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But by means of a large number of experiments upon a given material we
may obtain useful average or minimum values for it, and should in all cases
of practice keep the stresses well within such values; since, if the elastic
limit be exceeded (through miscalculation, or through subsequent increase
in the stress or decrease in the strength of the material) the structure
rapidly fails. The table, p 460, gives approximate average elastic limits
for a few materials. The elastic limit, as here defined, is sometimes called
the * true " elastic limit. Compare t 31.

27. Brittle materials, such as stones, cements, bricks, etc., can scarcely
be said to have an elastic limit; or, if they have, it is almost impossible to
determine it; since rupture, in such bodies, takes place before any stretch
can be satisfactorily measured.

28. A small permanent "set" (stretch) probably takes place in all
cases of stress even under very moderate loads; but ordinarily it first be-
comes noticeable at about the time when the elastic limit is exceeded.
The elastic limit is sometimes defined as that stress at which
the first marked permanent set appears.

29. The elastic ratio of a material is the quotient,
It is usually expressed as a decimal fraction.

The permissible working load of a material should be determined by its
elastic limit rather than by its ultimate strength. Hence, other things
being equal, a high elastic ratio is in general a desirable qualification; but,
on the other hand, it is possible, by modifying the process of manufacture,
to obtain material of high elastic ratio, but deficient in "body" or in resil-
ience i. e., in capacity to resist the effect of blows or shocks, or of sudden
application or fluctuation of stress. See If 34; also Iff 35 etc.

In the manufacture of steel, the elastic ratio is increased by increasing the
reduction of area in hammering or rolling, and the rate of increase of elastic
ratio with reduction of area increases rapidly as the reduction becomes very
great. Kirkaldy found t

for steel plates 1 inch thick, mean elastic ratio = 0.53

" " " M " " " " " = 0.53

" " " y?, " " " " " = 0.54

" M " " " " " = 0.61

* The U. S. Board appointed to test Iron, Steel, &c., found p, variation of
nearly 4000 Ibs. per square inch in the elastic limit of bars of one make of
rolled iron, prepared with great care and having very uniform tensile strength;
and, in another very carefully made iron, a difference of over 30 per cent,
between two bars of the same size. Report, 1881, Vol. 1, p. 31.

t Annual Report of the Secretary of the Navy, Washington, 1885, Vol. I,

. 499; and Merchant Shipping Experiments on Steel, Parliamentary Paper,
. 2897, London, 1881.



C2



460



STRENGTH OF MATERIALS.



SO. Elastic Moduli and Elastic Limits. Approximate averages.f
IS = elastic modulus, in millions of pounds per square inch ;
I = stretch or compression, in ins, in a length of 10 feet, under

a load of 1000 pounds per square inch.
= (10 X 12 X 1,000) +- (1,000,000 E) ;
s = stress at elastic limit, in thousands of pounds per square inch.





E


1


*e


Metals.


10 to 30


012 to 004


4 to 8


" " ordinarily


12 to 15


010 to 008


6 to 7




27 to 31


004


20 to 40


Steel structural*


" to "




34 to 38




8 to 10


015 to 012


5 to 7


** wire


12 to 16


0.010 to 0.007


14 to 18




10 to 14


012 to 009


6 to 7




10 to 14


012 to 009


8 to 12


Lead


8 to 10


150 to 120


1 to 1 2


Tin, cast


6 to 7


0.020 to 0.017


1.4 to 1.6


Bronzes


13 to 15


009 to 008


14 to 15




4 to 8


030 to 015


1 to 2


Masonry} 1


5 to 2


0.240 to 060


Art. 4 (b)


Wood f


1 5 to 2


080 to 060


5 to 7











31. Yield point. Commercial, Relative or Apparent Elas-
tic Limit. In testing specimens of iron and steel, it is commonly found that,
at a stress slightly exceeding the true elastic limit (If 26), the stretch begins
to increase without further increase of load. This point is usually called "the
yield point," or " the elastic limit" in commercial testing. The French Com-
mission on Methods of Testing the Materials of Construction called it the
" apparent elastic limit." The late Prof. J. B. Johnson (" The Materials of Con-
struction," New York, John Wiley & Sons, 1906, p. 19) applied the term, " rela-
tive or apparent elastic limit" to that point on the stress diagram at which the
rate of deformation is 50 per cent, greater than at points below the true elastic
limit.

