John C. (John Cresson) Trautwine.

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neither f g nor f c exceeds the elastic limit. When they exceed that limit, see
tI 21, 22, p 1122.

7. In equilibrium, the bending moment of the load (see p 474) is
balanced by the equal resisting moment of the couple composed of
the two equal hor forces, T and (7; these forces being the resultants respec-
tively of the tensile stresses in the steel and of the compressive stressest in
the cone.

8. The tensile stresses, f s , in the steel, are assumed to be uniformly
distributed over its entire cross section, a g ; and their resultant, T, is there-
fore taken as acting at the grav cen of the steel area; but the compres-
sive stresses, in the cone, in any cross sec, decrease uniformly! from a
max, f c , at the upper surf of the beam, to zero, at the neutral axis. Their
resultant, C, is therefore applied at a point distant kd/3 below the top of
the beam, kd being the distance from top of beam to neutral axis, and d the
distance from top of beam to grav cen of steel.

9. Value of "j." The lever arm, d', of the resisting
couple is therefore

d ' = jd = d kd/3 = d (1 Jb/3) (D

and we have

j = d'/d = 1 k/3 (2)

For approx values of j, see t 12.

10. Value of "It." From assumption 1, U 4 we have

e c /e s = k/(l k) (3)

From assumption 5, we have
Hence

J s = V*f = l^k '^s = "U *> '.' (4a)

For equilibrium, C = T; but

C = f c b k d/2 = e c E c b k d/2 (5)

and T = / 8 a, = f s p b d = e g E S p bd (6)

Hence, A; = 2 p s * = 2 p n - ;
e c E c k

or :

li = i/(p n) 2 + 2pn pn (7)

*See lit 15, 16, p 1118. ** See 1 13, p 1118. .

t Below the neutral axis, the cone is in tension, but its tensile stress is
neglected. See assumption 4, H 4, p 1115. t See Hi 21, 22.

Figs 2 and 3 are by Prof A. W. French, A S C E, Trans, Vol 56, '06,
pp 362, etc.



REINFORCED BEAMS.



1117



Scale of 100 jp
0.25 0.50 0.75 1.00 1.25 1.60




0.25 0.50 0.75

Scale of lOOp

Fig 2. For Working Stresses. (For ultimate stresses, see Fig 3.)
It = l/ (p n) 2 + 2 p n p n, j = d' / d,
f ft = unit stress in steel, f f = unit stress in cone at top of beam,
p = a s ja c = ratio of steel area to cone area,

[g , M C = resistg mom, based upon allowed value of f g , f c , resp,
M = resistg mom, actual.



n = E S /E C . Solid curves represent n = 10; dotted curves, n = 15.

Steel lines plotted f or n = 10; approx for n = 15.



1118 CONCRETE.

11. Hence the position of the neutral axis (given by k} de-
pends solely upon the ratio, p, of steel area to cone area, and upon the
ratio, n, of elasticity betw steel and cone. For approx values of k, see t 12.

12. Approximate values of j and k. See Fig 2.

when . and we have and

n = 10, p = 0.010: j = 0.88; k = 0.36;

p = 0.015: j = 0.86; A = 0.42;
n = 15, p = 0.010: j = 0.86; k = 0.42;

p = 0.015: j = 0.84; A; = 0.48.

13. When, as in reinfd cone, two widely diff materials are used in con-
junction, it usually happens that, owing to the impracticability of always
giving, to each, its ideal cross-sec area, one or the other is un-
avoidably and uneconomically subjected to less than its maxi-
mum allowable stress. Thus, with a given value of p = a s /a c ,
if we load the beam until either f g or f c reaches its max allowable limit, the
other (f c or / g respectively) will usually remain below its max allowable
limit. See U 19 /. Let F S and F C = respectively the max allowable
values of t' m and i' c .

