the effect of combined torsional and bending loads on a
TRANSCRIPT
Scholars' Mine Scholars' Mine
Masters Theses Student Theses and Dissertations
1966
The effect of combined torsional and bending loads on a channel The effect of combined torsional and bending loads on a channel
beam with one end restrained from warping beam with one end restrained from warping
David M. Schaeffer
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Recommended Citation Recommended Citation Schaeffer, David M., "The effect of combined torsional and bending loads on a channel beam with one end restrained from warping" (1966). Masters Theses. 5772. https://scholarsmine.mst.edu/masters_theses/5772
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THE EFFECT OF COMBINED TORSIONAL AND BENDING LOADS
ON A CHANNEL BEAM
WITH ONE END RESTRA NE FROM W R.PING
by
DAVID M. SCHAEFFER
A
THESIS
Submitted to the faculty of the
UNIVERSITY OF MISSOURI AT ROLLA
i n partia l fulfillmen t of the wor k required for the
Degree of
MASTER OF SCIENCE IN ENGINEERING MECHANICS
Rolla, Missouri
1966
Approved by
c;;;;t~
i i
PREFACE
This report is the result of research conducted in the
partial fulfillment of the requirements for the degree of
Master of Science in the field of Engineering Mechanics.
The work was performed at the University of Missouri
at Rolla under the guidance of Professor Karlheinz C. Muhl
bauer in the Department of Engineerlug fileclu.nlcs o1 wlllch
Professor Robert F. Davidson is the chairman.
At this time the author would like to thank his advis
or, Professor Muhlbauer, for his generous assistance in
completing the thesis. Also a word of thanks to Mr. Robert
Hackbarth for his aid in taking the data and programming
the equations to run in the computer.
1 1 1
ABSTRACT
Channel beams have been designed with the main purpose
of giving large resistance to bending while the torsional
strengths are known to be relatively small~ In this exper
iment, a cantilevered channel beam was loaded with a con
centrated load at the free end, first through the experi
mentally determined shear center and then through the cen
troid of the cross section~ For each loading condition 1
the strains were measured with the aid of SR-4 strain gages
placed at intervals along the length of the beam~ With the
aid of a computer, the strains were converted into longitu
dinal stresses and these stresses were compared to the the
oretically predicted values of longitudinal stresses~
The derivation of the torsional stress equation as
shown in the text Advanced Mechanics of Materials by Seely
and Smith is briefly compared to the derivation of the tor
sional stress equation as shown in the text Strength of
Materials by Timoshe~ko~ The two derivations are discussed
and, even though·the derivations are completely different 1
the values of longitudinal stress obtained by the use of
each equation agree very well with each other~
The longitudinal stresses calculated from the experi
mentally obtained values or strain agree very closely with
the theoretically predicted values· or longitudinal stress~
TABLE OF CONTENTS
LIST OF FIGURES
LIST OF SYMBOlS
• • • • • • • • • • • • • • • • • • •
• • • • • • • • • • • • • • • • • • •
SCOPE • • • • • • • • • • • • • • • • • • • • • • •
I. INTRODUCTION • • • • • • • • • • • • • • • • •
II~ REVIEW OF LITERATURE • • • • • • • • • • • • •
III~ GENERAL PREPARATION OF THE EXPERIMENT • • • •
IV~ EXPERIMENTAL DETERMINATION OF THE SHEAR CENTER ~ • • • • • • • ~ • • • • ~ ~ • • • • •
V~ OBSERVED VALUES OF LONGITUDINAL STRESSES WHEN
1v
Page v
vi
viii
1
4
8
.15
LOAD WAS APPLIED THROUGH THE SHEAR CENTER • • 19
VI~ LONGITUDINAL STRESSES IN CHANNEL BEAM WHEN THE TRANSVERSE LOAD PRODUCES TWISTING AS WELL AS BENDING -~ ~ • • • • • ~ • • • ~ ~ ~ • ~ • • •
B.
Theory According to Seely and Smith
Theory Developed by Timoshenko ~ • •
VII~ OBSERVED VALUES OF LONGITUDINAL STRESS IN
• • •
• • •
34
34
43
CHANNEL BEAM WHEN LOADED THROUGH THE CENTROID~ 46
VIII~ DISCUSSION OF LOADING THROUGH THE SHEAR CENTER ~ ~ ~ ~ • • ~ ~ • • • ~ ~ • ~ • • • • •
IX. DISCUSSION OF LOADING THROUGH THE CENTROID • •
X~ CONCLUSION • • • • • • • • • • • • • • • •. • • •
BmLIOGRAPHY • • • • • • • • • • • • • • • • • • • •
VITA . . • • • • • • • • • • • • • • • • • • • • • • • •
62
66
69
71
73
LIST OF FIGURES
FIGURE 1~ Fixed End of Channel Beam • • • • • • •
FIGURE 2~ Loading Apparatus • • • • • • • • • • • FIGURE 3~ General Arrangement of Level Bars • • • FIGURE . 4~ General Arrangement of Strain Gages • .•
FIGURE 5~ General View of Loading Apparatus • • • FIGURE 6~ Location of Shear Center Determined
from Test Data • • • • • • • • • • • •
FIGURE 7~ Transverse Cross Section of the Channel Beam • • • < • • • • • • • • • • • • • •
FIGURES 8-19~ Load Through Shear Center Versus
FIGURE 20~
FIGURE 21~
FIGURE 22~
FIGURE 23-32~
FIGURE
· Strain • • • • • • • • • • • • • • • •
Stress Versus Distance from Fixed End of Channel Beam • • • • ~ ~ ~ • ~ • • •
Load "P" Applied Through Centroid of Cross Section • ~ • • • • • • • • • • •
General Torsional Effect of Twisting Moment on Channel Beam • ~ • • • •
Load Through Centroid Versus Strain
• •
• •
v
Page 10
11
12
13
14
17
18
21-32
33
35
36
48-57
33~ Angle of Twist Versus Distance from Fixed End ~ ~ • • • • • • • • • • • • • 58
FIGURE 34-36~ Stress Versus Distance from Fixed End • • • • • • • • • • • • • • • • • • 59-61
FIGURE 37 ~ Distribution of Stress Across the Flange • • • • • • • • • • • • • • • • 64
FIGURE 38~ Distribution of Stress Across the Flange • • • • • • • • • • • • • • • • 67
a
A
b
B
c
e
E
G
h
I
J
L
M
m
p
s
LIST OF SYMBOLS
warping constant
integration constant
overall width of flange
integration constant
distance from centroid to outer fiber
perpendicular distance from the center line of the web to the shear center
modulus of elasticity
shearing modulus of elasticity
mean distance between the flanges
vi
moment of inertia of the entire cross section with respect to the centroidal axis parallel to the flanges
moment of inertia of one flange with respect to centroidal axis parallel to the web
moment of inertia of the entire cross section with respect to the centroidal axis parallel to the web
equivalent polar moment of inertia
length of beam .