Resilience.

32. The resilience of a bar, under a stress, s, is the work done, upon
the bar, in producing that stress, or, theoretically, the work which the bar
will do, in regaining its original shape, when relieved from stress. Usually
we are concerned with the elastic resilience, or that corresponding to
the stress, s e at the elastic limit.

33. Let

s e = the unit stress at the elastic limit ;
a = the section area of the bar ;
P e = a s e = the load corresponding to s e ;
L = the original length of the bar ;
I = its stretch, at the elastic limit ;
E = the elastic modulus.



*In rolled iron and steel, the elastic modulus is remarkably constant for all
grades. In wrought iron, the elastic limit depends chiefly upon the degree of
reduction of cross section in rolling; the smaller sizes having the higher elastic
limit. In steel, this effect is less marked.

t See 1HJ 25, 26.

Jin wood, "the extreme fiber stress at the true elastic limit (1J 26) of a beam
is practically identical with the compressive stress endwise of the material,"
table, p. 958. See discussion by S. T. Neely, in "Timber Physics," 1889 to 1898,
by Filibert Roth, House Document No. 181, 55th Congress. 3d Session, Wash-
ington, 1899, p. 374.



GENERAL PRINCIPLES. 461

The work has been done by the mean load, P e /2 = a s e /2, acting thru
the dist, I = L s e /E. Hence,

Resilience = K = P e 1/2 = a s e L s e /2 E = (//2 E) a L.

34. Here s//2 E is the

resilience modulus = resilience of a bar of unit section area and

unit Igth.

The resilience modulus of a material is a measure of its capacity for re-
sisting shocks or blows.

Suddenly applied loads.

35. Let a body, of weight, W, be suspended by a string, and let it just
touch the scale-pan of a spring balance, without depressing it. Now let
the string be cut with a pair of scissors.

36. At the moment of cutting, the spring has not been stretched; its
resisting stress, S, is therefore zero, and the net or resultant downward force,
acting upon the body, is F = W S = W = W.

37. Under the action of this force, the spring stretches, and S increases
proportionally with the stretch. Hence (W remaining constant) the re-
sultant downward or accelerating force, F, acting upon the body, decreases
until S = W, when F = W S = W W = 0.

38. The body, having thus far been constantly accelerated, (by a dimin-
ishing force, F), has constantly increased its velocity. Let h = the height
thru which it has now fallen, and let x be the point reached, at the end of h.

39. Beyond x (W remaining constant, while S continues to increase),
the moving body is acted upon by a constantly increasing, retarding up-
ward force, F = W S, which brings it to rest at a second point, z,
at the end of a second distance = h. Its total fall is therefore 2 h.

40. Let S max = the max value of S, or that at the end, z, of the fall,
2 h. Then, since S has increased proportionally with h, its mean value,
during the fall, 2 h, was S max/2; and the work done, during the entire
fall, 2 h, was 2 W h = (S max/2) 2 h = S max X h. Hence,

S max = 2 W.

41. At the end, z, of the fall, 2 h, the body, having come to rest, is acted
upon by an upward force, F = W S max = W 2W = W; and
(neglecting friction) the same performance is now repeated, but in the up-
ward direction, and so on indefinitely.

42. But losses of energy, due to air resistance and to internal
friction, render each oscillation less than its theoretical value ; and the body
therefore finally comes to rest at the point, x, midway of the fall, 2 h.

43. Thus (1 40), within the elastic limit, a load, suddenly applied
(tho without shock) produces temporarily a stretch nearly equal
to twice that which it could produce if applied gradually :
i.e., twice that which it can maintain after it comes to rest; and develops
temporarily, in the stretched body, a resisting stress = twice the
load.

