Unformatted text preview: LECTURE Third Edition SHAFTS: STATICALLY
INDETERMINATE SHAFTS
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering 7
Chapter
3.6 by
Dr. Ibrahim A. Assakkaf
SPRING 2003
ENES 220 – Mechanics of Materials
Department of Civil and Environmental Engineering
University of Maryland, College Park LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 1
ENES 220 ©Assakkaf Up to this point, the stresses in a shaft
has been limited to shearing stresses.
This due to the fact that the selection of
the element under study was oriented in
such a way that its faces were either
perpendicular or parallel to the axis of
the shaft (see Fig. 15) 1 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 2
ENES 220 ©Assakkaf Fig. 15 τ τ Axis of shaft LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 3
ENES 220 ©Assakkaf From our discussion of the torsional
loading on a shaft, we know this loading
produces shearing stresses τ in the
faces perpendicular to the axis of the
shaft.
But due to equilibrium requirement,
there are equal stresses on the faces
formed by the two planes containing the
axis of the shaft. 2 Slide No. 4 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion
– It is necessary to make sure that whether
the transverse plane is a plane of
maximum shearing stress and whether
there are other significant stresses induced
by torsion.
– Consider the following shaft (Fig. 16),
which is subjected to a torque T. Slide No. 5 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion
Fig. 16 t y y
(c) n τx y dA cos α τn t dA
α α A σn dA x
y τ xy
τyx dA sin α x τ yx
α τ yx (a) τ xy
(b) x 3 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 6
ENES 220 ©Assakkaf Other Stresses Induced By Torsion
– The stresses at point A in the shaft of Fig.
16a is analyzed.
– A differential element taken from the shaft
at point A and the stresses acting on
transverse and longitudinal planes are
shown in Fig. 16b.
– The shearing stress τxy can be determined
from
Tc
τ xy =
J LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 7
ENES 220 ©Assakkaf Other Stresses Induced By Torsion
– Let assume that differential element of Fig.
16b has length dx, height dy, and thickness
dz.
– If a shearing force Vx = τxy dx dy is applied
to the top surface of the element, the
equation of equilibrium ∑Fx = 0 then will
require application of an opposite shear
force V’x at the bottom of the element. 4 Slide No. 8 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion Vx = τ yx dx dz y Vx V y = τ xy dy dz Vy Vy α ∆z z ∆y
∆x x Vx
Fig. 17 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 9
ENES 220 ©Assakkaf Other Stresses Induced By Torsion
– If ∑Fx = 0 then requires application of an
opposite shear force V’x at the bottom of
the element, then it will the element
subjected to a clockwise couple.
– This clockwise couple must be balanced by
counterclockwise couple composed of Vx
applied to the vertical faces of the element. 5 Slide No. 10 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion
– The application of the equilibrium moment
equation ∑Mz = 0 gives τ yx (dx dz ) dy = τ xy (dy dz ) dx – From which the important result τ yx = τ xy (27) LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 11
ENES 220 ©Assakkaf Other Stresses Induced By Torsion
– If the equations of equilibrium are applied
to the freebody diagram of Fig. 16c (which
is a wedgeshaped part of the differential
element of Fig. 16b with dA being the area
of the inclined face), the following results
are obtained
+ ∑F =0 (28) t τ nt dA − τ xy (dA cos α ) cos α + τ yx (dA sin α )sin α = 0 6 Slide No. 12 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion
Fig. 16 y y t n τn t dA
τx y dA cos α A (c) α α x
y σn dA τ xy
τyx dA sin α x τ yx (a) τ xy α τ yx (b) x Slide No. 13 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion
t ∑F t =0 τ nt dA − τ xy (dA cos α ) cos α + τ yx (dA sin α )sin α = 0
From which τ nt = τ xy (cos 2 α − sin 2 α ) = τ xy cos 2α (29) n τn t dA
τx y dA cos α + y α α σn dA τyx dA sin α Fig. 16c 7 Slide No. 14 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion
– Likewise, if we take summation of forces in
the n direction (see Fig. 16c), then the
results would be + ∑F n =0 σ n dA − τ xy (dA cos α )sin α − τ yx (dA sin α ) cos α = 0 (30) Slide No. 15 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) ENES 220 ©Assakkaf Stresses in Oblique Planes Other Stresses Induced By Torsion
t ∑F t =0 σ n dA − τ xy (dA cos α )sin α − τ yx (dA sin α ) cos α = 0
From which σ n = 2τ xy sin α cos α = τ xy sin 2α (31) n τn t dA
τx y dA cos α + y α α σn dA τyx dA sin α Fig. 16c 8 Slide No. 16 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Maximum Normal Stress due to Torsion
on Circular Shaft
The maximum compressive normal stress
σmax can be computed from σ max = τ max = Tmax c
J (32) LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 17
ENES 220 ©Assakkaf Example 4
A cylindrical tube is fabricated by buttwelding a 6
mmthick steel plate along a spiral seam as
shown. If the maximum compressive stress in the
tube must be limited to 80 MPa, determine (a) the
maximum torque T that can be applied and (b) the
factor of safety with respect to the failure by
fracture for the weld, when a torque of 12 kN.m is
applied, if the ultimate strengths of the weld metal
are 205 MPa in shear and 345 MPa in tension. 9 Slide No. 18 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Example 4 (cont’d)
T
T Weld 300 60
150 mm 30 Fig. 18 Slide No. 19 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Example 4 (cont’d)
(a) The polar moment of area for the cylindrical
tube can be determined from Eq.14 as
J= π (r
2 4
o ) − ri 4 = 4
4
τ 150 150 − 6 − 2 2 2 6
4 = 14.096 × 10 mm The maximum torque can be computed from
Eq. 32 as
σ max = σ J 80 × 106 (14.096 × 10 −6 )
Tmax c
⇒ Tmax = max =
75 ×10 −3
J
c
3
= 15.036 × 10 N ⋅ m = 15.036 kN ⋅ m 10 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 20
ENES 220 ©Assakkaf Example 4 (cont’d)
(b)The normal stress σn and shear stress τnt on
the weld surface are given by Eqs. 30 and 29
as
Tc
12 ×103 (75 × 10 −3 )
sin 2α =
sin 2(60 0 ) = 55.29 MPa (T)
J
14.096 × 10 −6
Tc
12 × 103 (75 ×10 −3 )
τ nt = τ xy cos 2α = cos 2α =
cos 2(60 0 ) = −31.92 MPa
J
14.096 ×10 −6 σ n = τ xy sin 2α = LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 21
ENES 220 ©Assakkaf Example 4 (cont’d)
The factors of safety with respect to failure
by fracture for the weld are
345
σ ult
=
= 6.24
σ n 55.29
τ
205
FSτ = ult =
= 6.42
τ tn 31.92
FSσ = 11 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 22
ENES 220 ©Assakkaf Up to this point, all problems discussed
are statically determinate, that is, only
the equations of equilibrium were
required to determine the torque T at
any section of the shaft.