bending moment about the centroidal axis of the flange parallel to the web
bending moment about the centroidal axis of the entire cross section parallel to the flanges
distance from the shear center to the point of application of the load
vertical load
total stress in beam
t
v
X
y
z
a
'[
X
y
z 1'
m'
n'
1
z
vii
LIST OF SYMBOLS {Continued)
stress component due to bending about the centroidal axis parallel to the flanges
stress component due to bending about the centroidal axis parallel to the web
total resisting torque
resisting torque produced by shearing forces in flange
twisting moment produced by the pure torsional shearing stresses in the cross section
thickness of flange
lateral .shearing force !n flange
distance from fixed end of channel beam in inches
lateral deflection or the flanges
distance from web to neutral axis
angle of twist per unit length
total angle of twist
normal stress
shearing stress
body force
body force
body force
direction cosine
direction cosine
direction cosine
component of the surface force per unit area
component of the surface force per unit area
component of the surface force per unit area
viii
SCOPE
It is the purpose of this paper:
1. To investigate the applicability of using the tor
sional equation, which was derived for a cantilevered I
beam by Seely and Smith, for a cantilevered channel beam.
2. To check on the exactness of the warping constant
"a" used in the torsional equation which was derived for
an I beam.
3. To examine whether the flanges of the channel beam
bend about an axis through their centroid parallel to the
web of the channel at.the fixed end.
1
I. INTRODUCTION
Torsional stresses in a steel framed structure are
rarely serious enough to require design analysis. Many
times design engineers completely disregard stresses caused
by torsion. There are conditions, however, in which tor
sional loads will produce stresses of sufficient magnitude
to require torsional analysis of a structural member.
The shapes of rolled I beams and channel beams have
been designed primarily for providing large resistance to
bending in accordanae with the simple flexure theory for
beams; the torsional strength and stiffness of such sec
tions are known to be relatively small. In order to devel
op the bending resistance of I beams and·channel beams with
out permitting them to twist, it was first assumed that
transverse bending loads on these sections should be ap
plied through the centroid of the transverse cross sections
and parallel to. the web. In the case of the I beams this
assumption is correct, but when a channel beam is so loaded
it twists appreciably as it bends.
In order to cause a channel beam to bend without twist
ing, and to develop stresses in accordance with the simple
flexure formula for beams, the load must be applied in a
plane parallel to the web but at a considerable distance
from the centroid or the channel. The intersection or this
plane of loading with the neutral surface is called the
"axis of bending"; and the intersection of this axis with
2
a transverse cross section of the beam is called the "shear
center" for the section.
When I beams and channel beams are free from lateral
restraint, and hence are free to twist, and are loaded so
that the transverse bending loads do not pass through the
axis of bending, they will twist as they bend. If the beam
has a transverse cross section that remains plane and hence
does not warp as the beam twists, such as a fixed end can
tilever beam, this twisting will produce large additional
longitudinal stresses.
The problem of specific interest here is the study of
a beam having a thin-walled, open cross section subjected
to both bending and torsional loads with one end restrained
from warping.
The distribution of torsional stresses along the
length of the member and the torsional rigidity of the mem
ber depends on the end conditions~ The fixed end condition
is satisfied when rotation and warping of the cross section
at the end of the member is prevented. For the free end
condition, both rotation and warping of the cross section
are unrestrained. The unsupported end of a cantilever beam
illustrates this condition.
3
Various methods of analysis have been derived for de
termining the stresses at any point in a beam with one end
restrained from warping. They vary from analytical to em
pirical in nature and each with their own assumptions~
In order to determine the total stress condition of a
structural member at a point, the stresses due to torsion
and those due to plane bending are added algebraically. It
is imperative that the direction of the stresses be care
fully observed.
Seely and Smith in their text Advanced Mechanics of
Materials derived the stress equations for an I beam with
one end fixed. They concluded that these equations can
also be used for c,hannel and Z cross sections having the
same boundary conditions as the I beam. Timoshenko in his
text Strength of Materials derived equations for a channel
beam with one end fixed, but he used an approach which was
quite different from that used by Seely and Smith. These
two derivations will be discussed,and each will be compared
with the results of this thesis~
4
II~ REVIEW OF LITERATURE
In 1784, Coulomb1 developed the exact solution for
shearing stress in a circular cross section. He assumed
that the cross section of the bar remained plane and rota-
ted without any distortion during twisting. This same the-. 2
ory was used by Navier in 1864 to arrive at a solution for
the twisting of prismatical bars of noncircular cross sec
tions. Making the above assumption he arrived at the erro
neous conclusions that, for a given torque, the angle of
twist of bars is inversely proportional to the centroidal
polar moment or inertia or the cross section, and that the
maximum shearing stress occurs at the points most remote
from the centroid of the cross section.
In 1853, St. Venant3 solved the torsion problem of a
prismatical bar using the methods of the mathematical the
ory of elasticity. He made certain assumptions as to the
deformation of the twi.sted bar and showed that with these
assumptions he could satisfy the equations of equilibrium:
acrx aTX~ aT xz + X 0 -+ ay + = ax az
acr aTx~ aT __J_ + + ~z + y = 0 ay ax az
aaz 3Txz +
chyz + z - 0 -+ az ax ay
5
and the boundary conditions:
,;( =
y =
a 1 1 X
a m' y
+ T m' + T n' xy xz
+T 0 1 +T 1 1 yz xy
i =an' + T 1' + T m' z xz yz
Then from the uniqueness of the solution obtained by use of
the elasticity equations it follows that the assumptions
made at the start are correct and the solution obtained is
the exact solution of the torsion problem.
4 In 1909, Professor c. Bach published his work of a
steel .rolled channel beam simply supported and loaded with
two equal concentrated loads at the one third points. The
loads were applied through the centroid of the cross sec
tion. He measured the strains at the center of the beam
along each of the four edges, and found that the strain
along one edge of the top was much greater than that along
the other edge of the top. His experimental results indi
cated that the flexure formula gave values of stress in a
channel beam largely in error when.the channel was loaded
according to the conditions that had been assumed to make
the flexure formula applicable. He also applied the loads
through the web and concluded that the stresses found from
the measured strains were more nearly in accordance with
the flexure formula, but did not offer an explanation.
6
In 1921, R~ Maillart, A~ Eggenschwyler, and H~ Zimmer
man5 brought to the attention of engineers the location and
significance of the shear center for channels and some other
thin-walled sections. They obtained mathematical expres
sions for the additional longitudinal stress caused by the
twisting of a channel when the transverse loads on the chan-
nel do not pass through the axis of bending.