44. If the load be added in small instalments, each ap-
plied suddenly, then each instalment produces a small temporary stretch,
and afterward maintains a stretch half as great. Under the last small
instalment of load, the spring stretches temporarily to a length greater
than that which the total load can maintain, by an amount equal to half the
small temporary stretch produced by the sudden application of the last
small instalment.



DIAGONAL STRESSES IN BEAMS.



494 a



DIAGONAL STRESSES IN BEAMS.
Maximum Unit Stresses.

104. When a body (as a bolt) is under tensile (or comp) stress

only, the tendency of the body, as regards sections normal to the stress, is to
pull apart (or crush together) in the direction of the stress, or normally to the
section, and the entire stress acts normally upon the section; but, on planes
oblique to the stress, the stress is resolved into two components, one (n)
of tension (or comp) normal to the plane, and one (0 tangential to the
plane (shearing stress).

105. Under shearing 1 stress alone, the effect, upon a plane parallel
to & betw the 2 shearing forces, is pure shear; but, upon planes obliyue
to the forces, the shearing forces are resolved into (0 tangential or shearing
stresses, and (n) normal (tensile or comp) stresses.




Fig. 17.



1O6. Thus, Fig 17, let a bar, of length, L, and depth, D, be subjected to
a tension, S = S', in line with its hor axis, and to two pairs of forces, V=V'
and H = H', as shown; V and V constituting a right-hand vert shear, while
// and H' constitute a left-hand hor shear.

Suppose the bar divided by a section, as N N, F G or K M, and consider
the forces acting, in either case, upon the right-hand segment of the bar as
thus divided.

Upon the normal section, N N, the tension, S, and the hor shear, H, act
normally (S as tension, H as compression), and the vert shear, V, tangen-
tially (as shear); but, for an oblique section, F G or K M, we first resolve
each force, S, V and H, into two components, b and y, c and z, a and x,
respectively normal and parallel to the section, as shown by the force-triangles
on the right.* Then, summing these comps, algebraically, we obtain the
resultant forces, P n (normal) and P t (tangential or shearing), acting upon
the section in question. With the forces, S, V and H, as shown in Fig 17,
we have:

On sec F G,



On sec K M,



tension, = y + z x ;
right-hand shear, = a + c b ;
compression, = a + c b ;
right-hand shear, = y -f- z x.



107. If, now, we examine all possible planes cutting the body at a
given point, we shall find (1) one such plane upon which the resultant
unit tensile stress reaches its max; (2) another, normal to (1), upon which
the resultant unit comp stress reaches its max; and (3) two planes, normal
to each other & bisecting the right angles betw planes (1) & (2). Upon
the two planes last named, (3), the resultant unit shearing stresses reach
their max.

*In order that, for either force, S, V or H, the two force-triangles (for
the two sections, F G and K M) may be identical, and thus simplify the
figure, we take the two sections, F G and K M, normal to each other.



4945



STRENGTH OF MATERIALS.




Fig. 18.

1O8. Let Fig. 18 represent a small element in a bar under tensile &
shearing stresses; and let it be required to determine the positions of
these planes ami the corresponding max stresses. Let

s = the original normal (tensile or comp) unit stress ;
v ' " vertical (shearing) unit stress ;

= h = ' horizontal (shearing) unit stress ;

s = " max or min resultant normal unit stress ;
v = " max resultant shearing unit stress ;



angle betw s and s,



Then



s/2








/ o/9\2 _L- ^t 2






(2)


= s/2 + v =


*/ 9


+ y' (s/2) 2 + v 2


(3)










= 8/2 V. =


*/a


V (s/2) 2 + v 2 ....


C4)



s^max
Sp min

( tension ( + si P ^T, 63 max tension

Tf J I comp = min tension

1 ( + " " " comp

^comp | _ , tension = min comp-

1O9. Example. Let

8 = 2000 Ibs/sq inch, tension (not drawn to scale);

v = h = 1600 " / " " , shear ( " ' ).

Here v is left-handed, h right-handed. If this be reversed, the angle, A,
betw the resulting tension, s , & the hor, will be below the neut axis.



