Nov. 6, 1897.] 
FOREST AND STREAM. 
STB 
Weights for equal ultimate tensile strength, aluminum 1, 
steel 2. 
(fo) Elastic strength in tension.— Steel, 30,0001bs. per square 
inch. Aluminum, 28,0001bs. per square inch. Ratio 14 to 15, 
or 1 to 1.07. 
Cross section for equal elastic strength in tension, see Figs. 
6 and 8. 
Weights.— For equal elastic strengths in tension, alumi- 
num, 0.356, steel 1, or 1 to 2.8. 
2. In compression. 
(a) Ultimate strength of compression.— Steel, 60,000; alumi- 
num, 34,000. Ratio, 1 to 1.76. 
Weights.— For equal ultimate strengths of compression, 
alummum 1, steel 1.7, 
(b) Elastic strength in compression.— Steel, 30,0001bs. per 
square inch. Aluminum, 26,0001bs. per square inch. Ratio, 
1 to 1.15. 
Weights.— For equal elastic strengths of compression, alu- 
minum 1, steel 2.6. 
3. Ultimate strength in shear.— Steel, 45,0001bs. per square 
inch. Aluminum, 28,0001bs. per square inch. Ratio, 1 
to 1.6. 
Weights.— For equal ultimate strengths of shear, alumi- 
num 1, steel 1.875. 
Double riveted joints.— Steel, 45,0001bs. per square inch. 
Aluminum, 24,0001bs. per square inch. Ratio, 1 to 1.875. 
Summing up, for simple resistance, the comparison gives 
the following results: 
1. For ultimate tensile strength, mild steel is half again 
as strong as the best aluminum alloy, but at the same time 
it is three times as heavy; in consequence, the same strength 
can be furnished by aluminum with half the weight that 
steel would require, or, with the same weight, aluminum 
would furnish double the strength. 
2. For elastic strength in tension, aluminum is fourteen- 
flfteenths as strong as steel, and can supply the same elastic 
strength with but 0.86 of the weight required by steel; or, 
with the same weight, aluminum would furnish 2.8 times 
the strength. 
3. For ultimate strength of compression, steel is 1.76 times 
as strong as aluminum, but to furnish the same strength 
would be 1.7 times as heavy. 
4. For elastic strength of compression, aluminum is thir- 
teen fifteenths as strong as steel, and to furnish the same 
strength steel would be 2.6 times as henvy. 
5. For ultimate strength in shear, steel is 1.6 times as 
strong as aluminum, and would be 1.875 times as heavy. 
When the shear is that of a riveted joint, where the riveting 
is double, steel is 1.875 times as strong. The lower perform- 
I 
mm 
FIG 18. Sections for Equal Bending Moments. Fiber Under Com- 
pression Worked to Alternate Strenglb. 
FIG. 14. Sections for Equal Bending Moments. Fiber Under Com- 
pression Worked to Elastic Limit. 
ance of the double riveted joint would indicate for aluminum 
a less perfect cooperation among the rivets. As will be seen 
later, this is no doubt due to the very small elongation in 
the aluminum, which is no doubt the case for shear as it is 
for tension, causing the shearing more in detail, the rivets 
most strained giving way before there is sufficient elonga^ 
tion to permit the full concurrence of the rivets less strained. 
6. In sum, for simple resistance, from the standpoint of 
strength and weight alone, aluminum has pronounced ad- 
vantages over steel, advantages that are nearly double for 
ultimate resistance, and nearly treble for elastic resistance. 
This result emphasizes a striking feature of aluminum, 
namely, the very large proportion of its total strength that 
18 elastic. While this proportion for steel is half, for alum- 
inum it is seven-tenths. 
The consequences of this remarkable property will be seen 
below in comparison of resistance to dynamic forces; it will 
suffice for the moment to refer simply to the fact that the 
vast bulk of structural resistance must be elastic. 
B.— COMPARISON FOB COMPOTJKD EESISTAJJCE. 
For the present purpose, resistance to torsion need not be 
considered; moreover, there is lack of reliable data for the 
resistance of aluminum to torsion, and comparison would re- 
quire an assumption of strength based upon the resistance 
to shearing. The comparison will therefore be limited to 
beading. 
Recalling the formula, the bending moment, 
s being the stress per unit area of the fiber most strained I 
the moment of inertia of the cross section, and h the 
distance of the fiber most strained from the neutral axis 
which passes through the center of gravity of the cross 
section. 
Where the material has a different resistance for tension 
and compression, it is evident that with symmetrical cro=s 
bections, the fiber under the stress of the lesser kind of re- 
astance will give way first, and would impose the limit to 
the bending moment. 
It is also evident that the resistance of a given cross sec- 
tion for such a material is a maximum when the area is so 
distributed that the extreme fibers of tension and compres- 
sion are distant from the neutral axis in the direct ratio of 
their respective resistances. When so distributed, the bend- 
ing moment, M={8c+st) — • where sc is the maximum re- 
n. 
aistance to compression, st the maximum resistance to ten- 
sion, and h the entire depth of the beam. 