In 1925, Foppl and Huber6 also ran tests on a channel
beam. They measured angles of twist at various sections
for different lateral positions of the load to determine
the shear center. The results of the test agreed well with
the calculated position of the shear center.
In 1930, Seely, Putnam, and Schwalbe7 analyzed a num
ber of channel beams with loads through the shear center
and also through the centroid. Each channel was tested as
a horizontal cantilever beam with a vertical load applied
at the end. From the .results thus obtained the position
of the load that causes no twisting of the channel was eas
ily found, and the location of the shear center for each
channel section was thereby determined~ The effect of the
twisting of the channel on the longitudinal stresses at
different sections along the beam was also determined.
8 In 1956, Timoshenko studied the combination of bend-'
ing and torsion of a channel beam and his conclusion was
7
that an equation for the fiber stress in an I beam can also
be used for a channel if the quantity "a" is replaced by a
different quantity which is compatible with the channel
section. This value of "a" is a constant that has units
of length and depends upon the proportions of the beam.
In 1962, Seely and Smith9 concluded in their study on
I and channel beams that "a" could be used without modifi
cation for an I as well as for a channel beam for deter
mining the stresses at any point in the beam.
III. GENERAL PREPARATION OF THE EXPERIMENT
~he channel beam tested in this investigation was a 61
inch - 8.2 lb. channel. The channel beam was 10 feet long
and was tested as a horizontal cantilever beam with verti
cal loads applied at the free end. An eight inch plate
was welded between the flanges as shown in Figure 1. The
purpose of this was to increase the rigidity of the beam
so that no warping would take place at the fixed end where
the beam was clamped between the heads of a 300,000 lb.
compression testing machine.
At the free end of the beam a horizontal plate was
bolted to the beam, as shown in Figure 2, so loads could
be applied at various points along the plate. The angle
of twist was measured at four sections along the length of
the beam for varying lateral positions of the load. See
Figure 3 for general arrangement.
The loads at the free end were applied by means of a
mechanical screw jack. The jack was placed on a horizontal
plate that was free to move on a set of rollers to minimize
the possibility of a horizontal force developing as the ver
tical load was applied. The magnitude of load was deter
mined by means of a platform scale which was placed directly
under the mechanical screw jack. A "point" load was pro
duced by placing a ballbearing between the head of the jack
9
and the horizontal plate.
Loads were applied in 50 lb. increments, first with
the point of application at the shear center and then with
the point of application at the centroid. For each load
strains were measured at various sections along the edges
of the channel beam by the use of SR-4 strain gages. See
Figure 4. The strains were read for each load and, with
the aid of a computer, the stresses were calculated.
10
Fig. 1. Fixed End of Channel Beam
11
Fig. 2. Loading Apparatus
Level bars clamped to top flange at various sections; the change in
inclination of each bar was
measured by dial gages on
a 15-in.
12
gage length.
Uniform load of 4,000 psi
Horizontal plate -
p
Load applied at varying distances from the web.
Fig. 3. General Arrangement of Level Bars
13
83.75 11
'),.
RT Rosettes on top of beam > •
RB Rosettes on bottom of beam LB Linear gages LR Linear gages LL ,Linear gages
Fig. 4. General Arrangement·of Strain Gages
14
Fig. 5. General View of Loading Apparatus
15
IV. EXPERIMENTAL DETERMINATION OF THE SHEAR CENTER
The channel beam was tested as a horizontal cantilever
beam with a vertical load applied as shown in Figure 3.
One end of the beam was fixed by placing it between the
loading heads of' a Riehle testing machine and a load of
50,000 pounds was applied. The fixed end section of the
channel beam was thus maintained as a plane section free
from warping.
Vertical loads were applied to the horizontal plate
at distances of 0, 1, 2 and 3 inches from the back of the
web as shown in Figure 6. For each value of load, the
angle of rotation of the channel beam was measured by read
ing the inclination of the level bars. The change in ele
vation of these level bars was measured to 0.001 of an inch
by use of AMES dial gages.
The location of the shear center for the channel beam
was determined by plotting curves showing the angle of
twist at several sections along the beam, and the corres
ponding lateral positions of the load.
The value of"e'obtained from the experiment was 0.43
inches measured from the center of the web.
The shear center as determined from Seely and Smith's
mathematical equation, which considers the cross section
16
of the flanges and web to be rectangles, was found to be
0.64 inches. As one can see from Figure 7, the flanges and
web are not exactly rectangles.
The shear center was located so that the longitudinal
stresses could be calculated, when the channel beam was sub
jected to twisting. The value of"e 11determined from the ex
periment was used in calculating the longitudinal stresses.
.011
.010
.009
rJl • 008 ~ rn ..-l
~ .007 ~ ...
s:: ..-l .006 +-l rll
~ .p • 005 C+-1 0 (1) • 004
l"""f b()
~ < .003
.002
0 1 11
Distance of load
62.85 1bs
82.85 1bs
22.85 1bs
82.85 1bs
22.85 1bs
22.85 1bs
3" back or channel in
inches
Back of channel
17
I
Fig. 6. Location of Shear Center Determined from Test Data
. 20 11
t
.220"
6.0" I .212 11
i...:;;--Centroid
~ 2 . . 5 inches
i.93"
Fig. ·1· Transverse Cross Section ot the Channel Beam
18
V ~ OBSERVED VALUES OF LONGITUDINAL STRESSES WHEN LOAD WAS APPLIED THROUGH THE SHEAR CENTER
19
If the load on the channel beam is applied through the
experimental shear center, the channel beam bends without
twisting, as is assumed in the simple flexure theory. The
longitudinal stresses at any section are due to the hori-
zontal bending moment at a given section and are given by Mcr
the simple flexure formula S = :r where the axis of symme-
try is the neutral axis.
In order to determine the longitudinal stresses at any
section of the channel beam, the strains were measured at
various sections along the beam. See Figure 3. This was
accomplished by using four Budd Strain indicator units to
read the strain for each increment of loading. To avoid
the possibility of having error in the strain reading~,the
strains were read for each 50 lbs. of load. These values
of strain were plotted against the valuerof load and cor
rection lines were drawn parallel to these points to elim
inate residual initial strain readings. See Figures 8-19.
At a load of 250 pounds, the strains were determined
from the graphs and then used to determine the longitudinal
stresses.
Curves showing the relation between the longitudinal
stress in the flange and the distance from the fixed sec
tion of the beam are shown in Figure 20. The broken line
represents the value of stress obtained from the flexure
formula S = M;.
20
The reason for loading the beam through the shear cen
ter was to see if the loading apparatus gave values of
stress that conformed with the simple flexure formula. As
shown in Figure 20 the stresses did not conform with the
simple flexure formula and the reason for this will be sta
ted in the discussion.
Fig. 8 • Load Through Shear Center Versus Strain
SOOT-----------------------------------------------------------~
r I I //
~ 400 .0 r-i
lj ~ .,;
· ..