14. Moments. For resistg moms, based upon the max allowable values,
F a and F C , of f g and f c respectively, we have:

M 8 = Td' = F s a s jd = F s pjbd* ............................................ (8)

M c = C d' = C i d = F c b k d j d/2 = F c k j b d*/2 ................ ..(9)

For usual values, we may take (see H 12): / = %; A; = %,
k j = %. Hence, approx,

M s = 7 F 8 a s d/8;
M c = F c b d V6.

But the actual resisting mom, M, of the sec, in any given case,
can of course have but one value; and this is the less of the two values,
M s and M C . Since / b d 2 is common to both these values, M is determined
by whether F S p or F C k/2 is the greater.

15. Relation between f g , f c and p. Since C = T,orf c bk d/2
= f 8 p b d, we have:



From Eq (4a) we have:



/, n nk-

HenCe * = n/JV//
and p - */ e /2/ t - , ; 5 . , x = rW^ >T (11)



V
Usually, p ranges from 0.010 to 0.015. It is seldom < 0.005 or

16. Note that f g , f c and p cannot be arbitrarily selected. Given any two
of them, the third depends upon the two so given.

17. Value of M/bd 2. Let F g and F C be the max allowable values of
the. unit stresses, f s and f c , in steel and in cone respectively. Then, from
eqs (8) and (9), 1 14, we have (Fig 2, lower portion):

(nearly straight lines, for steel) (12)

M c /bd 2 = F c k j/2 = F c k (1 fc/3)/2;

(curved lines, for cone) (13)



REINFORCED BEAMS. 1119

The dotted and solid curved lines, for cone, represent n = 15 and n = 10,
respectively. The nearly straight lines, for steel, are plotted for n = 10,
but are sufficiently approx also for n = 15.

18. The upper portion of Fig 2 gives values of

Is. = 1/2 p n + (p n)' 2 p n,
(see 1 10) and of

j = 1 Jfc/3 = d'/d,

corresponding to given values of p, for n = 10 and n = 15. Note that /
varies but slightly with p.

Examples.
I. Investigation.

Required tbe resisting 1 moments, M a , M C and M.
19 a. Given a rectangular reinfd cone beam : b = 8";

d = 20"; a c = bd = 8 X 20 = 160 sq ins; n = E g /E c = 15. Let F g
= 16,000, and F C = 500 Ibs per sq inch, be the max allowable values of the
unit stresses, f g and f c , in steel and in cone respectively; and let P be the
value of p based upon these max allowable stresses.
F 8

Then FJF C = 32; ~- + 1 = 3.133; and, from Eq (11), H 15, we
nF c

have:

P - 32 X3.133 = - 004987 '

as given by the intersection, in Fig 2, of radial line, for f s = 16,000, with
dotted curve for f c = 500.

19 b. (Case 1) Reinforced with two round rods, %" diam;

a s = 2 . IT 0.375 2 = 0.884 sq ins;

p = a s /a c = 0.884/160 = 0.005525 > P;

pn = 15 X 0.0055 = 0.0825;

k = l/(pn)2 + 2 pn pn

= V 0.0825 2 + 0.1650 0.0825 = 0.3322;

d' = d j = d (1 k/3) = 20(1 0.1107) = 20 X 0.89 - 17.8 ins;

C = F c b k d/2 = 500 X 8 X 0.3322 X 10 = 13,288 Ibs;

M c = Cd f = 13,288 X 17.8 = 236,526 inch-lbs;

T = F g a s = 16,000 X 0.884 = 14,144 Ibs;

M s = Td' = 14,144 X 17.8 = 251,763 inch-lbs;

M = M c = 236,526 " " .

Notice that where, as in this case and in Case" 2, P < p, the mom,
M c , based upon the max allowable stress, F c in the cone, is the actual
mom, M. Where P > p, M B is the actual mom.

