37_ch 04 Mechanical Design budynas_SM_ch04

# 37_ch 04 Mechanical Design budynas_SM_ch04 - U M A = The...

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

106 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 4-70 M = FR sin θ QR (1 cos θ ), M Q =− R (1 cos θ ) F θ = Q cos θ + F sin θ , F θ Q = cos θ Q ( MF θ ) = [ FR sin θ QR (1 cos θ )] cos θ + [ R (1 cos θ )][ Q cos θ + F sin θ ] F r = F cos θ Q sin θ , F r Q =− sin θ From Eq. (4-33) δ = U Q ± ± ± ± Q = 0 = 1 AeE ² π 0 ( FR sin θ )[ R (1 cos θ )] d θ + R AE ² π 0 F sin θ cos θ d θ 1 AE ² π 0 [ FR sin θ cos θ FR sin θ (1 cos θ )] d θ + CR AG ² π 0 F cos θ sin θ d θ =− 2 FR 2 AeE + 0 + 2 FR AE + 0 =− ³ R e 1 ´ 2 FR AE Ans. 4-71 The cross section at A does not rotate, thus for a single quadrant we have
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: U M A = The bending moment at an angle to the x axis is M = M A F 2 ( R x ) = M A F R 2 (1 cos ) (1) because x = R cos . Next, U = M 2 2 E I ds = / 2 M 2 2 E I R d since ds = R d . Then U M A = R E I / 2 M M M A d = Q F M R F r F...
View Full 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