214 Allowable Shear Stress: Applying the torsion - Aerostudents

115 Pages·2010·16.41 MB·English

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•5–1. A shaft is made of a steel alloy having an allowable
shear stress of t = 12 ksi.If the diameter of the shaft is
allow
1.5 in., determine the maximum torque T that can be T
transmitted.What would be the maximum torque T¿ if a T¿
1-in.-diameter hole is bored through the shaft? Sketch the
shear-stress distribution along a radial line in each case.
Allowable Shear Stress:Applying the torsion formula
Tc
t = t =
max allow J
T (0.75)
12 =
p (0.754)
2
#
T = 7.95 kip in. Ans.
Allowable Shear Stress:Applying the torsion formula
T¿c
t = t =
max allow J
T¿ (0.75)
12 =
p (0.754 - 0.54)
2
# #
T¿ = 6.381 kip in. = 6.38 kip in. Ans.
T¿r 6.381(0.5)
tr=0.5 in = J = p (0.754 - 0.54) = 8.00 ksi
2
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5–2. The solid shaft of radius ris subjected to a torque T.
Determine the radius r¿ of the inner core of the shaft that r¿
resists one-half of the applied torque T 2 . Solve the
1 > 2 r
problem two ways:(a) by using the torsion formula,(b) by
finding the resultant of the shear-stress distribution.
T
Tc Tr 2T
a) t = = =
max J p r4 p r3
2
(T)r¿ T
t = 2 =
p (r¿)4 p(r¿)3
2

r¿ T r¿ 2T
Since t = t ; =
r max p(r¿)3 r apr3b
r
r¿ = = 0.841 r Ans.
1
24
r r¿
2
b) dT = 2p tr2 dr
L L
0 0
2r r¿r
dT = 2p t r2 dr
L L r max
0 0
2r r¿r 2T
dT = 2p r2 dr
L L r apr3b
0 0
T 4T r¿
= r3 dr
2 r4 L
0
r
r¿ = = 0.841r Ans.
1
24
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5–3. The solid shaft is fixed to the support at C and
subjected to the torsional loadings shown. Determine the
shear stress at points Aand Band sketch the shear stress on 10 kN(cid:2)m
volume elements located at these points. C
75 mm
A B 4 kN(cid:2)m
50 mm 75 mm
The internal torques developed at Cross-sections pass through point B and A are
shown in Fig.aand b,respectively.
p
The polar moment of inertia of the shaft is J = (0.0754) = 49.70(10-6) m4.For
2
point B,r = C = 0.075Thus,
B
T c 4(103)(0.075)
t = B = = 6.036(106) Pa = 6.04 MPa Ans.
B J 49.70(10-6)
From point A,r = 0.05 m.
A
T r 6(103)(0.05)
t = A A = = 6.036(106) Pa = 6.04 MPa. Ans.
A J 49.70 (10-6)
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#
*5–4. The tube is subjected to a torque of 750 N m.
75 mm
Determine the amount of this torque that is resisted by the
gray shaded section. Solve the problem two ways: (a) by
using the torsion formula,(b) by finding the resultant of the 100 mm
shear-stress distribution. 750 N(cid:2)m
25 mm
a)Applying Torsion Formula:
Tc 750(0.1)
t = = = 0.4793 MPa
max J p (0.14 - 0.0254)
2
T¿(0.1)
t = 0.4793 106 =
max A B p (0.14 - 0.0754)
2
#
T¿ = 515 N m Ans.
b)Integration Method:

r
t = t and dA = 2pr dr
acb max
dT¿ = rt dA = rt(2pr dr) = 2ptr2 dr
0.1m r
T¿ = 2ptr2 dr = 2p t r2 dr
L L max acb
0.075m
2pt 0.1m
= max r3 dr
c L
0.075m
2p(0.4793)(106) r4 0.1 m
=
0.1 c 4 d
0.075 m
#
= 515 N m Ans.
2
5–5. Thecopperpipehasanouterdiameterof40mmand
aninnerdiameterof37mm.Ifitistightlysecuredtothewall A
atAandthreetorquesareappliedtoitasshown,determine
theabsolutemaximumshearstressdevelopedinthepipe.
T c 90(0.02) 30 N(cid:2)m
t = max =
max J p2 (0.024 - 0.01854) 20 N(cid:2)m
= 26.7 MPa Ans..
80 N(cid:2)m
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5–6. The solid shaft has a diameter of 0.75 in. If it is F
subjected to the torques shown, determine the maximum E
shear stress developed in regions BCand DEof the shaft. D
The bearings at Aand Fallow free rotation of the shaft. C 25 lb(cid:2)ft
B 40 lb(cid:2)ft
(t ) = TBC c = 35(12)(0.375) = 5070 psi = 5.07 ksi Ans. A 20 lb(cid:2)ft
BC max J p2 (0.375)4 35 lb(cid:2)ft
T c 25(12)(0.375)
(t ) = DE = = 3621 psi = 3.62 ksi Ans.
DE max J p (0.375)4
2
5–7. The solid shaft has a diameter of 0.75 in. If it is F
subjected to the torques shown, determine the maximum E
shear stress developed in regions CDand EFof the shaft. D
The bearings at Aand Fallow free rotation of the shaft. C 25 lb(cid:2)ft
B 40 lb(cid:2)ft
T c
(t ) = EF = 0 Ans. A 20 lb(cid:2)ft
EF max J
35 lb(cid:2)ft
T c 15(12)(0.375)
(t ) = CD =
CD max J p (0.375)4
2
= 2173 psi = 2.17 ksi Ans.
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*5–8. The solid 30-mm-diameter shaft is used to transmit 300 N(cid:2) m 500 N(cid:2) m
the torques applied to the gears. Determine the absolute
maximum shear stress on the shaft. A
200 N(cid:2) m
Internal Torque:As shown on torque diagram.
#
C
Maximum Shear Stress:From the torque diagram T = 400 N m.Then,applying
max
torsion Formula. 400 N(cid:2) m
300 mm D
T c
tmabasx = mJax 400 mm B
400(0.015)
= = 75.5 MPa Ans.
p (0.0154) 500 mm
2
•5–9. The shaft consists of three concentric tubes, each
made from the same material and having# the inner and T (cid:3) 800 N(cid:2)m
outer radii shown.If a torque of T = 800 N mis applied to
the rigid disk fixed to its end,determine the maximum shear
stress in the shaft. 2 m ri (cid:3) 20 mm
r (cid:3) 25 mm
o
p p p
J = ((0.038)4 - (0.032)4) + ((0.030)4 - (0.026)4) + ((0.025)4 - (0.020)4)
2 2 2 r (cid:3) 26 mm
i
r (cid:3) 30 mm
J = 2.545(10-6) m4 o
t = Tc = 800(0.038) = 11.9 MPa Ans. ri (cid:3) 32 mm
max J 2.545(10-6) ro (cid:3) 38 mm
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5–10. The coupling is used to connect the two shafts T
together. Assuming that the shear stress in the bolts is
uniform,determine the number of bolts necessary to make
R
the maximum shear stress in the shaft equal to the shear
stress in the bolts.Each bolt has a diameter d.
r
nis the number of bolts and Fis the shear force in each bolt.