19 C. By Fig 2. The intersection of the vert line, on 100 p = 0.55,
with radial line for f s = 16,000 Ibs per sq inch, gives M g /bd 2 = 78.7; and
M s = 78.7 bd 2 = 78.7 X 8 X 20 2 = 251,840 inch-lbs; but the intersection
of vert line on 100 p = 0.55, with dotted curve (n = 15) for f c = 500 Ibs
per sq inch, gives M c /bd 2 = 74; and M = M C = 74 bd 2 = 73.9 X 8 X 20 2
= 236,480 inch-lbs.



1120 CONCRETE.

19 d. (Case 2) Reinforced with 3 round rods, I" diam;

a g = 3 if 0.5 2 = 2.356 sq ins;

p = a s /a c = 2.356/160 = 0.01473 > P;

pn = 15 X 0.01473 = 0.2209;

k = I/ (pn) 2 + 2 pn pn



= v 0.22 2 + 0.44 0.22 = 0.48;

d' = d] = d (1 fc/3) = 20 (1 0.16) = 20 X 0.84 - 16.8;
C = F c bkd/2 = 500 X 8 X 0.48 X 10 = 19,200 Ibs;
M c = Cd' = 19,200 X 16.8 = 322,560 inch-lbs;
T = F s a s = 16,000 X 2.356 = 37,696 Ibs;
M s = Td' = 37,696 X 16.8 = 633,293 inch-lbs;
M = M c = 322,560 " " .

19 e. By Fig 2. The intersection of the vert line on 100 p = 1.473,

with radial line for f s = 16,000 Ibs per sq inch, would give (on a sufficiently
accurate diagram) M s /b d * = 198; and M S = 198 b d 2 = 198 X 8 X 20 2 =
633,600 inch-lbs; but the intersection of vert line on 100 p = 1.473, with
dotted curve (n = 15) for f c = 500 Ibs per sq inch, gives M c /bd 2 = 101;
and M = M C = 101 & d 2 = 101 X 8 X 20 2 = 323,200 inch-lbs.

19 f. It will be noticed that, in these cases, an increase of 166.5 %,
in the amt of steel, has increased the resisting mom (which still

depends upon the cone) by less than 38 %; and the steel, in Case 2, is
stressed to only about 8,000 Ibs per sq inch or half the max allowable stress
(intersection of vert for 100 p = 1.473, with dotted curve for f c = 500, is
nearly intersected by radial line for f s = 8,000). See U 13.

19 g. In both cases, (1) and (2), the intersection of radial line for f g =
F g = 16,000, with dotted curve for f c = F C = 500, would give (on a suffi-
ciently accurate diagram) p = P = 0.004987; M/bd* = 71.5, and M =
71.5 bd 2 = 228,800 inch-lbs, the actual mom, for the given b and d, in the
ideal case where f s and f c = respectively F 9 and F C = 16,000 and 500.

II. Design.

20 a. Conversely, given the bending moment, 236,500
inch-lbs; F s = 16,000; F C = 500 Ibs per sq inch; whence P = 0.004987,
as before. Required b and d.

Let K and J = the values of k and of / respectively, corresponding to



Here we have

Pn = 15 X 0.004987 = 0.075;



K = (Pn)2 + 2 Pn Pn



= I/ 0.075 2 + 0.150 0.075 = 0.3193;
J =1 K/3 = 1 0.1064 = 0.8936;
,, 2 _ M 2M 2 X 236,500 ^ __

" F S PJ F c KJ 500 X 0.3193 X 0.8963

2O b. An infinite number of section areas, bd, giving the
same resisting moment, M , may be found from bd 2 .

2O c. Thus, in the example of H 20 a, with bd 2 = 3315, we may have
b d 2 d

6 552 23.5

8 414 20.3

10 331 18.2 etc, etc.



REINFORCED BEAMS.