H
f G)
~ 300 G) 0
H c:d G)
..c: m . 200
~ I I-V/ I _/ I 0 lRT Long.(- strain) 0 H
..c: 1 ffJ / 'f I ~ I I!J lRT 45° (- strain) .p
~ lOOJ L /.L I 1/ I VlRT Lat. (+strain)
0 50 150 200 250 300 350 -6 I Strain x 10 in in
1\)
~
Fig~ g. Load Through Shear Center Versus Strain
500~·------------------------------------------------------------
·u; 400' .J;l rl
~ H G)
~ 300 G) C)
~ ~ Vl -§, 200 ::s g ..c .p
'0 aS 100 .s
0
IZl
lj
50
~
-232
0 2RT Long.{~ strain)
~ 2RT 45° (- strain)
'f) 2RT Lat. { + strain)
150 200 250 300 350 -6 I Strain x 10 in in
I\) I\)
Fig~ 10. Load Through Shear Center Versus Strain
500 +-------------------------------------------------------~
• Ul 400 ,a r-i
s:: 'l""f
~ Q) +l
~ 300 (,)
~ aS Q) .c: Ul
~ 200 ::s I 0 ~ .c:
j lOJ
0
1./l I
11
50
//
/~
100 150 200 . 6
Strain x 10- in/in
I 0 3RT Long. (- strain)
[!] 3RT 45° (- strain)
V 3RT Lat. (+ strain)
300 350 1\)
VJ
Fig. 11. Load Through Shear Center Versus Strain
500------------------------------------------------------~
"oi 400 .a r-i
s:: ..-1
~ Q) .p s:: 300 Q) 0
~ as Q)
..c: I'll
~ 200 ::s I 0 ~
i lOJ
0
I Yl I
Jl~~ /
/
120
Strain x
I 0 4RT Long. (- strain)
c::J 4RT 45° (- strain)
~ 4RT Lat. (+ strain)
240 280 1\)
+="
Fig. 12. Load Through Shear Center -;;· :us Strain
500 ~-------------------------------------
·• co 4·oo .c .-1
~ .rf
~ Q)
.j.J
~ 300 Q) ()
H aS Q)
..c: al
§, 200 ::s I I II I I // I 0 5RT Long. (- strain) 0 H ..c:
Ill- ~-j lOJ l!l 5RT 45° (- strain)
\J 5RT Lat ~ ( + strain)
0 80 120 160 200 240
-6 I Strain x 10 in in
280 1\)
\J1
Fig. 13. Load Through Shear Center Versus Strain 500T-------------------------------------------------------~
~ 400 .0 n s:: .n ~ <1> .p s:: 300 <1> C)
~
"' ~ r1l
~ 200 ::s 2 ;::: .p
'0
"' .s 100
0
- ... o 40 80
0 6RT Long.(- strain)
8 6RT 45° (- strain)
V 6RT Lat. (+ strain)
12"0 160 200 240 -6 I Strain x 10 in in
2s-o 1\)
0\
Fig. 14. Load Through Shear Center Versus Strain 500T-------------------------------------------------------~
• 400
~ fQ .c r-t
t::
/ ori
H. Q)
~ 300
.. / 4> ()
H aS 4> ;.:::
/.{1 /
0 3RB 45° (+ strain) i 2001 / ~ 3RB Lat~ (- strain)
'{} 3RB Long. ( + strain) ;.:::· '/ .p
ro aS 100 .s
0 40 80 120 160 200 240 280
-6 I Strain x 10 in in 1\)
-'1
Fig. 15. Load Through Shear Center Versus Strain 500~----------------------------------------------------------~
oi 400 .0. r-f
s:: """ $.4 Q)
+> s:: 300 Q) ()
$.4 cd Q) ~ m
-§, 200 ::s f ~ +'
't:S aS 3 100
0·
+30 +74 40 80
o 4RB Long.(+ strain)
o 4RB 45° (+ strain)
zy 4RB Lat. (- strain)
120 160 200 -6 I Strain x 10 in in
240 280 1\) ())
_Fig. 16. Load Through Shear Center Versus Strain 500 \
oi 14-00 .0
I o Gage lLL (- strain) r-1
s:: m Gage lLR (- strain_)_ """ H G) .p
300 s:: (J) C,)
H aJ (J) .c: Cll
~ 200 ::s 0 H .c: .p
res aJ 100 .s
-253 0 240 280
Strain x in/in 1\)
\0
. Fig. 17. . Load Through Shea~ Center Versus Strain 500~--------------------------------------------------------------~
. . . . 11l 400 J:l
·r-4
s::: oM
~ Q)
+> s:: 300 a> <:.>
~ cd Q) ~ ell
fa 200 l :::s /I / I Q Gage 3LR (- strain) 0
J In 1£ I l!J Gage 2LR (- strain) ~ ..c: +> 'd
~ 100
0 40 Bo 120 160 200 24'0 280
-6 I Strain x 10 in in w 0
Fig. 18. ·Load Through Shear Center Versus Strain 500---------------------------------------------------------,
ol 400 .0 r-i
s:: '" H Q) ~
s:: 300 CD C)
H as CD ..c: til
~ 200j i' I r.l"/ I 0 Gage 3LL (- strain) ;:s I J 1// I 0 E!l Gage 2LL (- strain) H ..c: .p
'0 as .s 100
-128
0 40 80 120 160 200 240 280 -6 I Strain x 10 in in w ......
Fig. 19· Load Through Shear Center Versus Strain
soor-~------------------------------------------------
/ YJ
• 400 // .
l7l / ~-,0 r-i
~ 0 . ..-i /-~ S-1 Q)
+> 300 /0/'VJ s:: Q)
0
H '{/ «S Q)
..c:: Cll 200 J ~ I ~/-A/ I 0 Gage 2LB (+ strain)
..c::
~ / b()
~ 'f) Gage 1LB ( + strain) ::s
0 S.. C!l Gage 3LB ( + strain) .c /
+:>
~ 100 0 H
+76 +120 I +14 240 120 160 200
-6 Strain x 10 in/in
0 40 80 280 w 1\)
Fig. 2Q Stress Versus Distance from Fixed End of Channel Beam
~Boo 0 ..-; I '&... Br-1--A m m ~ 700vl ~
-~ ......... I I + 0 ·:6001 ~ ~ I c I D
Me m ~ ' r T p Pt -,:: 500 ..-;
Stress at edge A 0 l1l
Stress at edge B r1l 400 t!J (I) U Stress at center f.-1
.4-)
of flange r1l
n 300 - - -Theoretfcal stress ro ,::
...-1 'd 200 ~ +> ...-1 bO c 100 0 H
0 10 20 30 40 50 60 70 80 90 100 110 120
Distance from fixed end of channel beam in inches
w w
VI. LONGITUDINAL STRESS IN CHANNEL BEAM WHEN THE TRANSVERSE LOAD PRODUCES TWISTING AS WELL AS BENDING
A. Theory According to Seely and Smith
Let·a channel beam be loaded through the centroid.
