Unformatted text preview: D 3.01 Shaft AB consists of n homogeneous cylindrical elements. which
can be solid or hollow. Its end A is ﬁxed, while its end B is free. and it is sub—
. , . jetted to the loading shown. The length of element 1' is denoted by 1... its outer
"" " diameter by 0D,. its inner diameter by [0,, its modulus of rigidity by 6.. and I] = the torque applied to its right end by T,, the magnitude T, of this torque being
assumed to be positive if T. is observed as counterclockwise {mm end B and negative otherwise. (Note that ID, = 0 if the element is solid.) (a) Write a com puter pmgram that can be used to determine the maximum shearing stress in each element. the angle of twist of each element, and the angle of twist of the Elemenil entire shaft. (b) Use this program to solve Probs. 3.9. 3.35. 3.37. 3.150. and F02 Ema CU owe/ma Hemp”; Eli/7W: L on 10. G 5: 155432172 Fil/D {QMPU'I'I
J. = (2.732% ogre ~33») . . ' " 3.51.
‘l: k 5' SOLUTION 007L/29E c) F P/é‘Oﬁ/P‘GM UP 0474: Texaco 05 7= 7'? 7:
ﬂ/fo C(‘I/Ij 1:1"?? 7Tch. = '7” {og./2)/.{.
m4. = 742/4 {M545 a: We; 0F awe/g; S’s/937'} 57791.23»;
lav/m 63:0) [390.475 Tﬂ/ECJUM 7;“ [zEME/vf U Q: 63* (9’92"
PPﬁﬁ/gﬁrn OUT/’1‘?" Problem 3.9 and 335’
Element Maxim Stress Angle of Twist
(ups) (degrees) 1.0000 56.5884 2.5265
2.0000 36.6264 0.0887
Angle of twist: for entire shaft  3.4152 ° F Problem 3.37
, Element Maxim Stress Angle of Twist (MPal (degrees) 7—7 1.0000 33.9531 0.8314
2.0000 19.6483 0.7413
Angle of tuist [or entire shaft: . 1.5726 ° 1—] Problem 3.150 am! 3.15?
Element: Maximum Stress Angle or Twist
(ks‘i) (degrees) fl 1.0000 9.1266 3.5857
2.0000 8.5526 «3.0002
Angle of twist for entire shaft  0.5855 ° ...
View
Full Document
 Summer '10
 STAFF

Click to edit the document details