110. Then tan 2 A = -^ =



A = 29 C



/(s/2) 2



max
min



v r "
v, =



1/1000 2 + 1600 2 = 1887;

1000 + 1887 = 2887 (tension);
1000 1887 = 887 (comp).



8/2 +

111. In other words, we have, as resultants, (1) a max unit tension,
s max = 2887 Ibs/sq in, forming an angle, A = 29, with the axis of the
bar or with the direction of s ; (2) a min unit tension or max comp, s p min =
887 Ibs/sq in, normal to s p max; (3) a right-hand unit shear, V T = 1887
Ibs/sq in; and a left-hand unit shear, v. = 1887 Ibs/sq inch; the



DIAGONAL STRESSES IN BEAMS.



494 c



directions of the shearing stresses bisecting the right angles betw the max
normal stresses.

112. The max tension and compression, at any point, are called the
" principal stresses " for that point.

Horizontal and Vertical Shear in Beams.

See also pp 440 &c, 446 &c, 450 to 453, 478-9.

113. Let Fig. 19 represent the left half of a homogeneous beam, of

rectangular section; breadth, b, = 1 inch; depth, d, = 10 ins: span,!/, = 100
ins; with cen load, W,* of 200 Ibs; left reaction, R = W/2 = 100 Ibs. Weight
of beam neglected. The bendg mom, at cen of span, is M = RL/2 =
FPL/4* = 5000 inch-lbs; and the mom decreases uniformly,* from its max,
at cen of span, to zero at the supports. In the extreme upper & lower fibers,
the longitudinal unit stress, (^[ 10, p 468) s, = MT/I, where T = df2 =
dist from neut axis to extreme fibers = 5 ins; / = inertia mom of cross
section = bd*/12 = 1000/12. Hence, in Fig 19, s = 12X5 M/1000 =
0.06 M. Now s, being thus proportional to M, also decreases uniformly,*
from its max, at cen of span, to zero at the supports. Values of M and of
s, for the sections 0, a, b, c, d, e, are figured on the diagram.

Section












__


.


_.
r _.=== :== - H


~~ 1. '
>-


~_~_~Ji




\ S


W


H ,


L_


/ Neutral


/




/'


I /


"7


Axis /


"~ /


/S




Bf- 1
= 10

= ( 6


0" 2
00 20
U


o" a

00 30
li


0"
00 40
50 24


3" 9
00 50




\ff

X)




Fig. 19.



* Under a uniformly distributed load, the bendg mom, at cen of span,
is WL/8; and the bendg moms, M, and the resulting longitudinal unit
stresses, s, vary as the ordinates of a parabola, as indicated by the dotted
parabola, r m e, at top of Fig 19, which corresponds to a uniform load = 400
Ibs = 2 W, The unit shears, v, in a given hor section, then decrease uni-
formly, from a max, at the supports, to zero at the cen of the span. Com-
pare 3d and 4th figures, p 474.



494 d



STRENGTH OF MATERIALS.



114. The unit hor tensile and contp stresses, s, at the several pointa
in any vert section, are proportional to the dists of those points from

the neutral axis, as indicated by the diagram at each vert section, Fig 19.

115. In Fig 20, let n and g be two vert sections of this beam, such that,
at n and at g, the extreme unit fiber stresses are: mn = 15, and u g = 25,
respectively. Then the rectangular portion, n f, of the beam,
betw sections n V i>- is acted upon by a series of net or resultant forces,
ranging from compression, e g u g m < n = 25 ( 15) = 10, at
the top, to tension, = +10, at bottom, as indicated by the diagram, e k.

116. Suppose the piece nf to be divided into 10 hor strips of equal depth,
= 1 inch. Then the net unit stresses, s, acting at the tops and bottoms
of these strips, respectively, are those, (10, 8, 6, ... .6, 8, 10) figured
from e to k; and the mean stress, or (since depth of each strip = 6 = 1)

the force, acting upon each strip, is that (9, 7, 5 5, 7, 9)

figured betw g and /.

117. These forces are transmitted, from strip to strip, thru their
surfs of contact; and, in determining the shearing force, acting in the hor
plane betw any 2 strips, we regard the upper (or lower) strip as acted upon
by its own push or pull plus (algebraically) those of all the strips above (or
below) it.