1121



Scale of 1OO p

0.75 1.00 1.25



II




y>=steel area-T- concrete area
n _2?s \ n 10 for full curves

E c I jj=15 " dotted curves
EQ= Initial E for concrete



Steel lines-plotted for n10
Approximate for w=16 JP



700



TOO



JJOO






100



100







^ oc^tfe o/ 1<\O 1



.25 1.50 .1.75 2.00




0.25 0.50 0.75 J.OO

Scale oflOOp

Fig 3. For Ultimate Stresses. (For allowable stresses, see Fig 2.)



pn, j = d' / d,

f g = unit stress in steel, f c = unit stress in cone at top of beam,
p = a, g fa c = ratio of steel area to cone area,

resistg mom, based upon max allowed value of f s , f resp,



M



resistg mom, actual.



n = E S /E C . Solid curves represent n = 10; dotted curves, n = 15.

Steel lines for n = 10; approx for n = 15. E f = initial E for cone.
C7



1122



CONCRETE.



20 d. It can be shown (T & M, pp 175-6) that, with given M, given unit
stresses, and given unit prices, the cost of a reinfd cone beam, per unit of
length, varies inversely as d, directly as I/ 6, and directly as $ b/d .
Hence, for a given bd, the deeper the beam, the less is the cost; but practical
considerations (such as practical limits to reduction of 6, requirements as
to head room, etc) often limit the extent to which this economy can be carried
in practice.

21. Within the limit of allowable working stresses, Fig 2,
the stresses and deformations, in the several fibers, are taken (assumption 1,
U 4) as proportional to the dists of the fibers from the neutral axis, as repre-
sented by the shaded triangle in the small figure above the diagrams (said
triangle representing approx the lower portion of the parabolic area shown
in Fig 3); and we have, Eq (7), t 10,

li = l/ (pn)* + 2 pn pn.

22. For stresses exceeding the allowable worlig stresses,

up to the ult, Fig 3, assumption 1 is inadmissible, we must employ the entire
parabolic area, its vertex corresponding with the ult comp strgth of the
cone; and we have



It = y (3pn / 2) 2 + 3 pn 3 p ra / 2 (14)

Fig 3 gives values of j, k and M /b d 2 , for ult values of f s and f c .

23. Note that, for steel stresses, f s , not exceeding the usual elastic limit,
and with f c ultimate < 2000 Ibs per sq inch, the ult resistg mom in-
creases directly with the amount of reinfmt until this reaches
2 % or over. Thus, Fig 3, with f s = 30,000 Ibs per sq inch, f c ult < 2000,
and p = to 2 %, we have M /bd 2 = approx 25,000 p.

Tee Sections.

24. Tee sections. Fig 4. 6 = flange width; b' = stem width; t =
flange thickness; d = depth from top of flange to cen of steel; k d =
depth of neut axis; d r = j d = leverage of T and C.




Fig 4. Reinforced Tee Section. Theory.



25. When the tops of rectangular beams are connected by slabs, the
whole being placed at one time and properly bonded, all or a part of the
slab may be considered as a compression flange, in some respects
similar to those, composed of angles and plates, of steel plate girders.

26. The width of slab, b, Fig 4, which acts as flange, is sometimes
taken as the distance between beams, but should not exceed % of the span
of the beams. See Specifications, IfU 168-170.

27. Exact analysis of such a section is hardly possible, but it is believed
that the following method is reasonable and safe.

28. Determine the ratio, p = a s / c , of steel area to cone area as tho the
beam were rectangular, with depth = d, and width = the flange width, b.
With this value of p, determine the position of the neutral axis. If this
falls within the slab or just at its lower side, the resisting moment is found
exactly as with any rectangular section. See Case 1, If 19.

29. If the neutral axis falls below the bottom of the slab, the
position of the neutral axis will not be exactly given by the equation for
rectangular beams; but the difference will not be important.

30. The resisting moment is Cd' or Td', whichever is the less.,



REINFORCED BEAMS. 1123

31. Examples.

(1) Neutral axis within the slab.