This load may be resolved into an equal load through the
shear center and a twisting couple whose moment is Pm. See
Figure 21. The load through the shear center produces bend
ing without twisting, as stated in the preceding section,
and the bending moment due to this load is held in equili
brium at any section by a resisting moment produced by the
longitudinal stresses as given by the simple flexure formu-
la. The external twisting moment Pm also develops shearing
stresses that hold the external twisting moment in equili-
brium.
The shearing stresses producing this resisting moment
develop quite differently on sections near the restrained
end than on sections near the free end of the channel beam.
The sections near the free end of the beam can twist because
they are free to warp. When a channel section is free to
warp, the twisting moment does not appreciably affect the
longitudinal stresses in the beam near the free end but
merely produces shearing stresses on the section as shown
in Figure 22.
If a section is restrained from warping, as in the
35
Shear center
Fig. 21. Load "P" Applied Through Centroid of Cross Section
Lateral shear
Both lateral shear ahd torsional shear
Torsional shear
Fig. 22. General Torsional Effect of Twisting Moment on Channel Beam
36
37
case of the fixed end of a cantilever beam, the twisting
moment is transmitted by lateral shearing forces which ac
company the lateral bending of the flange; these stresses
are not negligible near the restrained section. This lat
eral bending of each flange causes a longitudinal tensile
stress along one edge and a compressive stress along the
other edge. These stresses must be added algebraically to
the longitudinal stress caused by the vertical bending load
to obtain the total longitudinal stress on the edge.
The relation between the angle of twist and the dis
tance from the fixed end of the beam (Fig. 33) shows clear
ly that the effect of restraining a section from warping
and hence from twisting extends a short distance "a" from
the restrained section. Beyond this section "a" the remain
der of the channel beam twists approximately a constant
value per unit length. At some distance from the restrained
end as shown in section b, Figure 22, the twisting moment
Pm is transmitted along the member in two ways. First,
by a twisting moment T1 , produced by the lateral shearing
forces at a distance (h) between the flange centroids.
This component of the total resisting moment has a magni
tude equal to Vh. Pure torsional shearing stresses are
also acting on the cross section. The resisting moment
resulting from this stress distribution is equal to GJ~, as
determined by use of the membrane analogy. The term G
represents the shearing modulus of elasticity of the mater
ial, J is an equivalent polar moment of inertia, and $ re
presents the angle of twist per unit of length. Since there
are two unknowns, V and ~' two equations are needed. They
are the equilibrium equation for moments about the axis of
twist
Pm = Vh + GJ~ (1)
The lateral shearing force (V) can be expressed in
terms of the unit angle of twlst (<P) by making use of the
elastic.curve equation for lateral bending of the flanges.
This second equation becomes
- M (2)
where M is the lateral bending moment in the flange, E is
the modulus of elasticity of the material, y is the lateral
deflection, and I is the moment of inertia of the entire y cross section of the beam with respect to the axis of sym-
1 metry in the web so that 2 IY closely approximates the value
of the moment of inertia of one flange.
Since small angles of twist are involved, the lateral
deflection of the flange can be expressed as
. h Y=2-& ( 3)
Differentiating equation 3 twice with respect to x gives
dB-and since dx = ~, equation 2 may be written
E~yh d~ = _ M dx
(4)
(5)
39
dM Since V = dx' equation 5, after both sides are differenti-
ated with respect to x, may be written as
(6)
Substituting this value of V into equation 1 gives
(7)
For convenience let
h J~~y a=21fJ (8)
Equation 7 may be written
d2 ¢ Pm a2-- ¢ =--dx2 , GJ
(9)
40
The solution of this second order, linear, differen
tial equation is obtained in the following manner.
(10)
First, obtain the complementary solution of equation 10 by
letting
The roots of equation 11 are
D = + 1 - a
(11)
Therefore the complementary solution of equation 10 is
¢c = A sinh ~ + B cosh ~ a a (12)
where A and Bare arbitrary constants.
The particular solution is obtained by assuming V is
a constant and substituting this into equation 9 to deter
mine the value of the constant. The result is
Pm (J>p = GJ (13)
Therefore, the complete solution of equation 9 is the sum
of the complementary and particular solutions. Therefore,
X X Prri (f) = A sinh a + B cosh a + G:f ( 14)
41
Equation 14 expresses the angle of twist per unit length as
a function of the distance from the fixed end of the sec-
tion.
The arbitrary constants, A and B, can be determined
by using two boundary conditions. They are
1) de
X = 0 dx= ~ = 0
2) X = L d2y
= 0 ~2
~he value of A anq B are determined and are substituted into
equation 14 which gives the angle of twist per unit length.
The result is
Pm[ cosh (L- x){a~ ¢ = OJ 1 - . cosh (L/a ~
The total angle of twist at the free end is
~= ~ [ J Pm L
0 dx = GJ L - a tanh a
(15)
(16)
The twisting moment T2 at any section of the beam is
obtained by substituting the value of ¢ from equation 15
into equation 1 which gives
GJm = Pm[l cosh(L- x~/aJ T2 = ~ - cosh(L/a (17)
The late~al bending moment M in the flange~ of the beam at
any section is obtained by taking the derivative of
42
equation 15 with respect to x and substituting it into equa
tion 5, which gives
M =.;. Pm a sinh(L- x~/a h cosh(L/a ( 18)
Assuming that the lateral bending of each flange is
in accordance with the flexure formula, and that each flange
has a rectangular cross section, the stress at the edge of
the flange is
32 = M 1/2 b If
Substituting equation 18 into equation 19,
3 _ Pm a [sinh(L - xj/a] lf2b 2 - 11 cosh(L/a J f
(19)
(20)
where If is the moment of inertia of the flange about its
centroidal axis, m is the distance from the shear center to
the application of the load, and b is the mean width of the
flange.
Therefore the longitudinal stresses in the edges of
the channel beam having one section restrained may be found
approximately by adding algebraically the stresses due to
pure bending and pure twisting. The total stress in the
edge of the beam is
(21)
= P(L - x)l/2h + ~ [sinh(L - x)/a] l/2b I h a cosh(L/a) j If
B. Theory Developed by Timoshenko
In the derivation for a channel beam by Timoshenko,
he found that the resisting torque produced by the shearing
forces in the flanges was
(22)
as compared to
(23)
from Seely and Smith.
