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the total energy values for static and dynamic conditions are identical. If the velocity is increased,
the impact values are considerably reduced. For further information, see Ref. 10.
10.6.6 Steady and Impulsive Vibratory Stresses
For steady vibratory stresses of a weight, W, supported by a beam or rod, the deflection of the bar,
or beam, will be increased by the dynamic magnification factor. The relation is given by
dynamic = Static x dynamic magnification factor
An example of the calculating procedure for the case of no damping losses is
»„ - S^ X i _ ( ^ ) 2
where a) is the frequency of oscillation of the load and O) n is the natural frequency of oscillation of
a weight on the bar.
For the same beam excited by a single sine pulse of magnitude A in./sec 2 and a sec duration,
then for t < a a good approximation is
S static (A/g) T 1 / a}\ "I
«<*-. = t J^/y [sin - - ^ (-) sin ^J
where A/g is the number of g's and o> is TT/a.
10.7.1 Definitions
TORSIONAL STRESS. A bar is under torsional stress when it is held fast at one end, and a force acts
at the other end to twist the bar. In a round bar (Fig. 10.23) with a constant force acting, the
straight line ab becomes the helix ad, and a radial line in the cross section, ob, moves to the
position od. The angle bad remains constant while the angle bod increases with the length of the
bar. Each cross section of the bar tends to shear off the one adjacent to it, and in any cross section
the shearing stress at any point is normal to a radial line drawn through the point. Within the
shearing proportional limit, a radial line of the cross section remains straight after the twisting
force has been applied, and the unit shearing stress at any point is proportional to its distance from
the axis.
TWISTING MOMENT, T, is equal to the product of the resultant, F, of the twisting forces, multiplied
by its distance from the axis, p.
RESISTING MOMENT, T r , in torsion, is equal to the sum of the moments of the unit shearing stresses
acting along a cross section with respect to the axis of the bar. If dA is an elementary area of the
section at a distance of z units from the axis of a circular shaft (Fig. 10.23£), and c is the distance
from the axis to the outside of the cross section where the unit shearing stress is r, then the unit
shearing stress acting on dA is (TZ/C) dA, its moment with respect to the axis is (TZ 2 Ic) dA, an
the sum of all the moments of the unit shearing stresses on the cross section is J (rz 2 /c) dA. In
Fig. 10.23 Round bar subject to torsional stress.
this expression the factor J z 2 dA is the polar moment of inertia of the section with respect to the
axis. Denoting this by 7, the resisting moment may be written rJIc.
THE POLAR MOMENT OF INERTIA of a surface about an axis through its center of gravity and
perpendicular to the surface is the sum of the products obtained by multiplying each elementary
area by the square of its distance from the center of gravity of its surface; it is equal to the sum
of the moments of inertia taken with respect to two axes in the plane of the surface at right angles
to each other passing through the center of gravity. It is represented by /, inches 4 . For the cross
section of a round shaft,
J = 1 X 32 TrJ 4 or V 2 TTr 4
For a hollow shaft,
/ - '/32TT(J 4 - d 4 )
where d is the outside and d 1 is the inside diameter, inches, or
J = y 2 7r(r 4 - r 4 )
where r is the outside and r l the inside radius, inches.
THE POLAR RADIUS OF GYRATION, k p , sometimes is used in formulas; it is defined as the radius of
a circumference along which the entire area of a surface might be concentrated and have the same
polar moment of inertia as the distributed area. For a solid circular section,
k 2 p = Vsd 2
For a hollow circular section,
k 2 = 1 X 8 (J 2 - d 2 )
10.7.2 Determination of Torsional Stresses in Shafts
Torsion Formula for Round Shafts
The conditions of equilibrium require that the twisting moment, T, be opposed by an equal resisting
moment, T r , so that for the values of the maximum unit shearing stress, r, within the proportional
limit, the torsion formula for round shafts becomes
T r = T=T-
if T is in pounds per square inch, then T r and T must be in pound-inches, / is in inches 4 , and c is in
inches. For solid round shafts having a diameter, d, inches,
J = 1 X 32 TrJ 4
c = 1 X 2 J
1 (~\T
T = 1 A 6 TTd 3 T
T= —-
TTd 3
For hollow round shafts,
7T(d 4 ~ d 4 )
c = l / 2 d
J =
and the formula becomes
167 1 J
TTr(J 4 - J 4 )
Tr(J 4 - d 4 }
or T=
The torsion formula applies only to solid circular shafts or hollow circular shafts, and then only
when the load is applied in a plane perpendicular to the axis of the shaft and when the shearing
proportional limit of the material is not exceeded.