Let b = 60 ins; b' = Sins; d = 20 ins; t = Sins; max allow-
able unit stresses, F C = 500, F g = 16,000 Ibs per sq in; E C =
3,000,000; E 8 = 30,000,000; n = 10. Let there be 3 round steel
rods, diam = 1 inch.
Then

3 X 0.785

P = 60 X 20 == - 02 '
k = V (pn) 2 + 2 p n p n

= ]/ (10 X 0.002) 2 + 2 X 10 X 0.002 10 X 0.002 = 0.18;
k d = 0.18 X 20 = 3.6 ins;

G = F c b k d/2 = 500 X 60 X 0.18 X 20/2 = 54,000 Ibs;
T = 3 X 0.785 F s = say 37,650 Ibs.

Using the smaller value (that for the steel) we have :
M = T d' = T (d d Jfc/3) = 37,650 (20 3.6/3) = 707,000 inch-lbs.

(2) Neutral axis below the slab.

Let 6 = 60 ins; b' = 10 ins; d = 30 ins; t = 4 ins; F C , F 8 , E C ,
E s and n as in Example (1); 6 round steel rods, diam = 1 inch. Then



P = N' Q = O- 0026 - and k = ' 2 ' fc <* = 0.2 X 30 = 6.

OU X oU

32. Since the comp unit stress, in the outer fibers of cone, is assumed to
be F c = 500 Ibs per sq inch, the stress, at the lower side of the slab, is 500
(k d t)/k d = 500 X 2/6 = 167; and the average stress, in the
slab, is 5 * 16 I = 333 lba per sq in>

33. The 2 inches of stem, which lie between the neutral axis and the
lower side of the slab, exert some comp resistance, but this is neglected,
with a small error on the safe side.

34. The position of the center of gravity of the compressive
forces in the slab may be found as for a trapezoid; but it is usual, safe, and
sufficiently approximate, to assume that it is at the cen of the slab, or, in
this example, at a distance of d t/2 = 30 -2 = 28 ins above the cen of
the steel. The mom of these forces is then M C = 333 X 60 X 4 X 28 =
2,238,000 inch-lbs; but the moment of the tensile resistance of the steel is
only M s = 6 X 0.785 X 16,000 X 28 = 2,110,000 inch-lbs; and this mom,
being the less of the two, is to be taken as the actual mom, M.

Shear.

35. Shear. In addition to the hor stresses, resisted by compression in
the cone and by tension in the longitudinal steel reinfmt, the vertical shear-
ing stresses require attention in relatively deep beams under heavy loads.

36. For the total shear, V, in any vert section, distant x from
a support, we have :

V = R W ............................................ (15)

where R = upward reaction at the support;

W = the total of any loads in the distance, x.

37. The vert shear is sometimes provided for by using a large safety
factor with the ult shearing strgth of cone, which is usually taken at from
500 to 800 Ibs per sq inch, while the working shearing stress is often
restricted to from 30 to 50 Ibs per sq inch. But see Stirrups, ^ 38, etc.



1124 CONCRETE.

Shear Reinforcement. Stirrnps.

38. Shear Reinforcement. Where the loading produces a shear-
ing stress exceeding the limit assumed for plain cone, the beam is often reiufd
by vert stirrups, which consist of rods, bent into the shape of a letter U,
and passing under the hor bars and up to near the top of the beam; or, in
the case of Tee beams (Fig 4), into the slab.

39. The distance between stirrnps is sometimes made such that,
within a hor length = d', there shall be an aggregate sectional area of vert
steel bars sufficient to carry the vert shear by means of the permissible unit
tension in the steel.

40. Example.

Consider the T beam of example (1) f 31, Fig 4; V = 8 ins; 6 = 60 ins;
d = 20 ins; d' = 20 k = 0.18; k d/3 = 20 1.2 = 18.8; safe mom of
resistce, M = 707,000 inch-lbs. Let span L = 20 ft = 240 ins. Then, for
a uniform load, we have JF = 8M/L = 8X 707,000/240 = 23,600 Ibs.

Shear at ends = W/2 = 11,800 Ibs.