■ _ In the cases of aluminum for ultimate resistance the re- 
sistance to tension being 40.0001bs. per square inch, and the 
resistance to compression 34,0001bs. per square inch the' 
symmetncal section would have its limit imposed by the 
fiber under compression, and the ratio of the resistance of 
Buch a section to the resistance of a section designed to give 
M 
a maximum resistance, if the moment of inertia is the 
34000 _34 
same, is 40000 + 34000 37- For elastic resistance the ratiols: 
2 
26000 
28000 + 26000 
26 
'27. 
With steel, however, the two resistances are the same and 
the symmetrical section realizes the maximum resistance. 
^7 27 
The ratio_U 1.09 and 11 =1.04 would represent the ad- 
34 26 
vantage to be derived from the best design for the alumi- 
num section, provided the moment of inertia did not change. 
It is evident, however, that the transference of metal from 
the side of tension to the side of compression will result in 
placing this metal nearer the neutral axis, and in conse- 
quence will reduce the moment of inertia. The two ratios, 
therefore, while remaining appreciably greater than unity, 
will be less than those indicated, 1.09 and 1.04, according to 
the form of the section. 
Bearing in mind this small but appreciable discrimination 
against aluminum, it will suflice, for the sake of simplicity, 
to compare only symmetrical sections. 
1. Comparison of usual sections designed for resistance 
other than bending. 
(a) The case of square cross sections designed for tension 
or compression. 
The ratio of bending moments, from equation (1), is: 
M' s' I' h 
M 
s I h' 
(2) 
the letters with the affixes denoting aluminum. 
If the side of the aluminum section is n times the side of 
the steel section, h'= nh, or ^ = - • 
h' n 
The areas of the cross sections being in the ratio = 112 and 
, • h'^ h'^ 
the radii of gyration squared in the ratio-^=ii2, the mo" 
h'* I' 
ments of mertia are in the ratio"^ = n* and Y = 
M 
s n s 
(3) 
Take first the case illustrated in Fig. 5, of (1) section de- 
^^SP|*?f°^ ^<l^^^ ultimate tensile strength, designating by S 
and b the two^ ultimate tensile strengths. 
, h'» S S 
h -b = h-iS or ^-or = g7 and n 
Substituting in equation (3), {—^^ 
Suppose first that the cross sections are worked to their 
limit of resistance to bending. It is the extreme compressed 
' nAmu alumintim sections which will break first when 
s =ci4,0001bs. per square inch. The fibers of the steel section 
resist equally for tension and compression, both attaining 
60,0001bs. per square inch, so that - = The two ulti- 
, , ., s 60000 
mate tensile strengths S and S' being 60,000 and 40,000 re- 
snectivPlv- =?°500«n^ ^'-34000 /60000U , 
spectively, y, ^^^^ and-^ » — x (^) ^= 1.04. 
Suppose next that the cross sections are worked only to 
the elastic limit of the most strained fibers. 
The limit for the aluminum cross section is again imposed 
bythe fibers of compression where s'=26000. 
For the steel section the limit being the same for both 
fibers, s = 30,000 ^' ^^^^ 
c ^ r.,-, . s 30000 
S and S being the same as before, 
M' 26000 ^ /60000\f 
M "30600 \40000^ 
■1.6. 
It does not suffice m general to know simply the amount 
of resistance to bending. It is also necessary to know the 
effect of the bending moment, the amount of deformation 
or deflection produced. It is therefore necessary to calcu- 
late the resistance to deformation, to know the stiffness 
i he measure of stiffness is the amount of bending or the de- 
flection pr^uced by a given force. The amount of deflec- 
tion, f = ^j., where E is the modulus of elasticity, and I the 
moment of inertia of the cross section as above, K being 
constant under the same conditions of bending moment and 
support. Therefore, the conditions being the same 
f E'. r ^' 
Taking tte modulus of elasticity of aluminum as that 
given by Hutte, 10,000,000. and the modulus of st^el as 
30,000,000, 3. As seen above, for the case in question, 
II , . . r 1 
2 25'- Substituting in equation 
-4 , wherein = 1.225, ~ 
f ^ 3 1 
(5),Y 2 075"" Willis, for square cross sections giv- 
ing equal ultimate strengths in tension, the deflection pro- 
ducea by the same force under similar conditions of bending 
is greater in the case of the aluminum section. The square 
bteel bar is one-third the stiffer. 
(2) . Sections designed for equal elastic strength in tension 
illustrated in Fig. 6. ' 
By a similar process of reasoning, 
M' 34000 / 30000 \f _ 
M 60000 \28000/ ' 
when worked to the ultimate limit. 
M'_ 26000 ^ /30000\f_ ^.q 
M 30000 \28000/ ' 
when worked to the elastic limit. 
£ 1.143 .381* 
(3) . Sections designed for equal ultimate strength of com- 
pression. 