Shearing Stress in Terms of Horsepower
If the shaft is to be used for the transmission of power, the value of T, pound-inches, in the above
formulas becomes 63,030//VAf, where H = horsepower to be transmitted and N = revolutions per
minute. The maximum unit shearing stress, pounds per square inch, then is
321 0007/
For solid round shafts:
r =
^91 OOOfjsl
For hollow round shafts:
r =
N(d -U 1 )
If T is taken as the allowable unit shearing stress, the diameter, d, inches, necessary to transmit
a given horsepower at a given shaft speed can then be determined. These formulas give the stress
due to torsion only, and allowance must be made for any other loads, as the weight of shaft and
pulley, and tension in belts.
Angle of Twist
When the unit shearing stress r does not exceed the proportional limit, the angle bod (Fig. 10.23)
for a solid round shaft may be computed from the formula
O = £-.
where 6 = angle in radians; / = length of shaft in inches; G = shearing modulus of elasticity of the
material; T = twisting moment, pound-inches. Values of G for different materials are steel,
12,000,000; wrought iron, 10,000,000; and cast iron, 6,000,000.
When the angle of twist on a section begins to increase in a greater ratio than the twisting moment,
it may be assumed that the shearing stress on the outside of the section has reached the proportional
limit. The shearing stress at this point may be determined by substituting the twisting moment at this
instant in the torsion formula.
Torsion of Noncircular Cross Sections
The analysis of shearing stress distribution along noncircular cross sections of bars under torsion is
complex. By drawing two lines at right angles through the center of gravity of a section before
twisting, and observing the angular distortion after twisting, it as been found from many experiments
that in noncircular sections the shearing unit stresses are not proportional to their distances from the
axis. Thus in a rectangular bar there is no shearing stress at the corners of the sections, and the stress
at the middle of the wide side is greater than at the middle of the narrow side. In an elliptical bar
the shearing stress is greater along the flat side than at the round side.
It has been found by tests 5 ' 1 1 as well as by mathematical analysis that the torsional resistance of
a section, made up of a number of rectangular parts, is approximately equal to the sum of the
resistances of the separate parts. It is on this basis that nearly all the formulas for noncircular sections
have been developed. For example, the torsional resistance of an I-beam is approximately equal to
the sum of the torsional resistances of the web and the outstanding flanges. In an I-beam in torsion
the maximum shearing stress will occur at the middle of the side of the web, except where the flanges
are thicker than the web, and then the maximum stress will be at the midpoint of the width of the
flange. Reentrant angles, as those in I-beams and channels, are always a source of weakness in
members subjected to torsion. Table 10.8 gives values of the maximum unit shearing stress r and
the angle of twist 6 induced by twisting bars of various cross sections, it being assumed that r is not
greater than the proportional limit.
Torsion of thin-wall closed sections, Fig. 10.24,
T = 2qA
q = rt
(10 .82)
2A2AG t
should be checked. When more than one cell is used 1 ' 1 2 or if section is not constructed of a single
where 5 is the arc length around area A over which r acts for a thin-wall section; shear buckling
material, 1 2 the calculations become more involved:
J = TTTt
§ dslt
Table 10.8 Formulas for Torsional Deformation and Stress
General formulas: B =• —-, T = — , where 9 =» angle of twist, radians; T = twisting moment, in.-lb;
L — length, in.; r — unit shear stress, psi; G — modulus of rigidity, psi; K 1 in. 4 ; and Q, in. 3 are func-
tions of the cross section.