With safe unit shearing stress = 50 Ibs per sq inch, we have safe shear
resistance of plain cone in section = 50 6' d' = 50 X 8 X 18.8 = 7,500 Ibs.

Under uniform load, this shear occurs at a dist, from the ends,

(11,800 7.500) L

2 X 11,800

From this point to the center of the span, the cone is able to care for the
shear, and no stirrups are there reqd. But see 1f^[ 41, 45.

Between this point and each support, let the stirrups be of % inch round
steel; aggregate cross section area of the two limbs of each stirrup = 0.22
sq inch.

Allowing 16,000 Ibs per sq in, one stirrup will sustain 16,000 X 0.22 =
3,520 Ibs.

The total shear, 11,800 Ibs, at the support, divided by 3520, gives 3.3
as the number of stirrups required, in 18.8 ins of length of beam;

-I Q O

or the spacing-, next~to the ends, should be = 5.5 ins.

d.p

Let the load, W, = 23,600 Ibs, be uniformly distributed. Then, at a

1 f) q

point 3 ft from the end, V = 1Q X 11,800 = 8260 Ibs; 8260/3520 =
2.35; and stirrup spacing* = 18.8/2.35 = 8 ins.

41. The spacing- may be made to vary uniformly betw these limits;
and it would be well for the vert reinft to extend beyond the theoretical
stopping point (3.6 ft from end; see If 40), by one or two stirrups spaced a
foot apart. See 1 45.

42. Let

A = aggregate vert cross sec area of hor rods, sq ins;
L = span, ft;

z = dist from end of beam to stirrup, ft;

S = aggregate cross section area reqd in the 2 limbs of the
stirrup, sq ins.

Then, when the stirrups are 1 ft apart,

a _ (i- 2 -^i) we)

(J. W. Schaub, E N, '03/Apr/16, p 348.)

43. In general, spacing betw stirrups > d'.

44. The cone, in each sec, has to act as a connecting medium between
the hor and the vert reinft. It is also subjected to comp forces, in transfer-
ring the shear from one stirrup to the next. The action here is complex,
and an ample safety factor should be used.

45. In order to provide against excessive loadings, which may come
temporarily upon the beams during construction it is advisable to use
stirrups, even where not actually required by the shearing stresses deter-
mined theoretically as above for the completed structure in use. The
etirrups being light, the cost of using them is principally for labor; so that,
if any are reqd, it is well to be liberal with them. See H 41.



^ff^

X*V*JB**A*ps.

C OF THi X

UNIVERSITY)

s y

x* S



REINFORCED BEAMS.



Unit Shear.



1125



46. Unit shear, v. In any hor section of a beam, Fig 5, under uniform
or central loading, the hor tensile or comp stresses increase from the ends,
where they are zero, toward the middle of the beam, where they are a max.
Hence, of any two vert plane sees, 1 and 2, the section, 2, nearer the cen of
the beam, will have the greater hor stresses, s.





:


.


F


-*-,




Neutral Axis \




FT 7 ' 4






T

. - ^-


B


d' )

Jr

-^-TVT.^^.


-




V'


se

'


.
i





Shear
Diagram.



Fig 5. Unit Shear.



47. Consider the forces acting upon the rectangular body, B, between the
two sections, 1 and 2.

48. At the left section, 1, the vert shear, V, coming from the left support,
pushes B upward; and the tension, T, of the steel pulls B horizontally toward
the left; while the total comp, C, acting at the cen of the comp forces, pushes
B toward the right.

49. At the right sec, 2, the vert shear, V, pushes B downward; while T'
and C' are in line with T and C respectively, but opposite to them. Note
that 2" > T, and C' > C. Let T' T = L

50. Let there be no load on B. Then V = F.* Since the vert
forces are distant by x, their moment = Vx = V'x* The mom of T' T
is (T' T) d' = td'. Hence, for equilibrium,

Vx = td'; or t = Vx/d' (17)

51. In a reinfd cone beam, Fig 5, we neglect the tensile strgth of the cone.
Hence, the diff, T' T = t, of tension, between sees 2 and 1, must be trans-
mitted, from the steel to the neut axis, by a total shear, = t, uniform* in
each hor sec; and, since the hor sec of the body, B, is 6 x, we have, for the
11 ii E t shear :

v = t/bx = Vx/d'bx = V/bd' = V'/bd f * (18)

Diagonal Stresses.

