21_ch 03 Mechanical Design budynas_SM_ch03

# 21_ch 03 Mechanical Design budynas_SM_ch03 -...

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

34 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 3-19 From Eq. (3-15) σ 3 (1 + 4 + 4) σ 2 + [1(4) + 1(4) + 4(4) 2 2 ( 4) 2 ( 2) 2 ] σ [1(4)(4) + 2(2)( 4)( 2) 1( 4) 2 4( 2) 2 4(2) 2 ] = 0 σ 3 9 σ 2 = 0 Roots are: 9, 0, 0 kpsi τ 2 / 3 = 0, τ 1 / 2 = τ 1 / 3 = τ max = 9 2 = 4 . 5 kpsi Ans. 3-20 (a) R 1 = c l FM max = R 1 a = ac l F σ = 6 M bh 2 = 6 2 l F F = σ 2 l 6 Ans. (b) F m F = ( σ m )( b m / b ) ( h m / h ) 2 ( l m / l ) ( a m / a ) ( c m / c ) = 1( s s ) 2 ( s ) ( s s ) = s 2 Ans. For equal stress, the model load varies by the square of the scale factor. 3-21 R 1 = w l 2 , M max | x = l / 2 = w 2 l 2 ± l l 2 ² = w l 2 8 σ = 6 M 2 = 6 2
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