Formula for K in 6 = —
Formula for Shear Stress
« = £
T =
r(d* - d! 4 )
K = 1/32TW 4 ~ <*1 4 )
K - 2/3 ^ 3
T = JI_
K ^TT 2
r== -^
-^ 2
T - ,a^d+^-n
K - ^TfJs [(I + 5 ) ~ l ] q _ 6-61
o — ai
7roi 3 bi3
6 V^
_ 2OT
T --^T
1 097 7
K = 2.696*
T = ±?j±
336 Vl- b4> ll
r _ Oa+ 1.86)7
a V
^ = 2^ 2 (a - <2) 2 (6 - <i) 2
r =
a<2 + Ui - t£ - tr
2/2(a - ^) (6 - ti)
r = 4 ^I
K = 0.140664
Table 10.8 (Continued)
General formulas: 6 — -^- , r •» — , where 6 «• angle of twist, radians; T «• twisting moment, in.-Ib;
L — length, in.; r «• unit shear stress, psi; G = modulus of rigidity, psi; K, in. 4 ; and Q, in. 9 are func-
tions of the cross section.
Formula for K in 9 — —
Formula for Shear Stress
r •» fillet radius
D — diameter largest inscribed circle
For all solid section* of irregular
A - ZA 1 + A 2 + t*U
for m th e maxi mum shea r 8tre M
K. — 06 3 F- — O 21 - f I — -^-\ "1
occurs at or very near one of the
I2a 4 /J
points where the largest inscribed
o \
92 c ' J
curvature of the boundary is alge-
— I
J 7
/f4 \ -i
circle touches the boundary, and
c ^
X 2 - «P - - 0.105? ( 1
L 3
— )
of these, at the one where the
b / n nj
n n _, r\
braically least. (Convexity rep-
« - - ^0.07 + 0.076 -J
A t a in t where the curvature is
regent 8 P 08 J 11 ^ concavity nega _
tive, curvature of the boundary.)
X « 2Xi -f X 2 -f ZaD 4
M \ T
positive (boundary of section
Xi — 06* J- — 0.21 - ( 1
—- ) J
straight or convex) this maximum
o \
I2o*/ J
stress is given approximately by:
T ~ K C
X 2 - V 3 cd 3
T " °~L C
a- -(0.15+ 0.1- r )
ti \
t - 6 if 6 <d
t - d if d < b
1+ £**
E 1+ -(S-:-!)]
16A *
X - Xi -f Xt + ctD 4
where D - diameter of largest in-
& \~]
scribed circle, r - r a da us of cur-
Xi - o63 ^- - 0.21 ^I- 72^JJ
vature of boundary at the point
(positive for this case), A •• area
o f tb e »<*«»•
/C 2 - cd» ri - O.J05- (\ - —^l
c \
a - - (0.07 + 0.076 f\
d \
Ultimate Strength in Torsion
In a torsion failure, the outer fibers of a section are the first to shear, and the rupture extends toward
the axis as the twisting is continued. The torsion formula for round shafts has no theoretical basis
after the shearing stresses on the outer fibers exceed the proportional limit, as the stresses along the
section then are no longer proportional to their distances from the axis. It is convenient, however, to
compare the torsional strength of various materials by using the formula to compute values of r at
which rupture takes place. These computed values of the maximum stress sustained before rupture
are somewhat higher for iron and steel than the ultimate strength of the materials in direct shear.
Computed values of the ultimate strength in torsion are found by experiment to be: cast iron, 30,000
psi; wrought iron, 55,000 psi; medium steel, 65,000 psi; timber, 2000 psi. These computed values of
twisting strength may be used in the torsion formula to determine the probable twisting moment that
will cause rupture of a given round bar or to determine the size of a bar that will be ruptured by a
given twisting moment. In design, large factors of safety should be taken, especially when the stress
is reversed as in reversing engines and when the torsional stress is combined with other stresses as
in shafting.
Fig. 10.24 Thin-walled tube.
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