It is often for torsionally loaded
members to be statically indeterminate
in real engineering applications. LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 23
ENES 220 ©Assakkaf When this occurs, distortion equations
involving angle of twist θ must written
until the total number of equations
agrees with the number of unknowns to
be determined.
A simplified angle of twist diagram will
often be of great assistance in obtaining
the correct equations. 12 Slide No. 24 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5
A steel shaft and aluminum tube are
connected to a fixed support and to a rigid
disk as shown in the figure. Knowing that
the initial stresses are zero, determine the
minimum torque T0 that may be applied to
the disk if the allowable stresses are 120
MPa in the steel shaft and 70 MPa in the
aluminum tube. Use G = 80 GPa for steel
and G = 27 GPa for aluminum. Slide No. 25 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5 (cont’d)
Aluminum 8 mm 76 mm Rigid disk 50 mm Steel 500 mm 13 Slide No. 26 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5 (cont’d)
– Freebody diagram for the rigid disk
Tal
T0 From statics,
T0 = Tal + Tst T st
– Deformation (39) Tal Lal Tst Lst
=
J alGal J st Gst θ al = θ st ⇒ (40) Slide No. 27 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5 (cont’d)
– Properties of the aluminum tube
Gal = 27 GPa 38 mm ri = 30 mm = 0.030 m
ro = 38 mm = 0.038 m 30 mm
76 mm J al = [(0.038) − (0.030) ] =
2 π 4 4 = 2.003 × 10 −6 m 4 14 Slide No. 28 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5 (cont’d)
– Properties of the steel tube
Gst = 80 GPa 25 mm c = 25 mm = 0.025 m
J st =
50 mm [(0.025) ] =
2 π 4 = 0.6136 × 10 −6 m 4 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 29
ENES 220 ©Assakkaf Example 5 (cont’d)
Substituting these input values in Eq. 40,
gives
Tal Lal Tst Lst
=
J alGal J st Gst Tal (0.5)
Tst (0.5)
=
−6
2.003 × 10 (27) 0.6136 ×10 −6 (80)
Tst = 0.908Tal
(41) 15 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 30
ENES 220 ©Assakkaf Example 5 (cont’d)
Let’s assume that the requirement τst is
less or to equal to 120 MPa, therefore
Tst = τ st J st
cst = 120 × 106 (0.6136 × 10 −6 )
= 2945 N ⋅ m
0.025 From Eq. 39, we have
Tst = 0.908Tal
2945 = 0.908Tal ⇒ Tal = 3244 N ⋅ m LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 31
ENES 220 ©Assakkaf Example 5 (cont’d)
Let’s check the maximum stress τal in
aluminum tube corresponding to Tal = 3244
N·m:
Tc
3244(0.038)
τ al = al al jal = 2.003 × 10 −6 = 61.5 MPa < 70 MPa OK Hence, the max permissible torque T0 is
computed from Eq. 39 as
T0 = Tal + Tst = 3244 + 2945 = 6189 N ⋅ m
= 6.2 kN ⋅ m 16 Slide No. 32 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 6
A circular shaft AB consists
of a 10inlong, 7/8 indiameter steel cylinder, in
which a 5in.long, 5/8in.diameter cavity has been
drilled from end B. The
shaft is attached to fixed
supports at both ends, and a
90 lb – ft torque is applied at
its midsection. Determine
the torque exerted on the
shaft by each of the
supports. Slide No. 33 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 6
• Given the shaft dimensions and the applied
torque, we would like to find the torque reactions
at A and B.
• From a freebody analysis of the shaft,
TA + TB = 90 lb ⋅ ft which is not sufficient to find the end torques.
The problem is statically indeterminate.
• Divide the shaft into two components which
must have compatible deformations,
φ = φ1 + φ2 = TA L1 TB L2
−
=0
J1G J 2G LJ
TB = 1 2 TA
L2 J1 • Substitute into the original equilibrium equation,
LJ
TA + 1 2 TA = 90 lb ⋅ ft
L2 J1 17 ...
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 Force, Shear Stress, τ yx, Statically indeterminate shafts

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