The difference between the two equations is that Tim
oshenko considered the additional stiffness of the web at
the fixed end where Seely and Smith imply that the shearing
stresses in the web at the fixed end is a function of the
angle of twist ~-
The total resisting torque was evaluated to be
T = Pm = GJ<b - f 1 + 1 -· EI h 2 ~ t h3Jd2~ 2 4! dx2
(24)
44 .
where
a2 = Eirh2 [ t 1hj
2GJ 1 + 4r (25)
as compared to
T = Pm = GJ~ ~Iyh2 Jd2~ 4 dx2 (26)
where
(27)
from Seely and Smith!
ln the derivation of the stress equation by Seely and
Smith, it is stated that Iy is the moment of inertia of the
entire cross section with respect to a centroidal axis par
allel to the web, and that Iy/2 closely approximates the
value of the moment of inertia of a flange cross section.
This statement is true if the contribution of the moment of
inertia of the web for an I beam is ignored, but it is not
true if the inertia of a channel beam about the centroidal
axis is considered.
Timoshenko considers the stiffness of the web at the
fixed end in his derivation of the stress equation for a
channel beam. He concludes that the stress equation derived
for an I beam can also be used for a channel if the quantity
a 2 given by ·.equation 25 is used for the warping constant.
Seely and Smith conclude that the stress equation for
45
an I beam can also be used for a channel beam. At the fixed
end of the I beam, they conclude that the entire twisting
moment is transmitted by means of the lateral shearing for
ces in the flange. This shearing force caused ench flange
extending a distance "a" from the fixed end to bend later
ally, thus producing a longitudinal stress at each edge.
Seely and Smith are very vague as to the statements
they make in their text and for this reason the author chose
to investigate the torsional stress induced in a channel
beam, and to verify their value of "a".
46
VII. OBSERVED VALUES OF LONGITUDINAL STRESS IN CHANNEL BEAM WHEN LOADED THROUGH THE CENTROID
Again the strains were measured for each 50 pound load
increment in the same mann0r as for the shear center. These
strains were plotted against each increment of vertical load
applied through the centroid as shown in Figures 23-32. At
a load of 250 pounds the strains were determined from the
r:;T'aphs :1.nd these strains were converted into stres:>~s wi t.h
the aid of a computer.
Curves of these stresses versus the distance from the
fixed end of the channel beam are shown in Figures 34-36.
Figure 34 is the observed value of longitudinal stress
versus Seely and Smith's theoretical value of longitudinal
stress.
Figure 35 is the observed value of longitudinal stress
versus Timoshenko's theoretical value of longitudinal stress.
Both Seely and Smith's and Timoshenko's torsional
stress equations were programed to be used in the computer
to determine the value of stresses at each edge of the beam
at 5 inch increments along the length of the channel beam.
The only difference between Seely and Smith's theore
tical stress equation and Timoshenko's is the warping con
stant "a". For the particular cross section of the channel
beam, "a" = 13.7 in. from Seely and Smith (Eq. 8) versus
14.1 in. from Timoshenko (Eq. 25).
47
Figure 36 shows the relati.onship between the observed
longi t11dinal ntreases and the theoretical value calculated
by using the value of "a" = 10 in. which was experimentally
determined from Figure 33.
450
. 400 t1l .0 .-1
s:: 350 ..-t
res ..-t 300 0 S... +> s:: Q) 250 0
..c: bO =' 200 0 H ..c: +> res 150 ro 3 loo
50
0 50
Fig. 23. Load Through Centroid Versus Strain
100 200 -6
Strain x 10 in/in
o lRT Long(- strain)
~ lRT Lat.(+ strain)
V lRT 45° (- strain)
-225 250 300
~ CX>
l7l .c r-1
s:: oM
'0 oM 0 H +> s:: Q) C)
..c:: bO ::s 0 H ..c:: +> '0 cd 0
....:!
F 24. Load Through Centroid Versus Strain 500~-----------
400
300
200
100 /
0 50 100
/0 ~ /e
/' 0 2RT Long.(- strain)
c:l 2RT 45°
TQ 2RT Lat.
-222
150 200 250 -6
Strain x 10 in/in
(- strain)
(+ strain)
300 350 +=" \0
Fig. 25. Load Through Centroid Versus Strain
500 I
·4ocl . I'll .0 r-i
s:: ..-1
't:S 300 ..-i
0 H ..j.)
s:: <1> C)
J 2001
't:S
100~ :j 0 ..:I
0
~I _/ ~' I ·-~
'
1/~~ c:::J 3RT 45 ( - strain) e 3RT LoJg.(- strain)
40 80
'V 3RT La ... ( + strain)
-123
120 160
-6 Strain x 10 in/in
-214
200 240 280 \J1 0
Fig. 26. Load Through Centroid Versus Strain 500~------------------------------------------------------------~
. 400 Ul .0 ,..-t
~ ..... 'tj
"ci 300 H +> ~ C1> 0
~ ::s 200 0 H
..c:: ..,_,
'tj aS s
100
0 40
[!)
I c::J
I I
[U
+51
80
~/ /~
120 160
0 4RT Long.(- strain) c::J 4RT Lat. (+ strain)
V 4RT 45° (- strain)
-167
200 240 -6 Strain x 10 in/in
280 \11 ......
Fig. 2~ Load Through Centroid Versus Strain
500~--------------------------------------------------------~
400 m .a &-i
s::: ..-f
'tj
'8 . 300 H .p s::: Q) C)
.c: til ::s 200 0 H .s:: .p
't:S . cd
0 H
100
~,
~
0 40 80
0 5RT Long.(- strain) \'J 5RT Lat.
m 5RT 45°
-120
120 160 200
-6 Strain x 10 in/in
(+ strain)
(- strain)
240 280 Vl r\)
Fig. 28. Load Through Centroid Vers:1s Strain
500~-------------------------------------------------------,
400 . rll
,.Q rl
s:::: ....-i
'0 300 ....-i 0 H .p s:::: (J) C)
..c
I 200 j # f" i I .