118. Thus, the 3d strip from the top is pushed to the left by a force of
_9 _7 _5 = 21, while the 4th strip, just below it, is pulled to the right
by a force of 9 + 7 + 5 + 3+1 1 3 = 21. Hence the surf betw
the 3d and 4th strips, sustains a counterclockwise shear of 21 ; which,
divided by the area, b I = I, of that surface, gives the unit shear in the
plane betw the 3d and 4th strips. With central load,* this unit shear is
uniform from each support to cen of span, where it changes sense (from plus
to minus, or vice versa) but is of the same intensity in the other half-span.
See 3d Fig, p 474.

119. In any vert section of the beam, let
V = the total shear

= " reaction of either support, minus the sum of all loads betw

that support and the section ;

7 = " inertia moment with respect to the neut axis;
b = " breadth; d = depth ;

a = " area above (or below) any given point in the section;
c = " dist from neut axis t9 grav cen of a;
M S = a c = static mom of a, with respect to the neut axis;
v = the unit vert shear = unit hor shear at a given point.

120. Then

M .



v = V



-f- & = V -z



IV



* See foot-note p 494 c.



DIAGONAL STRESSES IN BEAMS. 494 6

N d b d d? b
At the neut axis, M g ( = a c) = -^ ^ = -g .

Hence, at the neutral axis :

. _ v ^ = v 12cp = -

V 81 Sbd^ 2 bd

= X the mean vert shear in the cross section.
21

See also If 51 etc.

In f 115 we have taken diff in 8, betw n and g, Fig 20, = 10 = V I bd.

Hence.-f .f rf = f .^>=15 = f Xdiffin,

At neut ax, Fig 20, we have total hor shear = 9 + 7 + 5 + 3 + 1 = 25;
and dist nff = l = bl = total hor shear -r- unit shear = 25 / 15 = 1.666...;

and * max X 7-7^ = ( 30 x 1-666. . .) / 50 = 10 = diff in hor fiber stress,
Li I A

s, betw n and g.

121. At the left of Fig. 19 is a diagram showing the unit shears

in the several hor sections.

122. Let Fig 21 represent a small element of a body, of unit thickness,
normal to the paper, and acted upon by a right-hand vert shear, V = v D,
(where v = the unit vert shear, and D = the depth of the element) and by a
left-hand hor shear, // = h L (where h = the unit hor shear, and L = the
length of the element). For equilib of moments, we must have

V L = H D; orvDL = hLD; or v = h.
In other words,

unit vert shear = unit hor shear.




Fig. 21.

Maximum Unit Stresses in Beams.

123. The common theory of beams (pp 466 to 494, U If 1-103)
considers only the longitudinal tensile and eompressive forces

and the vert and hor shearing forces, due directly to the load and to
the upward reactions of the supports, and acting, at any point, upon vert
and hor planes passing thru such point; but, except in certain limited por-
tions of the beam, these stresses are not the maximum stresses act-
ing at such point; for they combine to form resultant diagonal stresses,
acting upon diagonal planes (passing thru the same point); and, upon some
of these diag planes, the resulting normal and tangential stresses are greater
than either of the original stresses.

124. The common theory is sufficiently well adapted to beams of
many kinds, and especially to steel beams, where the longitudinal forces
are resisted by the flanges, and the shears by the web; but in certain por-
tions of deep and heavily loaded beams, especially those of reinforced
concrete, the diagonal resultant or maximum stresses are the
ruling stresses, and must not be neglected.

125. In a beam, at top and bottom, we have, respectively, hor tensile
and corap stresses only, and, at the neut axis, shear (vert & hor 1 ) only;
but, at all other points, we have shear (vert & hor) acting conjointly
with hor stresses, either tensile or comp. At all points, these shearing and
longitudinal stresses may be resolved into components, normal &
tangl to any plane, at pleasure, as in the case of the bar or bolt, Fig 17.



494/



STRENGTH OF MATERIALS.