52. As a matter of fact, the longitudinal tensile stresses and the vert and
hor shearing stresses, combine to form, and are replaced by, diagonal
stresses; and reinfmt, against shear, is more rationally designed by deter-
mining, as nearly as may be, the directions and intensities of these resultant
diagonal stresses (See U 53), and so placing the reinfmt as best to resist them.

53. From "Maximum Unit Stresses in Beams," p 494 e, we have, in
homogeneous beams, for the angle. A, betw the neutral axis and the
resultant normal (tensile and comp) or "principal" stresses, s~, , at any point:



tan 2 A = 2 v/s;
and, for the max stress,



(19)



8/2 + l (8/2)2



(20)



where v = the unit vert or hor shear, and 8 = the unit hor tensile or comp
stress, at the given point.

* If there is a load, L, upon B (as, for instance, in the case of uniform
loading) we have V > V, and V V = L; and there are two couples of
vert forces, with moms, respectively: Vx and (V V) x', where x' = dist
from sec 1 to gravity center of L, Here we have, for sec 1, t/ = V'/b d'\
and, for sec 2, v = V/b d'.



1126 CONCRETE.

54. Bat, neglecting" the tensile strath of the cone, we

have, in beams with tension reinfmt of straight oars, and for points betw
the neutral axis and the steel, s = 0; whence :

tan 2 A _= oo; 2 A 90; A = 45;

s p = i/V = = v/bd' (21)

55. Hence, betw the neut axis and the steel, we should provide against
tensile unit stresses, s p = V/b d', acting in parallel directions form-
ing angles of 45 with the neut axis.

56. Other things being equal, this provision is preferably made by means
of rods, placed like the diag tension members of a Pratt
bridge truss, Figs 76, 86, 96, p 693, and forming angles of 45 with the hor.

57. Very commonly, the tension rods, at each end, in a hor dist about
equal the depth of the beam, are bent upward to form an angle of 45 or
thereabouts with the axis of the beams.

Adhesion. Seep 1111.

58. Unit of adhesion. Let

x = a given portion of the length of the beam;

t T' T = the increase, in total tension, T, in the steel, in the Igth, x;
V = the total vert shear in the cross section;
d' = the dist betw T and the cen of comp of the cone;
U = t/x = the bond stress, per unit of x;
m = the number of rods;
a = the circumference of one rod

= the circumferential contact area of one rod, per unit of x;
u = U/m a = the bond stress, per unit of a.

Then (see U 50), t d' = V x; t = V x/d'; U = t/x = V x/d' x = Vfd'; and
u = U/m a = V/d'ma (22)

59. For given values of the bond stress, U, per unit of length, and of the
bond stress, u, per unit of circumferential contact area, the product, m a
= U /u ( = total circumferential area per unit of length) in a given case,
is constant; but the cross sec area, weight and cost of the rods increase
as the square of a. Hence, for a given total adhesion, numerous small
rods are more economical than fewer larger rods; but there is, of
course, for each case, a practical limit to this economy.

Continuous beams.

60. Floor systems are usually composed of slabs and beams continuous
over supports; and, if the negative bending moments over the supports
(producing tension at top of beam) are amply provided for, by reinfmt near
the top, and if the supports are unyielding, or exactly equal in their yielding,
advantage is usually taken of the reduction in the positive bending moms
(at and near cen of span) due to continuity.

61. Where floor slabs are laid continuously over the supporting beams,
it is usual to assume WL/1Q wL 2 /10 as the max bending mom, where
L = span; W = total load on span; w = W/L = load per unit of L.


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