/ 0 ..:I 100
-73lf -76
0 40 80
o 6RT Long.(- strain)
D 6RT Lat. (+ strain)
"fJ 6RT 45° (- strain)
120 160 200 -6
Strain x 10 in/in
240 280 \J1 w
C1l .0 r-1
s:: ..-I
'0 ..-I 0 H .p s:: Q) Q
§ :::s 0 ~ .r:: .p
'tj m 0
...:I
Fig. 2~ Load Through Centroid Versus Strain .·,
500~----------------------------------------------------------~
400
300
200
l £~0 I I \/ 1LL (- strain)
1oo I L'~/ I I 0 1LR (- strain)
-190
0 50 100 150 200 250 300 350
-6 I Strain x 10 in in \Jl ..j:;'
500
400 . Ill ..0 r-1
~ ..-i
'0 300 ..-i
0 H ~ ~ Q) 0
~ 200 0 H
..c:
I ~
'0 ro 0 H 100
0
Fig. 30. Load Through Centroid Versus Strain
I v ~ Gage 3LR (- strain)
Gage 2LR (- strain) m
-40 -80 -120 -160 -200 -240 -6 I Strain x 10 in in
-280 \Jl \Jl
Fig. 31. Load Th~ough Centroid Versus Strain
500~----------------------------------~~--------------------~
400 . l1l .0 r-1
~ ..-1
'0 300 ..-1 0 H
.-1-)
~ (\) C)
..c! ~ 200 0 '/' j 'o H Gage 3RR (- strain) ..c! .-1-)
~\71 '0 I 'fJ Gage 2RR (- strain) m 0 H 100
-80 -120
0 -40 -80 -120 -160 -200 -240 -280 -6 I Strain x 10 in in
\.11 m
Fig. 32. Load Through Centroid Versus Strain 500T---------------------------~--------------~---.r-----------
. 400 l'1l ..0 r-f
s:: oM
't:l oM 0 300 H·
4-)
s:: Q.) 0
§ ~ 0 200 H .c 4-)
't:l J r/ ~ I I 0 3LB (+ strain) ro
t!l 2LB (+ strain) 0
100 ~ If ..:I
VI 1LB (+ strain)
+87 I +132
0 50 100 150 200 250 300 350 -6
Strain x 10 in/in \J1 ~
. 016
11.1
.0141
s:: ro :a . 012"1 ro H
!=1 • 010
~ I 11.1 ~ .008 ~
C+-1 I 0 .006 (1.) n bO s:: • 004 <
.0021
0
Fig. 33 . Angle of Twist Versus Distance from Fixed End
I "-.. "-.. Tangent curves
"' "'' ...-.250 lbs.
" ""'""' / ~200 lbs.
~ / ~ .-150 lbs.
6-in. 8.2-lbs. p
channel -~ _.,.,., ~ __.100 lbs.
~'/~~ -- 50 lbs.
-... -----~
I ---I
10 20 3) 40 50 60 Distance from fixed end of cantilever in inches
\]1 CX>
§ 8000 ..-i I'll I'll
~ 7000 ~ 0 0 6000
..-i I ttl p. ~ 5000
..-i
I'll
~ 4000 H
..p I ttl
r-1 3000 I ro s::: ..-i 'g 2000-1 ..p ..-i I bO § 1000 H
o·
Fig. 34. Stress Versus Distance from Fixed End
0 Stress in edge A t::l S:tress ip edge B
' - -Stress at center of flange
0 Corrected stress at edge A e Corrected stress at edge B y Corrected stress at center
c of flange Theoretical value of stress
B II A
-1~--
c .-n p
10 20 30 40 50 60 70 80 90 100 110 120 Distance from fixed end in inches
\11 \0
§ 8000
'" Ul Ul
~ 7000
~ 0 () 6000
'" Ul 0. s:: 5000
....-!
Cll
~ 4000 ~ ~ Ul
r-i 3000 m s:: '" 'g 2000 ~ ....-! bO s 1000
0
[:::1
B
c
10
Fig. 35. Stress Versus Distance from Fixed End
0 Stress in edge A
c:J Stress it: edge B VI Stress at center of flange e Corrected stress at edge A ~ Corrected stress at edge B
• Corrected stress at center .......... of flange
---Theoretical value of stress
A
D
20 30 40 50 60 70 80 90 100 110 120 Distance from fixed end in inches
0"1 0
§ Boo .,; t1l I'll i 7000
0 0 6000 .,; t1l Pc
s:: 500 or-i .
ril rn <1> 400 H .p rn
,..-f cd s:: .,; '0 ::3 .p or-i bO s:: ,s 100
0
p
10
Fig. 36. Stress Versus Distance from Fixed End
B
A
A
D
20 30 40 50 60
0 Stress in edge A m Stress in edge B V Stress at· center of flange e Corrected stress at edge A 21 Corrected stress at edge B
Y Corrected stress at center of flange
-Theoretical value of stress
70 80 90 100 110 120 Distance from fixed end in inches
0'\ .....
62
VIII. DISCUSSION OF LOADING THROUGH THE SHEAR CENTER
It was stated in the introduction that in order to
cause a channel beam to bend without twisting and to devel-
op stresses in accordance with the simple flexure formula
for beams, the load must be applied through the shear cen
ter or center of rotation~ This was accomplished by posi
tioning the vertical load at the free end of the beam so
that the inclination of the level bars (Figure 3) was zero~
For varying yalues of load the strains were read and graphs
were drawn as shown in Figures 8-19.
For a load of 250 pounds, the strains were determined
from the graphs, and these values of strains were placed in
a mathematical program so that strain could be converted
into stress at each gage. See Figure 4.
The curves showing the relation between the longitu
dinal stresses along the edges of the channel beam and the
distance from the fixed section of the beam are given in
Figure 20, when the load is applied through the shear cen
ter as found in Figure 6~ The broken lines represent the Me
value of stress obtained from the flexure formula S = ][•
Since the values of stress on either side of the chan
nel beam are approximately linear (Figure 20) it can be
stated that the loading apparatus must have applied a
63
transverse load to the channel beam at the same time the
vertical load was applied. This transverse load produces
additional compressive stresses beyond that of the theore-
tical value at gages lLR, 2LR and 3LR of magnitude 1300 psi,
760 psi and 440 psi respectively. At gages lLL, 2LL and
3LL the additional tensile stresses produced by this trans
verse load are 625 psi, 350 psi and 200 psi respectively.
If the possibility of a transverse load caused by misalign-
ment of the loading apparatus is assumed, and if the chan-
nel beam does bend about the 11 Y11 axis, then the transverse
load can be determined as follows.
At a distance of 60 inches from the fixed end(Fig. 20)
the value of stress at point A is SA = 3850 psi C where at
point B the value of stress is SB = 2700 psi C. It will
also be assumed that the stress variation across the flange
is linear and that it has a value of stress of 2700 psi C
at the left side and a value of 3850 psi C on the right side.
Plotting this stress distribution across the flange and de-
termining where this curve crosses the theoretical curve
Me 11 p 11 b d t i d s = ~, the transverse load can e e erm ne .
SB=2700 psi
760 psi
T-A=3850 psi
S=Mc/I =3090 si
l_ ~---1.-..1..-.1--'----L--
l-- 1.75"--1 Fig. 37. Distribution of Stress Across the Flange
390 = 760 z 1.75-z z = 0.59 inches
64
This value of z approximately equals the distance from
the back of the web to the centroid of the entire cross sec-
tion.
The transverse load that is necessary to produce this
stress distribution
Me P(L- x)(b - z) S = :r; = ry
P(ll3.25 - 60)(1.75 - .59) = 0. 7 760 psi
is P = 10.5 lbs.
65
Therefore a transverse force of 10.5 lbs. would com
pensate for the difference in the theoretical value and the
observed value of stress.