126. Thus, each element of the beam, Figs 22, 23, 24, is acted upon by
hor & vert forces (unit stresses), which, acting upon diagonal planes, are
resolved into diagonal components, and these components may be alge-

Section

- i



Top: =0; v r =s = s p mate; Cornp = s p maa;




Fig. 24.



braically summed into resultants ; but the original stresses vary in
intensity, and the resultant stresses both in intensity and in direction, from
point to point. For the directions and values of these resultant
stresses at their maxima, we have, from p 000,



V (s/2) 2 + v 2

8/2 v = s/2 i/(s/2) 2



(2)



where



original unit tensile or cpmp stress at the point ;
original (vert or hor) unit shear at the point.



The max normal stresses, s p , are called the principal stresses.

127. Applying these formulas at numerous points in the profile of the
beam, Fig 22, we are enabled to construct curves, Fig 23, showing the
directions of the stresses ; and to plot, as in Fig 24, for given points, the
directions and intensities of the stresses there acting. At any given
point, Fig 24, we have resultant normal and shearing stresses analogous to
those in Fig 18, p 4946; but, in the present Fig 24, owing to want of space,
only the max principal stress, max, is shown for each point selected.







128. In Fig. 23, the directions of the principal stresses, s p , are repre-
sented by the solid curves; those of the resultant shears, v f , by dotted
curves.


Of the solid curves
(principal stresses)


concave


horizontal
at cen of span


at 45
with


at 90
with


The tension curves are
The compression curves are


upward
downwd


below neut axis


neut axis


top of beam
bot "


above '





130. Where v = zero (viz: at any point in the vert cross section at
cen of span, and along the extreme upper and lower fibers), we have (U 126) :



The tensile and comp curves are normal to each other at their intersections.
129. Following any curve (concave upward) of normal tension,*
we find that,

(1) for its point of tangency with the hor (viz: at cen of span)
s max = tension = s ; s p min = comp = ;

(2) for the point where the curve crosses the neut axis (at 45)
s p max (tension) = s p min (comp) = v f = v (shear);

(3) above the neut axis, the tension becomes 8 p min, and continues
diminishing, as the direction approaches the vert, becoming zero at top,
where A = 90. Above the neut axis, for points in the same curve, the
compression (normal to the curve) is now a max, and increases from 8 p =
?V = v, at the neut axis, to 7J max (comp) = s, at top.

re v = z

id along tl
v r = s/2

s p max = s/2 + v f = s ; tan 2 A = ;
8 p min = s/2 v f = ; tan 2 A = 0.

181. The equation, tan 2 A = 0, gives either 2 A = or 2 A = 180;
i. e., A = 0, or A = 90; but we know that, at cen of span and along the
extreme upper and lower fibers, s p max is hor, or A = 0; and s p min is vert,
or A = 90.

132. Where s = zero (as at the neut surf and where bending mom
= zero), we have (If 126) : v r = v ; s max = s p min = j/v* = v;
tan 2 A = oo ; 2 A = 90 ; and A = 45.

133. Of the (dotted) shear curves. Fig 23, those of one set are tan-
gential to the neut axis and reach top & bottom of beam at angles of 45,
tending away from cen of span; while those of the other set are normal to
these and to the neut axis at their intersections, reaching top and bottom
of beam at 45, tending toward cen of span.



MOMENTS IW CONTINUOUS BEAMS.

See also HU 78, etc.

134. Figs 25 and 26 show positive and negative bending: moments
in two continuous beams, Fig 25 of two equal spans, and Fig 26
of three equal spans, resting freely upon their supports. Each span = 1.
Fig 26 (three spans) may be used, with sufficient approximation, for cases
where the spans are more numerous.

* Conversely for curves (concave downward) of normal compression.



494 A



STRENGTH OF MATERIALS.




-0.10








3


^ x


ij,j
















-


-.








J


'^


s*^




x xU












/














'/






m


\




s v


r.








2




/








\N


^












\


*








,




/










^\


-^r


1)









5




\


x


2 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

Online LibraryJohn C. (John Cresson) TrautwineConcrete → online text (page 2 of 23)