35 3-22 (a) Can solve by iteration or derive equations for the general case. Find maximum moment under wheel W 3 W T = W at centroid of W ’s R A = l x 3 d 3 l W T Under wheel 3 M 3 = R A x 3 W 1 a 13 W 2 a 23 = ( l x 3 d 3 ) l W T x 3 W 1 a 13 W 2 a 23 For maximum, dM 3 dx 3 = 0 = ( l d 3 2 x 3 ) W T l x 3 = l d 3 2 substitute into M , M 3 = ( l d 3 ) 2 4 l W T W 1 a 13 W 2 a 23 This means the midpoint of d 3 intersects the midpoint of the beam For wheel i x i = l d i 2 , M i = ( l d i ) 2 4 l W T i 1 ± j = 1 W j a ji Note for wheel 1: ± W j a = 0 W T = 104 . 4, W 1 = W 2 = W 3 = W 4 = 104 . 4 4 = 26 . 1 kip Wheel 1: d 1 = 476 2 = 238 in, M 1 = (1200 238) 2 4(1200) (104 . 4) = 20 128 kip
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/13/2011 for the course EML 3013 taught by Professor Shingley during the Fall '11 term at UNF.

Ask a homework question - tutors are online