This transverse load is caused by not having the cen
ter line of the web parallel to the line of action of the
load. This situation can be produced by the combination of
two things; first, by not placing the stiffened end between
the heads of the testing machine correctly, and second, by
not having the jack properly leveled. If the web of the
channel beam is initially inclined at an angle of 2.4 de
grees from the line of action of the load., the transverse
component of load would be approximately 10.5 lbs. when a
vertical load of 250 lbs. is applied to the free end.
66
IX. DISCUSSION OF LOADING THROUGH CENTROID
When a vertical load is applied through the centroid
of the cross section, the channel beam twists as it bends.
For each increment of load the inclination of the level
bars were read and the results are shown in Figure 33.
It was stated in the derivation for the stress equa
tion for an -I beam that ;..;ccLJons near the free end ol' the
I beam warp as they twist and that the longitudinal stress
es are only produced from the bending moment about the axis
of symmetry. For each curve of theoretical stress versus
the observed value of longitudinal stress, there is a dis
tinct difference between the curves near the free end. If
the stresses are observed a little closer, the stresses at
lLR, 2LR and 3LR are considerably higher than the theore
tical values, and the stresses at 1LL,2LL and 3LL are con
siderably lower than the theoretical values.
The author assumed that the same transverse load that
affected the shear center test was the cause for the error
when loading the channel beam through the centroid. From
Figures 34-36 the value of stresses at gage lLR, 2LR and
3LR are 6900 psi C, 3840 psi C and 2400 psi C respectively,
and stresses at gage lLL, 2LL and 3LL are 5700 psi C, 2640
psi C and 1320 psi C respectively. From theory, the value
67
of stresses at gage 2LL and 2LR, 3LL and 3LR should be 3090
psi C and 1700 psi C respectively.
If the transverse load of 10.5 lbs. is used to correct
the value of stresses at these points on the channel beam,
it is shown in Figures 34-36 that with these corrections
the observed value of longitudinal stress falls close to
the theoretical values near the free end.
At the fixed end of the channel beam the value of
stress at gage lLL and lLR should read 6455 psi C and 5485
psi C respectively with the correction factor applied to
these readings. It will be assumed that the stress distri-
bution across the flange is linear.
Plotting the stress distribution across the flange at
a distance of 6.125 inches from the fixed end the location
psi
Fig. 38. Distribution of Stress Across the Flange
68
of the neutral axis can be determined for lateral bending
of the flange. Subtracting the stress distribution proMe duced by S = ][ from SA and SB, the stress components pro-
duced by lateral bending on the right side and left side
are 700 psi and 270 psi respectively. By using proportions
the value of z can be determined.
285 = 700 z 1.75-z
z = 0.516 inches
This value of z falls very near to the distance from the
back of the web to the centroid of the cross section.
69
X~ CONCLUSION
The purpose of this thesis was threefold as was stated
in the scope. The first objective was to investigate the
applicability of using the torsional equation, which was
derived for a cantilevered I beam by Seely and Smith, for
a cantilevered channel beam. The author believes that Seely
and 'Smith's torsional equation can be used with a reasonable
degree of accuracy for a 6 inch - 8.2 lb. channel beam.
This same conclusion can be drawn for Timoshenko's torsion-
al stress equation since the two equations only differ by_a
numerical value for "a" which are approximately equal to
each other.
The second objective was to check on the exactness of
the warping constant "a" used in the torsional equation
which was derived for an I beam. The warping constant "a"
found experimentally falls very close to the value of "a"
derived by Seely and Smith and also by Timoshenko.
The third objective was to determine the axis about
which bending takes place when the torsional loads are ap
plied~ When the channel beam was loaded through the cen
troid, the observed values of longitudinal stress fell very
close to the theoretically predicted values by Seely and
Smith and by Timoshenko. See Figures 34J 36. If the stress Me
produced by S = j[J is subtracted from the stress calculated
70
from the strain readings at gages ILL and lLR, the results
would be the stress distribution across the flange due to
the torsional load only. It is shown in the discussion
that the neutral axis for the stress distribution across
the flange falls very close to the cent·roidal axis of the
entire cross section parallel to the web of the channel
beam.
1~ TIMOSHENKO, S •
2. NAVIER (1864)
BIBLIOGRAPHY
and J. N. GOODIER (1951) Theory of Elasticity. 2nd ed., McGraw-Hill Book Co., Inc., New York. p. 258-304, 228, 229. '
I I "Resume des lecons sur l 1application de la mecanique," 3rd ed., Paris, edited by St. Venant.
71
3. BORG, S~ F. and J. J. GENNARO (1960) Advanced Structure Analysis. D. Van Nostrand Company, Inc., Princeton, New Jersey. p. 218-242.
4. BACH, C. (1909) Zeit. d. Vereins deutscher Ingenieure. p. 382, 1790, 1910.
5. MAILLART, R. A. and A. EGGENSCHWYLER (1921) Schweiz, Bauz. Vol. 77, p. 195; (1920) Vol. 76, p. 266.
6. ZIMMERMAN, H. (1925) Bauingenieur. Vol. 6, p. 455.
7. SEELY, F~ B. and W. J. PUTNAM (July 1930) "The Torsional Effect of Transverse Bending Loads on Channel Beams", Engineering Experiment Station Bulletin, No. 211, University of Illinois.
8. TIMOSHENKO, S. (1956) Strength of Materials. Part I
9.
and Part II, 3rd ed., D. Van Nostrand Co. Inc., Princeton, New Jersey. p. 235-291.
SEELY, F. B. and J. O. SMITH (1952) Advanced Mechanics of Materials. 2nd ed., John Wiley and Sons, Inc., New York. p. 97-136, 266-294.
10. HIGDON, A., E~ H. OHLSEN and W. B. STILES (1960) Mechanics of Materials. John Wiley and Sons, Inc., New York. p. 210-212, 224-226.
11~ ADAMS, D. F. (1963) "An Analysis of the Torsional Rigidity of an Intermittently Stiffened I Beam", Department of Theoretical and Applied Mechanics, University of Illinois.
72
BIBLIOGRAPHY {Continued)
12. HEINS 1 C~ P~ and P. A~ Seaburg (1963) Torsional Analysis of Rolled Steel Sections. Bethlehem Steel Corporation, Bethlehem 1 Pa~
73
VITA
David M~ Schaeffer was born on May 28, 1942 in St~
Charles, Missouri. He received his primary education in
Portage des Sioux, Missouri and completed his secondary
education in St. Charles, Missouri. He has received his
college education from the Missouri School of Mines and
Metallurgy in Rolla, Missouri. He received a Bachelor of
Science Degree in Civil Engineering in June, 1964.
He has attended the University of Missouri at Rolla
since June, 1964 in pursuit of a Master of Science Degree
in Engineering Mechanics.