- Project Runeberg -  Machinists' and Draftsmen's Handbook /
270

(1910) Author: Peder Lobben - Tema: Mechanical Engineering
Table of Contents / Innehåll | << Previous | Next >>
  Project Runeberg | Catalog | Recent Changes | Donate | Comments? |   

Full resolution (JPEG) - On this page / på denna sida - Strength of Materials - Hollow round shafts - Square beams exposed to torsional stress

scanned image

<< prev. page << föreg. sida <<     >> nästa sida >> next page >>


Below is the raw OCR text from the above scanned image. Do you see an error? Proofread the page now!
Här nedan syns maskintolkade texten från faksimilbilden ovan. Ser du något fel? Korrekturläs sidan nu!

This page has never been proofread. / Denna sida har aldrig korrekturlästs.

27O STRENGTH OF MATERIALS.
Hollow Round Shafts.
In proportion to the amount of material used, around hollow
shaft has more torsional strength than a solid shaft of the same
diameter. This is because the fibers in any shaft exposed to
twisting stress only offer resistance to the load in proportion to
their stretch. Therefore, the fibers near the center are always
in position to offer less resistance than the fibers more remote
from the center.
The formula for torsional strength in round hollow shafts
will be
:
(
D* — d*\
\ D X m J
P — Ultimate breaking load in pounds applied at a leverage
of m feet.
D = Outside diameter of shaft in inches.
d = Inside diameter of shaft in inches.
m = Length of lever in feet.
c = Constant (same as for a solid shaft).
Square Beams Exposed to Torsional Stress.
The theoretical formula for twisting strength (on page 266)
will apply to square as well as round beams. The proportional
strength between a round and a square beam may, therefore, be
compared by using that formula. Let S represent the side of a
square beam and the polar moment of inertia is -} S*.
The distance from the center of the beam to the most
remote fiber in a square beam \s S \f l
/
2 , and, dividing the polar
moment of inertia by this distance, we have,
i ^4

*


= 0.23 S3
Let D represent the diameter of a round beam. The
polar moment of inertia is = 0.098 Z>4
The distance from the center to the most remote fiber in the
round beam is ]/
2 D. Dividing the polar moment of inertia by
this distance, we have
oms
^ = 0.196 D3
Suppose, now, that S and D are equal, for instance,
one inch ; the proportion in torsional strength between the two
beams must be 0.23 divided by 0.196, which equals 1.18.
Thus, for square beams, use the formulas given for round
beams, but multiply constant c, in Table No. 32, by 1.2, and

<< prev. page << föreg. sida <<     >> nästa sida >> next page >>


Project Runeberg, Sun Jan 5 23:50:27 2025 (aronsson) (download) << Previous Next >>
https://runeberg.org/lobben/0290.html

Valid HTML 4.0! All our files are DRM-free