When worked to the ultimate limit, - ' = 1.328 
M 
When worked to the elastic limit, — = 2 031 
M 
f'_ 1 
t 1.032* 
(4) . Sections designed for equal elastic strength of com- 
pression. 
When worked to the ultimate limit, M = 703 
M 
When worked to the elastic limit, = 1 073 
M 
f .441" 
(5) . Sections of equal weight. 
When worked to the ultimate limit, ~ = 2.94 
M 
When worked to the elastic limit. 
M 
f 3* 
_ ft. The case of rectangular cross sections designed for ten- 
sion plates of a given width. 
Refer to Figs. 7 and 8. 
The areas of the cross sections are bh' and bh respectively, 
b being the width; the radii of gyration squared are in the 
h'2 
ratio of~- Therefore, the moments of inertia are in the 
• h' h'^ h'^ 
ratio of _ X _ = and the bending moments are in the 
n n h'* 
ratio = 8 
h' 
h'2 a' 
X — = — X n- 
h^ s 
When worked to the ultimate limit, 
When worked to the elastic limit. 
■ 1.275 
= 1.94 
1 
M s h« '■ h' s " h^ r " (4) 
(1) . Sections designed for equal ultimate strength. Fig. 7. 
M' 
M 
M' 
M 
1' 
1 1.125" 
(2) . Sections designed for equal elastic strength. Fig. 8. 
When worked to the ultimate limit, — = .648 
M 
When worked to the elastic limit, ^' = .99 
' M 
f^ = _1 
t .408 
(3) . Sections designed for equal weight. 
M' 
M 
M' 
M 
?! _ 1 . 
ing <^o™Parison of sections designed for resistance to bend- 
a. Shapes having equal weight. 
(1). I Beams. Fig. 9. 
St A^ot^ ti^-^^*^^ I i« limited, being the 
I' 
When worked to the ultimate limit. 
When worked to the elastic limit, 
5.1 
= 7.8 
same for both sections 
Ratio of moments of inertia, 
- = 2.36 
When worked to the ultimate limit, ^ 
' M 1.43 
When worked to the elastic limit, ^'=905 
M ■ 
fl=JL_ 
f .78b" 
pr^StiS^ed dfr^eSon^ ^^-^ 
Ratio of moments of inertia, 
I' 
= 9.25 
J. 
When worked to the ultimate limit ^' 
'M 
When worked to the elastic limit ^' = 4 45 
'M 
f 
(2). Angle bars, /. Fig. 10. 
First, when the depth of / is the same. 
I' 
= 2.91 
3.08 
Ratio of moments of inertia, 
- = 4.61 
When worked to the ultimate limit, ^' = 1.56 
'M 
When worked to the elastic limit ^' = 24 ' 
' M 
f „ 1 , 
I 1.54 
Next, when the depth is proportioned 
Ratio of moments of inertia, 
When worked to the ultimate limit, — = 2.35 
M 
When worked to the elastic limit, ^ = 3 59 
iVJ. 
f ' _ 1 , 
I 2.3* 
h. Shapes giving equal bending moments. 
(1) . 1 Beams. Figs. 11 and 12. 
First, when designed from the ultimate limit of resist- 
ance. 
Fig. 11.— When the depth is the same, the ratio of cross 
sections is 1.67, the ratio of weights is 0.553 and *1 = _A_ 
1 59 * 
When the depth is proportioned, the ratio of cross sec- 
tions is 1.39, the ratio of weights is 0.46 and ^ = -J_ 
£ 82' " 
Next, when designed from elastic limit of resistance 
Fig. 12.— The depth being the same, the ratio of cross sec- 
tions is 1.11, the ratio of weights is 0.37 and ~ = ^ 
£ 38 
(2) . Angle bars Figs. 13 and 14. 
When designed'from the ultimate limit, the depth being 
the same, the ratio of cross sections is 1.74, the ratio of 
weights is 0.58 and — = 
1 .58 
When designed from the elastic limit, the depth being the 
same, the ratio of cross sections is 1.19, the ratio of weights 
is 0.396 audi- = JL, 
f .38 
The above comparisons for resistance to bending and for 
stiffness give the following results, namely: 
1. For bars of square section, giving the same ultimate 
tensile strength, the ultimate bending moment that each 
can resist is practically the same, the aluminum bar giving 
1.04 times the resistance of the steel bar, while the bending 
moment before the elastic limit is reached is half as large 
again for aluminum, the ratio being 1.6. On the other 
hand, the steel bar is the stiffer of the two, the deviation or 
flexion produced by the same force being but three-quarters 
tbat of the aluminum bar. Thus the aluminum bar giving 
the same ultimate tensile strength, an equal ultimate resist 
aace to bending, three halves as great an elastic resistance to 
binding, of three-quarters the stiffness, weighs half as much 
as the steel bar. 
2. For bars of square section, giving the same elastic 
strength m tension, the aluminum bar gives 0 6 the ultimate 
resistance to bending and 0.96 the elastic resistance to bend- 
ng, is 0.38 only as stiff, but weighs about a third as much as 
the steel bar, ^ 
SB.Q , 