T
T
T - nFR = 0; F =
nR
T
F 4T
t = = nR =
avg A (p)d2 nRpd2
4
Maximum shear stress for the shaft:
Tc Tr 2T
t = = =
max J pr4 pr3
2
4T 2T
t = t ; =
avg max nRpd2 p r3
2 r3
n = Ans.
Rd2
5–11. The assembly consists of two sections of galvanized
steel pipe connected together using a reducing coupling at B.
C
The smaller pipe has an outer diameter of 0.75 in.and an inner
diameter of 0.68 in., whereas the larger pipe has an outer
diameter of 1 in.and an inner diameter of 0.86 in.If thepipe is
tightly secured to the wall at C,determine the maximum shear B
stress developed in each section of the pipe when the couple
shown is applied to the handles of the wrench.
A
15 lb 6 in.
8 in.
Tc 210(0.375)
t = = = 7.82 ksi Ans.
AB J p (0.3754 - 0.344)
2
15 lb
Tc 210(0.5)
t = = = 2.36 ksi Ans.
BC J p (0.54 - 0.434)
2
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#
*5–12. The motor delivers a torque of 50 N mto the shaft
AB.This torque is transmitted to shaft CDusing the gears
at E and F.Determine the equilibrium torque T(cid:2) on shaft
CD and the maximum shear stress in each shaft. The
A
bearings B,C,and Dallow free rotation of the shafts. 50 mm
30 mm
Equilibrium: B C
E
a+©M = 0; 50 - F(0.05) = 0 F = 1000 N
E 35 mm 125 mm
a+©MF = 0; T¿ - 1000(0.125) = 0 T¿ D F
#
T¿ = 125 N m Ans.
Internal Torque:As shown on FBD.
Maximum Shear Stress:Applying torsion Formula.
T c 50.0(0.015)
(t ) = AB = = 9.43 MPa Ans.
AB max J p (0.0154)
2
T c 125(0.0175)
(t ) = CD = = 14.8 MPa Ans.
CD max J p (0.01754)
2
#
•5–13. If the applied torque on shaft CDis T¿ = 75 N m,
determine the absolute maximum shear stress in each shaft.
The bearings B,C,and Dallow free rotation of the shafts,
and the motor holds the shafts fixed from rotating.
A
50 mm
Equilibrium: 30 mm
a+©M = 0; 75 - F(0.125) = 0; F = 600 N B E C
F
a +©M = 0; 600(0.05) - T = 0 35 mm 125 mm
E # A T¿ D F
T = 30.0 N m
A
Internal Torque:As shown on FBD.
Maximum Shear Stress:Applying the torsion formula
T c 30.0(0.015)
(t ) = EA = = 5.66 MPa Ans.
EA max J p (0.0154)
2
T c 75.0(0.0175)
(t ) = CD = = 8.91 MPa Ans.
CD max J p (0.01754)
2
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5–14. The solid 50-mm-diameter shaft is used to transmit 250 N(cid:2) m
the torques applied to the gears. Determine the absolute
maximum shear stress in the shaft.
A 75 N(cid:2) m
325 N(cid:2) m
150 N(cid:2) m
B
500 mm C
D
The internal torque developed in segments AB,BCand CDof the shaft are shown 400 mm
in Figs.a,b and c.
500 mm
The maximum torque occurs in segment AB.Thus,the absolute maximum shear
stress occurs in this segment. The polar moment of inertia of the shaft is
p
J = (0.0254) = 0.1953p(10-6)m4.Thus,
2
T c 250(0.025)
t = AB = = 10.19(106)Pa = 10.2 MPa Ans.
A maxBabs J 0.1953p(10-6)
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5–15. The solid shaft is made of material that has an 15 N(cid:2)m
allowable shear stress of tallow = 10 MPa. Determine the 25 N(cid:2)m
required diameterof the shaft to the nearest mm.
A
30 N(cid:2)m
B 60 N(cid:2)m
C 70 N(cid:2)m
D
E
The internal torques developed in each segment of the shaft are shown in the torque
diagram,Fig.a.
Segment DEis critical since it is subjected to the greatest internal torque.The polar
p d 4 p
moment of inertia of the shaft is J = = d4.Thus,
2 a2b 32
d
70
T c a2b
t = DE ; 10(106) =
allow J p
d4
32
d = 0.03291 m = 32.91 mm = 33 mm Ans.
223